Squares, Midpoints, and Cosine

Let $\angle MAN$ be denoted by $\alpha$ and let $\angle MAB$ , which is trivially congruent to $\angle NAD$ , be denoted by $\beta$ .

By the cofunction property, $\cos \alpha = \sin (90 - \alpha)$ . Furthermore, note that $\sin (90 - \alpha) = \sin 2 \beta = 2 \cos \beta \sin \beta$ .

Now it remains to find $\cos \beta$ and $\sin \beta$ , which can be easily found to be $\frac{2}{\sqrt {5}}$ and $\frac{1}{\sqrt{5}}$ . Plugging this in, $\cos \angle MAN = \frac{4}{5}$ and our answer is $4 + 5 = 9$ .

There are other (easier) ways to solve this, such as using law of cosines, but I thought this way was pretty neat. Here's the law of cosines way now:

Without loss of generality, let the side of the square be $2$ . Then, by the Pythagorean theorem, $AM=AN=\sqrt{5}$ , and $MN = \sqrt{2}$ . By law of cosines, $2 = 5 + 5 - 10\cos \angle MAN$ . The fact that $\cos \angle MAN = \frac{4}{5}$ immediately follows.

Take a square whose coordinates are A(0,0), B(a,0), C(0,a) and D(a,a). MN= a sqrt(1/2) AM=AN= a sqrt(5/4) So, by cosine rule, we get 4/5