Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME Prove it United States

Problem:

Unit circle Ω\Omega has points XX, YY, ZZ on its circumference so that XYZXYZ is an equilateral triangle. Let WW be a point other than XX in the plane such that triangle WYZWYZ is also equilateral. Determine the area of the region inside triangle WYZWYZ that lies outside circle Ω\Omega.

Solution

Solution:

Answer: 33−π3\frac{3 \sqrt{3}-\pi}{3}

Let OO be the center of the circle. Then, we note that since ∠WYZ=60∘=∠YXZ\angle WYZ = 60^{\circ} = \angle YXZ, that YWYW is tangent to Ω\Omega. Similarly, WZWZ is tangent to Ω\Omega.

Now, we note that the circular segment corresponding to YZYZ is equal to 13\frac{1}{3} the area of Ω\Omega less the area of triangle OYZOYZ. Hence, our total area is
[WYZ]−13[Ω]+[YOZ]=334−13π+34=33−π3 [WYZ] - \frac{1}{3}[\Omega] + [YOZ] = \frac{3 \sqrt{3}}{4} - \frac{1}{3} \pi + \frac{\sqrt{3}}{4} = \frac{3 \sqrt{3} - \pi}{3}

Figure 1

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.