HMMT-Nov 2018 thm, Problem 7

Ben "One Hunna Dolla" Franklin is flying a kite KITE such that $I E$ is the perpendicular bisector of $K T$. Let $I E$ meet $K T$ at $R$. The midpoints of $K I, I T, T E, E K$ are $A, N, M, D$, respectively. Given that $[M A K E]=18, I T=10,[R A I N]=4$, find $[D I M E]$. Note: $[X]$ denotes the area of the figure $X$.

Solution: Let $[K I R]=[R I T]=a$ and $[K E R]=[T E R]=b$. We will relate all areas to $a$ and $b$. First, \[ [R A I N]=[R A I]+[I N R]=\frac{1}{2} a+\frac{1}{2} a=a \] Next, we break up $[M A K E]=[M A D]+[A K D]+[D E M]$. We have \[ \begin{aligned} & {[M A D]=\frac{A D \cdot D M}{2}=\frac{1}{2} \cdot \frac{I E}{2} \cdot \frac{K T}{2}=\frac{[K I T E]}{4}=\frac{a+b}{2}} \\ & {[A K D]=\frac{[K I E]}{4}=\frac{a+b}{4}} \\ & {[D E M]=\frac{[K T E]}{4}=\frac{b}{2}} \end{aligned} \] After adding these we get $[M A K E]=\frac{3 a+5 b}{4}$. We want to find \[ [D I M E]=2[I M E]=[I T E]=a+b=\frac{4}{5}\left(\frac{3 a+5 b}{4}\right)+\frac{2}{5} a=\frac{4}{5} \cdot 18+\frac{2}{5} \cdot 4=16 \] $\fbox{16}$.

Answer: 16