Tuesday, December 13, 2011

MGRE problem: wrong!

Even a teacher can get things wrong, and this time around, I did.

My answer to last week's Math Beast Challenge problem turns out to be incorrect. You'll recall that my answer was (D); MGRE's answer is (C): the quantities are equal. And they're right. But why? Because of one little fact about right triangles that I had missed: the triangle's altitude, drawn from the vertex of the two legs to a point on the hypotenuse, creates two right triangles that are geometrically similar to the large triangle. I should have realized this. Anyway, without further ado, here's part of MGRE's explanation for why (C) is correct:

This could be solved with the Pythagorean Theorem, as there are three right triangles in the figure: the small one on the left, the bigger one on the right, and the largest right triangle comprised of the other two. It should also be noted that these three triangles are similar triangles; that is, the three triangles have the same three angle measures.

For the largest triangle, a2 + b2 = c2 so by substitution, Quantity B = hc2. Now that Quantity B is more similar in form to Quantity A, we will compare.

Quantity A: abc
Quantity B: hc2

Divide both quantities by c. Dividing both quantities by the same positive number will not change the relative values; the larger quantity will still be larger. This comparison becomes

Quantity A: ab
Quantity B: hc

For similar triangles, the ratios of side lengths will be equal. For example, the ratio of the short leg to the hypotenuse will be the same in each triangle.

(short leg)/(hypotenuse) = a/c (from the largest triangle) = h/b (from the triangle on the right)

a/c = h/b

By cross-multiplying, we conclude that ab = hc and thus the two quantities are equal.

MGRE's explanation continues, but it's basically a plug-in-the-numbers approach. What bugs me is that I was obviously on the right track, but I stopped in my ruminations before I'd figured out the "similar right triangles" part. Had I done that, I'd have seen that the equality I had discovered for one case (45-45-90 triangles) must also obtain for all cases.

Live and learn, eh?


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