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Geometry Difficulty 5.6 AIME, harder Prove it United States

Problem:

Let $ABC$ be an equilateral triangle of side length $6$ inscribed in a circle $\omega$ . Let $A_1, A_2$ be the points (distinct from $A$ ) where the lines through $A$ passing through the two trisection points of $BC$ meet $\omega$ . Define $B_1, B_2, C_1, C_2$ similarly. Given that $A_1, A_2, B_1, B_2, C_1, C_2$ appear on $\omega$ in that order, find the area of hexagon $A_1A_2B_1B_2C_1C_2$ .

Solution

Solution:

Answer: $\frac{846 \sqrt{3}}{49}$

Let $A'$ be the point on $BC$ such that $2BA' = A'C$ . By law of cosines on triangle $AA'B$ , we find that $AA' = 2\sqrt{7}$ . By power of a point, $A'A_1 = \frac{2 \times 4}{2\sqrt{7}} = \frac{4}{\sqrt{7}}$ . Using side length ratios, $A_1A_2 = 2 \frac{AA_1}{AA'} = 2 \frac{2\sqrt{7} + \frac{4}{\sqrt{7}}}{2\sqrt{7}} = \frac{18}{7}$ .

Now our hexagon can be broken down into equilateral triangle $A_1B_1C_1$ and three copies of triangle $A_1C_1C_2$ . Since our hexagon has rotational symmetry, $\angle C_2 = 120^\circ$ , and using law of cosines on this triangle with side lengths $\frac{18}{7}$ and $6$ , a little algebra yields $A_1C_2 = \frac{30}{7}$ (this is a 3-5-7 triangle with an angle $120^\circ$ ).

The area of the hexagon is therefore $\frac{6^2 \sqrt{3}}{4} + 3 \times \frac{1}{2} \frac{18}{7} \frac{30}{7} \frac{\sqrt{3}}{2} = \frac{846 \sqrt{3}}{49}$

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