[ Law of Cosines $\quad]$ [ Similar auxiliary triangles ]

Points $K$ and $M$ are located on side $BC$ and height $BP$ of an acute triangle $ABC$, respectively.

Find the area of the equilateral triangle $AMK$, given that $AP=3, PC=11/2, BK: KC=10: 1$.
Prove that the line $B P$ divides the segment $A K$ in the ratio $3: 5$.

## Solution

Let $F$ and $T$ be the projections of point $K$ onto $A C$ and $B P$ respectively. Then $P F=10 / 11, P C=5, K T=P F=5$. Let $Q$ be the intersection point of $A K$ and $B P$. From the similarity of triangles $A P Q$ and $A F K$, it follows that $A Q: A K=A P: A F=$

Let $A K=K M=A M=x$. Then $A Q=3 x / 8$. By the cosine theorem, $M Q^{2}=A M^{2}+A Q^{2}-2 A M \cdot A Q \cos 60^{\circ}=$ $x^{2}(1+9 / 64-3 / 8)=49 / 64 x^{2}$.

Since $K T=5$ is the height of triangle $M K Q$ and $S_{M K Q}=5 / 8 S_{A M K}$, then $M Q \cdot K T=5 / 4 S_{A M K}$, or

$\frac{35 x}{8}=\frac{5 x^{2} \sqrt{3}}{16}$. From this, $x=\frac{14}{\sqrt{3}}$.

Therefore, $S_{A M K}=\frac{\sqrt{3}}{4} x^{2}=\frac{49}{\sqrt{3}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f7f9d90261bfc416bd71g-23.jpg?height=441&width=774&top_left_y=477&top_left_x=642)

Answer

$\frac{49}{\sqrt{3}}$.

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