2023 SSMO Team Round Problems/Problem 3

Problem

Let $ABC$ be a triangle such that $AB=4\sqrt{2}, BC=5\sqrt{2},$ and $AC=\sqrt{82}.$ Let $\omega$ be the circumcircle of $\triangle ABC$ . Let $D$ be on the circle such that $\overline{BD} \perp \overline{AC}.$ Let $E$ be the point diametrically opposite of $B$ . Let $F$ be the point diametrically opposite $D$ . Find the area of the quadrilateral $ADEF$ in terms of a mixed number $a\frac{b}{c}$ . Find $a+b+c$ .

Solution

Note that $\Delta{ABC}$ is right with the right angle at $B$ . This means that $AC$ is the diameter of the circle. We can divide quadrilateral $ADEF$ into $\Delta{DEF}$ and $\Delta{FAD}$ , both of which are right triangles. Mark the intersection point between BD and AC as G. We can use the fact that $\Delta{ABC}$ and $\Delta{AGB}$ are similar to find that $BG=\frac{20\sqrt{2}}{\sqrt{41}}$ , so $BD=\frac{40\sqrt{2}}{\sqrt{41}}=FE$ by symmetry. Then, $FD=\frac{9\sqrt{2}}{\sqrt{41}}=FE$ by the Pythagorean Theorem, so the area of $\Delta{DEF}$ is $\frac{360}{41}$ . Since $DA=FA=4\sqrt{2}$ by symmetry, $FA=4\sqrt{2}$ , so the area of $\Delta{FAD}$ is $20$ . This means that the area of the entire quadrilateral equals $\frac{360}{41}+20=28\frac{32}{41}$ , so the answer is $28+32+41=\boxed{101}$ .

~alexanderruan

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2023 SSMO Team Round Problems/Problem 3

Problem

Solution