AMC 10 · 2002 · #23

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
pythagorean-theoremisosceles-trianglesystems-of-equations convert-to-algebra ↑ Prerequisites: pythagorean-theorem
📏 Long solution 💡 3 insights
Problem
Four points A, B, C, D sit on one line in that order, with the two outer gaps equal (AB = CD) and the middle gap BC = 12. A point E off the line is the same distance from B and from C: BE = CE = 10. The perimeter of the big triangle AED is exactly twice the perimeter of the small triangle BEC. Find the length AB.

Pick an answer.

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

AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

A clear picture is the whole game here. Once you plot the four points and E, the equal distances BE = CE (and AB = CD) make the figure mirror-symmetric, so E sits directly above the midpoint of BC — and of AD. Dropping that vertical altitude (Tool #1) splits each triangle into right triangles, which is exactly what the Pythagorean theorem needs. The perimeter condition can't be used until we name the unknown, so we let AB = x (Tool #4) and write both AD and the slant side AE in terms of x. That gives two expressions for the same length AE; setting them equal (Tool #13) collapses to one equation for x. Tool #7 (Identify Subproblems) keeps the two jobs — find the altitude, then find AE — separate and clean.

1STEP 1

Draw it and drop the altitude

Plot A, B, C, D left to right with E above. BE = CE = 10 makes BEC isosceles, so E sits over the midpoint M: BM = 6, height EM = 8.

BM = 6, EM = √(10² - 6²) = √(64) = 8
2STEP 2

Name AB and use the perimeter fact

Small perimeter 10 + 10 + 12 = 32, so AED must total 64. Let AB = x: AD = 2x + 12, and symmetry gives AE = DE, so AE = 26 - x.

perim(BEC)=32 → perim(AED)=64; AD = 2x+12; AE = (64-(2x+12))/2 = 26 - x
3STEP 3

Measure AE from the right triangle

AB = CD makes M the midpoint of AD too, so the same height 8 drops there. With AM = x + 6, right triangle AME gives AE² = (x+6)² + 64.

AM = x + 6, AE² = (x+6)² + 8² = (x+6)² + 64
4STEP 4

Set the two AE's equal and solve

Set them equal: (26 - x)² = (x+6)² + 64. The x² cancels, leaving 676 - 52x = 12x + 100, so 64x = 576 and AB = 9, choice (D).

(26-x)² = (x+6)² + 64 → 676 - 52x = 12x + 100 → 64x = 576 → x = 9 (D)
Answer
9
Plug x = 9 back in. Then AM = 9 + 6 = 15 and EM = 8, so AE = √(15² + 8²) = √(225 + 64) = √(289) = 17 — a clean 8-15-17 right triangle, and it matches 26 - x = 26 - 9 = 17. The big triangle then has AD = 2(9) + 12 = 30 and two slant sides of 17, giving perimeter 30 + 17 + 17 = 64, which is exactly twice the small triangle's 32. Everything lines up, so (D) is correct. The half-integer choices like 15/2 or 17/2 are traps for anyone who mishandles the factor of two.
💡Key takeaway

Drop the altitude to split the figure into right triangles, write the slant side two ways — once from the doubled perimeter and once from the Pythagorean theorem — and make them agree to pin down AB = 9.

  • Draw it and drop the altitude
  • Name AB and use the perimeter fact
  • Measure AE from the right triangle
  • Set the two AE's equal and solve

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