AMC 10 · 2009 · #13

Grade 8 geometry-2d
pythagorean-theoreminteger-pythagorean-triplesarea-triangles identify-subproblemscasework ↑ Prerequisites: pythagorean-theoremarea-triangles
📏 Medium solution 💡 3 insights
Problem
A triangle has two known sides and a known altitude to the third. Two different triangles fit. Find the sum of the two possible third sides.

Pick an answer.

(A)
15
(B)
16
(C)
17
(D)
18
(E)
19
How to solve
Strategy Draw a Diagram

Tool #1 (Draw a Diagram) is primary because the altitude is the only object in the problem that is not yet drawn, and drawing it turns one slanted triangle into two right triangles. Tool #7 (Identify Subproblems) then handles those two right triangles separately: each one has a hypotenuse and the shared leg 12, so each gives up its own horizontal piece by the Pythagorean theorem. Those two numbers come out completely forced — which means the entire problem is not arithmetic at all, but the placement question: where on line BC can the foot of the altitude actually sit? Tool #2 (Make a Systematic List) answers that by enumerating every position instead of assuming one, which is what turns "two possible values" into something proved rather than taken on faith. Tool #6 (Guess and Check) then builds each candidate triangle and measures its altitude, because listing a possibility is not the same as showing it exists. Tool #3 (Eliminate Possibilities) closes by reading the answer list, which quietly confirms that no single triangle can be the whole story.

1STEP 1

Drop the altitude and name its foot

The altitude splits the picture into two right triangles.

AD = 12, ∠ ADB = ∠ ADC = 90°
2STEP 2

Pythagoras fixes both pieces

Each fixes its own piece of the base.

BD = √(13² - 12²) = 5, DC = √(15² - 12²) = 9
3STEP 3

List every way the points can sit

The foot can land on either side, giving two lengths.

B ∈ {-5, 5}, C ∈ {-9, 9} → BC = |C - B| ∈ {4, 14}
4STEP 4

Build both triangles and check them

Both triangles really exist, so both count.

BC = 14: 1/2(5)(12) + 1/2(9)(12) = 84 = 1/2(14)(12) BC = 4: 1/2(9)(12) - 1/2(5)(12) = 24 = 1/2(4)(12)
5STEP 5

Add the two lengths

Adding gives 18, choice (D).

14 + 4 = 18 → (D)
Answer
18
The first triangle is the well-known 13-14-15 triangle, whose area is 84; its altitude to the 14-side is (2 · 84)/14 = 12 exactly, matching the given number on the nose. The second, 4-13-15, is legal but barely so: 4 + 13 = 17 only just clears 15, which is precisely the picture of an extremely obtuse triangle whose altitude foot has slid off the end of the base — the geometry and the arithmetic agree about which triangle is the strange one. There is also a size window that catches any stray value: with two sides 13 and 15, the third side must satisfy 15 - 13 < BC < 15 + 13, that is 2 < BC < 28. Both 14 and 4 sit inside that window, and no third candidate was dropped, since the enumeration of foot positions was exhaustive. Finally, 18 sits comfortably inside the range of the choices, 15 through 19, so nothing about the size is suspicious — but the real assurance is that both triangles were built and measured, not merely named.
💡Key takeaway

The foot of an altitude does not have to land between the two corners of the side — check whether it can fall outside, because that outside picture is a whole second answer.

  • Drop the altitude and name its foot
  • Pythagoras fixes both pieces
  • List every way the points can sit
  • Build both triangles and check them
  • Add the two lengths