Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Find the answer

Equilateral triangle $ABC$ has circumcircle $\Omega$ . Points $D$ and $E$ are chosen on minor arcs $AB$ and $AC$ of $\Omega$ respectively such that $BC=DE$ . Given that triangle $ABE$ has area 3 and triangle $ACD$ has area 4, find the area of triangle $ABC$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

A rotation by $120^{\circ}$ about the center of the circle will take $ABE$ to $BCD$ , so $BCD$ has area 3. Let $AD=x, BD=y$ , and observe that $\angle ADC=\angle CDB=60^{\circ}$ . By Ptolemy's Theorem, $CD=x+y$ . We have $4=[ACD]=\frac{1}{2} AD \cdot CD \cdot \sin 60^{\circ}=\frac{\sqrt{3}}{4} x(x+y)$ $3=[BCD]=\frac{1}{2} BD \cdot CD \cdot \sin 60^{\circ}=\frac{\sqrt{3}}{4} y(x+y)$ By dividing these equations find $x: y=4: 3$ . Let $x=4t, y=3t$ . Substitute this into the first equation to get $1=\frac{\sqrt{3}}{4} \cdot 7t^{2}$ . By the Law of Cosines, $AB^{2}=x^{2}+xy+y^{2}=37t^{2}$ The area of $ABC$ is then $\frac{AB^{2} \sqrt{3}}{4}=\frac{37}{7}$

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