Two isosceles triangles $A B C(A B=B C)$ and $M N P(M P=N P)$ are similar and positioned such that points $M$, $N$, and $P$ lie on sides $A B$, $B C$, and $C A$ respectively. Find the ratio $\frac{M P}{A B}$, if $\frac{N C}{B N}=2$, and the angle $B A C$ is equal to $\operatorname{arctg} 4$.
Apply the Law of Sines.

## Solution

Let

$$
\alpha=\angle B A C, \beta=\angle M N B, A B=B C=a, P M=P N=x
$$

$$
\begin{gathered}
B N=\frac{a}{3}, N C=\frac{2 a}{3}, M N=2 x \cos \alpha, \angle M P N=180^{\circ}-2 \alpha \\
\angle M P A=2 \alpha-\beta, \angle B M N=2 \alpha-\beta
\end{gathered}
$$

In triangle $B M N$ by the Law of Sines

$$
\frac{B M}{\sin \beta}=\frac{2 x \cos \alpha}{\sin 2 \alpha} \Rightarrow B M=\frac{x \sin \beta}{\sin \alpha}
$$

In triangle АМР by the Law of Sines

$$
\frac{A M}{\sin (2 \alpha-\beta)}=\frac{x}{\sin \alpha} \Rightarrow A M=\frac{x \sin (2 \alpha-\beta)}{\sin \alpha}
$$

Since

$$
A M+B M=a=\frac{x(\sin (2 \alpha-\beta)+\sin \beta)}{\sin \alpha}=2 x \cos (\alpha-\beta)
$$

then $a=2 x \cos (\alpha-\beta)$.

In triangle $P N C$ by the Law of Sines

$$
\frac{x}{\sin \alpha}=\frac{2 a}{3 \sin \beta} \Rightarrow a=\frac{3 x \sin \beta}{2 \sin \alpha}
$$

Therefore,

$$
\cos (\alpha-\beta)=\frac{3 \sin \beta}{4 \sin \alpha}, \text { or } \cos \alpha \cos \beta+\sin \alpha \sin \beta=\frac{3 \sin \beta}{4 \sin \alpha} .
$$

Since $\sin \alpha=\frac{4}{\sqrt{17}}$ and $\cos \alpha=\frac{1}{\sqrt{17}}$, then

$$
\cos \beta=-\frac{13}{16}, \sin \beta=\frac{16}{5 \sqrt{17}}
$$

Therefore,

$$
\frac{x}{a}=\frac{3}{2} \cdot \frac{\sin \beta}{\sin \alpha}=\frac{5}{6}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_e1837d72990d2467a29dg-30.jpg?height=466&width=535&top_left_y=2008&top_left_x=767)

Answer

## Hint

Replace the given triangle $A B C$ with a right triangle $A B_{1} C\left(\angle C=90^{\circ}\right)$.

## Solution

Suppose that a rectangle $K L M N$ with diagonal $L N$ equal to the given segment $a$ is inscribed in the given triangle $A B C$. Let vertices $K$ and $N$ be on side $A C$, and vertices $L$ and $M$ be on sides $A B$ and $B C$ respectively.

Draw a line through vertex $B$ parallel to $A C$ until it intersects at point $B_{1}$ with the perpendicular to the base $A C$ passing through vertex $C$. Let the line $L M$ intersect the hypotenuse $A B_{1}$ and the leg $B_{1} C$ of the right triangle $A B_{1} C$ at points $P$ and $Q$ respectively. If $F$ is the projection of point $P$ on $A C$, then the rectangle $P Q C F$ is equal to the rectangle $L M N K$, since

$$
\frac{L M}{A C}=\frac{B L}{A B}=\frac{B_{1} P}{A B_{1}}=\frac{P Q}{A C}, P F=L K
$$

Therefore, the diagonals of these rectangles are respectively equal.

From this, the following construction method follows. Construct a right triangle $A C B_{1}\left(\angle A C B_{1}=90^{\circ}\right)$ with leg $B_{1} C$ equal to the height $B D$ of the given triangle $A B C$. With center at point $C$ and radius $a$, equal to the given diagonal, draw a circle. Through the point of intersection of this circle with the hypotenuse $A B_{1}$, draw a line parallel to $A C$. The points of intersection of this line with sides $A B$ and $B C$ are the vertices of the desired rectangle.

Depending on whether the height of the triangle $A B_{1} C$, dropped from $C$, is less than, equal to, or greater than the given segment, the problem will have two solutions, one, or none.

![](https://cdn.mathpix.com/cropped/2024_05_06_e1837d72990d2467a29dg-31.jpg?height=458&width=774&top_left_y=1343&top_left_x=639)

Problem