 # Problem of the Week Problem D and Solution Which Term is Which?

## Problem

In $$\triangle PQR$$, $$\angle PRQ = 90^{\circ}$$. An altitude is drawn in $$\triangle PQR$$ from $$R$$ to $$PQ$$, intersecting $$PQ$$ at $$S$$. A median is drawn in $$\triangle PSR$$ from $$P$$ to $$SR$$, intersecting $$SR$$ at $$T$$. If the length of the median $$PT$$ is $$39$$ and the length of $$PS$$ is $$36$$, determine the length of $$QS$$.

Note: An altitude of a triangle is a line segment drawn from a vertex of the triangle perpendicular to the opposite side. A median is a line segment drawn from a vertex of the triangle to the midpoint of the opposite side.

## Solution

Since $$T$$ is a median in $$\triangle PSR$$, $$ST = TR$$. Let $$ST=TR = a$$. Let $$PR = b$$, $$QS = c$$, and $$QR = d$$. The variables and the given information, $$PS = 36$$ and $$PT = 39$$, are shown in the diagram. Since $$\triangle PST$$ contains a right angle at $$S$$, \begin{aligned} ST^2 &=PT^2-PS^2\\ a^2&=39^2-36^2\\ &=225\end{aligned} Then, since $$a>0$$, $$a=15$$ follows. Thus, $$SR=2a=30$$.

Since $$\triangle PSR$$ contains a right angle at $$S$$, \begin{aligned} PR^2&=PS^2+SR^2\\ b^2&=36^2+30^2\\ &=2196\end{aligned} Then, since $$b>0$$, $$b=\sqrt{2196}$$ follows.

We will now use $$a=15$$ and $$b=\sqrt{2196}$$ in the three solutions that follow.

Solution 1

In $$\triangle PSR$$ and $$\triangle PRQ$$, $$\angle PSR=\angle PRQ = 90^{\circ}$$ and $$\angle SPR =\angle QPR$$, a common angle. So $$\triangle PSR$$ is similar to $$\triangle PRQ$$. It follows that \begin{aligned} \frac{PS}{PR}&=\frac{PR}{PQ}\\ \frac{36}{\sqrt{2196}}&=\frac{\sqrt{2196}}{36+c}\\ 1296+36c&=2196\\ 36c&=900\\ c&=25\end{aligned}

Thus, the length of $$QS$$ is $$25$$.

Solution 2

Since $$\triangle RSQ$$ contains a right angle at $$S$$, $$QR^2=QS^2+SR^2=c^2+30^2=c^2+900$$.
Therefore, $$d^2=c^2+900$$.

Since $$\triangle PQR$$ contains a right angle at $$R$$, $$PQ^2=PR^2+QR^2$$. Therefore, $$(36+c)^2=(\sqrt{2196})^2+d^2$$, which simplifies to $$1296+72c+c^2=2196+d^2$$. This further simplifies to $$c^2+72c=900+d^2$$.

Substituting $$d^2 = c^2 +900$$, we obtain $$c^2+72c=900+c^2+900$$. Simplifying, we get $$72c=1800$$ and $$c=25$$ follows.

Thus, the length of $$QS$$ is $$25$$.

Solution 3

Position $$\triangle PQR$$ on the $$xy$$-plane so that $$PQ$$ lies along the $$y$$-axis, and altitude $$SR$$ lies along the positive $$x$$-axis with $$S$$ at the origin. Then $$P$$ has coordinates $$(0, 36)$$, $$T$$ has coordinates $$(15,0)$$, and $$R$$ has coordinates $$(30,0)$$.
Since $$Q$$ is on the $$y$$-axis, let $$Q$$ have coordinates $$(0,b)$$ with $$b<0$$. Notice that $\text{slope }PR=\dfrac{36-0}{0-30}=\dfrac{-6}{5}\text{ and slope }QR=\dfrac{b-0}{0-30}=\dfrac{b}{-30}$

Since $$\angle PRQ=90^{\circ}$$, $$PR\perp QR$$, and so their slopes are negative reciprocals of each other. That is, $$\dfrac{b}{-30}=\dfrac{5}{6}$$, and so $$b=-25$$.

It then follows that the coordinates of $$Q$$ are $$(0,-25)$$. Thus, the length of $$QS$$ is $$25$$.