[Law of Cosines] [Law of Sines]

In a right triangle $KLM$, a segment $MD$ is drawn connecting the vertex of the right angle to point $D$ on the hypotenuse $KL$ such that the lengths of segments $DL, DM$, and $DK$ are distinct and form a geometric progression in the given order with a common ratio of $\sqrt{2}$, and $DL = 1$. Find the measure of angle $KMD$.
Denote $\angle M K L$ and find $\cos \alpha$, applying the cosine theorem to triangle $M D K$. Then apply the sine theorem to triangle $M K D$.

## Solution

Since the lengths of segments $D L, D M$ and $D K$ form a geometric progression with a common ratio of $\sqrt{2}$, then

$$
D M=D L \cdot \sqrt{2}=\sqrt{2}, D K=D M \cdot \sqrt{2}=2
$$

Let $\angle M K L=\alpha$. Then $K M=K L \cos \alpha=3 \cos \boldsymbol{\alpha}$. By the cosine theorem,

$$
M D^{2}=K M^{2}+K D^{2}-2 \cdot K M \cdot K D \cdot \cos \alpha, \text { or } 2=9 \cos ^{2} \alpha+4-2 \cdot 6 \cos ^{2} \alpha,
$$

from which we find that $\cos \alpha=\sqrt{\frac{2}{3}}$. Then $\sin \alpha=\frac{1}{\sqrt{3}}$. Applying the sine theorem to triangle $M K D$, we find that

$$
\sin \angle K M D=K D \cdot \frac{\sin \alpha}{M D}=\frac{2 \sin \alpha}{\sqrt{2}}=\sqrt{2} \sin \alpha=\sqrt{\frac{2}{\sqrt{3}}}
$$

Since angle $K M D$ is acute, then

$$
\cos \angle K M D=\sqrt{1-\frac{2}{3}}=\frac{1}{\sqrt{3}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_8027e864227276d12944g-37.jpg?height=586&width=426&top_left_y=1153&top_left_x=816)

Answer

$\arccos \frac{1}{\sqrt{3}}$.