2002 AMC 10A Problem 23

Below is the professionally curated solution for Problem 23 of the 2002 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 10A solutions, or check the answer key.

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Concepts:isosceles triangle Pythagorean Theorem perimeter

Difficulty rating: 1660

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 $AB.$

$\dfrac{15}{2}$

$8$

$\dfrac{17}{2}$

$9$

$\dfrac{19}{2}$

Solution:

Let $M$ be the midpoint of $BC.$ Since $BE=CE,$ $EM\perp BC$ and $EM=\sqrt{10^2-6^2}=8.$ By symmetry $AE=ED;$ write $AB=CD=x$ and $AE=ED=y.$

The perimeter condition gives $2y+(2x+12)=2(10+10+12)=64,$ so $x+y=26.$ Also $y^2=EM^2+(x+6)^2=64+(x+6)^2.$

Substituting $y=26-x,$ $(26-x)^2=64+(x+6)^2,$ which simplifies to $676-52x=100+12x,$ so $64x=576$ and $x=9.$

Thus, the correct answer is D.

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