# 1966 AHSME Problems/Problem 38

## Problem

In triangle $ABC$ the medians $AM$ and $CN$ to sides $BC$ and $AB$, respectively, intersect in point $O$. $P$ is the midpoint of side $AC$, and $MP$ intersects $CN$ in $Q$. If the area of triangle $OMQ$ is $n$, then the area of triangle $ABC$ is:

$\text{(A) } 16n \quad \text{(B) } 18n \quad \text{(C) } 21n \quad \text{(D) } 24n \quad \text{(E) } 27n$

## Solution

Construct triangle $\triangle ABC$ with points $M,N,P$ being the midpoints of sides $\overline{CB}, \overline{AB}, \overline{AC}$, respectively. Proceed by drawing all medians. Then draw all medians (so draw $\overline{AM}, \overline{BP}, \overline{CN}$). Next, draw line $\overline{PM}$ and label $\overline{PM}$'s intersection with $\overline{CN}$ as the point $Q$. From the problem, the area of $\triangle QMO$ is $n$, but by vertical angles we know that $\angle QOM = \angle AOM$. Furthermore, since line $\overline{PM}$ is drawn from the midpoint of $\overline{AC}$ to the midpoint of $\overline{CB}$, we know that $\overline{PM}$ is parallel to $\overline{AB}$ (via SAS similarity on triangles PCM and ABC). From these parallel lines we know that $\angle PMO = \angle OAB$ which indicates that $\triangle QOM \sim \triangle ANO$. The linear ratio from $\triangle QOM$ to $\triangle ANO$ is 1:2 because line segment $\overline{OM}$ is one half of line segment $\overline{AO}$ since $\overline{AO}$ and $\overline{OM}$ make up the median $\overline{AM}$. Thus the area ratio is 1:4. So $\triangle ANO$ has area $4n$. Since $\triangle ONB$ has the same height and base as $\triangle ANO$ we know that the area of $\triangle AOB = 8n$. The medians form 3 triangles each with area $1/3$ of the total triangle (these triangles are $\triangle AOB, \triangle COB, \triangle COA$). Thus since $\triangle AOB = 4n \underset{\text{multiply by 3}}{\implies} \triangle ABC = 24n$ $\fbox{D}$.

- LJ