In the drawing, $AM=MN=NC$ and $\frac {BP}{PC}=\frac {5}{3}$, if the area of the gray region is 8, whats the area of $\triangle ABC$?
I saw that $\triangle ABN$ and $\triangle BNC$ have the same height but their bases are in ratio $2:1$, and the sum of the areas of those triangles gives the final answer, but i don't know how to continue and how to use the area of the gray triangle. I think i'm close.
Any hints?
Best Answer
Add an auxiliary segment through P parallel to AC intersecting BN at Q and AB at S. Call the intersection of BN and MP R. Let $h_1$ be the altitude of triangle QRP
Let $h_2$ be the altitude of triangle MRN
Let $h_3$ be the altitude of triangle PMC
Let $h_4$ be the altitude of triangle BSP
Let $h_5$ be the altitude of triangle BAC
$h_3$ = $h_1$ + $h_2$
$h_5$ = $h_4$ + $h_3$
Triangle BQP is similar to triangle BNC. $\frac {BP}{BC}=\frac {5}{8}$, so $\frac {QP}{NC}=\frac {QP}{MN}=\frac {5}{8}$. Triangle QRP is similar to triangle NRM. So $h_1$ = $\frac {5}{8}$$h_2$
$h_3$ = $h_1$ + $h_2$ = $\frac {5}{8}$$h_2$ + $h_2$ = $\frac {13}{8}$$h_2$
Triangle BSP is similar to triangle BAC, so $h_4$ = $\frac {5}{8}$$h_5$
$h_5$ = $h_4$ + $h_3$ = $\frac {5}{8}$$h_5$ + $h_3$
$\frac {3}{8}$$h_5$ = $h_3$, so $h_5$ = $\frac {8}{3}$$h_3$ = $\frac {13}{3}$$h_2$
Area of RMN = 8 = $\frac {1}{2}$(MN)$h_2$
Area of BAC = $\frac {1}{2}$(3MN)$h_5$ = $\frac {1}{2}$(3MN)$\frac {13}{3}$$h_2$ = 13($\frac {1}{2}$(MN)$h_2$) = (13)(8) = 104