# Problem of the Week Problem E and Solution Medians

## Problem

In $$\triangle ABC$$, $$\angle ABC = 90^{\circ}$$. A median is drawn from $$A$$ to side $$BC$$, meeting $$BC$$ at $$M$$ such that $$AM=5$$. A second median is drawn from $$C$$ to side $$AB$$, meeting $$AB$$ at $$N$$ such that $$CN=2\sqrt{10}$$.

Determine the length of the longest side of $$\triangle ABC$$.

Note: In a triangle, a median is a line segment drawn from a vertex of the triangle to the midpoint of the opposite side.

## Solution

Since $$AM$$ is a median, $$M$$ is the midpoint of $$BC$$. Then $$BM=MC$$. Let $$BM=MC=y$$.
Since $$CN$$ is a median, $$N$$ is the midpoint of $$AB$$. Then $$AN=NB$$. Let $$AN=NB=x$$.

Since $$\angle B=90^{\circ}$$, $$\triangle NBC$$ is a right-angled triangle. Using the Pythagorean Theorem, \begin{align*} NB^2+BC^2&=CN^2\\ x^2+(2y)^2&=(2\sqrt{10})^2\\ x^2+4y^2&=40\tag{1}\end{align*}

Since $$\angle B=90^{\circ}$$, $$\triangle ABM$$ is a right-angled triangle. Using the Pythagorean Theorem, \begin{align*} AB^2+BM^2&=AM^2\\ (2x)^2+y^2&=5^2\\ 4x^2+y^2&=25 \tag{2}\end{align*}

Adding equations $$(1)$$ and $$(2)$$, we get $$5x^2+5y^2=65$$ or $$x^2+y^2=13$$.

The longest side of $$\triangle ABC$$ is the hypotenuse $$AC$$. Using the Pythagorean Theorem, \begin{aligned} AC^2&=AB^2+BC^2\\ &=(2x)^2+(2y)^2\\ &=4x^2+4y^2\\ &=4(x^2+y^2)\end{aligned}

Since $$x^2 + y^2 =13$$, we have $$AC^2 = 4(13)$$. And since $$AC>0$$, $$AC=2\sqrt{13}$$ follows.

Therefore, the length of the longest side of $$\triangle ABC$$ is $$2\sqrt{13}$$.

Note: The solver could have instead solved a system of equations to find $$x=2$$ and $$y=3$$, and then proceed to solve for the longest side. The above approach was provided to expose the solver to alternate way to think about the solution of this problem.