In the diagram, square $P Q R S$ has side length 2. Points $M$ and $N$ are the midpoints of $S R$ and $R Q$, respectively. The value of $\cos (\angle M P N)$ is
(A) $\frac{4}{5}$
(B) $\frac{\sqrt{2}}{2}$
(C) $\frac{\sqrt{5}}{3}$
(D) $\frac{1}{3}$
(E) $\frac{\sqrt{3}}{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-034.jpg?height=360&width=374&top_left_y=262&top_left_x=1339)
The side length of the square is 2 and $M$ and $N$ are midpoints of sides.

Thus, $S M=M R=Q N=N R=1$.

Using the Pythagorean Theorem in $\triangle P S M$, we get $P M=\sqrt{P S^{2}+S M^{2}}=\sqrt{2^{2}+1^{2}}=\sqrt{5}$ since $P M>0$.

Similarly, $P N=\sqrt{5}$.

Using the Pythagorean Theorem in $\triangle M N R$, we get $M N=\sqrt{M R^{2}+N R^{2}}=\sqrt{1^{2}+1^{2}}=\sqrt{2}$ since $M N>0$.

Using the cosine law in $\triangle P M N$, we get

$$
\begin{aligned}
M N^{2} & =P M^{2}+P N^{2}-2(P M)(P N) \cos (\angle M P N) \\
2 & =5+5-2(\sqrt{5})(\sqrt{5}) \cos (\angle M P N) \\
2 & =10-10 \cos (\angle M P N) \\
10 \cos (\angle M P N) & =8 \\
\cos (\angle M P N) & =\frac{8}{10}=\frac{4}{5}
\end{aligned}
$$

ANswer: (A)