2002 AMC 10A Problem 23

Let $AB = CD = x$ and let $M$ be the midpoint of $BC$. Since $EB = EC$, the segment $EM$ is perpendicular to the line, with $BM = MC = 6$ and $EM = \sqrt{10^2 - 6^2} = 8$.

Triangle $BEC$ has perimeter $10 + 10 + 12 = 32$, so triangle $AED$ has perimeter $64$. By symmetry about $EM$, $AE = DE$, and $AM = x + 6$, so
$$AE = \sqrt{(x+6)^2 + 8^2}.$$
The perimeter condition reads
$$2\sqrt{(x+6)^2 + 64} + (2x + 12) = 64 \quad\Longrightarrow\quad \sqrt{(x+6)^2 + 64} = 26 - x.$$
Square both sides:
$$x^2 + 12x + 100 = 676 - 52x + x^2 \quad\Longrightarrow\quad 64x = 576 \quad\Longrightarrow\quad x = 9.$$

Check: $AM = 15$, $EM = 8$, $AE = 17$, and $17 + 17 + 30 = 64$.

The answer is $\boxed{\textbf{(D)}\ 9}$.

An isosceles apex sitting over a line invites the altitude to the base; it produces a right triangle with a known leg ($8$) for every point on the line, so all the slanted lengths become Pythagorean expressions in one variable. The $x^2$ terms cancel after squaring because both sides are quadratic with leading coefficient $1$, leaving a linear equation. Symmetry ($AE = DE$) halves the work.

Guess a Pythagorean triple: the altitude is $8$, and $8$-$15$-$17$ is the natural triple with leg $8$ other than $6$-$8$-$10$. If $AM = 15$ then $AE = 17$, $AD = 30$, and the perimeter $17 + 17 + 30 = 64$ is exactly twice $32$. Hence $AB = 15 - 6 = 9$.

• Setting the perimeter of AED to twice 32 but forgetting that AD = 2x + 12 contains the two unknown segments, or squaring without isolating the square root.
• Taking $AM = x$ instead of $x + 6$, which yields a different (non-answer) value and signals an error only if the check is done.