AMC 10 · 2013 · #15

Grade 7 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

area-trianglesratio-proportion convert-to-algebrabound-inequality-then-enumerate ↑ Prerequisites: area-triangles

📏 Medium solution 💡 2 insights

Problem

A triangle has two sides of length 10 and 15. In a triangle, each side has an altitude drawn to it (the perpendicular height when that side is the base). The altitude drawn to the third side equals the average of the altitudes drawn to the sides of length 10 and 15. Find the length of the third side.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

The altitudes are never given as numbers, so the useful move is to name the one quantity that ties them all together: the triangle's area. That is Tool #4 (Introduce a Variable) — call the area S. The single triangle has one area, and S = 1/2 (base)(height) for each of the three bases, so every altitude can be written as 2S/side. Once each altitude is an expression in S, Tool #13 (Convert to Algebra) turns the words "is the average of" into an equation. The reason this works so cleanly is that the unknown area S appears in every term and cancels, leaving a short equation in x alone.

1STEP 1

Name the shared area

One triangle has one area whichever side is the base — call it S and write area = half base times height for all three sides.

S = 1/2 · 10 · h₁₀ = 1/2 · 15 · h₁₅ = 1/2 · x · h_x

2STEP 2

Write each altitude

Divide each copy by its own base: every altitude becomes twice the area over that side, giving $\frac{2S}{10}$ , $\frac{2S}{15}$ , $\frac{2S}{x}$ .

h₁₀ = 2S/10 = S/5, h₁₅ = 2S/15, h_x = 2S/x

3STEP 3

Translate "average"

Turn the sentence straight into algebra: the third altitude equals the sum of the other two divided by 2.

h_x = (h₁₀ + h₁₅)/2

4STEP 4

Cancel the area

Substitute, and the factor 2S divides out of every term, leaving $\frac{1}{x}$ as half of $\frac{1}{10}+\frac{1}{15}=\frac{1}{6}$ .

2S/x = 1/2(2S/10 + 2S/15) → 1/x = 1/2(1/10 + 1/15) ; 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6

5STEP 5

Solve for the side

Half of $\frac{1}{6}$ is $\frac{1}{<span class="hl-ask">12</span>}$ , so equal reciprocals force the third side to be 12 — choice (D).

1/x = 1/2·1/6 = 1/12 → x = 12 → (D)

Answer

Plug in a concrete area to test. Give the triangle area S = 30: then h₁₀ = (2 · 30)/10 = 6 and h₁₅ = (2 · 30)/15 = 4, whose average is 5. The altitude to a side of length 12 would be h_x = (2 · 30)/12 = 5 — it matches the required average exactly. The sides 10, 12, 15 also pass the triangle inequality (10 + 12 ＞ 15), so a real triangle exists. Both checks confirm x = 12.

💡Key takeaway

A triangle's area is the same whichever side you call the base, so each altitude equals twice the area divided by its side — name the area once, watch it cancel, and a clean equation for the missing side is all that is left.

Name the shared area
Write each altitude
Translate "average"
Cancel the area
Solve for the side

Name the shared area

Write each altitude

Translate "average"

Cancel the area

Solve for the side

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