In triangle $K L M$, the ratio of the radii of the circumscribed and inscribed circles is 3. The inscribed circle touches the sides of triangle $K L M$ at points $A, B$, and $C$. Find the ratio of the area of triangle $K L M$ to the area of triangle $A B C$.

#
Let $O$ be the center of the circle inscribed in triangle $KLM$. Then point $O$ lies inside triangle $ABC$ and

$$
S_{\triangle \mathrm{ABC}}=S_{\triangle \mathrm{BOC}}+S_{\triangle \mathrm{AOC}}+S_{\triangle \mathrm{AOB}}.
$$

Next, apply the Law of Sines and formulas for the area of a triangle.

## Solution

Let the vertices $A, B$, and $C$ of triangle $ABC$ lie on the sides $ML, KM$, and $KL$ of triangle $KLM$ respectively; $O$ be the center of the circle with radius $r$ inscribed in triangle $KLM$, and $R$ be the radius of the circumcircle of this triangle. Denote

$$
ML=a, KM=b, KL=c, \angle LKM=\alpha, \angle KLM=\beta, \angle LMK=\gamma
$$

Then

$$
\angle BAC=90^{\circ}-\frac{\alpha}{2}<90^{\circ}, \angle ABC=90^{\circ}-\frac{\beta}{2}<90^{\circ}, \angle ACB=90^{\circ}-\frac{\gamma}{2}<90^{\circ}.
$$

Since all angles of triangle $ABC$ are acute, the center $O$ of the circumcircle of triangle $ABC$ lies inside this triangle. Therefore,

$$
S_{\triangle \mathrm{ABC}}=S_{\triangle \mathrm{BOC}}+S_{\triangle \mathrm{AOC}}+S_{\triangle \mathrm{AOB}}=
$$

$$
\begin{gathered}
=\frac{1}{2} \cdot OB \cdot OC \cdot \sin \left(180^{\circ}-\alpha\right)+\frac{1}{2} \cdot OA \cdot OC \cdot \sin \left(180^{\circ}-\beta\right)+\frac{1}{2} \cdot OA \cdot OB \cdot \sin \left(180^{\circ}-\gamma\right)= \\
=\frac{1}{2}(r \cdot r \cdot \sin \alpha+r \cdot r \cdot \sin \beta+r \cdot r \cdot \sin \gamma)=\frac{1}{2} r^{2}(\sin \alpha+\sin \beta+\sin \gamma)
\end{gathered}
$$

Since

$$
S_{\Delta \mathrm{KLM}}=\frac{1}{2} \cdot KL \cdot KM \cdot \sin \angle LKM=\frac{1}{2} bc \sin \alpha
$$

then

$$
\begin{gathered}
\frac{S_{\triangle ABC}}{S_{\triangle KLM}}=\frac{\frac{1}{2} r^{2}(\sin \alpha+\sin \beta+\sin \gamma)}{\frac{1}{2} bc \sin \alpha}=\frac{r^{2}}{bc} \cdot\left(1+\frac{\sin \beta}{\sin \alpha}+\frac{\sin \gamma}{\sin \alpha}\right)= \\
=\frac{r^{2}}{bc} \cdot\left(1+\frac{b}{a}+\frac{c}{a}\right)=\frac{r^{2}(a+b+c)}{abc}
\end{gathered}
$$

Since

$$
S_{\Delta \mathrm{KLM}}=\frac{a+b+c}{2} \cdot r \text{ and } S_{\Delta \mathrm{KLM}}=\frac{abc}{4R}
$$

then

$$
a+b+c=\frac{2 S_{\Delta KLM}}{r}, \text{ and } abc=4 S_{\Delta \mathrm{KLM}} \cdot R
$$

Thus,

$$
\frac{S_{\triangle ABC}}{S_{\triangle KLM}}=\frac{r^{2} \cdot \frac{2 S_{\triangle KLM}}{r}}{S_{\triangle KLM} \cdot 4R}=\frac{r}{2R}=\frac{1}{6}
$$

Therefore, $\frac{S_{\triangle KLM}}{S_{\triangle ABC}}=6$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-47.jpg?height=692&width=738&top_left_y=1658&top_left_x=660)

## Answer

6.00

## Answer

$40^{\circ}$.

The diagonals of the convex quadrilateral $ABCD$ intersect at point $E, BC=CD, AC$ is the angle bisector of $\angle A, \angle BCD=160^{\circ}, \angle CED=130^{\circ}$. Find the angle $\angle ABD$.