## [

 Law of Sines In triangle $ABC$, it is known that $\angle A=60^{\circ}, \angle B=45^{\circ}$. The extensions of the altitudes of triangle $ABC$ intersect the circumcircle of the triangle at points $M, N, P$. Find the ratio of the areas of triangles $ABC$ and $MNP$.

#
The angles of triangle $M N P$ are $60^{\circ}, 90^{\circ}, 30^{\circ}$.

## Solution

Let the extensions of the altitudes of triangle $A B C$, drawn from vertices $A, B$, and $C$, intersect the circumcircle of the triangle at points $M, N$, and $P$ respectively. In triangle $M N P$,

$$
\begin{aligned}
\angle M & =\angle N M P=\angle N M A+\angle P M A=\angle N B A+\angle A C P= \\
& =\left(90^{\circ}-\angle A\right)+\left(90^{\circ}-\angle A\right)=30^{\circ}+30^{\circ}=60^{\circ}
\end{aligned}
$$

Similarly, $\angle P=30^{\circ}$ and $\angle N=90^{\circ}$.

Let $R$ be the radius of the circumcircle of triangle $A B C$. Then

$$
\begin{gathered}
M N=2 R \sin \angle P=2 R \sin 30^{\circ}=R \\
N P=2 R \sin \angle M=2 R \sin 60^{\circ}=R \sqrt{3} \\
S_{\triangle \mathrm{MNP}}=\frac{1}{2} M N \cdot N P=\frac{R^{2} \sqrt{3}}{2} \\
A C=2 R \sin \angle B=2 R \sin 45^{\circ}=R \sqrt{2} \\
B C=2 R \sin \angle A=2 R \sin 60^{\circ}=R \sqrt{3} \\
S_{\triangle \mathrm{ABC}}=\frac{1}{2} A C \cdot B C \sin \angle C=R^{2} \cdot \frac{\sqrt{6}}{2} \cdot \sin 75^{\circ}
\end{gathered}
$$

Therefore,

$$
\frac{S_{\triangle A B C}}{S_{\triangle M N P}}=\sqrt{2} \sin 75^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5827e7e7a03e1b3b311bg-51.jpg?height=432&width=506&top_left_y=1&top_left_x=776)

## Answer

$\sqrt{2} \sin 75^{\circ}=\frac{\sqrt{3}+1}{2}$