2023 SSMO Team Round Problems/Problem 3
Problem
Let be a triangle such that and Let be the circumcircle of . Let be on the circle such that Let be the point diametrically opposite of . Let be the point diametrically opposite . Find the area of the quadrilateral in terms of a mixed number . Find .
Solution
Note that is right with the right angle at . This means that is the diameter of the circle. We can divide quadrilateral into and , both of which are right triangles. Mark the intersection point between BD and AC as G. We can use the fact that and are similar to find that , so by symmetry. Then, by the Pythagorean Theorem, so the area of is . Since by symmetry, , so the area of is . This means that the area of the entire quadrilateral equals , so the answer is .
~alexanderruan