I help some students read about Computer Science Education Week and "Hour of Code," which takes place the first week in December. Every time I see a coding lesson, I can't help but think back to the old charter school and coding Mondays. I believe that the coding teacher briefly mentioned "Hour of Code," but I didn't post much about what coding Mondays were like in December 2016.
I also assist some students with some math, including adding simple dollar amounts. I keep saying that I'm going to use Number Talks tricks to help these students out, and I never do -- the students end up just using the standard algorithm. One student does use ten-frames to add single-digit numbers -- ten-frames being a staple of Common Core.
Another student is supposed to use the standard algorithm to add a pair of two-digit numbers -- but the numbers are labeled "TouchPoints," dots that I've only seen in special ed classes:
I'm not quite sure how traditionalists would react to something like TouchPoints. On one hand, we see that TouchPoints clearly don't interfere with the standard algorithm at all. On the other, they arguably interfere with memorization. I've only seen special ed classes use them, but as the link implies, there's no reason that grade-level lower elementary students can't use them as well. At any rate, TouchPoints aren't clearly associated with Common Core in the way that ten-frames are.
And I keep wanting to introduce Putnam questions to whichever math students I teach this week -- but of course, a special ed class is the last place where I want to talk about something as advanced as the Putnam exam. (If I were in that Algebra I class from three weeks ago, I would have given a Putnam question!)
But that doesn't prevent me from posting a Putnam question here on the blog, of course. Here's a link to the Geometry question that I'd like to discuss, Question A6:
Suppose that 
, 
, 
, and 
are distinct points, no three of which lie on a line, in the Euclidean plane. Show that if the squares of the lengths of the line segments 
, 
, 
, 
, 
, and 
are rational numbers, then the quotient
![\[\frac{\mathrm{area}(\triangle ABC)}{\mathrm{area}(\triangle ABD)}\]](https://latex.artofproblemsolving.com/3/9/0/390f5a4a417c52a9ffd75564ac2acc815ea1babc.png)
is a rational number.
, , , and are distinct points, no three of which lie on a line, in the Euclidean plane. Show that if the squares of the lengths of the line segments , , , , , and are rational numbers, then the quotientis a rational number.
Once again, Dr. Kent Merryfield is no longer with us to post these questions. Instead, the poster CantonMathGuy provides us with this question, and he also posts a solution.
Notice that this question is A6 -- typically the most difficult question in the morning (A) session. One poster remarks that he finds this question easier than A5, but that's not saying much (as A5 has Calculus). I myself attempted to solve it and struggled mightily.
Let's discuss CantonMathGuy's solution here. This is a coordinate proof -- that is, he begins by placing the four points on a coordinate plane. Notice that AB is the base of both triangles, and so it's handy to place A and B on the x-axis. In fact, here are our coordinates:
A(-1, 0)
B(1, 0)
C(x, y)
D(u, v)
Not only do we place A and B on the x-axis, but at these two particular points. This makes it even easier to find the area of the triangles -- the base of Triangle ABC is AB = 2 and its height, the distance between C and x-axis, is in fact y. So its area is just (1/2)(2)(y) or y. Likewise, the area of Triangle ABD is v.
You might point out that nowhere in the problem does it state AB = 2, only that AB^2 is rational. But actually, we can perform a dilation -- exactly the dilation that makes AB = 2. Since AB^2 = 4 is also rational, this dilation will keep all the other squares rational as well.
In general, in a coordinate proof, we're allowed to set one variable equal to 1 (or any other value). If we'd written A(-w, 0) and B(w, 0), we could have chosen x = 1 or v = 1 instead of w = 1. The coordinate proofs in Chapter 11 of the U of Chicago text don't do this, but for us, setting w = 1 makes the proof easier.
We know that the squares of all the lengths are rational. This means that we can use the Distance Formula to find all the distances. Since we're not interested in the distances per se but their squares, we can just leave out the square root symbol:
AB^2 = 4
AC^2 = (x + 1)^2 + y^2 = x^2 + 2x + 1 + y^2
BC^2 = (x - 1)^2 + y^2 = x^2 - 2x + 1 + y^2
AD^2 = (u + 1)^2 + v^2 = u^2 + 2u + 1 + v^2
BD^2 = (u - 1)^2 + v^2 = u^2 - 2u + 1 + v^2
CD^2 = (x - u)^2 + (v - y)^2 = x^2 - 2xu + u^2 + y^2 - 2yv + v^2
The left-hand sides are rational, and so the right-hand sides are rational.
We complete the proof by using the closure properties of the rationals. The sum, difference, product, and quotients of (nonzero) rationals are themselves rational. For example, we have:
AC^2 - BC^2 = 4x is the difference of rationals, hence itself rational.
AC^2 + BC^2 = 2x^2 + 2 + 2y^2 is the sum of rationals, hence itself rational.
And if 4x is rational, then x is rational. (It's the quotient of the rationals 4x and 4.) Then x^2 is rational. (It's the product of the rationals x and x.) And in the second line, 2y^2 is rational. (It's the difference of some already known rationals.) Then y^2 is rational. (It's the quotient of the rationals 2y^2 and 2.) Notice that at this point, we cannot conclude that y is rational. (The square root of a rational might be irrational.)
We follow the same steps with AD^2 and BD^2 and conclude that u and v^2 are rational (but not v).
We haven't looked at the CD^2 line yet, so let's do so now:
CD^2 = (x - u)^2 + (v - y)^2 = x^2 - 2xu + u^2 + y^2 - 2yv + v^2
The left-hand side is rational, and five of the six terms on the right-hand side are known rationals. Therefore the last term, -2yv, is rational, as is yv (the quotient of the rationals -2yv and -2).
This still isn't enough to conclude y or v are rational. But notice that y/v must be rational, since it's the quotient of the known rationals yv and v^2.
Now let's go back to what we're asked to prove. The areas of the two triangles are y and v, but we aren't asked to prove that either is rational -- we need to show that their ratio is rational. Fortunately, we already proved that their ratio y/v is rational. This concludes the proof. QED
(Note: C and D could be on opposite sides of line AB, the x-axis. Technically, the areas of the triangles aren't y and v but |y| and |v|, since y and v could be negative. But that doesn't matter for the proof, since the absolute value of a rational number is rational.)
Here is the original version of proof as written by CantonMathGuy:
Set 
, 
, 
, and 
. Note
![\[\mathbb{Q} \ni AC^2 - BC^2 = (x_C + 1)^2 - (x_C - 1)^2 = 4x_C\]](https://latex.artofproblemsolving.com/f/4/6/f461454420fcd9117cbe433850915bc10d3406b7.png)
so 
; now note
![\[\mathbb{Q} \ni AC^2 = (x_C + 1)^2 + y_C^2\]](https://latex.artofproblemsolving.com/a/6/b/a6bd393c1cf2124531e872597c2d42eaa162a3a6.png)
so 
. Likewise 
. Now it follows that
![\[\mathbb{Q} \ni CD^2 = (x_C - x_D)^2 + (y_C - y_D)^2\]](https://latex.artofproblemsolving.com/1/5/8/158d601a3651b956a12876b4b445b67e6ed7b083.png)
and so 
. Thus 
and thus
![\[\frac{[ABC]}{[ABD]} = \left|\frac{y_C}{y_D}\right| = \left|\frac{y_Cy_D}{y_D^2}\right| \in \mathbb{Q}.\]](https://latex.artofproblemsolving.com/9/b/2/9b28c627a096ef4bec9a48c3115a31e7f704668f.png)
Recall that the (blackboard) bold Q stands for "rational." He uses subscripts to distinguish between the coordinates of C and D -- instead I follow another poster in this thread and use (x, y) and (u, v).
, , , and . Noteso ; now noteso . Likewise . Now it follows thatand so . Thus and thusRecall that the (blackboard) bold Q stands for "rational." He uses subscripts to distinguish between the coordinates of C and D -- instead I follow another poster in this thread and use (x, y) and (u, v).
Suppose we want to present this proof to high school Geometry students. There are two places where they are likely to struggle understanding -- the use of (-1, 0) and (1, 0) for the coordinates of a and b, and the whole issue with "rationals."
In any coordinate proof, we're allowed to set one of the variables equal to 1. For example, consider setting up coordinates for a rectangle centered at the origin. The vertices are (a, b), (-a, b), (-a, -b), and (a, -b). We can set either a = 1 or b = 1 without disturbing any possible coordinate proof involving this rectangle, since we can imagine enlarging or shrinking (via a dilation) the rectangle until a or b becomes 1. But we can't set both a and b to be 1 at the same time, since this would make the rectangle into a square, and an arbitrary rectangle can't suddenly become a square via a mere dilation. And besides, we can tell them that by letting AB = 2, the areas of the two triangles are simply y and v (or their absolute values).
As for rational numbers, students do learn about rationals and irrationals in eighth grade, and they're aware of such irrationals as sqrt(2) and pi. But these are rational expressions involving variables, which might be too abstract for high school students to consider. They should at least be aware that the Distance Formula contains a square root, which is the obvious source of any possible irrationals in this problem.
The importance of rationals to this proof is closure -- the fact that the sum, difference, product, and quotient of any two rationals is rational (excluding division by 0). We might try to explain this to high school students by writing certain expressions in boxes -- the ones proved to be rational.
Then if we add two terms in boxes, the sum goes in a box. Likewise if we subtract, multiply, or divide two terms in boxes. The known rationals 1, 2, 3, and 4 all go in boxes. If all but one term of a polynomial is in a box and the polynomial itself is equal to something in a box, then the last term goes in a box. (Color-coding all rationals instead in, say, red, might be easier. But then it's tricky to explain why yv or y/v are colored red without y or v themselves being red. Instead, we can write yv and y/v in a single box without individual boxes for y or v.)
I admit that I never came up with this proof on my own (though I wish I did). My first attempt was to note that since all the squares of sides are rational, perhaps we should start with an expression involving the squares of sides -- the Law of Cosines. From c^2 = a^2 + b^2 - 2ab cos C, we notice that since a^2, b^2, and c^2 are all rational, 2ab cos C must be rational as well. But that doesn't help us -- especially since the desired heights y and v can be found in terms of the sine of C, not the cosine. So our high school Geometry students will find this problem even more different. This is problem A6 after all, not A1.
Well, let's take a look at Question A1:
Find all ordered pairs 
of positive integers for which
![\[\frac{1}{a} + \frac{1}{b} = \frac{3}{2018}.\]](https://latex.artofproblemsolving.com/2/3/2/232bffeaf3bc1927e2cde4db377f8936d50e1671.png)
This problem is all about adding fractions, which students first learn in fifth grade. If I were subbing in a regular math class below Geometry, I'd introduce this question (possibly even as low as fifth grade if we restate the problem without formal variables).
of positive integers for whichThis problem is all about adding fractions, which students first learn in fifth grade. If I were subbing in a regular math class below Geometry, I'd introduce this question (possibly even as low as fifth grade if we restate the problem without formal variables).
One solution jumps out at me. Obviously 3/2018 = 1/2018 + 2/2018, and 2/2018 reduces to 1/1009, so this gives our first solution, (2018, 1009). For any solution (a, b), we have another solution (b, a), so this leads to our second solution, (1009, 2018). But we're asked to find all the solutions. Not only must we find all of them, but we must also prove that there are no others.
Let's think about how to add fractions for a moment. We must first find a common denominator -- the least common multiple of a and b in this case. Since the sum has 2018 in the denominator, we might guess that the LCM of a and b is 2018 -- in which case a and b are factors of 2018.
Almost every year, the Putnam mentions the current year in the problem. (It's the Putnam equivalent of Pappas problems with the date as the answer.) This year there are two such problems, namely A1 and B6. And more often than not, factoring that number will be helpful. Thus every Putnam participant should memorize the factors of the current year before taking the test.
Clearly 2018 = 2 * 1009. As it turns out 1009 is prime, so we are done factoring. Next year is a multiple of three -- so if you wish to get a head start on next year, 2019 = 3 * 673, with 673 a prime. (As it turns out, the factorization of 2019 is relevant in this year's A1 as well.)
If I were to show this to students from Math 5 to Algebra I, that 2018 in the denominator is intimidating. So it's instructive to consider a similar problem with a smaller denominator. I suggest 14, because like 2018, fourteen is twice a prime. Also, like 2018, fourteen is one less than a multiple of 3 -- that is, it's equivalent to 2 mod 3. (That three will turn out to be significant is hinted at by the 3 in the numerator.)
So let's try solving 1/a + 1/b = 3/14. Once again, the obvious solutions (7, 14) and (14, 7) jump out at us. But since 14 is so much smaller, a trial-and-error process is suggested. Notice that 3/14 - 1/a = 1/b, and so we might try plugging in different unit fractions for 1/a to see whether another unit fraction 1/b appears:
1/2, 1/3, and 1/4 are all larger than 3/14, so 3/14 - 1/a would be negative. It's given that b must be positive.
3/14 - 1/5 = 1/70 *
3/14 - 1/6 = 1/21 *
3/14 - 1/7 = 1/14 *
3/14 - 1/8 = 5/56
3/14 - 1/9 = 13/126
3/14 - 1/10 = 8/70
And we can stop here, since 8/70 is clearly larger than 1/10 (= 7/70). Indeed, it suffices to consider only values of a such that 1/a exceeds half of 3/14 -- if 1/a were less than half, then 1/b would be more than half, and so it would have already as a possibility for 1/a (as a and b are interchangeable).
We've found three solutions -- (5, 70), (6, 21), (7, 14). Of course, (14, 7), (21, 6), and (70, 5) are solutions. This means that we've found all six solutions.
I found out that these solutions generalize for all denominators following the same pattern as 14 and 2018 -- even numbers that are equivalent to 2 mod 3. Besides the obvious solutions like (7, 14) or (1009, 2018), the analogs of (5, 70) and (6, 21) are found by taking the next multiple of 3 and dividing by 1. Then both that number and the next integer work for a. For 14 we obtain (14 + 1)/3 = 5 so a = 5 and a = 6 works. For 2018, well, we already wrote 3 * 673 earlier, so a = 673 and a = 674 are solutions. For these solutions, 1/a is just barely smaller than the fraction, and so 1/b ends up being very tiny (that is, b is much larger than a). In the case of 2018:
(1009, 2018), (673, 1358114), (674, 340033)
Don't forget to reverse these, so that there are six solutions. (If the denominator is equivalent to 1 mod 3 rather than 2 mod 3, then only four of these six solutions generalize.)
I was able to find the six solutions, but I couldn't prove that they were the only solutions. As it turns out, the proof given by CantonMathGuy involves rearranging the original equation:
1/a + 1/b = 3/2018
But CantonMathGuy doesn't explain how to rearrange the equation. Instead, it's up to other posters in the thread, who use something called "SFFT" (Simon's Favorite Factoring Trick), named after a prolific poster named Simon Rubenstein.
The first thing we tell Algebra I/II students to do is clear fractions by multiplying:
3ab - 2018a - 2018b = 0
As it turns out, we must multiply both sides by 3 again in order for SFFT to work:
(3^2)ab - (3*2018)a - (3*2018)b = 0
Now here's the factoring trick -- the left-hand side can almost be factored by grouping, except the last term appears to be missing. So add that missing term to both sides:
(3^2)ab - (3*2018)a - (3*2018)b + 2018^2 = 2018^2
(3a - 2018)(3b - 2018) = 2018^2
"Simon" refers to this as "completing the rectangle." Ordinarily, doing so is useless, since we no longer have the zero on the right-hand side needed for the Zero Product Property. There are only two cases when this is helpful:
- when completing the square (since we don't need 0 on RHS -- just take the square root of both sides)
- in a Diophantine equation where the solutions are integers
And this is indeed the second case. The equations gives the solutions in terms of factors of 2018^2 -- so now there's a strong limit on how many solutions there can be. There can't be more solutions of the equation than factors of 2018^2. And we can set 3a - 2018 and 3b - 2018 equal to those factors to find solutions.
I'll let CantonMathGuy take it from here:
Rewrite it as 
; then it follows that
![\[\{3a - 2018, 3b - 2018\} \in \big\{ \{1, 2018^2\}, \{4, 1009^2\}, \{1009, 4036\} \big\},\]](https://latex.artofproblemsolving.com/5/1/6/51692a25a80db3931b15a2da3ce8cd4ec7b1747d.png)
giving six solutions
![\[\{a, b\} \in \big\{ \{673, 673 \cdot 2018\}, \{674, 337 \cdot 1009\}, \{1009, 2018\} \big\}.\]](https://latex.artofproblemsolving.com/c/1/2/c1253ef1a05db746354c0dca0f8d4b6d6e9922bd.png)
(The other three factorizations of 
do not give integers for 
and 
.)
; then it follows thatgiving six solutions(The other three factorizations of do not give integers for and .)I was right that the factorization of 2018 mattered -- well 2018^2, not 2018, but obviously 2018 and its square have the same prime factors.
One poster in this thread remarked that this might be the easiest Putnam in years. On the other hand, I struggled right out of the gate with A1, since I'd never heard of SFFT, and it never would have occurred to me to factor this way. I did figure out A2 without looking at the answer -- but I still had to look up a certain matrix formula (which I couldn't do if I were a real college student taking the test.)
As a math teacher, it's always humbling to find problems that are too difficult to me. Many of my students feel about basic Algebra I and Geometry the same way I feel about the Putnam. Neither of us can believe how easy Geometry class/Putnam is for others when we struggle so much.
Lesson 7-2 of the U of Chicago text is called "Triangle Congruence Theorems." As you already know, this is one of the most important lessons in the entire text.
This is what I wrote last year about today's lesson:
And so we finally reach Lesson 7-2 of the U of Chicago text, the Triangle Congruence Theorems. I will be able to demonstrate how SSS, SAS, and ASA follow from the definition of congruence in terms of isometries.
Let's start with ASA, since as I said earlier this week, we can use the same proof of ASA directly out of the U of Chicago text. Here is the proof as given in the text:
ASA Congruence Theorem:
If, in two triangles, two angles and the included side of one are congruent to two angles and the included side of the other, then the triangles are congruent.
Proof:
Given
Think of reflecting Triangle A'B'C' over line DE. Applying the Side-Switching Theorem to Angle C'DF, the image of Ray A'C' is Ray DF. Applying the Side-Switching Theorem to Angle C'EF, the image of Ray B'C' is Ray EF. This forces the image of C' to be on both Ray DF and Ray EF, and so the image of C' is F. Therefore the image of Triangle A'B'C' is Triangle DEF.
So if originally two angles and the included side are congruent (
Now as we said earlier, the U of Chicago text uses the Isosceles Triangle Theorem to prove SAS, when we instead want to use SAS to prove the Isosceles Triangle Theorem. But as it turns out, we can write a proof of SAS that's not much different from the ASA proof given above. In this proof, we will discuss more in detail how we perform the first isometry -- the one that maps Triangle ABC to the position A'B'C', from which we can perform the final reflection.
SAS Congruence Theorem:
If, in two triangles, two sides and the included angle of one are congruent to two sides and the included angle of the other, then the triangles are congruent.
Proof:
Given
Next, we must map the whole segment
Now it could be the case that C" is already F -- in which case, we'd already be done. (Euclid erroneously made the assumption that C" is always F.) This is why the U of Chicago text always draws the case where C" is not F, in hopes that one final reflection, over line DE, will map C" to F.
So far, this part of the proof isn't particular to SAS. All of the congruence theorems will begin with this same isometry -- reflect A to D, then rotate DB' to DE. (By the way, it's possible to dispense with the rotation and use only reflections in the proof. Instead of rotating, use the angle bisector of B'DE as the mirror. Then by the Side-Switching Theorem, Ray DB' maps to DE.)
Notice that I have written C" double-primed. This is because C mapped to C' under the first reflection and then C' maps to C" under the rotation (or second reflection). We would have to write a third prime for the final reflection, to show that C'" is F. (The U of Chicago text shows only a single isometry mapping
Now here's the part of the proof that's particular to SAS. Instead of using isosceles triangles as in the U of Chicago proof, we notice that just as in the ASA proof, since Angles CAB (which has been moved to CDE) and FDE are given to be congruent, the Side-Switching Theorem once again tells us that the reflection over line DE maps Ray DC to Ray DF, and thus segment
So if originally two sides and the included angle are congruent (
When presenting this proof in class, we can start with this SAS proof so that students can see how to perform the opening reflection and rotation. Then when we get to ASA and the other proofs, we can just say "there exists an isometry" mapping ABC to the reflection of DEF, so that the proof only needs to discuss the final reflection.
Now let's go for SSS. I mentioned earlier that the proof in the U of Chicago text creates a kite -- and the properties of kites ultimately go back to isosceles triangles. I wrote that we can avoid the Isosceles Triangle Theorem by using the Converse to the Perpendicular Bisector Theorem instead. Here is the new proof:
SSS Congruence Theorem:
If, in two triangles, three sides of one are congruent to three sides of the other, then the triangles are congruent.
Proof:
Given
We are given that
So line DE is the perpendicular bisector of
But Lesson 7-2 contains another congruence theorem -- AAS. The U of Chicago, like most texts, prove AAS using ASA plus the Triangle Sum Theorem. I like the following proof of AAS:
AAS Congruence Theorem:
If, in two triangles, two angles and a non-included side of one are congruent respectively to two angles and the corresponding non-included side of the other, then the triangles are congruent.
Proof:
Given
We know that Angle ACB (same as DCE) is mapped to Angle DC'E, and as reflections preserve angle measure, these angles are congruent. And we are given that Angles ACB and DFE are congruent, so this tells us that DC'E and DFE are congruent. But this isn't sufficient to identify the images of any rays, since we don't know whether the vertices of the angles, C' and F, are the same point yet.
So let's try an indirect proof -- assume that C' and F are not the same point. This may be tough to visualize, so try drawing a picture. We already know that C' lies on Ray DF, so we can draw C' to be any point on Ray DF other than F. It doesn't matter whether C' is between D and F or on the opposite side of F from D -- both will lead to the same contradiction.
After we label the two angles known to be congruent, DC'E and DFE, we notice something about the diagram we've drawn. We see that lines C'E and FE are in fact two lines cut by the transversal DF, and the two angles DC'E and DFE turn out to be corresponding angles that are congruent. Thus, by the Corresponding Angles Test, lines C'E and FE are parallel! And so we have two parallel lines that intersect at E, a blatant contradiction. So the assumption that C' is not F must be false, and so C' is exactly F. QED
Like previous indirect proofs involving parallel lines, this can be converted into a direct proof if we use the U of Chicago definition of parallel. Then C'E and FE are parallel lines with E in common, so they are identical line -- that is, C' lies on FE. Then just as in the ASA proof, C' lies on both DF and FE, so C' is exactly F.
I like this proof as it has the same flavor as the SAS, ASA, and SSS proofs. Now I have an alternate proof of HL that avoids AAS and uses only theorems that have been proved previously on the blog so far.
HL Congruence Theorem:
If, in two right triangles, the hypotenuse and a leg of one are congruent to the hypotenuse and a leg of the other, then the two triangles are congruent.
Proof:
Given
Now since CAB (same as CDE) and FDE are right angles, line DE is perpendicular to
I was wondering whether there's a proof of AAS that avoids TEAI (or the Corresponding Angles Test, which can also be proved as a result of TEAI). As the U of Chicago uses AAS to prove HL and we've already been reversing many of the proofs in the text, I'm wondering whether we might possibly use HL to prove AAS (drawing in altitudes in order to generate right triangles) -- but I wasn't able to find such a proof.
Dr. Randall Holmes, a math professor at Boise State, also tried to find a proof of AAS that avoids the TEAI, but he could not find such a proof. Two years ago, he wrote:
http://math.boisestate.edu/~holmes/math311/M311S13announcements.html
"AAS proof note: I'm convinced that there is no way to prove AAS without using the exterior angle theorem, which makes it less attractive as a test proof (because of the need for cases – but see that I actually handle the cases quite compactly below). By the way, the ASA proof does not need cases, because the application of the Angle Construction Postulate in it does not depend on the position of the new point in the same way the application of the Exterior Angle theorem in the AAS proof does."
We can easily why we need two cases in our proof of AAS using TEAI. Recall that in our above proof of AAS, the image C' could either be on the segment
Now Dr. Holmes isn't merely a geometer -- he's also a set theorist. Set theory ultimately goes back to the mathematician Georg Cantor. (Yes, the same Cantor for whom the Cantor dust is named. But no, he is not one of the mathematicians whose biography is given in Mandelbrot's book.) Now as it turned out, Cantor's original theory led to contradictions (for example, a set containing all sets was problematic for Cantor). Ever since then, set theorists have been trying to find new theories that avoid the contradiction of the Cantor's theory.
Now here's where Holmes comes in -- two years ago, he completed a proof that an alternate set theory, called New Foundations, allows for a set of all sets without contradictions. This theory is complicated -- recall that I once called the Axiom of Choice the set theorists' Parallel Postulate. Well, the Axiom of Choice is not even compatible with New Foundations.
Thus Holmes is undoubtedly an expert at proofs and determining which theorems can be proved using which axioms or postulates. So if someone like Holmes is unable to come up with a proof of AAS without using TEAI (or a theorem derived from TEAI), then who am I even to try?
And so today I post worksheets for SAS, ASA, and SSS, but not AAS yet. We'll get to HL next week since it's not until Lesson 7-5 of the U of Chicago text. I'm still holding out hope that I can find a proof of AAS from HL by the time we get to Lesson 7-5 (but considering what Holmes wrote above, don't count on it).
Here are the worksheets for today. I've actually written a worksheet for ASA last year, but I couldn't post it due to the computer problems I had last Thanksgiving. I'd written it using the prime-notation from the U of Chicago text, where we begin with some isometry mapping Triangle ABC to A'B'C' as we prepare for the final reflection. The new worksheets for SAS and SSS that I created this year do not refer to Triangle A'B'C' -- instead we abuse the name ABC again.
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