x^2 + 6y = 4x - y^2 + 12 is a circle with area = ?pi square units.
This is the sort of question that goes with Lesson 11-3 of the U of Chicago text, except that none with completing the square appear in that lesson. Since completing the square does appear on the SBAC, we should do this problem:
x^2 + 6y = 4x - y^2 + 12
x^2 - 4x + y^2 + 6y = 12
x^2 - 4x + 4 + y^2 + 6y + 9 = 12 + 4 + 9
(x - 2)^2 + (y + 3)^2 = 25
Notice that our circle is centered at (2, -3), and that r^2 = 25. There's no need to find r (even though it's easy), since the desired circle area is pi r^2. So we just plug in 25 for r^2. Thus the desired area is 25pi square units -- and we're already supplied the factor of pi. Therefore the number that ? represents is 25 -- and of course, today's date is the 25th.
Questions like this straddle the border between Algebra II and Geometry. Yesterday's Pappas problem asks for the slope of a line parallel to the line whose equation is given. Slopes of parallel lines appear in Lesson 3-4 of the text as well as in Algebra I. Yet I made the arbitrary decision to highlight today's Pappas problem here on the blog, but not yesterday's.
This goes back to what Stewart writes in Chapter 6 of his book -- the Cartesian plane blurring the algebraic-geometric distinction. Oh, and speaking of Stewart's book...
Chapter 8 of Ian Stewart's The Story of Mathematics is called "The System of the World: The Invention of Calculus." Here's how it begins:
"The most significant single advance in the history of mathematics was calculus, invented independently around 1680 by Isaac Newton and Gottfried Leibniz."
As the title implies, this chapter is all about Calculus. Stewart calls it "the system of the world" due to its applications in science. He tells us that Calculus is all about functions:
"The procedure is usually specified by a formula, assigning to a given number x (possibly in some specific range) an associated number f (x)."
The author now goes on to explain the two main operations -- differentiation and integration. I won't take the time here to explain what these are, or attempt to write an integral sign in ASCII. Instead, let's pick up Stewart as he explains why Calculus is needed:
"But the main stimulus towards calculus came from physics -- the growing realization that nature has patterns."
In particular, these patterns involved the motions of the sun, moon, and planets. The author describes two models -- Ptolemy's, which was based on circles (and epicycles), and that of Hipparchus, which was based on spheres:
"Hipparchus's model was not terribly accurate, compared to observations, but Ptolemy's model fitted the observations very accurately indeed, and for over a thousand years it was seen as the last word on the topic."
According to Ptolemy, the sun orbits the earth. But during the Renaissance, some astronomers began to challenge this view -- and this quickly became an issue of God versus science:
"But observations can -- and did -- debunk the view of the Earth as the centre of the universe. And this caused a huge fuss, and got a lot of innocent people killed, sometimes in hideously cruel ways."
One of these heretics was Copernicus, who tried to simplify the Ptolemaic epicycles:
"Copernicus realized that if all these epicycles were transferred to the Earth, only one of them would be needed."
Another major astronomer during this period was Kepler, who wanted to describe the six planets known to him at the time:
"Kepler wondered why there were six planets. He noticed that six planets leave room for five intervening shapes, and since there were exactly five regular solids, this would explain the limit of six planets."
Stewart describes how this solids fit into Kepler's model:
"Between the spheres, nestling tightly outside one sphere and inside the next, he placed the five solids, in the order Mercury - Octahedron - Venus - Icosahedron - Earth - Dodecahedron - Mars - Tetrahedron - Jupiter - Cube - Saturn. The numbers fitted reasonably well, especially given the limited accuracy of observations at the time."
Kepler also discovered his Laws of Planetary Motion. The laws state:
(i) Planets move round the Sun in elliptical orbits.
(ii) Planets sweep out equal areas in equal times.
(iii) The square of the period of revolution of any planet is proportional to the cube of its average distance from the Sun.
"The most unorthodox feature of Kepler's work is that he discarded the classical circle (allegedly the most perfect shape possible) in favour of the ellipse."
At this point, Stewart begins describing the invention of Calculus, a new branch of math. Calculus was used to solve several unrelated problems:
"These included calculating the instantaneous velocity of a moving object from the distance it has travelled at any given time, finding the tangent to a curve, finding the length of a curve, finding the maximum and minimum values of a variable quantity, finding the area of some shape in the plane and the volume of some solid in space."
It was Gottfried Leibniz who discovered the connection between two of these problems -- finding the tangent to a curve, and finding areas and volumes:
"The latter boiled down to finding a curve given its tangents; the former problem was exactly the reverse. Leibniz used this connection to define what, in effect, were integrals, using the abbreviation, omn (an abbreviation of omnia, the Latin word for 'all')."
OK, now this I can write in ASCII: omn x^2 = x^3/3. Stewart tells us that Leibniz went on to discover many more concepts of Calculus:
"Unfortunately, his writings were fragmented and virtually unreadable. The other creator of calculus was Isaac Newton."
Of course, Newton was working with his new theory of gravity. He was trying to prove that a sphere of a given mass acted, as far as gravity is concerned, like a point of the same mass:
"In 1686, he succeeded in filling the gap, and the Principia saw the light of day in 1687. It contained many novel ideas."
But the Principia Mathematica didn't actually contain any Calculus. Indeed, Newton originally didn't publish his ideas about Calculus:
"Among his earliest work on the topic was a paper titled On Analysis of Equations with an Infinite Number of Terms, which he circulated to a few friends in 1669."
In these early works, instead of limits as a variable approached zero, Newton used something that he called fluxions. Several decades later, Bishop George Berkeley criticized these fluxions:
"The method was about fluxions, not numbers. The mathematicians sought refuge in physical analogies -- Leibniz referred to the 'spirit of finesse' as opposed to the 'spirit of logic' -- but Berkeley was perfectly correct."
It would take centuries before limits were given a rigorous foundation. We return to the battle between Leibniz and Newton regarding the invention of Calculus -- and, much to the dismay of our British author Stewart, the English got left behind:
"When Leibniz published his work in 1684, some of Newton's friends took umbrage -- probably because Newton had been pipped to the publication post and they all belatedly realized what was at stake -- and accused Leibniz of stealing Newton's ideas."
Thus much of mathematical physics was developed by continental Europeans, not the English. The author continues by describing differential equations, and how they were used to calculate the motion of an object under the influence of gravity:
"The formula tells us that the graph of height y against time t is an upside-down parabola. This is Galileo's observation."
Stewart concludes the chapter with the idea that nature depends not as much on the values of variables, but on their rates of change:
"It was a profound insight, and it created a revolution, leading more or less directly to modern science, and changing our planet forever."
Sidebars in this chapter include biographies of Kepler, Galileo, and Newton, what Calculus did for them, and what Calculus does for us.
Lesson 5-1 of the U of Chicago text is called "Isosceles Triangles." In the modern Third Edition of the text, isosceles triangles appear in Lesson 6-2. But the new Lesson 6-2 expands beyond the old Lesson 5-1 to discuss scalene triangles as well (which don't appear until Lesson 13-7 in the old text).
This is what I wrote last year about today's lesson:
Section 5-1 of the U of Chicago text covers isosceles triangles. Now the most important result of this section is, of course, the Isosceles Triangle Theorem, which states that if a triangle has two equal sides, then the angles opposite them are equal. But this first theorem in this lesson is something called the Isosceles Triangle Symmetry Theorem. Here's a statement of the theorem and its proof as given by the U of Chicago:
Isosceles Triangle Symmetry Theorem:
The line containing the bisector of the vertex of an isosceles triangle is a symmetry line for the triangle
Given: Isosceles triangle ABC with vertex angle A bisected by m
Prove: m is a symmetry line for triangle ABC
Proof:
There are three things given. Each leads to conclusions that are used later in the proof. First, since m is an angle bisector, because of the Side-Switching Theorem, when ray AB is reflected over m, its image is ray AC. Thus B' is on ray AC. Second, it is given that A is on the reflecting line, so A' = A. Hence, since reflections preserve distance, AB' = AB. Third, it is given that triangle ABC is isosceles with vertex angle A, so AB = AC. Now put all of these conclusions together. By the Transitive Property of Equality, AB' = AC. So B' and C are points on ray AC at the same distance from A, and so B' = C. By the Flip-Flop Theorem, C' = B. So, by the Figure Reflection Theorem, the reflection image of triangle ABC is triangle ACB, which is the sufficient condition for the symmetry of the triangle to line m. QED
Now, when a pre-Common Core teacher first sees this proof, this is not what is expected. Such a teacher expects to give the point where BC and m intersect the name D. Then one tries to prove the triangles ABD and ACD congruent. We have AB = AC by the definition of isosceles, angles BAD and CAD equal by definition of angle bisector, and AD = AD by the Reflexive Property of Equality. And so by SAS, the two triangles and congruent, and so angles B and C are equal by CPCTC.
But we must note that in Chapter 7, we use isometries to prove SAS as a theorem. So we could have the Isosceles Triangle Theorem derived from SAS, which in turn is derived using isometries. Or we can skip the middleman and just derive the Isosceles Triangle Theorem from isometries. This is definitely in the line of Common Core thinking:
CCSS.MATH.CONTENT.HSG.CO.A.3
Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself.
In this section we found a reflection that maps an isosceles triangle to itself -- namely the reflection over the line containing the angle bisector of the vertex angle. Later on in Chapter 5, we'll find the isometries that map the other figures mentioned in that standard (rectangles, etc.) to themselves.
A few things about the traditional proofs. First of all, these proofs usually involve adding a new point D, where the angle bisector of BAC intersects side BC. But, according to Dr. Hung-Hsi Wu, we can't be sure that these even intersect at all without a Crossbar Axiom. But in the U of Chicago, we don't need it because the point D is never mentioned in the proof.
The most common proof of the Isosceles Triangle Theorem uses an angle bisector and SAS. Notice that we also could have used a median and SSS. But the SAS proof is usually better because some texts use the Isosceles Triangle Theorem to prove SSS. (In fact, Wu uses ITT to prove SSS.) So a proof of ITT using SSS would be circular.
Euclid himself proves the Isosceles Triangle Theorem as his Proposition I.5:
http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI5.html
Euclid's proof also uses SAS (which is his Proposition I.4), but his proof is more complicated than any text appearing in a high school text because he was also concerned with the exterior angles at B and C, not just the interior angles of the triangle. According to Euclid, the Isosceles Triangle Theorem was the Pons Asinorum, or bridge of SSA --- well, actually SSA spelled backwards. In a way, we cross a bridge when we enter Chapter 5, as here the somewhat more difficult proofs of geometry begin.
I must mention one of my favorite proofs of ITT here, although it's also based on SAS and so not appropriate for the students right now. According to the above link, it can be attributed to the Greek geometer Pappus, who lived about 500 years after Euclid. Here is the proof given directly from the above link:
The two triangles BAC and CAB have two sides equal to two sides, namely side BA of the first triangle equals side CA of the second triangle, and side AC of the first triangle equal to side AB of the second, and the contained angles are equal, namely angle BAC of the first triangle equals angle CAB of the second, therefore, by I.4, the corresponding parts of the two triangles are equal, in particular, the angle B in the first triangle equals the angle C of the second. QED
In two-column form, the Pappus proof would look like this:
Given: AB = AC
Prove: Angles B and C are equal.
Proof:
Statements Reasons
1. AB = AC 1. Given
2. Angle A = Angle A 2. Reflexive Property of Equality
3. AC = AB 3. Symmetric Property of Equality
4.
5. Angle B = Angle C 5. CPCTC
Notice that the closest I could get to a triangle symbol in ASCII is the caret with a strikethrough. I was considering writing an angle symbol as an underlined slash, but I decided that simply writing out the word "Angle" was less confusing.
We see that in the Pappus proof, triangle ABC is congruent to itself -- but under a new name ACB. In a way, this is exactly what symmetry is -- a figure is symmetrical if there is a nontrivial congruence between the figure and itself (under a different name). Triangle ABC is always congruent to triangle ABC, but it's only congruent to ACB if the triangle is isosceles. So we know that there is an isometry from ABC to ACB -- but we don't know that it's a reflection yet, much less a reflection over the line m containing an angle bisector of angle A, although this can be deduced. Notice that the line m never appears in the Pappus proof -- the only proof that doesn't require an auxiliary line.
Now let's return to the U of Chicago proof. This proof is on the longish side -- indeed, many of the heavy-duty theorems have long proofs. It's instructive to cut out an isosceles triangle with the bisector of A drawn in, and ask the students where B would land if the triangle were folded along that angle bisector. Chances are that the students will say that it would land on C. But this doesn't work unless the triangle is isosceles. For the scalene triangle DEF drawn on the U of Chicago page right below the Isosceles Triangle Symmetry Theorem, E will not land on F if the triangle is folded along the bisector of angle D. So we need to prove that it works for isosceles triangles.
This proof contains a common trick in isometry-based proofs -- to prove that the image of B is C, we first show that the image of B lies on ray AC, then show that the image is a point that's the correct distance from A. And the point at just the right distance from A is exactly C.
But there's something missing from this lesson in the U of Chicago. We have the Isosceles Triangle Theorem, but there's no mention of its converse. The text even states, "The Isosceles Triangle Theorem is useful in proofs in which you must go from equal sides to equal angles," but what if one wants to go from equal angles to equal sides?
The most common proof is based on AAS. Once again, we draw the angle bisector of angle A (which intersects BC at D). We are given that angles B and C are equal, and angle BAD = angle CAD by definition of angle bisector and AD = AD by the Reflexive Property. So this gives us ABD and ACD, congruent by AAS, and AB = AC by CPCTC.
In Euclid, the converse is Proposition I.6. His proof is an indirect proof that is not usually given in high school classes:
http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI6.html
There is also a Pappus-style proof of this converse. We are given that angles B and C are equal, we have BC = CB by the Reflexive Property, and we have that angles C and B are equal. And so once again, triangles ABC and ACB are congruent, this time by ASA, and so AB = AC by CPCTC.
But is there a proof based on reflections, similar in style to the U of Chicago's proof of the forward Isosceles Triangle Theorem? Here's what such a proof may look like. Instead of letting m be the bisector of angle A, let it be the perpendicular bisector of BC:
Given: Triangle ABC with equal angles B and C, m perpendicular bisector of
Prove: m is a symmetry line for triangle ABC
Proof:
We obtain that the reflection image of B is C immediately, as the reflecting line m is the perpendicular bisector of
But there is a gap in this proof here. How do we know that the image of angle B is not an angle of the same vertex and measure as angle C, but on the opposite side of
This problem appeared in our Alternate Interior Angles Test proof as well -- and it occurs in any proof where we have equal angles and want to map one to the other with an isometry. The solution is the Plane Separation Postulate, which tells us that line BC divides the plane into two half-planes, and we want to prove that angle B and its image each has a side in the same half-plane.
To prove this, suppose we have a point P in the coordinate plane and wish to reflect it over, let's say, the y-axis. So of course P and P' are on opposite sides of the y-axis. But notice that P and P' must be on the same side of the x- (not y-, but x-) axis. This is because line PP', by definition of reflection, is perpendicular to the reflecting line, the y-axis, and the x-axis is also perpendicular to the y-axis. So by the Two Perpendiculars Theorem, line PP' must be parallel to the x-axis, and so the line can hardly intersect x-axis if it is parallel to the x-axis! (Of course if P lies on the x-axis then so does P'.) This tells us that the while the reflection image of a half-plane whose boundary is the mirror must be the opposite half-plane, its reflection image in any mirror perpendicular to the boundary is itself. Any line perpendicular to the boundary line of a half-plane is a symmetry line of that half-plane.
Notice that the Isosceles Triangle and Perpendicular Bisector Theorems are ultimately related, in that the latter is used to prove the converse of the former. Moreover, the converse of the latter can be used to prove the former as well! In fact, my proof of the Converse of the Perpendicular Bisector Theorem actually borrowed portions of the proof of the Isosceles Triangle Theorem. This means that some steps of the proof are redundant. We can shorten the Isosceles Triangle Theorem proof by using the Converse of the Perpendicular Bisector Theorem, as follows. Notice that now m must be the perpendicular bisector of
Given: Isosceles triangle ABC with m the perpendicular bisector of
Prove: m is a symmetry line for triangle ABC
Proof:
First, it is given that A is on the reflecting line, so A' = A. Second, it is given that triangle ABC is isosceles with vertex angle A, so AB = AC. Since A is equidistant from B and C, by the Converse of the Perpendicular Bisector Theorem, A lies on the perpendicular bisector of BC, which is m. Third, by the definition of reflection, we have B' = C and C' = B. So, by the Figure Reflection Theorem, the reflection image of triangle ABC is triangle ACB, which is the sufficient condition for the symmetry of the triangle to line m. QED
Returning to 2019, today is an activity day. When this school year began, I planned on finding activities from two particular sources:
- old activities from past years on the blog
- the Exploration questions from the current U of Chicago lesson
- In the past, never posted any activities for Lesson 5-1.
- The Exploration questions for Lesson 5-1 are very important.
CCSS.MATH.CONTENT.HSG.CO.C.10
Prove theorems about triangles. Theorems include: measures of interior angles of a triangle sum to 180°; base angles of isosceles triangles are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point.
[emphasis mine]
Well, today we cover base angles of isosceles triangles in the main lesson. But the key here is on the medians of triangles and how they are concurrent. That doesn't appear in the U of Chicago text at all, except in this Exploration question.
In the past, I squeezed in the Exploration question as a bonus, but now I'm creating an activity worksheet for this question. Still, this is what I wrote in the past about medians:
In the exercises, I just had to include Exploration Question 21 as a bonus question, since it refers to the centroid or point where the medians intersect, which is explicitly mentioned in Common Core. I know that this is an exploration question, so nothing is actually proved yet. Earlier I wrote that the actual proof can wait until second semester, but after comparing the U of Chicago with Wu, I see that the proof is better given during first semester. In fact we may be able to give a proof very soon.
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