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2002 AMC 10A Problem 23

2002 AMC 10A Problem 23AMC 10 Step by Step

Name: Date: October 28, 2026

Triangles: Area & Pythagorean · difficulty 4 of 5 · about 6 min · position P21-25

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 $AB$ .

A) $15/2$
B) $8$
C) $17/2$
D) $9$
E) $19/2$

Answer

Key insight

The line from E down to the middle of BC has length 8, so AE is the long side of a right triangle with legs 8 and AB + 6.

Solution

Point $E$ is the same distance from $B$ and $C$ , so it sits directly above the middle of $BC$ . A line straight down from $E$ gives right triangles, and right triangles let us find the slanted lengths.

1. The height of $E$ . Let $M$ be the midpoint of $BC$ , so $BM = MC = 6$ . Triangle $EMB$ has a right angle at $M$ and long side $EB = 10$ , so
$EM = \sqrt{10^2 - 6^2} = \sqrt{64} = 8.$
(It is the $6$ - $8$ - $10$ right triangle.)

The perimeters. Triangle $BEC$ has perimeter $10 + 10 + 12 = 32$ , so triangle $AED$ has perimeter $64$ .

3. Name the unknown. Let $AB = CD = x$ . Then
- $AD = x + 12 + x = 2x + 12$ ,
- $AM = x + 6$ , and by the same picture on the other side, $DM = x + 6$ , so $AE = DE$ .

4. Write the perimeter condition. $AE + DE + AD = 64$ means $2 \cdot AE + 2x + 12 = 64$ , so
$AE = 26 - x.$

5. Use the right triangle $AME$ . Its legs are $AM = x + 6$ and $EM = 8$ , and its long side is $AE$ :
$(x + 6)^2 + 8^2 = (26 - x)^2.$

6. Solve. Multiply out both sides:
$x^2 + 12x + 36 + 64 = 676 - 52x + x^2.$
The $x^2$ cancels on both sides, leaving $64x = 576$ , so $x = 9$ .

Check. $AM = 15$ and $EM = 8$ give $AE = 17$ (the $8$ - $15$ - $17$ right triangle). Perimeter: $17 + 17 + 30 = 64$ . Correct.

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

Why this works

A point that is equally far from two points on a line sits right above their midpoint, so dropping a line straight down creates right triangles with a known height. Every slanted length then comes from the Pythagorean theorem with one unknown. The $x^2$ terms cancel because both sides of the equation are squares of something with a single $x$ , so the equation ends up easy.

Alternative approach

Guess with right triangles that have whole-number sides. The height is $8$ , and besides $6$ - $8$ - $10$ , the familiar right triangle with a side $8$ is $8$ - $15$ - $17$ . Try $AM = 15$ , $AE = 17$ : then $AD = 30$ and the perimeter is $17 + 17 + 30 = 64$ , exactly twice $32$ . So $AB = AM - BM = 15 - 6 = 9$ .

The trap

Forgetting that AD is AB + 12 + CD, so AB appears twice in the perimeter, or squaring before getting the square root alone on one side.

Concise solution

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 $\boxed{\textbf{(D)}\ 9}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Algebra, Chapter 11: Special Factorizations (§11.2, §11.3, §11.5)
Introduction to Geometry, Chapter 19: Problem Solving Strategies in Geometry (§19.1)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)

Techniques

Add construction lines/points (drop altitudes, extend segments, connect centers) · Exploit symmetry to reduce work or pair up objects