[ Law of Cosines ]

Find the sum of the squares of the distances from a point $M$, taken on the diameter of a certain circle, to the ends of any chord parallel to this diameter, if the radius of the circle is $R$, and the distance from the point $M$ to the center of the circle is $a$.
Apply the Law of Cosines.

## Solution

Let $O$ be the center of the circle. $AB$ is an arbitrary chord parallel to the given diameter. Denote $\angle BOM = \varphi$. Then $\angle AOM = 180^\circ - \varphi$.

By the Law of Cosines in triangles $BOM$ and $AOM$, we find that

\[
\begin{gathered}
BM^2 = a^2 + R^2 - 2aR \cos \varphi \\
AM^2 = a^2 + R^2 - 2aR \cos (180^\circ - \varphi) = a^2 + R^2 + 2aR \cos \varphi
\end{gathered}
\]

Adding these equations term by term, we get

![](https://cdn.mathpix.com/cropped/2024_05_06_ae6ec2c916c776dd1759g-11.jpg?height=400&width=460&top_left_y=2258&top_left_x=799)

Answer

$2(a^2 + R^2)$

The bisector $AE$ of angle $A$ divides the quadrilateral $ABCD$ into an isosceles triangle $ABE (AB = BE)$ and a rhombus $AEC D$. The radius of the circle circumscribed around triangle $ECD$ is 1.5 times the radius of the circle inscribed in triangle $ABE$. Find the ratio of the perimeters of these triangles.

## Hint

Denote $\angle BAE = \angle EAD = \angle ECD = \angle AEB = 2\alpha$ and set up an equation in terms of $\alpha$.

## Solution

Let $AE = x$, $\angle BAE = 2\alpha$. Then $DC = EC = x$, $\angle EAD = \angle ECD = \angle AEB = 2\alpha$. Let $r$ be the radius of the circle inscribed in triangle $ABE$. Then $r = (x / 2) \cdot \tan \alpha$. Let $R$ be the radius of the circle circumscribed around triangle $ECD$. Then

\[
R = \frac{EC}{2 \cdot \sin \angle EDC} = \frac{x}{2 \cdot \sin (90^\circ - \alpha)} = \frac{x}{2 \cdot \cos \alpha}
\]

According to the problem,

\[
\frac{R}{r} = \frac{3}{2}, \text{ i.e., } \frac{x}{2 \cdot \cos \alpha} = \left(\frac{x}{2} \cdot \tan \alpha \right) \cdot \frac{3}{2}
\]

Then

\[
\tan \alpha \cdot \cos \alpha = \frac{2}{3}, \sin \alpha = \frac{2}{3}, \cos \alpha = \frac{\sqrt{5}}{3}, \cos 2\alpha = \frac{1}{9}
\]

In triangle $ABE$, $AB = BE = \frac{x}{2 \cdot \cos 2\alpha} = \frac{9x}{2}$. Therefore, its perimeter is $\frac{9x}{2} + \frac{9x}{2} + x = 10x$.

In triangle $ECD$, $DE = 2x \cdot \sin \alpha = \frac{4x}{3}$. Therefore, its perimeter is $x + x + \frac{4x}{3} = \frac{10x}{3}$.

Thus, the ratio of the perimeters is $\frac{10x}{\frac{10x}{3}} = 3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ae6ec2c916c776dd1759g-12.jpg?height=458&width=809&top_left_y=1323&top_left_x=630)

## Answer

3.