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
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
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
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.
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.
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