In parallelogram $A B C D(A B \| C D)$, diagonal $B D=a$, and $O$ is the point of intersection of the diagonals. Find the area of the parallelogram if $\angle D B A=45^{\circ}, \angle A O B=105^{\circ}$.

## Answer

$S=\frac{1+\sqrt{3}}{4} a^{2}$.

Circles with radii 3 and 6 have centers at points $O_{1}$ and $O_{2}$, respectively, and touch externally at point $A$. A common external tangent and a common internal tangent are drawn to the circles. These tangents intersect at point $B$, and $L$ is the common point of the external tangent and the circle with radius 6. Find the radius of the circle inscribed in quadrilateral $A B L O_{2}$.
Prove that $B A$ is the height of the right triangle $O_{1} B O_{2}$, drawn from the vertex of the right angle $B$.

## Solution

Let $K$ be the point of tangency of the common external tangent with the circle of radius 3. Since $B O_{1}$ and $B O_{2}$ are the bisectors of angles $A B K$ and $A B L$, then $\angle O_{1} B O_{2}$, which means that $B A$ is the height of the right triangle $O_{1} B O_{2}$, drawn from the vertex of the right angle, so

$$
B A=\sqrt{A O_{1} \cdot A O_{2}}=3 \sqrt{2}
$$

Then $B L=B A=3 \sqrt{2}$ as segments of tangents drawn from the same point to the circle. Since $B L + A O_{2} = B A + L O_{2}$, a circle can be inscribed in the quadrilateral $A B L O_{2}$. The center $O$ of this circle lies on the diagonal $B O_{2}$, since $B O_{2}$ is the bisector of the angles at vertices $B$ and $O_{2}$ of this quadrilateral. Let $P$ and $Q$ be the points of tangency of the desired circle with segments $B A$ and $A O_{2}$, respectively, and let $r$ be the desired radius. Then

$$
\begin{aligned}
S_{\triangle \mathrm{ABO}_{2}}= & S_{\triangle \mathrm{AOO}_{2}}+S_{\triangle \mathrm{AOB}}=\frac{1}{2} \cdot A O_{2} \cdot O Q+\frac{1}{2} \cdot A B \cdot O P= \\
& \frac{1}{2} \cdot 6 \cdot r+\frac{1}{2} \cdot 3 \sqrt{2} \cdot r=\frac{1}{2} \cdot r(6+3 \sqrt{2})
\end{aligned}
$$

On the other hand,

$$
S_{\triangle \mathrm{ABO}_{2}}=\frac{1}{2} \cdot A O_{2} \cdot A B=\frac{1}{2} \cdot 6 \cdot 3 \sqrt{2}=\frac{1}{2} \cdot 18 \sqrt{2}
$$

From the equation $\frac{1}{2} \cdot r(6+3 \sqrt{2})=\frac{1}{2} \cdot 18 \sqrt{2}$, we find that

$$
r=\frac{6}{1+\sqrt{2}}=6(\sqrt{2}-1)
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-31.jpg?height=503&width=700&top_left_y=1935&top_left_x=679)

## Answer

$6(\sqrt{2}-1)$

Submit a comment