2002 AMC 10A Problems/Problem 23

Problem 23

Points $A,B,C$ and $D$ lie on a line, in that order, with $AB = CD$ and $BC = 12$ . Point $E$ is not on the line, and $BE = CE = 10$ . The perimeter of $\triangle AED$ is twice the perimeter of $\triangle BEC$ . Find $AB$ .

$\text{(A)}\ 15/2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 17/2 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 19/2$

Solution

First, we draw an altitude to $BC$ from $E$ . Let it intersect at $M$ . As $\triangle BEC$ is isosceles, we immediately get $MB=MC=6$ , so the altitude is $8$ . Now, let $AB=CD=x$ . Using the Pythagorean Theorem on $\triangle EMA$ , we find $AE=\sqrt{x^2+12x+100}$ . From symmetry, $DE=\sqrt{x^2+12x+100}$ as well. Now, we use the fact that the perimeter of $\triangle AED$ is twice the perimeter of $\triangle BEC$ .

We have $2\sqrt{x^2+12x+100}+2x+12=2(32)$ so $\sqrt{x^2+12x+100}=26-x$ . Squaring both sides, we have $x^2+12x+100=676-52x+x^2$ which nicely rearranges into $64x=576\rightarrow{x=9}$ . Hence, AB is 9 so our answer is $\boxed{\text{(D)}}$ .

Simpler Solution

Let $M$ be the foot of the altitude from $E$ to $BC.$ Then $MB=MC=6$ because $\triangle BEC$ is isosceles. By the Pythagorean triple $(6,8,10)$ the altitude is $8.$ Since $(8,15,17)$ is the only primitive Pythagorean triple with leg $8,$ we test $AE=DE=17,AM=DM=15.$ Since $2(10+10+12)=(17+17+2\cdot 15)$ this works, giving us $AB=15-9=\boxed{\text{(D)}\ 9}.$

~dolphin7

2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2002 AMC 10A Problems/Problem 23

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Problem 23

Solution

Simpler Solution

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