Let A(1,0), B(6,2) and C(32,6) be the vertices of a triangle ABC.

Question

Let $A(1,0),\ B(6,2)$ and $C\left(\dfrac32,6\right)$ be the vertices of a triangle $ABC$. If $P$ is a point inside the triangle $ABC$ such that the triangles $APC, APB$ and $BPC$ have equal areas, then the length of the line segment $PQ$, where $Q$ is the point $\left(-\dfrac76,-\dfrac13\right)$, is \_\_\_\_\_\_\_\_.

Answer: 5

Accepted Answer

Since the three sub-triangles have equal areas, $P$ is the centroid of $\triangle ABC$:
$$P = \left(\frac{1+6+\frac32}3,\frac{0+2+6}3\right) = \left(\frac{17}6,\frac83\right)$$

$$PQ = \sqrt{\left(\frac{17}6+\frac76\right)^2+\left(\frac83+\frac13\right)^2} = \sqrt{4^2+3^2}=5$$

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