AMC 10 · 2025 · #21

Grade 11 geometry-2d
law-of-cosinestrigonometric-ratiossupplementary-angles bound-inequality-then-enumerateconvert-to-algebra ↑ Prerequisites: area-triangles
📏 Long solution 💡 4 insights
Problem
Two triangles have the same area but are not congruent. Each of them has one side of length 8 and one side of length 9, and in each the remaining third side has whole-number length. Find the sum of those two third sides.

Pick an answer.

(A)
20
(B)
22
(C)
24
(D)
26
(E)
28
How to solve
Strategy Introduce a Variable

Two sides are pinned at 8 and 9, so the only freedom left in each triangle is the angle squeezed between them. Naming that angle (Introduce a Variable) turns both the area and the third side into expressions in one unknown, and a quick sketch of the two triangles (Draw a Diagram) shows why one must be squat and the other tall. Equal area then forces a relation between the two angles, the Law of Cosines converts that relation into a single equation with no angle left in it, and what remains is a small integer hunt: bound the third side with the triangle inequality (Eliminate Possibilities) and scan the surviving candidates (Make a Systematic List) to prove exactly one pair works, then push that pair back through the Law of Cosines (Guess and Check) to confirm it really builds a triangle.

1STEP 1

Name the angle between 8 and 9

The included angle decides everything.

sides 8, 9 and included angle θ ⟶ triangle determined; 0° < θ < 180°
2STEP 2

Equal area forces equal sines

Equal areas force equal sines.

1/2 · 8 · 9sinθ₁ = 1/2 · 8 · 9sinθ₂ ⟹ sinθ₁ = sinθ₂
3STEP 3

Non-congruent makes them supplementary

Not congruent means they are supplementary.

θ₁ ≠ θ₂, sinθ₁ = sinθ₂ ⟹ θ₂ = 180° - θ₁
4STEP 4

Law of Cosines on both triangles

The cosine law gives two formulas differing only in sign.

s₁² = 8² + 9² - 2 · 8 · 9cosθ₁ = 145 - 144cosθ₁, s₂² = 145 - 144cos(180° - θ₁) = 145 + 144cosθ₁
5STEP 5

Add the equations to kill the angle

Adding kills the angle: the squares sum to 290.

s₁² + s₂² = (145 - 144cosθ₁) + (145 + 144cosθ₁) = 290
6STEP 6

Bound each third side

The triangle inequality traps each side between 2 and 16.

9 - 8 < s < 9 + 8 ⟹ 1 < s < 17 ⟹ s ∈ {2, 3, …, 16}
7STEP 7

Scan for two squares summing to 290

The only squares summing to 290 are 11 and 13.

290 ≡ 2 (mod 4) ⟹ s₁, s₂ both odd; 11² + 13² = 121 + 169 = 290
8STEP 8

Confirm the pair and add

Verified and added, that is 24.

cosθ₁ = (145 - 121)/144 = 1/6 ∈ (-1, 1), s₁ + s₂ = 11 + 13 = 24
Answer
24
Both triangles should have the same area, and they do: with cosθ₁ = 1/6, sinθ₁ = √(35)/6, so each area is 36 · √(35)/6 = 6√(35) ≈ 35.5. The 8-9-11 triangle is the acute-hinge one and the 8-9-13 triangle the obtuse-hinge one, matching the picture of one squat and one stretched triangle. Both third sides sit comfortably inside the allowed band from 2 to 16, and the sum 24 lands exactly on choice (C). A sanity note on scale: 24 is a bit above 2 × 12, which is what you would expect since the two third sides straddle the balanced value √(145) ≈ 12.04.
💡Key takeaway

When two sides are fixed, the angle between them is the only dial — equal area means the two triangles use an angle and its supplement, and since cosine flips sign between them, the two third sides squared always add up to the same fixed number.

  • Name the angle between 8 and 9
  • Equal area forces equal sines
  • Non-congruent makes them supplementary
  • Law of Cosines on both triangles
  • Add the equations to kill the angle
  • Bound each third side
  • Scan for two squares summing to 290
  • Confirm the pair and add