[ [ Inscribed Quadrilaterals] $]$

The lateral sides of a trapezoid are equal to 3 and 5. It is known that a circle can be inscribed in the trapezoid. The midline of the trapezoid divides it into two parts, the ratio of the areas of which is 5:11. Find the bases of the trapezoid.
Solve the system of equations with two unknowns - the lengths of the bases of the trapezoid.

## Solution

Let $x$ and $y$ be the lengths of the bases of the trapezoid. Then

$$
x+y=3+5=8
$$

Therefore, the midline is $\frac{x+y}{2}=4$. The second equation of the system is found from the ratio of the areas:

$$
\frac{x+4}{y+4}=\frac{5}{11}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-35.jpg?height=386&width=646&top_left_y=2039&top_left_x=706)

## Answer

1 and 7. Send a comment

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-35.jpg?height=115&width=1109&top_left_y=2695&top_left_x=0)

Lines $P A$ and $P B$ are tangent to a circle with center $O$ ( $A$ and $B$ are the points of tangency). A third tangent to the circle intersects lines $P A$ and $P B$ at points $X$ and $Y$. Prove that the measure of angle $X O Y$ does not depend on

## Hint

The angle between the radii drawn to the points of tangency complements the angle between the tangents to $180^{\circ}$.

## Solution

Let $\angle A P B=\alpha$. Let $M$ be the point of tangency of the circle with line $X Y$. Then

$$
\begin{gathered}
\angle X O Y=\frac{1}{2} \angle A O M+\frac{1}{2} \angle B O M=\frac{1}{2}(\angle A O M+\angle B O M)= \\
=\frac{1}{2} \angle A O B=\frac{1}{2}\left(180^{\circ}-\alpha\right)=90^{\circ}-\frac{\alpha}{2} .
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-36.jpg?height=108&width=1575&top_left_y=1098&top_left_x=1)

Difficulty: $3+$

Grades: 8,9

Two circles touch each other externally at point $C$. The radii of the circles are 2 and 7. A common tangent to both circles, passing through point $C$, intersects another common tangent at point $D$. Find the distance from the center of the smaller circle to point $D$.

## Hint

Apply the property of the altitude of a right triangle dropped to the hypotenuse.

## Solution

Let $O_{1}$ and $O_{2}$ be the centers of the smaller and larger circles. The angle $\angle O_{1} D O_{2}$ is the angle between the bisectors of adjacent angles, so $\angle O_{1} D O_{2}=90^{\circ}$.

From the right triangle $O_{1} D O_{2}$, we find that

$$
O_{1} D^{2}=O_{1} C \cdot O_{1} O_{2}=2 \cdot 9
$$

Therefore, $O_{1} D=3 \sqrt{2}$.

Let $O_{1}$ and $O_{2}$ be the centers of the smaller and larger circles. The angle $\angle O_{1} D O_{2}$ is the angle between the bisectors of adjacent angles, so $\angle O_{1} D O_{2}=90^{\circ}$.

From the right triangle $O_{1} D O_{2}$, we find that

$$
O_{1} D^{2}=O_{1} C \cdot O_{1} O_{2}=2 \cdot 9
$$

Therefore, $O_{1} D=3 \sqrt{2}$.

Let $O_{1}$ and $O_{2}$ be the centers of the smaller and larger circles. The angle $\angle O_{1} D O_{2}$ is the angle between the bisectors of adjacent angles, so $\angle O_{1} D O_{2}=90^{\circ}$.

From the right triangle $O_{1} D O_{2}$, we find that

$$
O_{1} D^{2}=O_{1} C \cdot O_{1} O_{2}=2 \cdot 9
$$

Therefore, $O_{1} D=3 \sqrt{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-37.jpg?height=403&width=606&top_left_y=-1&top_left_x=729)

## Answer

$3 \sqrt{2}$.

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