2024 SSMO Accuracy Round Problems/Problem 4

Problem

Right triangle $ABC$ has a right angle at $C$ and hypotenuse $1$ . Let points $D$ and $E$ lie on $AC$ such that $\angle BDC=\angle BEC=45^{\circ}$ . $A,D,C,$ and $E$ are colinear in that order. Given that $EA=13DA$ , the area of $\triangle ABC$ can be expressed as $\frac{m}{n}$ for relatively prime $m$ and $n$ . Find $m+n$ .

Solution

Since $BDC$ and $BEC$ are isosceles right triangles, we have $DC = BC = EC$ . Since $AD+DC+CE = EA = 13AD,$ we have $DC+CE = 12AD \implies DC = CE = 6AD.$ From here, we see that $AC = AD+DC = 7AD.$ By the Pythagorean Theorem, $AC^2+BC^2 = AB^2 = 1\implies (7AD)^2+(6AD)^2 = 1\implies AD^2 = \frac{1}{85}.$ The area of triangle $ABC$ is $\frac{AC\cdot BC}{2} = 21AD^2 = \frac{21}{85}\implies 21+85 = \boxed{106}.$

~SMO_Team

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2024 SSMO Accuracy Round Problems/Problem 4

Problem

Solution