Let $M$, $N$, and $P$ be the midpoints of sides $\overline{TU}$, $\overline{US}$, and $\overline{ST}$ of triangle $STU$, respectively. Let $\overline{UZ}$ be an altitude of the triangle. If $\angle TSU = 60^\circ$ and $\angle STU = 45^\circ$, then what is $\angle ZNM + \angle ZMN + \angle PNM + \angle PMN + \angle NZM + \angle NPM$ in degrees?

Stanry May 1, 2024

#1**+1 **

I think this is more of a trick question

First off, I graphed the triangles. Note, please excuse my awful drawing skills.

Alright, now we have a graph, we can easily visualulize the angles needed.

We honestly don't need to do much math if we notice one major fact.

If we rearrange the order of the angles we need to find a bit, \(\angle ZNM + \angle ZMN + \angle NZM + \angle PNM + \angle PMN+ \angle NPM\) , we notice one big factor.

The angles \(\angle ZNM + \angle ZMN+ \angle NZM\) are the three angles of the triangle ZMN, meaning they simply add up to 180 degrees.

The same goes for the angles \(\angle PNM + \angle PMN + \angle NPM\). They are the 3 angles to triangle PMN, meaning they also add up to 180 degrees.

This means that \(\angle ZNM + \angle ZMN + \angle PNM + \angle PMN + \angle NZM + \angle NPM = 180 + 180 = 360\).

360 degrees is the final answer!

Thanks! :)

NotThatSmart May 1, 2024