2024 AMC 10A Problem 14

Attempt Problem 14 of the 2024 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 10A solutions, or check the answer key.

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14.

One side of an equilateral triangle of height 2424 lies on line .\ell. A circle of radius 1212 is tangent to \ell and is externally tangent to the triangle. The area of the region exterior to the triangle and the circle and bounded by the triangle, the circle, and line \ell can be written as abcπ,a\sqrt{b} - c\pi, where a,a, b,b, and cc are positive integers and bb is not divisible by the square of any prime. What is a+b+c?a + b + c?

7272

7373

7474

7575

7676

Answer: D
Concepts:tangent linespecial right trianglesectorkite
Difficulty rating: 1660
Solution:

The equilateral triangle has side 163.16\sqrt3. Put \ell on the xx-axis with base vertex V=(163,0)V = (16\sqrt3, 0); the slanted side lies on 3x+y=48.\sqrt3\,x + y = 48. The circle sits on ,\ell, has radius 12,12, and touches that side externally, so its center is O=(203,12).O = (20\sqrt3, 12). Let T=(203,0)T = (20\sqrt3, 0) be its tangency point on ,\ell, and let PP be the tangency point on the slanted side. The two tangent lengths from VV satisfy VT=VP=43,VT = VP = 4\sqrt3, so kite VTOPVTOP has area 4312=483.4\sqrt3 \cdot 12 = 48\sqrt3. The angle at VV is 120,120^\circ, so the removed sector has angle 6060^\circ and area 16π(12)2=24π.\tfrac16 \pi (12)^2 = 24\pi. The region has area 48324π,48\sqrt3 - 24\pi, giving a+b+c=48+3+24=75.a + b + c = 48 + 3 + 24 = 75. Therefore, the answer is D.

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