The incircle of triangle $ABC$ with radius 1 touches its sides $AB, BC$, and $AC$ at points $K, M$, and $N$ respectively. It is known that $\angle MKN = \angle ABC = 45^{\circ}$. Find the sides of triangle $ABC$.

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From the theorem about the angle between a tangent and a chord, it follows that

$$
\angle M N C=\angle M K N=45^{\circ}, \angle N M C=\angle M K N=45^{\circ} .
$$

Therefore, $\angle A C B=90^{\circ}$. This means that triangle $A B C$ is a right triangle, and since $\angle A B C=45^{\circ}$, this triangle is isosceles. Let $O$ be the center of the inscribed circle of triangle $A B C$. Denote $A N=A K=x$. Then

$$
A B=2 x, A C=x+1, A C=A B \sin 45^{\circ}=x \sqrt{2}
$$

From the equation $x+1=x \sqrt{2}$, we find that

$$
x=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1
$$

Therefore,

$$
A B=B C=2+\sqrt{2}, A C=2+2 \sqrt{2} .
$$

## Answer

$2+\sqrt{2}, 2+\sqrt{2}, 2+2 \sqrt{2}$.

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