$\triangle A B C$ is an isosceles triangle in which $A B=A C=10$ and $B C=12$. The points $S$ and $R$ are on $B C$ such that $B S: S R: R C=1: 2: 1$. The midpoints of $A B$ and $A C$ are $P$ and $Q$ respectively. Perpendiculars are drawn from $P$ and $R$ to $S Q$ meeting at $M$ and $N$ respectively. The length of $M N$ is
(A) $\frac{9}{\sqrt{13}}$
(B) $\frac{10}{\sqrt{13}}$
(C) $\frac{11}{\sqrt{13}}$
(D) $\frac{12}{\sqrt{13}}$
(E) $\frac{5}{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-141.jpg?height=295&width=488&top_left_y=197&top_left_x=1274)
$\triangle A B C$ is an isosceles triangle in which $A B=A C=10$ and $B C=12$. The points $S$ and $R$ are on $B C$ such that $B S: S R: R C=1: 2: 1$. The midpoints of $A B$ and $A C$ are $P$ and $Q$ respectively. Perpendiculars are drawn from $P$ and $R$ to $S Q$ meeting at $M$ and $N$ respectively. The length of $M N$ is
(A) $\frac{9}{\sqrt{13}}$
(B) $\frac{10}{\sqrt{13}}$
(C) $\frac{11}{\sqrt{13}}$
(D) $\frac{12}{\sqrt{13}}$
(E) $\frac{5}{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-237.jpg?height=331&width=572&top_left_y=1016&top_left_x=1335)

## Solution

The triangle $A B C$ is isosceles and we start by drawing a perpendicular from $A$ to $B C$ to meet at $D$, the midpoint of $B C$. This makes $B D=6$ and using 'Pythagorus' in $\triangle A B D \quad A D=8$. If we join $P$ to $Q, P$ to $S$ and $Q$ to $R$ we note that because $B P=P A=5$ and $B S=S D=3$ we conclude that $P S \| A D$ and $\angle P S B=90^{\circ}$. This implies that

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-237.jpg?height=344&width=569&top_left_y=1541&top_left_x=1339)
$P S=4$. We use symmetry to also conclude that $Q R=4$ and $Q R \| A D$. This implies that $P S=Q R=4$ and $P S \| Q R$. $P Q R S$ is thus a rectangle and $P Q=6$ since $P Q=S R=6$.

To make the calculations easier we place all the information on the diagram.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-237.jpg?height=347&width=528&top_left_y=2041&top_left_x=1357)

Now, $S Q^{2}=4^{2}+6^{2}=52$ so $S Q=\sqrt{52}$. In $\triangle S Q R$ we calculate the value of $N R$ by using the fact
that it has an area of 12. ( $\triangle S Q R$ has an area of 12 because it is one-half the area of the rectangle $P Q R S$ which itself has area 24 .)

Therefore, $\quad \frac{1}{2}(\sqrt{52})(N R)=12$

$$
N R=\frac{24}{\sqrt{52}}=\frac{24}{2 \sqrt{13}}=\frac{12}{\sqrt{13}}
$$

Using 'Pythagoras', $\quad N R^{2}+N Q^{2}=Q R^{2}$.

Substituting, $\quad\left(\frac{12}{\sqrt{13}}\right)^{2}+N Q^{2}=4^{2}$.

Therefore, $\quad N Q^{2}=16-\frac{144}{13}$

$$
\begin{aligned}
& N Q^{2}=\frac{208-144}{13} \\
& N Q^{2}=\frac{64}{13}
\end{aligned}
$$

And, $N Q=\frac{8}{\sqrt{13}}$ and $M S=\frac{8}{\sqrt{13}}$ because $N Q=M S$.

Therefore, $M N=2 \sqrt{13}-2\left(\frac{8}{\sqrt{13}}\right)=\frac{26}{\sqrt{13}}-\frac{16}{\sqrt{13}}=\frac{10}{\sqrt{13}}$.