HMMT-Nov 2011 guts, Problem 25

Let $X Y Z$ be an equilateral triangle, and let $K, L, M$ be points on sides $X Y, Y Z, Z X$, respectively, such that $X K / K Y=B, Y L / L Z=1 / C$, and $Z M / M X=1$. Determine the ratio of the area of triangle $K L M$ to the area of triangle $X Y Z$.

Solution: First, we note that \[ [K L M]=[X Y Z]-[X K M]-[Y L K]-[Z M L] \] Then, note that \[ \begin{gathered} {[X K M]=\frac{X K}{X Y} \cdot \frac{X M}{X Z} \cdot[X Y Z]=\frac{B}{B+1} \cdot \frac{1}{2} \cdot[X Y Z]} \\ {[Y L K]=\frac{Y L}{Y Z} \cdot \frac{Y K}{Y X} \cdot[X Y Z]=\frac{1}{C+1} \cdot \frac{1}{B+1} \cdot[X Y Z]} \\ {[Z M L]=\frac{Z M}{Z X} \cdot \frac{Z L}{Z Y} \cdot[X Y Z]=\frac{1}{2} \cdot \frac{1}{C+1} \cdot[X Y Z]} \end{gathered} \] Consequently, \[ \begin{aligned} A & =\frac{[K L M]}{[X Y Z]} \\ & =1-\frac{B}{B+1} \cdot \frac{1}{2}-\frac{1}{C+1} \cdot \frac{1}{B+1}-\frac{C}{C+1} \cdot \frac{1}{2} \\ & =\frac{B+C}{(B+1)(C+1)(2)} \end{aligned} \] If we solve our system of equations for $A, B, C$, we get that $A=\frac{1}{5}$. $\fbox{\frac{1}{5}}$.

Answer: \frac{1}{5}