Learning through Teaching - An Example
by rrusczyk, Jun 17, 2006, 3:21 PM
#14 of the second AIME this year was:
In triangle ABC, AB = 13, BC = 15, and CA = 14. Point D is on BC with C D= 6. Point E is on BC such that
. Given that BE = p/q, where p and q are relatively prime positive integers, find q.
Had I seen this problem as little as 2 years ago, I probably would have tried a little geometry, maybe finding the somewhat clever angle bisector + Menelaus's Theorem approach. But I probably would have gotten stuck and then just trigged it out with a little help from Stewart's Theorem. But because I've been teaching brilliant students for the last two years, I was able to solve this with elementary geometry in just a few minutes. I used basic tools we teach even in our Intro Geometry class.
The problem statement reminded me of the Angle Bisector Theorem, which tell us that if W is on YZ such that XW is an angle bisector of XYZ, then XY/YW = XZ/ZW.
Thinking of the Angle Bisector Theorem made me think of how to prove the Theorem, since I hoped to use the same approach from that proof in this problem. We prove the Angle Bisector Theorem by creating similar triangles (the equal angles and the ratios in the Angle Bisector Theorem are our tip-off to look for similar triangles). Parallel lines make similar triangles, so we draw a line through Z parallel to XY. Extending our angle bisector gives us our similar triangles.
We have XYZ ~ VZW, and VZ = ZX, so XY/YW = ZV/ZW = XZ/ZW.
So, in our AIME problem, I tried the same thing. I add a line through C parallel to AB, and I extend the cevians AD and AE to meet it as shown.
Voila! Similar triangles galore. Specifically, we have
$\begin{eqnarray*} \triangle ABE &\sim&\triangle XCE \\ \triangle ABD &\sim &\triangle YCD \\ \triangle AXC &\sim & \triangle YAC \end{eqnarray*}$
We know a lot of the lengths of the sides of these triangles - see if you can finish the problem from here!
My main purpose in relating this story isn't to talk about the Angle Bisector Theorem, or the power of cleverly adding a parallel line to a geometry problem. My main purpose is to relate the power of teaching as a learning tool. I simply would not have ever found this clever solution had I not taught the Angle Bisector Theorem proof several times in the last year. Each time I taught it, I preached to the students the power of hunting for similar triangles, and adding parallels in order to create them. The lesson sunk in to me, at least!
Teach. No activity, not even intense studying, will make you learn a subject as thoroughly as preparing yourself to teach it will.
In triangle ABC, AB = 13, BC = 15, and CA = 14. Point D is on BC with C D= 6. Point E is on BC such that
. Given that BE = p/q, where p and q are relatively prime positive integers, find q.Had I seen this problem as little as 2 years ago, I probably would have tried a little geometry, maybe finding the somewhat clever angle bisector + Menelaus's Theorem approach. But I probably would have gotten stuck and then just trigged it out with a little help from Stewart's Theorem. But because I've been teaching brilliant students for the last two years, I was able to solve this with elementary geometry in just a few minutes. I used basic tools we teach even in our Intro Geometry class.
The problem statement reminded me of the Angle Bisector Theorem, which tell us that if W is on YZ such that XW is an angle bisector of XYZ, then XY/YW = XZ/ZW.
Thinking of the Angle Bisector Theorem made me think of how to prove the Theorem, since I hoped to use the same approach from that proof in this problem. We prove the Angle Bisector Theorem by creating similar triangles (the equal angles and the ratios in the Angle Bisector Theorem are our tip-off to look for similar triangles). Parallel lines make similar triangles, so we draw a line through Z parallel to XY. Extending our angle bisector gives us our similar triangles.
We have XYZ ~ VZW, and VZ = ZX, so XY/YW = ZV/ZW = XZ/ZW.
So, in our AIME problem, I tried the same thing. I add a line through C parallel to AB, and I extend the cevians AD and AE to meet it as shown.
Voila! Similar triangles galore. Specifically, we have
$\begin{eqnarray*} \triangle ABE &\sim&\triangle XCE \\ \triangle ABD &\sim &\triangle YCD \\ \triangle AXC &\sim & \triangle YAC \end{eqnarray*}$
We know a lot of the lengths of the sides of these triangles - see if you can finish the problem from here!
My main purpose in relating this story isn't to talk about the Angle Bisector Theorem, or the power of cleverly adding a parallel line to a geometry problem. My main purpose is to relate the power of teaching as a learning tool. I simply would not have ever found this clever solution had I not taught the Angle Bisector Theorem proof several times in the last year. Each time I taught it, I preached to the students the power of hunting for similar triangles, and adding parallels in order to create them. The lesson sunk in to me, at least!
Teach. No activity, not even intense studying, will make you learn a subject as thoroughly as preparing yourself to teach it will.
February 2012
January 2012