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2024 AMC 10A Problem 14

Problem 14 of 25IntermediateGeometry

One side of an equilateral triangle of height 2424 lies on line ℓ.\ell. A circle of radius 1212 is tangent to ℓ\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 ℓ\ell can be written as ab−cπ,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?

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Solution

The equilateral triangle has side 163.16\sqrt3. Put ℓ\ell on the xx-axis with base vertex V=(163,0)V = (16\sqrt3, 0); the slanted side lies on 3 x+y=48.\sqrt3\,x + y = 48. The circle sits on ℓ,\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 ℓ,\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 43⋅12=483.4\sqrt3 \cdot 12 = 48\sqrt3. The angle at VV is 120∘,120^\circ, so the removed sector has angle 60∘60^\circ and area 16π(12)2=24π.\tfrac16 \pi (12)^2 = 24\pi. The region has area 483−24π,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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Tagged: tangent line · special right triangle · sector · kite

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