A circle of radius 2 passes through the midpoints of three sides of triangle $ABC$, in which the angles at vertices $A$ and $B$ are $30^{\circ}$ and $45^{\circ}$, respectively.

Find the height drawn from vertex $A$.
Let $K, L$ and $M$ be the midpoints of sides $AB, BC$, and $AC$ respectively. Then triangle $ABC$ is similar to triangle $LMK$ with a similarity coefficient of 2.

## Solution

Let $K, L$ and $M$ be the midpoints of sides $AB, BC$, and $AC$ respectively, and let $AP$ be the desired height. Triangle $ABC$ is similar to triangle $LMK$ with a similarity coefficient of 2. If $R$ is the radius of the circumcircle of triangle $KLM$, then $ML = 2R \sin \angle MKL = 4 \sin 105^\circ = 4 \cos 15^\circ$. Let $LH$ be the height of triangle $KLM$. From the right triangle $MHL$, we find that $LH = ML \sin \angle KML = 4 \cos 15^\circ \sin 45^\circ = 2(\sin 60^\circ + \sin 30^\circ) = \sqrt{3} + 1$.

Therefore, $AP = 2LH = 2 + 2\sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c5c78c6462ff090b828g-24.jpg?height=474&width=623&top_left_y=212&top_left_x=723)

Answer

$2 + 2\sqrt{3}$.