Van Brummelen Chapter 7: Areas, Angles, and Polyhedra

Table of Contents

1. Van Brummelen Chapter 7: Areas, Angles, and Polyhedra
2. Exercise 7.2
3. Exercise 7.3
4. Exercise 7.7
5. Exercise 7.8
6. Exercise 7.12
7. Exercise 7.13
8. Exercise 7.14
9. Catching Up
10. Conclusion: One California!

Van Brummelen Chapter 7: Areas, Angles, and Polyhedra

Before we begin, I like to point out that earlier, my June 12th post contained errors -- not on the Van Brummelen portion, but in trying to prove "a line and its translation image are parallel." I believe that after editing the post, I finally have a valid proof of this theorem. But the lemmas needed to prove are a bit awkward (including the "Three Perpendiculars Theorem"), so while the proof is correct, it still might not be something to present in a high school class. I'll save a further discussion of the proof and its viability in the classroom for my next post, not today's.

Chapter 7 of Van Brummelen's Heavenly Mathematics is called "Areas, Angles, and Polyhedra." Here is how it begins:

"The first goal of trigonometry -- to solve any triangle given some information about its sides and angles -- has been accomplished, so it is at this point that most textbooks stop. This is a pity, because while the straightforward practical work has been completed, a wealth of mathematical pleasures that might have spurred a lot of curiosity lies just around the corner."

And according to the author, one such math pleasure is spherical area. He begins by finding the area of a two-sided spherical polygon -- a lune:

"Its area is easy to find, since the ratio of the angle theta between the two great circles to 360 degrees is equal to the ratio of the area of the lune to the surface area of the sphere. In the standard unit sphere the area is 2tau r^2 = 2tau, so:

area of lune/2tau = theta/360.

Thus, measuring theta in degrees,

area of lune = tau theta/180."

Notice that we haven't quite reached Pi Approximation Day yet, and so we're still measuring circles and spheres in terms of tau, not pi.

Van Brummelen proceeds to find the area of a spherical triangle. He informs us that the first person to discover this formula was the 17th century French mathematician Albert Girard. At first he didn't publish this formula -- instead, it was published three years later by Bonaventura Cavalieri. That's right -- it's the same Cavalieri for whom Cavalieri's Principle is named. Cavalieri's Principle can be used to prove the sphere volume (but traditionalists seem to hate it for some reason).

Let's follow the author as he derive's Girard's formula. (In past years, I posted Legendre's derivation, which is more or less the same.)

Derivation:
Girard's idea is simple: extend all three sides of the given triangle into great circles and consider the triangles and lunes that result. In figure 7.2 the original triangle is ABC, and the ' symbols represent antipodal points. The triangle may be extended in three different ways to form lunes (colunar triangles): extend the sides departing from A all the way to the antipodal point A', forming ABA'C; or extend the sides from B; or extend the sides from C. If we add these three lunes together, we get:

2 Triangle ABC = (Triangle ABC + Triangle A'BC + Triangle AB'C + Triangle ABC').

By symmetry we can replace Triangle A'BC with Triangle AB'C'; then the four triangles grouped in parentheses will form the hemisphere at the front of figure 7.2, with area tau. But the areas of the three lunes, considered separately, may be found using the area formula we derived a few moments ago. So:

2 Triangle ABC + tau = (tau/180)(A + B + C),

which simplifies to:

Triangle ABC = (tau/360)(A + B + C - 180).

So Van Brummelen tells us that the area of a spherical triangle is proportional to the amount by which the sum of its angles exceeds 180. (As usual, if we assume the unit sphere and radian measure, then the constant of proportionality is 1.) He refers to this amount as the spherical excess and denotes it by the value 2E. The author keeps us in suspense what that factor of 2 is for.

And of course, he generalizes this to spherical polygons with more than three sides:

Area = (tau/360)[(Sum of triangle's angles) - n * 180]
         = (tau/360)[(Sum of polygon's angles) + 360 - n * 180]
         = (tau/360)[(Sum of polygon's angles) - (n - 2)180]

I argue that this formula applies to the lune as well. Since a lune has two sides, (n - 2)180 = 0, and so this becomes:

Area = (tau/360)(Sum of lune's angles).

Now a lune has two angles, each of measure theta, and so:

Area = (tau/360)(2theta)
         = tau theta/180

just as before.

At this point, Van Brummelen takes a surprising turn, apparently away from spheres altogether. I've discussed some of these related topics on the blog in past years.

He begins by describing a convex polyhedron, with its vertices, edges, and faces. And he reveals the famous formula relating these parts -- Euler's polyhedral formula, V - E + F = 2:

"Much of the notation and form of the algebra that we use today in calculus, such as functions and exponentials, was formulated by Euler."

I've mentioned Euler's polyhedral formula on the blog before. It appears in an exercise in Lesson 9-7 of the U of Chicago text, but on the day we reached that lesson (January 26th), I posted an activity instead and never mentioned the formula at all.

According to the author, there are at least 19 ways to prove Euler's formula:

"Perhaps the most common approach uses graph theory, an area of mathematics sometimes attributed (mostly falsely) to Euler. The story goes that the citizens of Konigsberg enjoyed weekend strolls through their town, which is located along a river with two islands (figure 7.6)."

And we might as well cut Van Brummelen off here, since we already know exactly what he's about to describe here -- the Bridges of Konigsberg problem. For the first few years of this blog, the Bridges of Konigsberg was my first day of school activity. It appears in Lesson 1-4 of the U of Chicago text -- and so once I switched to the digit pattern, it became a 14th day of school activity instead. Most recently, I discussed the Bridges of Konigsberg in my September 5th post.

So let's skip to the author's proof of V - E + F = 2, which is what we're really discussing here:

Proof:
The proof of V - E + F = 2 using graph theory works as follows. Remove one face of the polyhedron, and stretch out the edges to produce a graph on a flat surface (so that the edges from the missing face form the outer boundary). Since one face has been removed, we must show that V - E + F = 1 for this graph. For each face that is not a triangle, add an edge to the graph by joining two non-adjacent vertices; continue doing this until only triangles remain. Each time we add an edge, E and F increase by one, leaving V - E + F unchanged. Now choose a triangle on the outer boundary of the graph. If only one of the triangle's edges is on the boundary, then remove that edge. Then E decreases by one and F decreases by one, leaving V - E + F unchanged. If two of the triangle's edges are on the boundary, remove them and their shared vertex. This action decreases E by two, V by one, and F by one, again leaving V - E + F unchanged. Repeat these steps until all that remains is a single triangle. For that triangle, V - E + F = 3 - 3 + 1 = 1, and the theorem is proved. QED

Van Brummelen continues:

"Oddly, however, the first rigorous proof of Euler's polyhedral formula came from an entirely different and seemingly unrelated corner of mathematics: spherical trigonometry."

In fact, the mathematician who came up with this proof is none other than Adrien-Marie Legendre -- the author of the book we used to learn spherical geometry in past years before I ended up purchasing the Van Brummelen book. Legendre's book is described thusly:

"In various versions and translations it swept through Europe and America, in many cases replacing Euclid's Elements, and became the standard geometry text for over a century. It contains the first proof that pi^2 [uh, make that tau^2/4] is irrational, as well as the first proof that V - E + F = 2."

OK, I admit it's awkward to use tau^2/4 here. As it turns out, the first circle-related constant that was proved irrational was neither pi nor tau, but the square of pi. It follows from the irrationality of pi^2 that pi, and hence tau, is irrational as well. Let's just keep pi^2 for historical reasons here.

Anyway, Legendre's proof of V - E + F = 2 was not one of the theorems we previously covered. So let's reproduce it via Van Brummelen here. We begin by projecting the polyhedron outwards onto an enclosing sphere.

Proof:
Assuming that the sphere has a unit radius, its surface area is 2tau. But we can find the surface area another way, by adding together the areas of each face of the spherical polyhedron. We happen to have a formula for these areas. In our notation,

sum (or sigma) (tau/360)[(Sum of polygon's angles) - (n - 2)180] = 2tau.

Cleaning up a bit and expanding out the sum produces:

(Sum of all angles) - sigma n * 180 + 2F * 180 = 720.

But the angles encompass each of the vertices, so the sum of the angles is just V * 360. And since every edge is counted as part of exactly two polygons, sigma n = 2E. So:

V * 360 - 2E * 180 + 2F * 180 = 720.

Canceling 360 brings us to our goal:

V - E + F = 2. QED

At this point, Van Brummelen describes the five regular polyhedra -- namely the tetrahedron, cube, octahedron, dodecahedron, and icosahedron. That these are the only regular polyhedra was known as early as Euclid, but Van Brummlen uses V - E + F = 2 to prove it:

Proof:
Let m be the number of sides in a face of a regular tetrahedron, and let n be the number of faces that meet at each vertex. Then the number of edges E is equal to mF/2, since each edge is the side of twofaces; and E is also equal to nV/2, since each edge touches two vertices. From V - E + F = 2 and E = mF/2 = nV/2, a little algebra gets us to:

V = 4m/(2(m + n) - mn), E = 2mn/(2(m + n) - mn), and F = 4n/(2(m + n) - mn).

The denominator of this expression must be positive, so 2(m + n) > mn, or (dividing through by 2mn) 1/m + 1/n > 1/2. But both m and n must be greater than 2. A bit of plugging and chugging quickly reveals that the only possible pairs m, n that satisfy the inequality are 3, 3 (tetrahedron); 3, 4 (octahedron); 4, 3 (cube); 3, 5 (icosahedron); and 5, 3 (dodecahedron). QED

At this point, Van Brummelen writes about Johannes Kepler and his model of the solar system. This is based on the radii of inscribed and circumscribed spheres:

"Imagine a regular polygon enclosed within a unit sphere, meeting the polyhedron at the vertices. Then a sphere inscribed within the polyhedron will contact the polyhedron at the center of each face."

Here is how the author calculates the radius, r, of this inscribed sphere:

Derivation:
Figure 7.12 illustrates one face (portrayed here as a triangle, but it could be a square or pentagon) of a regular polyhedron, with O at the center, OA = 1, and since the inscribed sphere touches the polyhedron at C, OC = r. Let E be the end of the perpendicular dropped from C to the midpoint of AB. Imagine OC and OE extended to C' and E' on the circumscribed sphere, forming the right-angled spherical triangle AC'E'. Then since Angle OCA is right,

r = cos AOC = cos AC'.

But from the second of the identities in the first column of Napier's Rules,

cos AC = cot AC'E' cot C'A'E'.

Now, imagine dropping perpendiculars from C to every side of the face we're considering, not just AB. We end up with 2m identical angles around c', each equal to Angle AC'E'. So Angle AC'E' = 360/2m = 180/m. Similarly, if we imagine arcs drawn from A to the centers of each of the faces containing A as a vertex, we end up with 2n identical angles around A each equal to Angle C'AE'. But Angle C'AE' = 360/2n = 180/n. We are left with the pleasingly compact relation:

r = cot(180/m)cot(180/n).

Now Van Brummelen tells us how Kepler ordered his planets and polyhedra: Mercury, icosahedron, Venus, octahedron, Earth, dodecahedron, Mars, tetrahedron, Jupiter, cube, Saturn. The average error between the real planet distances and Kepler's was about 3%.

Now Van Brummelen describes how he calculates i.

Derivation:
In figure 7.14 one face of the polyhedron is drawn as a pentagon (although, as before, a square or a triangle are also possible). The adjoining face below edge AB (not drawn) has DE on its surface. C is the center of the face above AB perpendicularly onto AB. Segment DE is the result of doing the same thing to the face below AB. So the inclination between the two faces is i = Angle CED, and Angle CEO = i/2.

As before, the spherical trigonometry arises by "popping" Triangle ACE outwards onto the circumscribed sphere, producing the spherical right triangle AC'E'. From the second identity in column II of Napier's Rules,

cos C'AE' = sin AC'E' cos C'E'.

But if we consider (as before) the same sorts of constructions drawn on all faces with a vertex at A, we see that Angle C'AE' = 360/2n = 180/n. And considering arcs drawn from c' to every vertex and the midpoint of every edge of the visible face, we see that Angle AC'E' = 360/2m = 180/m. Finally, C'E' = Angle COE; but Angle OCE is right, so C'E' = 90 - i/2. Putting all of these results into our Napier's Rule formula above, we get:

cos(180/n) = sin(180/m)cos(90 - i/2).

A bit of algebraic cleanup provides a pleasing formula for the inclination between two faces:

sin(i/2) = cos(180/n)/sin(180/m).

Van Brummelen now proceeds to calculate the length of an edge:

Derivation:
We are now ready to determine a, the length of a side of a regular polyhedron inscribed in a unit sphere. We can reuse figure 7.14. Consider Triangle ACE; we notice that:

CE/AE = cot ACE = cot(180/m).

But AE = a/2, so CE = (a/2)cot(180/m). On the other hand, from Triangle CEO we have:

r = CE tan CEO = CE tan(i/2).

Combining these two results gives us r = (a/2)cot(180/m)tan(i/2). But we already know that r = cot(180/m)cot(180/n). Setting these two equations equal to each other and solving for a brings us home:

a = 2 cot(180/n)cot(i/2).

Finally, the author calculates the volume of each Platonic solid:

Derivation:
This problem turns out to be relatively simple. Connect the center of each sphere to each of the vertices of the polyhedron. This splits the polyhedron into F pyramids with the faces as bases. The volume of each pyramid is (1/3)(area of base)(height). The height of the pyramid is just the radius r of the inscribed sphere. The area of the base, i.e., a face of the polyhedron, may be found by joining the face's central point to each of the vertices. We see that the area of the base is (ma^2/4)cot(180/m). By combining this information, we arrive at:

Volume = (mFra^2/12)cot(180/m).

Van Brummelen divides by 2tau/3 -- the volume of the unit sphere -- to find the proportion of the sphere filled by the polyhedron. He summarizes all of this in the following table (figure 7.13):

Solid                 m  n  r                                                            i              a               Volume
Tetrahedron      3   3  1/3                                                        70.529    1.633         0.5132 (12.3%)
Cube                 4   3  sqrt(3)/3 = 0.5774                                90           2sqrt(3)/3  1.5397 (36.8%)
Octahedron       3   4  sqrt(3)/3 = 0.5774                                109.471  sqrt(2)       1.3333 (31.8%)
Dodecahedron  5   3  ((sqrt(5)+1)sqrt(6))/(6sqrt(5-sqrt(5)))  116.565  0.7136      2.7852 (66.5%)
Icosahedron      3   5  = 0.7947 (dodeca/icosa are the same)  138.190  1.052        2.5362 (60.5%)

The author concludes:

"The tabulations in figure 7.13 reveal a surprising fact: although the icosahedron appears to adhere most closely to its circumscribed sphere, the dodecahedron actually fills about 10% more of the sphere's volume. Sometimes, appearances can be deceiving."

This reminds us slightly of the Isoperimetric Theorems of Lessons 15-8 and 15-9. We found out that if a polygon is inscribed in a circle, then the polygon with more sides is closer to the circle's area. The same ought to hold for polyhedra inscribed in a sphere, yet the dodecahedron beats the icosahedron. I would point out that a dodecahedron is made up of pentagons, but on the other hand, an icosahedron is made up of triangles. The pentagon beats the triangle (in the 2D case), so that "outweighs" the fact that there are only 12 pentagons vs. 20 triangles.

Exercise 7.2

Verify that a spherical equilateral triangle with 60-degree angles has no area, and that the largest possible triangle is a hemisphere.

The correct formula for us to use is:

Triangle ABC = (tau/360)(A + B + C - 180).

In the first case, all three angles are 60 degrees, so this gives us:

Triangle ABC = (tau/360)(A + B + C - 180)
                       = (tau/360)(60 + 60 + 60 - 180)
                       = (tau/360)0
                       = 0.

A triangle with three 60-degree angles is Euclidean, and we already know that a spherical triangle becomes more Euclidean as it shrinks down to nothing. Hence the triangle has no area.

As for the largest possible triangle, we let all three angles have their greatest possible value, 180:

Triangle ABC = (tau/360)(A + B + C - 180).
                       = (tau/360)(180 + 180 + 180 - 180)
                       = (tau/360)(360)
                       = tau.

This is exactly half of the surface area of the sphere, 2tau. A triangle with three 180-degree angles looks straight on the sphere. Therefore the largest triangle is a hemisphere. QED

Exercise 7.3

The state of Colorado is close to a spherical rectangle, ranging from 37N to 41N latitude and from 102.05W to 109.05W. Find Colorado's area.

(There are two reasons why Colorado is not quite a spherical rectangle. Firstly, its northern and southern boundaries are actually circles of latitude rather than great circle arcs, i.e., they bend slightly to the left as you walk eastward along them. Secondly, surveying errors led to some small irregularities in the legal borders.)

Actually, there's a third reason why Colorado isn't a spherical rectangle. Simply speaking, it's because there are no spherical rectangles. Indeed, we can use the formula to find the area of a rectangle:

Area = (tau/360)[(Sum of polygon's angles) - (n - 2)180]

         = (tau/360)[4(90) - (4 - 2)180]

         = (tau/360)[360 - 360]

         = (tau/360)0

         = 0.

So just like a 60-degree equilateral triangle, a 90-degree spherical rectangle has an area of zero -- and of course, the area of Colorado isn't zero. Therefore Colorado isn't a spherical rectangle. Any figure whose angle measures add up to the Euclidean sum (180 for a triangle, 360 for a quadrilateral, and so on) must have an area of zero (because the "spherical excess" 2E over the Euclidean sum is zero).

How exactly does Van Brummelen expect us to solve this problem? After all, the fact that two of the sides aren't lines (great circles) means that Colorado isn't even a polygon, much less a rectangle -- yet the formulas he provides us only work for polygons.

I assume the author expects us to find the area of a polygon whose vertices match Colorado's. After all, he doesn't expect us to find the exact area of the state -- in his "secondly" above, the borders are irregular, anyway, so we can never get an exact area.

I notice that this problem can be solved similarly to Exercise 6.13 from my July 12th post. Instead of two Canadian cities, we join the NW and NE corners of Colorado to the North Pole. Two sides of this triangle are 49 degrees (90N - 41N) and the angle at the pole is 7 degrees (109.05 - 102.05). Then we use Napier's first analogy:

tan(1/2)(NW + NE)/cot(1/2)7 = cos(1/2)(49 - 49)/cos(1/2)(49 + 49)

tan(1/2)(NW + NE)/cot 3.5 = 1/cos 49

tan(1/2)(NW + NE) = 1/(cos 49 tan 3.5)

tan(1/2)(NW + NE) = 24.92

(1/2)(NW + NE) = 87.7

But notice that this is an isosceles triangle, so by the Isosceles Triangle Theorem the angles at NW and NE corners are congruent -- that is, each angle is 87.7 degrees. Also, notice that these angles are interior angles of the North Pole triangle -- that is, they are exterior to Colorado. The actual NW and NE corners of Colorado are 180 - 87.7 = 92.3 degrees.

Now let's find the SW and SE corners of Colorado:

tan(1/2)(SW + SE)/cot(1/2)7 = cos(1/2)(53 - 53)/cos(1/2)(53 + 53)

tan(1/2)(SW + SE)/cot 3.5 = 1/cos 53

tan(1/2)(SW + SE) = 1/(cos 53 tan 3.5)

tan(1/2)(SW + SE) = 27.17

(1/2)(SW + SE) = 87.89

These angles are interior to both the North Pole triangle and the state, so the corners are 87.89.

Now let's find the sum of all four angles of Colorado and plug it into the area formula:

Area = (tau/360)[(Sum of triangle's angles) - n * 180]

Area = (tau/360)[(92.3 + 92.3 + 87.89 + 87.89) - 360]

Area = (tau/360)[0.38]

Area = 0.0066

which of course, assumes that the earth is a unit sphere. All the way back in Chapter 1, Van Brummelen gives a value of 6371 km -- and this is like a scale factor. The Fundamental Theorem of Similarity tells us that we must multiply by the square of this scale factor to find the area:

Area = 0.0066(6371)^2 = 268931 km^2.

A Google search reveals the measured area of Colorado as 269837 km^2, so our calculated answer, which is close to 270000 km^2, is reasonable.

Exercise 7.7

Prove the following analogy from Breitschneider:

sin(1/2)E cos(1/2)(A - E)/sin(1/2)A = sin(1/2)(s)sin(1/2)(s - a)/cos(1/2)a,

where s is the half perimeter of the triangle. [Casey 1889, 47] (Enterprising readers may look up a hint and Breitschneider's other seven analogies on page 47 of Casey's book, available online, all derivable from Delambre's analogies.)

Well, what are we waiting for? We're definitely "enterprising," aren't we? And this is a blog, so it's easy enough for me to link to Casey's "golden age" text right here:

https://books.google.com/books?id=VYwLAAAAYAAJ&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false

And here is Casey's hint from the above link:

For example, [equation #]1 is obtained by subtracting both sides of the equation:

cos(1/2)(B + C)/sin(1/2)A = cos(1/2)(b + c)/cos(1/2)a from unity.

This, by the way, is Delambre's third analogy, and subtracting from "unity" means subtracting from 1, so now we may begin the proof.

Proof:
1 - cos(1/2)(B + C)/sin(1/2)A = 1 - cos(1/2)(b + c)/cos(1/2)a.

Let's rewrite each side with a common denominator:

(sin(1/2)A - cos(1/2)(B + C))/sin(1/2)A = (cos(1/2)a - cos(1/2)(b + c))/cos(1/2)a.

Now we can use the sum-to-product trig identities:

cos alpha - cos beta = -2sin((alpha + beta)/2)sin((alpha - beta)/2)
sin alpha - sin beta = 2 cos((alpha + beta)/2)sin((alpha - beta)/2).

The numerator on the RHS is already set up for the difference of cosines. The numerator on the LHS is sine minus cosine, so we convert the cosine to a sine to set up the difference of sines:

(sin(1/2)A - cos(1/2)(B + C))/sin(1/2)A = (cos(1/2)a - cos(1/2)(b + c))/cos(1/2)a
(sin(1/2)A - sin(90 - (1/2)(B + C)))/sin(1/2)A = (cos(1/2)a - cos(1/2)(b + c))/cos(1/2)a
(sin(1/2)A - sin(1/2)(180 - B - C))/sin(1/2)A = (cos(1/2)a - cos(1/2)(b + c))/cos(1/2)a.

Now we use the two formulas on the left and right sides:

2cos(1/4)(A-B-C-180)sin(1/4)(A+B+C-180)/sin(1/2)A = -2sin(1/4)(a+b+c)sin(1/4)(a-b-c)/cos(1/2)a.

Now we can simplify the sums inside the parentheses. Notice that a+b+c is just the perimeter of the triangle, so 1/2 of this is the semiperimeter. And 1/4 of this is the, um, "quarter-perimeter"? It doesn't matter, since we already see (1/2)s in our goal identity. Now (1/2)(a -b-c) isn't a semiperimeter -- instead, it's a - s, side a minus the semiperimeter. This must be negative, and so we use the fact that sine is an odd function to remove the - in front of the right side and make it s - a, which is positive:

2cos(1/4)(A-B-C-180)sin(1/4)(A+B+C-180)/sin(1/2)A = 2sin(1/2)(s)sin(1/2)(s - a)/cos(1/2)a.

All we need to do now is clean up the LHS. Inside the sine we have (1/4)(A+B+C-180), but notice that A+B+C-180 already has the name -- 2E, the spherical excess. Van Brummelen still hasn't explained why the excess 2E has that 2 there, but notice the similarity between the semiperimeter s on the RHS and the semi-excess E on the LHS. Likewise, we find that (1/2)(A-B-C-180) is A - E:

2cos(1/2)(A - E)sin(1/2)(E)/sin(1/2)A = 2sin(1/2)(s)sin(1/2)(s - a)/cos(1/2)a.

Dividing both sides by 2 and rearranging the LHS so that sine appears before cosine gives us the analogy that we're trying to prove:

sin(1/2)E cos(1/2)(A - E)/sin(1/2)A = sin(1/2)(s)sin(1/2)(s - a)/cos(1/2)a. QED

Exercise 7.8

If E, E_A, E_B, and E_C denote the half spherical excesses of a spherical triangle and its three colunar triangle and its three colunar triangles respectively, show that E + E_A + E_B + E_C = 180, and hence that the sum of the areas of these triangles is equal to half the area of the sphere.

Proof:
Well, we already know what E, the half-excess of Triangle ABC is. By definition:

E = (1/2)(A + B + C - 180).

Now let's find E_A, the half-excess of the colunar triangle A'BC. The interior angles of the new triangle at B and C are the exterior angles of the original triangle -- that is, 180 - B and 180 - C. And the angle at A' is equal to the original angle at A (as the two angles of a lune are congruent):

E_A = (1/2)(A + 180 - B + 180 - C - 180)
E_A = (1/2)(A - B - C + 180).

Likewise, we find E_B and E_C:

E_B = (1/2)(-A + B - C + 180)
E_C = (1/2)(-A - B + C + 180).

Let's line up all four equations together once again:

E     = (1/2)(  A + B + C - 180)
E_A = (1/2)(  A - B - C + 180)
E_B = (1/2)(-A + B - C + 180)
E_C = (1/2)(-A - B + C + 180).

Adding up all four equations, we see two A's and two -A's, which are eliminated. Similarly, two B's cancel two -B's, and two C's cancel two -C's. All that's left are 3 180's and a -180:

E + E_A + E_B + E_C = (1/2)(2 * 180)
E + E_A + E_B + E_C = 180,

which is exactly what we wanted to prove. And hence the total excess of these triangles is 360 (since E is the half-excess), and therefore the total area is tau -- wich is half the area of a sphere (as the surface area of the whole sphere is 2tau). QED

Hmm, perhaps this sheds more light on why the excess is defined to be 2E rather than E. The value of E for the whole sphere is 360, while the total excess for the entire sphere is 720. This is because the area of the sphere isn't tau, but 2tau. And if you think about it, a lune of angle theta has E = theta, while its excess is the sum of its two angles, each of measure theta, so it's 2theta.

Recall that the excess of a lune is simply the sum of its angles. Officially, it's the sum of its two angles minus (n - 2)180, but since n = 2, we find that (n - 2)180 = 0. For all other polygons, the excess equals the sum of its angles minus the nonzero value of (n - 2)180. (Notice what impact some of these equations has on the tau vs. pi debate. Tau is convenient when we notice that E = tau for an entire sphere. On the other hand, many of today's formulas contain 180's, which can be converted to pi radians.)

Exercise 7.12

If a dodecahedron and an icosahedron were each described about a given sphere, the sphere described about these polyhedra will be the same. [Todhunter/Leathern 1907, 216]

We notice in the chart (figure 7.13) that r for the dodecahedron and icosahedron are equal -- indeed, I didn't feel like typing out that exact radical expression in ASCII twice, so I just wrote it once. So anyway, Van Brummelen (or Todhunter) never actually uses the word "prove" in this problem, but I doubt he wants us to write "they're the same because figure 7.13 says so." Let's assume that he wants us to prove it.

Proof:
The formula for r is:

r = cot(180/m)(180/n).

Let's review what the m and n stand for again. Now m stands for the number of sides on each face -- so m = 5 for the dodecahedron (pentagons) and m = 3 for the icosahedron (triangles). And n stands for the number of faces that meet at a vertex. All we need to do is look at each Platonic solid to see that n = 3 for the dodecahedron and n = 5 for the icosahedron. So this gives us:

dodecahedron: r = cot(180/5)cot(180/3) = cot 36 cot 60,
icosahedron: r = cot(180/3)cot(180/5) = cot 60 cot 36,

which are obviously equal. Thus completes the proof -- oh, someone might nitpick here. We just found the radius of the inscribed sphere if both circumscribed spheres have radius 1, but we're asked to find the radius of the circumscribed spheres if both inscribed spheres have the same radius. Well then, just let both inscribed spheres have radius 1, then both circumscribed spheres have radius 1/r, which is tan 60 tan 36 . Now that really completes the proof. QED (It's because of this fact that the dodecahedron and icosahedron are said to be dual polyhedra.)

By the way, that radical expression on the chart appears because the trig ratios of 60 and 36 do have exact radical values. Trig students know that tan 60 = sqrt(3) and cot 60 = tan 30 = sqrt(3)/3, but we don't ordinarily teach the exact trig values of 36 or 54 degrees.

Let's try to find the value of r from the chart, which is cot 60 cot 36 = tan 30 tan 54. Now we know that tan 30 = sqrt(3)/3. Here's a website that derives the exact value of tan 54 (as well as some other trig ratios of multiples of 18 degrees):

http://www.math-only-math.com/exact-value-of-tan-54-degree.html

This link finds that tan 54 = (sqrt(5)+1))/sqrt(10 - 2sqrt(5)). This gives us:

tan 30 tan 54 = sqrt(3)/3 * (sqrt(5)+1))/sqrt(10 - 2sqrt(5))
                      = (sqrt(5)+1)sqrt(3)/(3sqrt(10 - sqrt(5))).

Apparently, Van Brummelen multiplied the numerator and denominator by sqrt(2) to obtain the value in his chart:

r = ((sqrt(5)+1)sqrt(6))/(6sqrt(5-sqrt(5))).

Exercise 7.13

A regular octahedron is inscribed in a cube so that the corners of the octahedron are at the centers of the faces of the cube: show that the volume of the cube is six times that of the octahedron. [Todhunter/Leathen 1907, 216]

Proof:
Notice that if we have a sphere, then the cube is circumscribed about the sphere while the octahedron is inscribed. (The cube and the octahedron are also dual polyhedra.)

This is a volume problem, so let's look at the volume formula:

Volume = mFra^2/12 cot(180/m).

And recall that we need to find radius r, inclination i and side length a before we can find the volume:

r = cot(180/m)cot(180/n)
sin(i/2) = cos(180/n)/sin(180/m)
a = 2cot(180/n)cot(i/2).

For the octahedron, we have m = 3 (triangles) and n = 4 (triangles at each vertex):

r = cot(180/3)cot(180/4)
r = cot 60 cot 45
r = tan 30
r = sqrt(3)/3.

sin(i/2) = cos(180/4)/sin(180/3)
sin(i/2) = cos 45/sin 60
sin(i/2) = sqrt(2)/sqrt(3).

a = 2cot(180/4)cot(i/2)
a = 2cot 45 cot(i/2).

Fortunately cot 45 is just 1, but what about cot(i/2)? We could use rounded values (or just get it from the figure 7.13 table), but since the goal statement to the proved contains an exact integer, we likely need to keep all figures exact. We know sin(i/2) and we need to find cot(i/2).

sin(i/2) = sqrt(2)/sqrt(3)
cos(i/2) = sqrt(1 - 2/3) = sqrt(1/3) = sqrt(3)/3
cot(i/2) = cos(i/2)/sin(i/2)
             = sqrt(3)sqrt(3)/(3sqrt(2))
             = sqrt(1/2) = sqrt(2)/2
a = 2sqrt(2)/2 = sqrt(2).

And now we can find the volume:

Volume = mFra^2/12 cot (180/m)
V = (3)(8)(sqrt(3)/3)(2)(1/12) cot(180/3)
V = (48/12)(sqrt(3)/3)^2
V = 4(1/3)
V = 4/3.

So the volume of the inscribed octahedron is 4/3 cubic units. But we can't find the volume of the circumscribed cube using the equations or the chart, since these are only set up for polyhedra that are inscribed (within the unit sphere)!

But then again, it's just a cube circumscribed about the unit sphere, so it should be easy. The side length of the cube equals the diameter of the sphere, which is 2. Hence its volume is 8 cubic units.

And now we can simply find the ratio:

Volume of cube/Volume of octahedron = 8/(4/3) = 8(3/4) = 6. QED

Notice that the figure 7.13 chart gives the volume of the octahedron as 1.3333. We could have made the educated guess that 1.3333 is just 4/3 rounded, and since the cube volume is easily found to be 8, we obtain the ratio as 6. But once again, this is a proof -- merely assuming that 1.3333 in a chart is actually 4/3 isn't exactly a valid proof.

Exercise 7.14

Derive precise expressions (that is, containing no decimal approximations) for the volumes of the regular polyhedra in terms of their side length a.

(Dang it -- no decimals, so we can't just use the figure 7.13 chart!)

Well, let's do the easy one first -- the volume of a cube of side length a is just a^3 (and besides, we already used this formula in the previous exercise). It took us just three seconds to find this formula.

Oh, Van Brummelen probably expects us to use the equations:

r = cot(180/m)cot(180/n)
sin(i/2) = cos(180/n)/sin(180/m)
a = 2cot(180/n)cot(i/2)
V = mFra^2/12 cot(180/m).

Well, let's check our cube formula from by using these equations. For a cube, m = 4, n = 3:

r = cot(180/4)(cot 180/3)
r = cot 45 cot 60
r = sqrt(3)/3.

sin(i/2) = cos(180/3)/sin(180/4)
sin(i/2) = cos 60/sin 45
sin(i/2) = sqrt(2)/2
i/2 = 45
i = 90 (as if we didn't already know that the angles in a cube are right angles).

a = 2cot(180/3)cot(90/2)
a = 2cot 60 cot 45
a = 2sqrt(3)/3.

V = (4)(6)(sqrt(3)/3)(4/3)(1/12)cot(180/4)
V = (96/108)sqrt(3)
V = 8sqrt(3)/9.

And we might check that a^3 = (2sqrt(3)/3)^3 = 8sqrt(27)/27 = 8sqrt(3)/9 = V, but then again, just because V = a^3 for one particular value of a, it doesn't mean that V = a^3 holds for all such values.

It's implied by the Fundamental Theorem of Similarity that the volume of any Platonic solid with side a is Ca^3 for some constant C. We just showed that for the cube, C = 1. In general, we'll find a and V for one particular solid (the one inscribed in the unit sphere, since that's what the equations are set up for) and then find the constant C = V/a^3.

Volume of Cube: V = a^3.

We might as well do the octahedron next, since we already did all the work in the previous problem:

a = sqrt(2)
V = 4/3
C = V/a^3
C = (4/3)/sqrt(2)^3
C = (4/3)(sqrt(2)/4)
C = sqrt(2)/3.

Volume of Octahedron: V = (sqrt(2)/3)a^3.

Let's try the tetrahedron, which appears to be the easiest. Here m = n = 1:

r = cot(180/3)cot(180/3)
r = cot^2(60)
r = tan^2(30)
r = 1/3.

sin(i/2) = cos 60/sin 60
sin(i/2) = cot 60
sin(i/2) = tan 30
sin(i/2) = sqrt(3)/3.

cos(i/2) = sqrt(1 - 1/3) = sqrt(2/3) = sqrt(6)/3
cot(i/2) = sqrt(6)/sqrt(3) = sqrt(2).

a = 2 cot 60 sqrt(2)
a = 2 tan 30 sqrt(2)
a = 2(sqrt(3)/3)sqrt(2)
a = 2sqrt(6)/3.

Actually, let's save ourselves a step here. The formula for V is:

V = mFra^2/12 cot(180/m)

but we're trying to find C = V/a^3, not V itself. So let's just divide this formula by a^3:

C = mFr/(12a) * cot(180/m)
C = (3)(4)(1/3)/(12 * 2sqrt(6)/3) cot 60
C = 1/(2sqrt(6))(sqrt(3)/3)
C = sqrt(3)/(6sqrt(6))
C = sqrt(2)/12.

Volume of Tetrahedron: V = (sqrt(2)/12)a^3.

For the dodecahedron, let's use the fact that we already calculated the exact value of r above:

r = ((sqrt(5) + 1)sqrt(6))/(6sqrt(5 - sqrt(5))).

Don't forget that m = 5, n = 3 for the dodecahedron.

sin(i/2) = cos(180/n)/sin(180/m)
sin(i/2) = cos 60/sin 36

Fortunately, sin 36 is one of the values provided at the above (Math Only Math) link:

sin(i/2) = (1/2)/(sqrt(10 - 2sqrt(5))/4)
sin(i/2) = 2/sqrt(10 - 2sqrt(5))

cos(i/2) = sqrt(1 - 2/(5 - sqrt(5))
cos(i/2) = sqrt(((5 - sqrt(5))/10)

cot(i/2) = sqrt(((5 - sqrt(5))/10)/(2/sqrt(10 - 2sqrt(5))
cot(i/2) = (sqrt(5) - 1)/2)
cot(i/2) = phi (that is, lowercase phi = 0.618...)

The manipulation of radicals here is very tedious, and I don't really want to write it out in ASCII. I was surprised when it turned out to be something as simple as phi. (Hint: multiply top and bottom by sqrt(2), combine to make a common denominator, then multiply top and bottom by conjugate.)

a = 2cot(180/3)phi
a = 2phi cot 60
a = 2phi tan 30
a = 2phi sqrt(3)/3

C = (5)(12)((sqrt(5) + 1)sqrt(6))/(6sqrt(5 - sqrt(5)))/(12(2phi)sqrt(3)/3) * cot(180/5)
C = (5)((sqrt(5) + 1)sqrt(6))/(6sqrt(5 - sqrt(5)))(sqrt(3)/(2phi))* (sqrt(5) + 1)/sqrt(10 - 2sqrt(5))

(Notice that cot 36 = tan 54, the value given at the above link.)

C = sqrt(5)(sqrt(5) + 1)^2/2(sqrt(5) - 1)^2
C = (1/4)(15 + 7sqrt(5))

Volume of Dodecahedron: V = (1/4)(15 + 7sqrt(5))a^3.

Now it's time for the hardest one of all, the icosahedron. Fortunately, the r is the same:

r = ((sqrt(5) + 1)sqrt(6))/(6sqrt(5 - sqrt(5))).

Don't forget that m = 3, n = 5 for the icosahedron.

sin(i/2) = cos(180/n)/sin(180/m)
sin(i/2) = cos 36/sin 60.

Fortunately, cos 36 is one of the values provided at the above (Math Only Math) link -- (1 + sqrt(5))/4 or Phi/2, which we'll use to make it simpler:

sin(i/2) = (Phi/2)/(sqrt(3)/2)
sin(i/2) = Phi/sqrt(3).

cos(i/2) = sqrt(1 - (1 + Phi)/3) (since Phi^2 = 1 + Phi)
cos(i/2) = sqrt(2 - Phi)/sqrt(3)
cos(i/2) = phi/sqrt(3) (that's right -- Phi^2 + phi^2 = 2)

cot(i/2) = (sqrt(2 - Phi)/sqrt(3))/(Phi/sqrt(3))
cot(i/2) = phi/Phi (which also equals 1 - phi, but let's keep it as phi/Phi)

a = 2cot(180/5)(phi/Phi)
a = 2(sqrt(5) + 1)/sqrt(10 - 2sqrt(5)) * (sqrt(5) - 1)/(sqrt(5) + 1) (multiplying phi/Phi by 2/2)
a = 2(sqrt(5) - 1)/sqrt(10 - 2sqrt(5))

C = (3)(20)((sqrt(5) + 1)sqrt(6))/(6sqrt(5 - sqrt(5)))/(12 * 2(sqrt(5) - 1)/sqrt(10 - 2sqrt(5)))cot(180/3)
C = (5)((sqrt(5) + 1)sqrt(6))/(6sqrt(5 - sqrt(5))) * (sqrt(10 - 2sqrt(5)))/(2(sqrt(5) - 1)) * sqrt(3)/3)
C = (5sqrt((5) + 1))/(6sqrt((5) - 1)
C = (5/12)(3 + sqrt(5))

Volume of Icosahedron: V = (5/12)(3 + sqrt(5))a^3.

I double-checked and triple-checked the values of r, i, and a with the chart, while I checked the values of C with the Wolfram website:

http://mathworld.wolfram.com/PlatonicSolid.html

I actually found the dodecahedron more difficult than the icosahedron (mainly because the part that needed the conjugate appears squared -- see the above).

Catching Up

OK, there are other things going on in the world besides Van Brummelen's questions. Regarding the traditionalists, Barry Garelick posted earlier today:

https://traditionalmath.wordpress.com/2018/07/20/levels-of-understanding/

This post is titled "levels of understanding." Garelick provides an example of a problem that to him demonstrates many levels of understanding -- a complex fraction problem (fortunately with no radicals, as opposed to all those radicals I had to simplify for Van Brummelen today). As usual, SteveH discusses more on complex fractions, "invert & multiply," and "proving" basic facts.

I have more I'd like to say about this, but I've already devoted this post to Van Brummelen -- it's been my plan to post today, and I'd already begun working out some of the problems a few days before Garelick decided to post. I'm not sure when my next traditionalists post will be.

Meanwhile, I'm continued to be fascinated with the summer school classes. The students have just completed the second of three B-session weeks.

The top student from Algebra 1A is currently earning a grade of 95% in Algebra 1B. On his most recent test, "Quadratic Functions," he earned a score of 84. The best student on this particular test earned a score of 96 -- she's also has a current grade of 95%. These are currently the top two students in Algebra 1B -- a third student also as an A (or A-) as the current grade, but is falling behind -- she hasn't reached the quadratic functions test yet.

On the other side, the bottom student from 1A has a high D (or D+) as the current grade, despite scoring only 52 on the quadratic functions test. These aren't the lowest scores on the recent test -- one student scored 16 while another has a 28 score. These are the only two students currently failing.

The top student from 1A is right on pace, having completed 67% of the 1B class (note that this is basically the 2/3-mark). But the bottom student from 1A is ahead of schedule -- he's completed 85% of the 1B class, including a score of 65 on the next test (on "Solving Quadratic Equations").

In fact, three students are listed as having completed 100% of the class! Two of those students currently have B's (including scores of 84 and 76 on the last unit test, "Quadratic Formula and Modeling"), but the third currently has a D (50 on the last unit test). I wonder why that third student is moving ahead on the material if he can't get better than a D. It would make far more sense for him to slow down, review the material, and retake the tests to earn a C or better. (Technically, no one has completed 100% of the course yet, since the district-mandated final must be given on the last day).

Conclusion: One California!

Let me end this post by noting that judges have removed the controversial Three Californias initiative from the ballot. It had already been assigned a number on the ballot (Proposition 9), and now that proposition will be vacant. Keep in mind that I wasn't going to vote for Prop 9 anyway, and the chances of it passing were always remote.

Meanwhile, Year-Round DST remains on the ballot as Proposition 7. So far, I haven't heard much about its prospects of passing.

Traditional Word Problems

Table of Contents

1. Barry Garelick on Traditional Word Problems
2. Wayne Bishop on Number Bases
3. Anna on Homeschooling
4. SteveH on Governing Equations
5. Conclusion

Today is a traditionalists' debate post -- and it's just in time, too. Our favorite traditionalist, Barry Garelick, posted on Friday the 13th. And this post has drawn eight comments, including one from our favorite commenter, SteveH.

https://traditionalmath.wordpress.com/2018/07/13/a-traditional-approach-to-a-traditional-word-problem-in-algebra/

Unlike most Garelick posts, this one doesn't begin with a link to an article, but rather a traditional word problem that's on his mind:

I saw this problem a few days ago and it got me thinking how I would present it in my algebra class: 

“An airplane can fly 550 miles with the wind in the same amount of time it takes to fly 425 miles against the wind. Find the speed of the wind, if the airplane is flying at a constant speed of 195 miles per hour.”

After his students set up an equation to solve, Garelick writes about a common problem that many traditionalist teachers have -- reducing the students' dependence on calculators:

Big numbers, so before they can ask whether they can use calculators, the first thing I ask the students is whether we can whittle down 550 and 425 by dividing both sides by a common factor? Dividing by 5 two times in a row (or 25 once) yields:  22/(195+w) = 17(195-w).

They still want to use their calculators.  I forbid their use for this problem. 

“Let’s just write it out without doing any multiplying,” I tell them.

And after his students have solved the problem, Garelick concludes:

This type of approach helps develop number sense, which some claim traditionally taught math doesn’t do a good job developing. I present it here as an example of what I consider a rich problem.  It’s not the problem that is rich so much as the way it can be used to address different strategies that may be employed within the same problem.

OK, so there are several issues addressed in this post. First of all, Garelick tells us that this is a traditional word problem. Traditional word problems include not just rate-time-distance problems as in this example, but age problems and mixture problems as well.

The relationship between word problems and the Common Core is a bit complex. The Common Core emphasizes mathematical modeling -- converting real-world language to math. Yet traditional word problems such as distance, age, and mixture problems are rare in Common Core math.

For example, back in my May 23rd post, I mentioned the following word problem -- Question 15 from the SBAC Practice Exam:

A store sells new and used video games. New video games cost more than old video games. All used video games cost the same. All new video games also cost the same.

Omar spent a total of $84 on 4 used video games and 2 new video games. Sally spent a total of $78 on 6 used video games and 1 new video game. Janet has $120 to spend.

Enter the number of used video games Janet can purchase after she purchases 3 new video games.

Why is Garelick's distance problem considered "rich," but not the SBAC price problem? Conversely, why does the Common Core approve of this price problem, but not Garelick's distance problem? We will try to discover the answers to these questions throughout this post.

Garelick's method of solving this without a calculator is interesting. Notice that in one step, to divide by 25, he divides by 5 twice. I'd argue that division by 25 is interesting in itself, because we can think of 25 as a quarter. So we can get the students thinking in terms of money -- how many quarters does it take to make $5.50 or $4.25? Otherwise I agree with his steps -- especially the way he avoids multiplying anything by 195.

According to Garelick, this question helps develop number sense -- and the idea that division by 5 twice is equivalent to division by 25 once contributes to this development.

But here's something similar to which traditionalists might object. Instead of dividing by 25, just divide by 5 twice -- therefore instead of dividing by 4, just divide by 2 twice. I found an example of this on YouTube -- the "Half Half Strategy":

As we can plainly see here, Carrie Susong -- the lady in the video -- divides 36 by 4 using this "Half Half Strategy." I can't read Garelick's mind, my feeling is that he'd criticize this division method as a waste of time. So why would Garelick object to dividing 36 by 2 twice yet have no problem with dividing 550 by 5 twice?

That's an easy once -- 36/4 = 9 (and the related 9 * 4 = 36) are basic one-digit math facts that students should have memorized, whereas 550/25 is always a multi-step division problem. Garelick and other traditionalists would especially complain if Susong were a teacher who forces her students to show the work for "Half Half," and awards zero credit to the kid who divides 36/4 = 9 in a single mental step because he can retrieve the multiplication 9 * 4 = 36 in one second. (By the way, nothing in this video suggests that Susong is such a teacher, but you know what I'm talking about here.)

The "Half Half Strategy" of division by 4 might have been common in the old days when everyone used Roman Numerals. But now that we use a simpler numeration system, traditionalists expect students to multiply and divide single digits in one step.

Wayne Bishop on Number Bases

Even though SteveH is the commenter we follow the most closely, let's look at some of the points made by others in this comment thread, in the order they posted. We begin with Wayne Bishop.

Wayne Bishop:

My only criticism/suggestions would be on the integer simplifications. Simple tests for common factors (the divisors of 10) and the sum of the digits divisible by 3 should be known and is usually enough (it makes the 39 much more obvious in this example) and reduction by multiples of 1001 for 7, 11, and 13 simultaneously for big numbers is kinda cute.

Hmm, I'm not quite sure how knowing that 39 is a multiple of three makes 195/39 easier -- except, of course, for the additional knowledge that 195 is also a multiple of three.

Wayne Bishop:

Beyond knowing and using these “tricks”, the “whys” of these simple tests are interesting and worth exploring in an algebra class. Even more, examining their equivalents in other base notation adds another level of understanding.

In my last post, I wrote a little about number bases (in discussing the Pappas quaternary problem for Friday the 13th). It's interesting that Bishop, presumably a traditionalist (since I assume that anyone who comments on the Garelick blog, especially to express agreement with him, is a traditionalist), would advocate spending time in a math class on number bases.

And of course, Bishop is writing about the divisibility rules. For decimal, he describes both the divisor rule (for 2 and 5) and the omega rule for 3 (which works for 9 as well).

He also mentions an obscure rule for 7, 11, and 13. This is technically a cube alpha rule (based on 1001, one more than the cube of the base). Sometimes I wonder whether such a rule is worth it, since for 7, we must memorize three-digit 143 multiples of 7. Well, since Bishop uses the rule, we might as well use it too.

To those of us familiar with divisibility rules, it may be strange to use cube alpha for 11, since 11 can be found using alpha itself. Well, Bishop's cube alpha allows us to resolve three primes, including a prime smaller than 11 (namely 7). Since 11 isn't an important (small) prime, we might as well use one rule to resolve all three primes.

Bishop suggests that we should examine another base in our classes. Suppose we take up Bishop's suggestion -- so which base would be the best "starter base" to introduce our students.

Well, consider that my knowledge of number base terminology (cube alpha, SPD, {c} default dozenal, etc.) comes from the Dozens Online website. The members of that forum would obviously recommend that dozenal be the base we introduce to students. The only problem with dozenal is that the two extra digits might confuse students. If we were to choose dozenal, I'd recommend using the digits introduced in the Schoolhouse Rock video "Little Twelvetoes" -- dek (X), el (E), doe (10). I might as well post the video here:

{c} (default dozenal)

Here are the relevant divisibility rules similar to the ones Bishop uses for decimal. The divisor rule applies to 2, 3, 4, 6, and omega applies to el (E). For Bishop's cube alpha (1001) rule, we find that 1001 = 7 * 11 * 17 in dozenal, and so this rule also applies to three primes, 7, 11, and 17. If we're going to have cube alpha, then we might as well have square alpha (SPD), which gives us two more primes since 101 = 5 * 25. So the resolvable primes are 2, 3, 5, 7, E, 11, 17, 25. But 17 and 25 are quite large, with two skipped primes 15 and 1E.

But should dozenal indeed be the base we introduce to students? If we want to avoid transdecimal digits (digits greater than 9) yet have a base as close to decimal as possible, then we should choose base 9 (nonary). If we wish to avoid odd bases, then base 8 (octal) is the best choice. Furthermore, some might consider teaching number bases to be a waste of time, but if we choose octal, then we can at least tie this to computer programming.

Another computer-related base, hexadecimal, suffers from needing more digits than dozenal. Binary, the pure computer base, is well below the human scale. Base 2 (binary) has several interesting divisibility rules, but if the purpose of teaching the number base is to compare it to decimal, then binary becomes a poor choice.

I'm actually in a base 4 (quaternary) mood right now because of the recent Pappas question. Indeed, there was a second Pappas question this month that I didn't mention:

?10 (base four) = 52 (base ten)

The correct digit is 3 -- and of course, this problem is from July 3rd.

{4} (default quaternary)

Imagine that we're teaching a math class, and that following Bishop's suggestion, we've chosen quaternary as the starter base to introduce to our students. Here are some questions we might ask:

1. In decimal, the divisor (last-digit) rule applies to two, five, and ten. To what numbers does the divisor (last-digit) rule apply in quaternary?

This one should be easy -- in quaternary, the divisor rule applies only to two (2) and four (10).

2. Notice that Bishop doesn't mention the rules for regular numbers, like four in decimal. Regular numbers have divisibility rules where we must check the last several digits (for four in decimal, we must check the last two digits for divisibility by four).

Even the regular test for four is a bit tricky. The next regular number, eight, requires checking the last three digits, but it might be doable if we piggyback on the rule for four. Sixteen, unfortunately, is too difficult to resolve in decimal.

On the other hand, regular tests in a base like quaternary are more important and much easier. To test for divisibility by eight (20), the last two digits must be -00 or -20. To test for divisibility by sixteen (100), the last two digits must be -00. For the next two powers of two (200 and 1000), we check the last three digits. If they're -000 or -200, the number is divisible by 200, and in the former case -000, the number is also divisible by 1000, and so on. Compare this to the difficult task for checking the fifth and sixth powers of two in decimal.

3. In decimal, the omega (sum-of-digits) rule applies to three and nine. Does the omega rule apply to three (3) in quaternary? Does it apply to nine (21)?

Students should be able to check that the omega rule applies to three but not nine. A counterexample for nine is nine itself -- nine is written as 21, and 2 + 1 = 3, which clearly isn't a multiple of nine.

10 (since we're still counting in quaternary). Why or why not?

As for "why," it's said that regular rules "extend upward," while omega rules "extend downward." So the regular rule in quaternary (and theoretically in decimal) works for any power of two. But the omega rule only works for the omega number (one less than the base) and any of its factors. In decimal, nine is the omega, and three inherits nine's omega properties. But in quaternary, three is the omega, so no other number inherits three's omega properties. The fact that nine is 3^2 doesn't mean anything in quaternary, just as 3^3 doesn't have a simple omega rule in decimal.

11. Describe an alpha rule for five (11) in quaternary.

Notice that Bishop doesn't give the alpha rule for eleven in decimal -- instead, eleven is one of three primes available via cube alpha. But five is a more important prime than eleven. Thus in quaternary, we should resolve five (11) using just alpha, not cube alpha.

The alpha rule involves adding and subtracting alternate digits. For example:

21: 2 - 1 = 1, therefore 21 isn't a multiple of 11.
313: 3 - 1 + 3 = 11, therefore 313 is a multiple of 11.
3001: 3 - 0 + 0 - 1 = 2, therefore 3001 isn't a multiple of 11.

12. To what numbers does the cube alpha (1001) rule apply?

We factor 1001 is quaternary: 1001 = 11 * 31. Thus cube alpha applies to the primes five (11) as well as thirteen (31). In practice, cube alpha will be used for 31 only, since 11 is resolved by alpha itself (and it's not worth using cube alpha for an extra prime much larger than the alpha, if all we want to resolve is the alpha).

13. To what numbers does the cube omega rule apply?

In quaternary, cube omega is 333, one less than the cube of the base 1000. In decimal, cube omega only helps to resolve two large numbers, but cube omega is more useful in quaternary. We notice this by factoring 333 = 3 * 3 * 13. So not only does cube omega provide us an additional prime seven (13), but we can also do nine (3^2 = 21) -- which, as you recall, is unavailable with just omega. (In decimal, cube omega likewise gives us an extra power of three, 3^3, and a prime larger than 3^3.)

Bonus: Is there a divisibility rule for eleven (23) in quaternary?

The answer is no, if we allow divisibility tests up to cube alpha (as Bishop does). The number eleven is resistant to intuitive divisibility tests -- we can show that unless eleven is resolved by divisor, omega, or alpha, it's resolved only by fifth-power omega/alpha. Fifth powers are too large to work with in any base except 2 (binary).

But as I mentioned in my last post, quaternary is easily convertible to binary -- one quaternary digit equals two binary digits. Then we apply fifth-power alpha to the binary number to resolve eleven.

I consider this to be a "bonus" question -- I wouldn't mention a rule for eleven with students unless it's an honors class. If we do give a rule for eleven, then we have divisibility rules for all two-digit numbers in quaternary. Of course, 100 is regular, 101 is square alpha, and 102 = 2 * 21, which we resolve as even + 21 (from cube omega). So the first prime we can't resolve is nineteen (103).

{a} (default decimal)

When exactly would we teach these number bases anyway? Since Bishop mentions number bases in connection with Garelick's word problems, perhaps there should be a unit on traditional word problems and number bases. Students learn about divisibility rules, then use these rules in some  calculations similar to those mentioned in this post, then reinforce the divisibility rules by comparing decimal to other number bases. In Algebra I, such a word problem unit can occur late in the first or early in the second semester. This lesson on number bases might be an interesting one to teach just before/after Thanksgiving/winter break.

Another possible time to teach it is during the lesson on factoring. An interesting question is:

There are square alpha, cube omega, and cube alpha rules? Why isn't there a square omega rule?

The reason is factoring. Square omega would be one less than the square of the base -- that is, in base b, square omega is b^2 -1. But this factors as (b - 1)(b + 1), which is omega * alpha. So any prime available in square omega would be available in omega or alpha.

Sometimes, though, alpha rules are given as square omega -- for example, a frequently given rule for 11 in decimal is to add the last two digits to the rest of the number. The assumption is that anything omega is easier than alpha (where we must remember to alternate signs).

There is one case where square omega is relevant -- odd bases. If b is an odd base, then both b - 1 and b + 1 are even. One of these is a multiple of four (and possibly of a higher power of two as well). If without loss of generality, say this is b - 1. Then b + 1, which is also even, provides one more factor of two than is available using omega only:

• In quinary, omega resolves 4, while square omega resolves 8.
• In nonary, omega resolves 8, while square omega resolves 16.
• In pentadecimal (base 15), alpha resolves 16, while square omega resolves 32.

I think back to my former middle school classes. It might have been interesting to replace one of the Illinois State projects with a different number base for each grade. (And unlike Illinois State projects, number bases would be Bishop-approved). One idea might be to teach the same number base as the grade level -- senary in sixth grade, septimal in seventh grade, octal in eighth grade. Senary might be a good starter base for sixth grade (with divisor rules for 2, 3, omega rules for 5, alpha rules for 7, regular rules for 4, 8, 9, forget about 11). Septimal adds the challenges of an odd base plus a more complicated rule required for 5. In eighth grade, we can discuss the relationship between octal and computers/binary (and perhaps even mention converting to binary for 11).

If we continue this pattern, then we teach nonary in Algebra I. Since nonary is an odd base, we might discuss the connection to difference of squares (divisibility rule for 16). And according to this pattern, the number base for Geometry would be decimal -- but Geometry is all about shapes and figures, not calculations, so there'd be no need for number bases at all. By Algebra II, students would be ready for transdecimals with undecimal (base 11), but by this point it's better just to skip to dozenal. (Of course, traditionalists might prefer octal in Algebra I, since that's their recommended class for eighth grade.)

By the way, I was continuing to think about number bases and Mocha/Atari music. You might think -- is anyone really going to perform the divisibility test for 11 right in the middle of a song just to determine whether a note's Degree is amber (11-limit) or not?

If we stick to the 7-limit, we notice that among the bases that are powers of two, octal resolves 7 much more easily (in the omega) than quaternary or hexadecimal can. (It's also the only base that resolves 19, via the cube alpha, for khaki notes.) The problem is that octal doesn't resolve 5 as easily as the other bases. But an argument can be made that 7 is more important than 5, even in 5-limit music (on the Atari, since we found the importance of 28, and hence 7, for the notes in 16-bit).

Anna on Homeschooling

The most prolific commenter in this thread for once isn't SteveH -- it's Anna. Let's see what Anna has to say in this thread:

Anna:

Excellent question, Mr. Garelick. My daughter actually had a question similar to this for her algebra class this year. Once your students get to ratio and rates, they could tackle one like this:

“A plane flies from Alphaville to Betaville and then back to Alphaville. When there is no wind, the round trip takes 4 hours and 48 minutes, but when there is a wind blowing from Alphaville to Betaville at 100 miles per hour, the trip takes 5 hours. How many miles is the distance from Alphaville to Betaville?”

I don't need to post all of Anna's work -- her answer is 1200 miles. Here's her second comment:

Anna:
It was fun to listen to my son (who is in pre-Algebra) discuss and argue concepts with her on this question. He couldn’t see why the tail wind/head wind didn’t cancel each other out and my daughter had to reason through the problem herself to explain it.

The way I think about this is, since traveling into the wind slows us down, we must spend more time in the headwind than in the tailwind -- which only slows us down even more. After all, if the wind speed had been 250 mph then it would have taken twice as long to get from Betaville to Alphaville as it would in still air. In other words, that one way trip would have taken just as long as the round trip in still air (that is, 4:48). There's no way that the tailwind can make up that time (since even with a tail wind, the trip from Alphaville to Betaville takes more than 0:00). And if the wind speed were more than 250 mph, the one-way trip would take longer than the round trip in still air.

On Square One TV, there was a Mathnet case where George Frankly, one of the main characters, is on trial for robbing a bank. The case against him hinged on whether a plane could fly from the mainland to the island where Frankly is staying, to bring him back to the mainland in time to rob the bank. But it's a windy day. In still air the trip would be 30 minutes each way, and with a tailwind, the plane arrives on the island in 15 minutes. But Kate Monday, Frankly's partner, determines that the return trip to the mainland would take close to an hour -- too late for Frankly to have robbed the bank, thus clearing his name. (They eventually reveal that an impostor robbed the bank.)

OK, that's enough on that. Let's get to Anna's third and final comment:

Anna:
We finally pulled the plug on the public school and decided to go the homeschool route. They jettisoned Geometry in 8th grade 2-3 years ago and in the fall, they will no longer offer separate math courses. It will be 7th grade math, 8th grade math. Middle schoolers will be bussed to the HS if they want Algebra.

She isn't the first traditionalist (again, I assume that everyone who comments on Garelick's traditional math blog is a traditionalist) who is a former public school parent before homeschooling. And the implication is obvious -- if only the public schools would teach math (or other subjects) the way she'd prefer it to be taught, her children would still be at the public school.

For example, she intimates that if the district would make it easy for students to take Algebra I and Geometry in eighth grade, then her children would still attend. (Notice that a compromise presents itself to me immediately -- teach Algebra I to some eighth graders and bus only the Geometry eighth graders to high school, since there would be much fewer of them.)

No, her district isn't San Francisco. She explains that her family lives in Washington State. But Washington, like California, is an SBAC state. This strongly suggests a reason for her district dropping eighth grade Algebra I -- the SBAC that all eighth graders must take at the end of the year is based on Common Core 8, and so it discourages eighth grade Algebra I.

I've mentioned a solution to this many times before -- just as ninth graders in Algebra I don't have to take the math SBAC, neither should eighth (or seventh) graders have to take it. We can even come up with an incentive for schools to place students in Algebra I -- for school accountability purposes, eighth graders in Algebra I are counted as taking the SBAC and earning the same score as they did on the seventh grade test. And so it's good for schools to place students who earn a 4 (or possibly a 3) on the seventh grade SBAC in eighth grade Algebra I, where they don't have to take the test. They're guaranteed to be counted as earning another 4 (or 3), with no possibility of the score dropping.

In fact, I was thinking about state accountability incentives. Before we had the Common Core, California had the CST, and schools were given API scores. The API was determined by assigning each student a score and then taking the average over all students:

• Advanced = 1000
• Proficient = 875
• Basic = 700
• Below Basic = 475
• Far Below Basic = 200

Notice that raising a student from FBB to BB was worth 275 points, while raising a student from Proficient to Advanced was worth only 125 points. And so schools had an incentive to help the lower students more than the higher students. (And this was even in the era when California encouraged schools to place eighth graders in Algebra I.)

Most traditionalists prefer that schools focus on the smart students. And so they'd reverse the differences between the five levels:

• Advanced = 1000
• Proficient = 725
• Basic = 500
• Below Basic = 325
• Far Below Basic = 200

In the old days, the goal for all schools is to attain an API of 800 (majority Proficient). With this scale, we might set the goal at 600, which can be attained by either having a near-majority Proficient or a significant number of Advanced students.

Anna continues to describe another problem with her district -- the textbooks:

Anna:

For middle school, they’ve just adopted Big Ideas by Larson and will change the HS curriculum to that series I think. I previewed the text and it is very heavy on linear algebra to the detriment of other concepts and has a strong focus on discovery. No partial fraction decomposition is presented, it is weak on ratios and rates, barely covers completing squares and factoring, and doesn’t touch optimization.

It's ironic that Anna would mention the Big Ideas text here. My district uses the Big Ideas text as an eighth grade Algebra I text, since it includes eighth grade SBAC material as well. Thus it's strange that her district would adopt the publisher of the ideal eighth grade Algebra I text and yet not offer eighth grade Algebra I class.

Anna reveals her ideal Algebra I text -- Dolciani. OK, that's settles it -- she really is a traditionalist:

Anna:
My oldest started with the 1988 Dolciani Algebra text and then moved to Art of Problem Solving Algebra. I think the AoPS material is great BUT students may need extra psets for review and practice. The online AoPS courses moved very fast. They covered the basics plus complex numbers, exponentials and logs, arithmetic and geometric series (finite and infinite), optimization with linear and quadratic inequalities, and special functions.

Recall that I, as a young student, used a Dolciani Algebra I textbook. This is back in 1993-94, so it's possible that our class used the 1988 text. Meanwhile, I've linked to the Art of Problem Solving website, but that was for college math contests with local Professor Kent Merryfield. Apparently, AoPS is also used by traditionalist parents disillusioned with public schools.

Notice that Anna doesn't distinguish between Algebra I and II here. So Anna criticizes Big Ideas because it "doesn't touch optimization." I assume that by "optimization," she means what the U of Chicago Algebra II text calls "linear programming" (Lessons 5-8 and 5-9 of that text), especially since she mentions optimization in connection with inequalities later on (but some of those were quadratic inequalities, which explains why she avoids the phrase "linear programming"). But I don't recall seeing "optimization" in the Dolciani text either -- the first time I'd seen linear programming is in that very U of Chicago text (which indeed was the Algebra II text I used as a young student).

And "partial fractions" don't appear in Dolciani either. The first time I've ever seen partial fraction decomposition was in AP Calculus BC (since the sole purpose of partial fractions is to make certain integrals easier). And so unless I'm misunderstanding what she's saying, Anna criticizes Big Ideas for lacking lessons that don't appear in her beloved Dolciani Algebra I text (based on my own memories of that same text).

Here is Garelick's reply to Anna:

Garelick:
I didn’t care for the Big Ideas series; I skipped the discovery lessons and just focused on the direct instruction lessons. So I gave up on it and use Dolciani’s algebra textbook from 1962. I was able to get them for very low cost via Amazon. The students like them a lot better than Big Ideas, which I find interesting.

Once again, Garelick likes to point out that his students actually prefer the direct instruction lessons of Dolciani to the discovery lessons of Big Ideas. Again, this sounds incredible, but I must respect the fact that Garelick is referring to his real students. Sometimes I wonder whether or not he's talking about real students. (Well, he did say that they kept reaching for a calculator -- OK, that definitely sounds like real students!)

SteveH on Governing Equations

Well, I guess I saved the "best" for last. Let's see what SteveH has to say:

SteveH:

Back when I taught college math and computer science, I realized that when I prepared to explain how to do a problem, I consolidated all of my analysis into a final (and clear) process to explain. My error was thinking that my final solution process was helpful to them when they approached a new problem – as if they should be able to get immediately to that level of problem understanding. They might have understood exactly what I did, but that didn’t mean that it would be helpful to them on a new problem. I evolved into giving them a more general process that starts with finding the governing equation, like D=RT, and then defining unique and simple equations, like

550 = R1 * T1
and
425 = R2 * T2

This isn't the first time that SteveH has mentioned "governing equations," but this is the first time that I notice his full explanation.

I've heard of setting up tables or charts containing such information, but SteveH likes to jump directly into setting up equations.

He's also criticized the Common Core for trying to avoid equations. For some problems, such as the Pythagorean Theorem, even Common Core texts recommend using a^2 + b^2 = c^2 -- the governing equation -- since there is no other reasonable way to solve such a problem. But for other problems, particularly proportions, the Core goes out of its way to avoid equations.

SteveH:
I told them to define proper variables and write down legal equations. Let the math do the thinking for them. This is also psychologically good because they can quickly get partial credit even if they somehow get confused about wind being completely additive and subtractive to rate. Is it exactly? I can imagine my son questioning that. They might cancel out, but what cancels out might not be the free wind speed.

SteveH claims that writing equations means not having to think about it. But if this is true, why does the appearance of an equation in a book cut its sales in half? I'd argue is because readers don't want to think about equations -- that is, equations force readers to think more, not less.

SteveH:
Math will explain it all to you, but that doesn’t seem to be the “understanding” many educators are looking for. They seem to think that it’s a concept/gut level approach – which can get you into big trouble.

And of course by "many educators," SteveH includes proponents of the Common Core. To SteveH, writing equations should replace having to explain one's own answers on the Common Core.

Conclusion

Earlier in today's post, I asked the question:

Why is Garelick's distance problem considered "rich," but not the SBAC price problem? Conversely, why does the Common Core approve of this price problem, but not Garelick's distance problem?

Maybe that's the reason -- Garelick's distance problem is best explained by equations, while the Core's price problem is explained by discourse. Each type of word problems reflects one of the two views of what "understanding" really is.

Van Brummelen Chapter 6 The Modern Approach: Oblique Triangles

Table of Contents

1. Van Brummelen Chapter 6 The Modern Approach: Oblique Triangles
2. Exercise 6.4
3. Exercise 6.5
4. Exercise 6.13
5. Exercise 6.16
6. Another Error on the Pappas Calendar?
7. A Line and Its Translation Image Are Parallel?
8. Conclusion

Van Brummelen Chapter 6 The Modern Approach: Oblique Triangles

Chapter 6 of Glen Van Brummelen's Heavenly Mathematics is called "The Modern Approach: Oblique Triangles." Here's how it begins:

"So far spherical trigonometry hasn't looked much like the plane theory we learned in high school. However, the parallels often lie just below the surface."

In this chapter, Van Brummelen writes about applying spherical trig to oblique triangles -- that is, triangles that aren't right triangles. In plane trig, we generally solve oblique triangles by using the Laws of Sines and Cosines. The author has already discussed the spherical Law of Sines, and so this chapter is all about the spherical Law of Cosines.

But first, he writes about the familiar planar cosine law. He gives the formula:

c^2 = a^2 + b^2 - 2ab cos C

and points out the apparent link to the Pythagorean Theorem. He then asks, who was the first person to discover the Law of Cosines:

"I've been asked this question before, and it sounds like the answer should be a simple fill-in-the-blank."

Van Brummelen makes the argument that the discoverer of the cosine law was none other than Euclid, despite his lacking a cosine ratio. He points out a certain proposition in Book II of The Elements, and since we've linked to Euclid before, we might as well do so again here.

Proposition 12.

In obtuse-angled triangles the square on the side opposite the obtuse angle is greater than the sum of the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle.

Proposition 13.

In acute-angled triangles the square on the side opposite the acute angle is less than the sum of the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle, namely that on which the perpendicular falls, and the straight line cut off within by the perpendicular towards the acute angle.

Van Brummelen decides to focus on II.13, the acute angle case. By the way, he points out that some people refer to Book II as "geometric algebra," and indeed, David Joyce (the author of the above links) does refer to it as such.

Now Van Brummelen explains that II.13 really is the Law of Cosines:

"In Triangle ABC with an acute angle as C and a perpendicular dropped from A onto BC (defining D), c^2 = a^2 + b^2 - 2AD * BC. But BC = a and AD = b cos C, so what Euclid is 'really' saying is: c^2 = a^2 + b^2 - 2ab cos C."

And the author provides Euclid's full proof, written in modern algebraic notation:

Proof:
Let x = CD, which we remember is equal to b cos C. Then Euclid asserts:

a^2 + x^2 = 2ax + (a - x)^2,

which we may verify with a little algebra, as long as we close our ears to the historians' howls of protest. (Euclid himself appeals to a previous geometric theorem at this point.) We add DA^2 to both sides:

a^2 + x^2 + DA^2 = 2ax + (a - x)^2 + DA^2.

Apply the Pythagorean Theorem to both sides; we get:

a^2 + b^2 = 2ax + c^2.

Finally, a bit of rearrangement takes us to:

c^2 = a^2 + b^2 - 2ax,

which is what we wanted to prove. QED

In fact, we see that Joyce writes something similar under the link for II.12, the obtuse angle case:

"This conclusion is very close to the law of cosines for oblique triangles, since AD equals –b cos A, the cosine of an obtuse angle being negative. Trigonometry was developed some time after the Elements was written, and the negative numbers needed here (for the cosine of an obtuse angle) were not accepted until long after most of trigonometry was developed. Nonetheless, this proposition and the next may be considered geometric versions of the law of cosines."

Notice that while most high school Geometry classes nowadays don't teach or prove the Law of Cosines, a few do. This version of Euclid's proof is indeed suitable for an Honors Geometry course.

Now Van Brummelen turns next to the spherical analog of this law. In his figure 6.2, he draws the same diagram on the spherical, adding a new variable h = AD, the altitude of the triangle that divides it into two right triangles. Here is the author's derivation:

This time Pythagoras looks a bit different:

cos b = cos h cos x and cos c = cos h cos (a - x)

Solving both expressions for cos h and setting them equal to each other has the advantage of removing the reference to the undesirable h:

cos b/cos x = cos c/cos (a - x).

From here we solve for cos c, apply the cosine subtraction law, and let the algebra run its course:

cos c cos x = cos b(cos a cos x + sin a sin x)
cos c = cos a cos b + sin a cos b tan x.

To get rid of the tan x term we apply identity four in the first column of the Napier's Rules theorems [from my Fourth of July post -- dw]. This takes us directly to the spherical Law of Cosines:

cos c = cos a cos b + sin a sin b cos C.

Van Brummelen points out that just as the planar Law of Cosines is a generalization of the planar Pythagorean Theorem, the spherical law generalizes its respective Pythagoras. The cos a cos b part comes from spherical Pythagoras, while by the locality principle, sin a and sin b approach a, b. The only thing from the planar law is the 2 -- make that the negative 2. But it's easy to see where the -2 went -- recall that cos c doesn't approach c^2, but 1 - (1/2)c^2. So the -1/2 factor swallows up the -2 we expect to see in the Law of Cosines.

Now Van Brummelen demonstrates applications of the Law of Cosines. He points out that many texts use the Law of Cosines to compute distances on common sea routes. In his first example, students are asked to find the distance traveled from Queenstown/Cobh, Ireland (51.78N, 8.18W) to Sandy Hook, New York Harbor (40.47N, 74.13W). Notice that this Sandy Hook is in New Jersey, and so it has nothing to do with the Sandy Hook school shooting in Connecticut (the two Sandy Hooks are about a hundred miles apart). This text is from the "Golden Age of Textbooks" in 1895, seventeen years before the the voyage of the Titanic along this route (which is itself a century before the tragedy at the other Sandy Hook).

As the author explains, the key to this problem is to join both New York (Y in figure 6.3) and Queenstown (Q) to the North Pole (N). Then YN = 90 - 40.47 = 49.53 and QN = 90 - 51.78 = 38.22, while the angle at N is the difference between the two longitudes, 65.95. We now apply the Law of Cosines, letting the C in the formula be the angle at the North Pole, and we find:

cos QY = cos 38.22 cos 49.53 + sin 38.22 sin 49.53 cos 65.95.

Thus QY = 45.43, which we multiply by 60 to get 2726 nautical miles, or (using the 1.15078 conversion factor) 3137 statute miles (that is, what we ordinarily call "miles"). The author points out that in practice, the ship would have traveled 2783 to 2889 miles, since Newfoundland and Nova Scotia get in the way of the true great circle route.

Van Brummelen's next example is the ASA case. He points out that in Euclidean ASA cases, we don't use the Law of Cosines -- instead, we use the fact that the angles of a triangle add up to 180 to find the third angle and then use the Law of Sines. The problem is that the angles of a triangle don't add up to 180, or any fixed value, in spherical geometry. Instead, the author uses:

Polar Duality Theorem: The sides of a polar triangle are the supplements of the angles of the original, and the angles of a polar triangle are the supplements of the sides of the original.

We discussed this theorem back in my Eid-al-Fitr post on June 15th, as well as in our discussion of spherical geometry from Legendre in years past. Anyway, Van Brummelen applies this to the Law of Cosines to obtain the "Law of Cosines for Angles."

cos(180 - C) = cos(180 - A)cos(180 - B) + sin(180 - A)sin(180 - B)cos(180 - c)
cos C = -cos A cos B + sin A sin B cos c.

Here's the author's example -- suppose we fly from Vancouver to Edmonton on a great circle route, measuring the distance we travel as well as the headings at departure and arrival. From this information alone, an application of the Law of Cosines for Angles gives us the difference in longitude between two cities, and the values of both latitudes follow immediately. The distance from Vancouver to Edmonton (figure 6.4) is VE = 507 statute miles, or 440.9 nautical miles, which corresponds to 7.35 degrees. We leave Vancouver with a heading 50.7 east of north, and arrive in Edmonton with a heading 58.22 east of north. Joining our two cities to the North Pole as before, we have Angle V = 50.7 and Angle E = 121.78. Apply the Law of Cosines for Angles, letting c be the journey from Edmonton:

cos N = -cos V cos E + sin V sin E cos VE.

This gives us Angle N = 9.6, the difference in longitude between Vancouver and Edmonton. Now that we have all three angles, we calculate the two latitudes using the Law of Sines; Vancouver works out to 49.3, and Edmonton to 53.6.

Van Brummelen tells us that with the Law of Cosines for Angles, we can even solve the AAA case -- which is impossible in plane trig. But, as warns us, not even on the sphere can we avoid the ambiguous case of SSA. Here's his example:

"A ship leaves Honolulu (latitude 21.31N) traveling towards Tokyo (latitude 35.7N) on a great circle route with a heading of 60.5 west of north. What will be the length of the voyage, in miles?"

From figure 6.5 we see that we are in an SSA situation. Unfortunately, there are two endpoints that satisfy the givens. The first is X, a spot in the middle of the Pacific Ocean. If you extend BX past X, it will eventually reach its northernmost point, and then start heading slightly southward, crossing the 35.7N latitude circle again at Tokyo.

The mathematical face of this ambiguity appears immediately when we use the Law of Sines to find Angle A: sin A = sin a sin B/sin b = 0.9985, so A = 86.84 or 180 - 86.84 = 93.16. We can tell from our diagram that we want the smaller angle 86.64, but if the ambiguity had passed without notice, we might have run out of fuel at X, with no land in sight.

We now know a, b, A, and B; but finding c and C proves surprisingly awkward. It is possible to use the Law of Cosines (rearranging the letters in the diagram appropriately), but only by changing sin c into sqrt(1 - cos^2 c), eventually requiring us to solve a quadratic equation for cos c. A cleaner approach would be nice.

It's at this point when Van Brummelen suddenly brings up the feud between Newton and Leibniz -- and I already mentioned this in my Fourth of July post. So let's just skip to the point he's making -- two mathematicians developed the "cleaner approach." One is Napier, and the other is the French astronomer Jean-Baptiste Delambre. Here the author derives the first of "Delambre's analogies":

Derivation (from Van Brummelen):
Delambre's analogies are usually demonstrated disappointingly through algebraic manipulation of various known identities. It is not easy to approach them geometrically, but at least one textbook (Isaac Todhunter's classic) gives it a go. The argument, based on Delambre's original demonstration, begins with Triangle ABC in figure 6.6. We begin by bisecting AB at M and drawing a perpendicular upward [towards V]. Next, bisect BCP to form CV. Drop arcs perpendicular to CP and CB, defining P and Q respectively. Finally, join AV and BV. We show first that Triangles AVP and BVQ are equal, by matching three elements of the triangles. They both contain a right angle; since MV is perpendicular to AB we know that AV = BV; finally, since Angle BCP was bisected by CV we know that PV = QV. So the two triangles are equal.

[In a footnote, the author realizes that he just "proved" the triangles "congruent" by SSA -- and it was his own students who told him that he was an, um, backwards-SSA, for making this mistake. The fact that these are right triangles is of no help, since HL is also invalid in spherical geometry. He leaves it to readers to figure out how to patch this flaw -- now I'm wondering, is the U of Chicago's "SsA" theorem valid in spherical geometry. This would require that the corresponding sides opposite the right angles be longer than the other given congruent sides -- that is, AV > PV and BV > QV. These inequalities certainly appear true from the diagram.]

This fact allows us to label the angles at A and B as we have done in the diagram. The angles at the base of the original triangle are Angle A = y + x and Angle B = y - x; a little algebra yields x = (A - B)/2 and y = (A + B)/2. We are in a similar algebraic situation with respect to sides a and b of the original triangle: a = BQ + CQ and b = AP - CP, but BQ = AP (since triangles AVP and BVQ are equal) and CQ = CP, so BQ = (a + b)/2 and CQ = (a - b)/2. Turning to the angles at the top of the diagram, we know that Angle AVB = PVQ since both are composed of AVQ and equal angles. Cutting these angles in half, we have Angle AVM = PVC.

We're in the home stretch of the proof. Consider the two right-angled triangles AVM and PVC, with equal angles at V. Apply Geber's Theorem [Fourth of July post] to both and bring the results together:

sin VAM cos AM = cos AVM = sin PCV cos CP.

Substituting each of the angles for the values from the elements of the original triangle and rearranging the terms, we finally arrive at Delambre's first analogy:

sin(1/2)(A + B)/cos(1/2)C = cos(1/2)(a - b)/cos(1/2)c.

Here are the other three analogies:

sin(1/2)(A - B)/cos(1/2)C = sin(1/2)(a - b)/sin(1/2)c,
cos(1/2)(A + B)/sin(1/2)C = cos(1/2)(a + b)/cos(1/2)c, and
cos(1/2)(A - B)/sin(1/2)C = sin(1/2)(a + b)/sin(1/2)c.

The remaining Napier's analogies are found by dividing Delambre's analogies by one another:

tan(1/2)(A + B)/cot(1/2)C = cos(1/2)(a - b)/cos(1/2)(a + b),
tan(1/2)(A - B)/cot(1/2)C = sin(1/2)(a - b)/sin(1/2)(a + b),
tan(1/2)(a + b)/tan(1/2)c = cos(1/2)(A - B)/cos(1/2)(A + B), and
tan(1/2)(a - b)/tan(1/2)c = sin(1/2)(A - B)/sin(1/2)(A + B).

The author uses Napier's third analogy to solve the Honolulu-to-Tokyo problem:

tan(1/2)(68.69 + b54.3/tan(1/2)c = cos(1/2)(86.84 - 60.5)/cos(1/2)(86.84 + 60.5)

from which we have tan(1/2)c = 0.5317, and so c = 28 degrees, equal to 1680 nautical miles or 1933 statute miles. We can find C easily using any relevant identity.

Van Brummelen ends the chapter as follows:

"There is plenty more to spherical trigonometry than we've seen so far, but this concludes the basic theory needed for solving triangles. In the remaining three chapters we'll see some special topics and applications."

Exercise 6.4

A ship sailing on a great circle from Ceylon to Madagascar crosses the meridian of 79E longitude bearing S50W. After sailing 2060 nautical miles farther, it crosses the meridian of 52E longitude. Find its latitude and the bearing of its course at this point and its latitude at the first point. [Brink 1942, 41].

(Yes, this is definitely a 1942 question, since "Ceylon" is now known as Sri Lanka.)

Let's set up the triangle. We'll use C for "Ceylon" and M for Madagascar, and as usual we'll label the North Pole N. Here are the givens: the angle at C is 130 (since that's 50 degrees west of south, not north) and CM is 2060/60 = 34+1/3 degrees. The angle at N is the difference between longitude, 27.

This is an AAS situation. We must first use the Law of Sines to find NM and then Napier's analogies to find the remaining parts:

sin NM/sin C = sin CM/sin N
sin NM/sin 130 = sin(34+1/3)/sin 27
sin NM = sin(34+1/3)sin130/sin 27
sin NM = 0.95168
NM = 72.1 or NM = 180 - 72.1 = 107.9

Hmm, so there are two possible answers. I suspect that the answer is 107.9, because Madagascar is in the southern hemisphere. The actual latitude is 90 - 72.1 = 17.9S.

Now we use Napier's analogies to find M and NC. We could use any of the analogies, but let's follow Van Brummelen and use the third analogy:

tan(1/2)(107.9 + 34.3)/tan(1/2)NC = cos(1/2)(130 - 27)/cos(1/2)(130 + 27)
tan(1/2)NC = tan 71.1 cos 78.5/cos 51.5
tan(1/2)NC = 0.93587
(1/2)NC = 43.1
NC = 86.2

So the latitude of Ceylon is 90 - 86.2 = 3.8N.

Now all that remains is to find the angle at M. Van Brummelen tells us that we can use any formula at this point, so let's use the Law of Cosines for Angles:

cos M = -cos N cos C + sin N sin C cos NC
cos M = -cos 27 cos 130 + sin 27 sin 130 cos 86.2
cos M = 0.59574
M = 53.4

This means that the bearing from Madagascar back to Ceylon is N53.4E. The ship is now heading away from Ceylon, so the bearing at Madagascar is S53.4W.

Therefore our answers are Madagascar 17.9S, Ceylon 3.8N, final bearing S53.4W.

I decided to check my answers by looking up the latitudes of Sri Lanka and Madagascar. I find that the latitude of Sri Lanka is about 7N and the latitude of Madagascar is about 19S. So our answers are fairly close, though I'd prefer to be a bit closer for Sri Lanka.

Also, I wonder whether there's a real way to assert that Madagascar is closer to 18S than 18N, which can't be determined from sin NM = 0.95168. If we assumed that Madagascar is at 17.9N, then we obtain 43.6N for the latitude of Ceylon (which is way off) and S35.6W for the final bearing. All of the identities are satisfied for this second triangle. After all, AAS is not a valid congruence theorem in spherical geometry -- therefore AAS, like SSA, leads to an ambiguous case.

Exercise 6.5

Charles A. Lindbergh flew his plane The Spirit of St. Louis on the great circle route from New York (40.75N, 73.97W) to Paris (48.83N, 2.33E). He left New York at 7:52 AM (Eastern Standard Time) on May 20th, 1927, and arrived at Paris the next day at 5:24 PM (Eastern Standard Time). What was his average speed for the flight? [Rosenbach/Whitman/Moskovitz 1937, 332]

Once again, we place N on top of the world, Y at the Big Apple, and P at the City of Light. The angle at N is 73.97 + 2.33 = 76.3, NY = 90 - 40.75 = 49.25, and NP = 90 - 48.83 = 41.17.

Let's use the Law of Cosines:

cos YP = cos NY cos NP + sin NY sin NP cos N
cos YP = cos 49.25 cos 41.17 + sin 49.25 sin 41.17 cos 76.5
cos YP = 0.60779
YP = 52.57 degrees

This equals 3154.2 nautical miles or 3629.8 statute miles. The famous transatlantic flight lasted for 33 hours and 32 minutes (33+32/60 hours), and so the average speed is 108.2 mph. A modern jet travels about four or five times as fast as Lindbergh's flight, but notice that this problem comes from a text written a mere decade after the famous flight.

There's no need to calculate either Angle Y or Angle P, since the question doesn't ask for the heading at either point.

Exercise 6.13

Napier's analogies were sometimes used as a method of solving geographical problems. Suppose that we know the latitudes of Edmonton (53.6N) and Vancouver (49.3N), as well as their difference in longitude (9.6). The Law of Cosines is the obvious choice here, but instead use Napier's analogies to determine the headings of a great circle path between the cities, and then use some other identity to determine the distance.

Well, let's look at the first two identities again:

tan(1/2)(A + B)/cot(1/2)C = cos(1/2)(a - b)/cos(1/2)(a + b),
tan(1/2)(A - B)/cot(1/2)C = sin(1/2)(a - b)/sin(1/2)(a + b).

The first equation gives the value of A + B, while the second provides the value of A - B. So we can use these two to find E + V and E - V, and then these two equations are solved easily for E and V:

tan(1/2)(E + V)/cot(1/2)9.6 = cos(1/2)(40.7 - 36.4)/cos(1/2)(40.7 + 36.4)
tan(1/2)(E + V) = cos 2.15/(cos 38.55*tan 4.8)
tan(1/2)(E + V) = 15.217
(1/2)(E + V) = 86.24

tan(1/2)(E - V)/cot(1/2)9.6 = sin(1/2)(40.7 - 36.4)/sin(1/2)(40.7 + 36.4)
tan(1/2)(E - V) = sin 2.15/(sin 38.55*tan 4.8)
tan(1/2)(E - V) = 0.71689
(1/2)(E - V) = 35.64

(1/2)(E + V) = 86.24
(1/2)(E - V) = 35.64
E = 121.88
V = 50.6

Since in the original problem, we're traveling from Vancouver to Edmonton, we read the heading at Vancouver as N50.6E, and the heading at Edmonton becomes 180 - 121.88 = N58.12E.

We must use some other identity to find the distance between the two cities -- and Van Brummelen specifically tells us not to use the Law of Cosines. So let's use the Law of Sines instead:

sin VE/sin 9.6 = sin 36.4/sin 50.6
sin VE = sin 9.6 sin 36.4/sin 50.6
sin VE = 0.12806
VE = 7.36 degrees = 441 nautical miles = 508 statute miles

We double-check the answers with figure 6.4 from the book. In that figure, V = 50.7, E = 121.78, and VE = 507 miles, so our answers are reasonable.

Exercise 6.16

(a) Show that cot a sin b = cos b cos C + cot A sin C.

Proof:
We mimic Van Brummelen's derivation of the Law of Cosines. We start by dividing ABC into two triangles and then applying the ten Chapter 5 formulas to these two triangles.

Notice that the statement to be proved starts with cot a sin b. We don't know a difference formula for cotangent, but we do know one for sine. Along with the fact that B doesn't appear in the goal formula, this indicates that we should drop the perpendicular from B to form Triangles ABD and CBD. We'll let CD = x and AD = b - x and choose the formula involving sin(b - x) to use the difference formula (h, again, equals the height, which this time is BD):

sin(b - x) = tan h cot A
sin b cos x - cos b sin x = tan h cot A

Applying this same formula to Triangle CBD gives:

sin x = tan h cot C

Dividing these two formulas gives:

sin b/tan x - cos b = cot A/cot C

We apply another formula to CBD involving tan x:

cos C = tan x cot a
tan x = cos C/cot a
1/tan x = cot a/cos C

We substitute this in to eliminate tan x:

sin b cot a/cos C - cos b = cot A/cot C

Multiplying by cos C gives:

sin b cot a - cos b cos C = cot A sin C (since cos C/cot C = cos C tan C = sin C)
cot a sin b = cos b cos C + cot A sin C

which is exactly what we want to prove. QED

(b) Use the above result (with the variables rearranged appropriately) in conjunction with Triangles AVP and CVP in the derivation of Delambre's first analogy (figure 6.6) to demonstrate Napier's second analogy.

Proof:
Let's begin by repeating the information from figure 6.6, which I can see, but you can't.

From earlier:
The argument, based on Delambre's original demonstration, begins with Triangle ABC in figure 6.6. We begin by bisecting AB at M and drawing a perpendicular upward [towards V]. Next, bisect BCP to form CV. Drop arcs perpendicular to CP and CB, defining P and Q respectively. Finally, join AV and BV. We show first that Triangles AVP and BVQ are equal, by matching three elements of the triangles. They both contain a right angle; since MV is perpendicular to AB we know that AV = BV; finally, since Angle BCP was bisected by CV we know that PV = QV. So the two triangles are equal. This fact allows us to label the angles at A and B as we have done in the diagram. The angles at the base of the original triangle are Angle A = y + x and Angle B = y - x; a little algebra yields x = (A - B)/2 and y = (A + B)/2. We are in a similar algebraic situation with respect to sides a and b of the original triangle: a = BQ + CQ and b = AP - CP, but BQ = AP (since triangles AVP and BVQ are equal) and CQ = CP, so BQ = (a + b)/2 and CQ = (a - b)/2.

(We'll cut it off there, since that's all the given information we need to solve the problem.)

We're instructed to apply the formula from (a) to Triangles AVP and CVP, so let's do so:

cot VP sin CP = cos CP cos CPV + cot VCP sin CPV
cot VP sin AP = cos AP cos APV + cot VAP sin APV

First of all, from the diagram, there's a right angle at P. So CPV and APV are right angles, which means cos CPV = cos APV = cos 90 = 0 and sin CPV = sin APV = sin 90 = 1:

cot VP sin CP = cot VCP
cot VP sin AP = cot VAP

Now let's plug in the values from the info above -- CP = CQ = (a - b)/2 and AP = BQ = (a + b)/2. In the diagram we see that Angle VAP = x, so we obtain VAP = x = (A - B)/2 from above.

That leaves us with Angle VCP. From the diagram, BCP and BCA (which is C from the original triangle ABC) form a linear pair, so BCP = 180 - C. And from the info above, ray CV bisects BCP, and so VCP = (180 - C)/2. Now we can fill in the equations:

cot VP sin (1/2)(a - b) = cot (1/2)(180 - C)
cot VP sin (1/2)(a + b) = cot (1/2)(A - B)

Of course, there's still one value we didn't substitute into the equation -- VP. But that doesn't matter, because we're just going to divide the equations:

sin(1/2)(a - b)/sin(1/2)(a + b) = cot(1/2)(180 - C)/cot(1/2)(A - B)
sin(1/2)(a - b)/sin(1/2)(a + b) = tan(1/2)(A - B)/tan(1/2)(180 - C)

But notice that tan(1/2)(180 - C) = tan(90 - (1/2)C) = cot (1/2)C, so we have:

sin(1/2)(a - b)/sin(1/2)(a + b) = tan(1/2)(A - B)/cot(1/2)C,

which, as you may have noticed, is exactly the formula we set out to prove, Napier's second analogy:

tan(1/2)(A - B)/cot(1/2)C = sin(1/2)(a - b)/sin(1/2)(a + b). QED

Another Error on the Pappas Calendar?


Tomorrow on her Mathematics Calendar 2018, Theoni Pappas writes:

1, 2, 3, 10, 11, 12, ____, 100, 101, 102, ...

It appears that Pappas intends this to be the sequence of natural numbers base 4 -- after all, the list begins 1, 2, 3, and then skips up to 10. And so the next value in the sequence after 12 should be 13 -- and of course, tomorrow's date is the thirteenth (albeit in decimal -- triskaidekaphobes beware on Friday the 13th!).

{4} (default quaternary)

But the next number after 13 isn't 100 -- it's 20. The sequence should have read:

1, 2, 3, 10, 11, 12, _13_, 20, 21, 22, 23, 30, 31, 32, 33, 100, 101, 102, ...

Thus assuming that quaternary is intended here, Pappas makes another error. Of course, it's possible that Pappas doesn't intend this sequence to be the quaternary naturals. But I performed a OEIS search and no sequence in the database begins with 1, 2, 3, 10, 11, 12, 13, 100.

{a} (default decimal)

The closest I found is:

https://oeis.org/A007961

This is the "square base" with a ones place, fours place, nines place, 16's place, and so on. But even then, we don't skip directly from 13 (seven) to 100 (nine) -- we still need 20 (eight).

Also, notice that any list of numbers can be the start of a sequence. We could take the Pappas list of ten numbers:

1, 2, 3, 10, 11, 12, _13_, 100, 101, 102, ...

and find an interpolating polynomial of degree nine that passes through these points. The Wolfram online calculator finds the following polynomial:

23x^9/2340 - 27x^8/80 + 12967x^7/1890 - 28081x^6/360 + 587903x^5/1080 - 1731947x^4/720

     + 5413807x^3/810 - 501803x^2/45 + 6300671x/630 - 3600

By the way, the next number in the sequence after 102 is 7543. But seriously, I doubt this is what Pappas has in mind.

As I wrote back in my May 25th post, number bases is one of my favorite topics. Base four, while superior to base two, is still considered a bit too small to be human scale.

But it's precisely because four is so small that certain SPD divisibility tests that are impractical in human scale bases become usable in quaternary. We see that 4^3 = 64, and so its neighbors 63 and 65 provide SPD tests for 7 and 9 (cube-omega) as well as 13 (cube-alpha), and all of these require memorizing at most ten multiples. We know that 11 resists being resolved by SPD, but on May 25th I mentioned how 11 can be resolved in binary -- and we can use that same test in quaternary by first converting it to binary (two digits at a time). This means that we have divisibility rules for all numbers up to the square of the base 4.

Even though base 4 isn't human scale, its square 16 is (albeit on the upper end of) human scale. So these divisibility tests are available in hexadecimal by first converting it two quaternary (two digits at a time). Of course, all of these tests are esoteric -- the three most common primes 2, 3, and 5 all have intuitive tests in both 4 and 16.

One interesting place where these divisibility rules might come into play is Mocha/Atari music, if the degree is written in hexadecimal (which is logical, since this is on a computer). Then knowing whether a number is divisible by 7, 11, 13, or 17 will help us determine whether the color of the note is red, amber, ocher, or umber. That's all I'm going to say about this -- don't worry, this won't turn into yet another music post.

A Line and Its Translation Image Are Parallel?

By now, I wanted to post a complete proof of "a line and its translation image are parallel" -- the proof of which I made the Fermat-like announcement last year ("I have the proof, but I can't fit it within the margins of this blog!").

Note: This is a correction of an earlier error.

To complete the proof, we begin with two important lemmas about reflections:

Line Perpendicular to Mirror Theorem:
If a line is perpendicular to the mirror, then it is invariant (that is, its image is itself).

Line Parallel to Mirror Theorem:
If a line is parallel to the mirror, then its image is also parallel to the mirror.

I've proved the first lemma in years past, so let's repeat the proof:

Proof (Line Perpendicular to Mirror):
We're given that l, the line, is perpendicular to m, the mirror. Our goal is to show that if P lies on l, then P', the reflection image of P, also lies on l.

Let M be a point where l and m intersect, and measure MP. Then choose P' to be the point on l at a distance MP from M, but on the other side of M. Then m is the perpendicular bisector of PP', which, by definition of reflection, means that P' is the image of P. Thus P' lies on P, and therefore l is its own reflection image. QED (Line Perpendicular to Mirror)

Proof (Line Parallel to Mirror):
This is a new proof, since I don't like my previously posted proofs. Instead, we use the U of Chicago definition of parallel -- two lines are parallel if they have no point in common or are identical. Thus the statement l is parallel to l' is equivalent to if l and l' have a point in common, then l and l' are the same line."

So let l | | m, and let l' be the reflection image of l across m. So our goal is to prove l | | l' -- that is, we show that if l and l' have a point P in common, then they are the same line:

Thus P lies on l, and P lies on l'. The question is, where does P', the image of P, lie? Well, we know that P' must lie on l', since l' contains the image of every point on l (definition of image of a line).

Case I:
P and P' are distinct points on l'. By definition of reflection, m is the perpendicular bisector of PP', so that m is perpendicular to l'. Therefore by Line Perpendicular to Mirror, l' is invariant. But to say that l' is invariant is to say that l itself is invariant. (You might ask, if T is an isometry, why does the fact that T(T(l)) = T(l) imply that T(l) = l? Well, according to the Reflection Theorem, we can show that if T(x) = T(y) then it follows that x = y. It follows from the idea that all isometries are invertible, since these are all one-to-one correspondences. The betweenness and collinearity clauses imply that they are invertible for lines as well as points -- no two distinct lines can ever have the same image.) So l and l' must be the same line.

Case II:
P and P' are the same point (known to be on both l and l'). But the only fixed points (that is, points such that P = P') of a reflection are points on the mirror. Thus l and m have a point in common -- but we are given that l | | m, so l and m are the same line -- and so l', the reflection image of l, must also be that same line. QED (Line Parallel to Mirror)

Notice that this proof reads more easily as an indirect proof -- we assumed that l is not parallel to l' and derived a contraction, therefore l is parallel to l'. The assumption I make here is that the students haven't learned about indirect proofs yet.

An important result based on these two lemma is the idea that if P is a fixed point of a certain isometry -- either a reflection of the composite of two reflections in distinct mirrors -- then P must lie on (both of) the relevant mirror(s). This is clear if there is only one mirror. If there are two mirrors, then suppose the first mirror maps P to some point P'. Then either P = P' and the point must lie on the mirrors, or else the second mirror would have to map P' back to P. But the two distinct mirrors can't both be the perpendicular bisector of PP'. Therefore P must be a fixed point of both reflections -- that is, it must lie on both mirrors.

Now let's try the main proof:

Theorem:
A line and its translation image are parallel.

Proof:
Let l be the preimage and l' its image. Our goal is to show l | | l'. Let m and m' be the mirrors of the two reflections whose composite is the translation. By definition of translation, m | | m'.

Again, our goal is to show l | | l' -- that is, if some point P lies on both l and l', then l and l' must be the same line. We follow the same pattern as the Line Parallel to Mirror Theorem -- we know that P', the translation image of P, must lie on l' since l' contains the translation image of every point on l.

Now consider a line through P perpendicular to the first mirror m. By Line Perpendicular to Mirror, this line is invariant with respect to the first mirror. But any line perpendicular to the first mirror must be perpendicular to the second mirror (which is where this proof fails in hyperbolic geometry). So this same line is also invariant with respect to the second mirror. Thus the line is invariant with respect to both mirrors, hence it's invariant with respect to the composite translation. By the definition of invariant, if P lies on this line, then so does P'.

(Notice that P and P' must be distinct. As the composite of two reflection, a translation can have a fixed point only if it lies on both mirrors -- but the mirrors are parallel, so no point can lie on both of the mirrors. Thus a translation has no fixed points. If P lies on one of the mirrors, it can't lie on the other, and so PP' is still invariant with respect to that other mirror.)

That is, l', a line containing both P and P', is invariant. And so by the same reasoning as in the Line Parallel to Mirror proof, l is invariant -- that is, l and l' are the same line. QED

You might ask, how exactly would be prove the Perpendicular to Parallels Theorem -- the one step that fails in hyperbolic geometry? In some old posts, I actually assumed Perpendicular to Parallels as a postulate, but this is clumsy. We prefer to use the parallel postulate that appears in most Geometry texts, namely Playfair's Postulate. (Through a point not on a line, there exists at most one line parallel to the given point.)

As it turns out, we can prove Perpendicular to Parallels directly from Playfair:

Proof:
Let a, b, t be lines such that a | | b and a perpendicular to t. We can say that t intersect a at A, and let t intersect b at B. (That t and b even intersect at all derives from Playfair -- if t | | b, then t and a are two distinct lines through A parallel to b, when Playfair tells us that there's at most one such line.)

Now let m be the perpendicular bisector of AB. Consider the reflection across m. This reflection maps A to B and t to itself by Line Perpendicular to Mirror. And since a | | m, by Line Parallel to Mirror, it maps a to a line parallel to a through B. By Playfair, at most one such line exists -- and that line must be exactly b. Thus our reflection maps t to itself and a to b.

Reflections preserve angle measure. Since the angle between t and a is 90, therefore the angle between their images t and b is also 90. QED

And from "a line is parallel to its translation image," we can finally prove the Corresponding Angles Consequence ("if two parallel lines are cut by a transversal,...") It turns out that this is proved exactly like Perpendicular to Parallels, except with a translation mapping a to b rather than a reflection (namely the translation that maps A to B, which leaves t invariant). This was our original goal -- to prove Corresponding Angles using transformations rather than assuming it as a postulate. It turns out that we need a reflection for the perpendicular case (one mirror), and then that case plus a translation is used for the general case (two mirrors).

My idea was that all of these proofs, while invalid in Euclidean geometry, would be valid in both Euclidean and spherical geometry. Some of these theorems (such as Perpendicular to Parallels) are vacuously true in spherical geometry, while others have a possible interpretation in spherical geometry (for example, the Case II of Line Parallel to Mirror shows us that if a line and its image are distinct and intersect, then the mirror must intersect there as well, in both geometries). But in no case did I use a postulate that holds in Euclidean but not spherical geometry.

Or did I? Consider the idea that if PP' is invariant and both P and P' lie on l', then l' is invariant. This appears to assume that there exists at most one line through P and P' (namely l'), but this fails in spherical geometry. But this is built into Line Perpendicular to Mirror -- it holds in spherical geometry that mirror m reflects P to P' and infinitely many lines pass through P and P', then every single such line is invariant with respect to m (because every such line is perpendicular to m).

We also note that Perpendicular to Parallels uses the Two Perpendiculars assumption (to show that a must be parallel to m). It seems silly to worry about this (since the theorem is vacuous unless parallel lines exist), but still, it's possible to avoid this:

Three Perpendiculars Theorem:
If a, b, c are all perpendicular to t, then a, b, c are either all parallel or all concurrent.

(Theorem is indeed "natural" in that it holds in both Euclidean and spherical geometry. Of course, "parallel" corresponds to Euclidean and "concurrent" to spherical geometry, but nowhere in the theorem or its proof is this stated directly, hence it's natural.)

Proof:
Let A, B, C be the points where a, b, c (respectively) intersect t. Without loss of generality, our goal is to show that if a and b have a point P in common, then P must lie on c as well.

Let m be the perpendicular bisector of AB. This reflection leaves t invariant and maps A to B. Since reflections preserve angle measures, a (perp. to t at A) is mapped to b (perp. to t at B). And since a intersects its image b at P, the mirror m must pass through P as well (from that above argument about spherical Case II), and as reflections preserve distance, PA = PB.

Now let n and o be the perpendicular bisectors of AC and BC respectively. Both reflections leave t invariant, and map A to C and C to B respectively. Since reflections preserve angle measure, the reflections map a to c and c to b respectively. The composite isometry maps a to b -- and since isometries preserve distance and PA = PB, the composite isometry maps P to itself.

Since P is a fixed point of an isometry that's the composite of two reflections, we conclude (from earlier in this post) that P must lie on both mirrors n and o. Thus the reflection over n, which maps a to c, must map PA (on a) to PC (on c). Therefore P lies on c. QED

The sole purpose of Three Perpendiculars Theorem is to take a proof that would ordinarily use Two Perpendiculars and apply it to natural geometry, without using a single postulate that distinguishes between Euclidean and spherical geometry.

But if we need an awkward Three Perpendiculars Theorem just to prove that a line and its translation image are parallel, then we should just forget about it. I'll use this -- my 750th post -- to declare this for future reference. Unless I make a sudden insight, this is probably as far as I'll go with this idea.

There are more interesting problems that I'd like to solve with transformations anyway. For example, I've mentioned how graphing linear equations is always a dark spot in eighth grade math and Algebra I classes -- students tend to struggle with it. Can transformations be used to make this unit easier? We know the Core's suggestion that similarity/dilations be used to prove the existence of slope -- but does this insight make graphing easier for students to learn? And does this make related lessons in Math 8 or Algebra I easier, such as solving systems of equations?

Conclusion

Notice how much of our algebraic manipulation in today's post involves solving simple systems of equations (where x + y and x - y are known). I'm continuing to monitor how the students are doing in the B sessions of the summer school classes (which I was never scheduled to teach). The first unit test of Algebra 1B is indeed on systems of equations (which are sometimes covered in first semester, but is covered in second semester for us).

The top student from A session (with his grade of 87%) earned 90 on the first unit test. His overall grade so far in B session is 93%.

The bottom student from A session (with his grade of 59%) earned 65 on the first unit test. His overall grade so far in B session is 70.

It's notable that so far, both students are finding systems of equations easier than graphing. As it turns out, today the famous math teacher Sarah Carter writes one of her final Algebra I blog posts -- on systems of equations:

https://mathequalslove.blogspot.com/2018/07/systems-of-equations-and-inequalities.html

(Oh how I'd looked up her blog earlier in anticipation of summer school, only to wind up not teaching summer school!) Carter writes:

And, now the same for elimination. By far, students definitely preferred elimination over substitution.

Next, we addressed solving systems of equations by both substitution and elimination. Last year, I had these as a single skill of solving systems of equations algebraically. As a result, most students chose the method that they preferred and used that method every single time. This meant that I had a lot of students that became really good at solving by elimination but were rubbish at solving by substitution.

This year, I decided to make separate skills for solving by substitution and solving by elimination.
Hmm, well, the top student scored 90 on both quizzes, while the bottom student actually scored slightly higher on substitution (70) than on "linear combinations" -- which is the name that Edgenuity gives to elimination (68). But overall, Carter's observations seem to hold, with higher scores on elimination than substitution. This includes the only student who failed substitution with 50 and went on to earn 68 on elimination. Indeed, no one failed elimination, while four had perfect scores -- twice as many as substitution.

In today's post, we used elimination since our questions gave values to x + y and x - y.

Meanwhile, in the sports world, many Southern Californians have been talking about the signing of the NBA free agent LeBron James with the Los Angeles Lakers. However, the defending champions from Northern California, the Golden State Warriors, have also gotten stronger.

As a general rule, fans from Northern California and Southern California don't cheer for each other's teams, even though we're the same state -- for now, that is. If the long-shot ballot proposition passes, the Warriors would play in Northern California while the Lakers would play in California. (The third state, Southern California, would have no NBA teams.)