The circle inscribed in triangle $ABC$ touches side $AB$ at point $M$, with $AM=1, BM=4$. Find $CM$, given that $\angle BAC=120^{\circ}$.
Let the circle inscribed in triangle $ABC$ touch side $AC$ at point $K$. Denote $CK=x$ and use the Law of Cosines to form an equation in terms of $x$.

## Answer

## $\sqrt{273}$.

Let the circle inscribed in triangle $ABC$ touch sides $AC$ and $BC$ at points $K$ and $N$ respectively. Denote $CK=x$. Then

$$
CN=CK=x, AK=AM=1, BN=BM=4, AC=x+1, BC=x+4.
$$

By the Law of Cosines,

$$
BC^{2}=AB^{2}+AC^{2}-2 AB \cdot AC \cos 120^{\circ}, \text{ or } (x+4)^{2}=25+(x+1)^{2}+5(x+2)
$$

From this equation, we find that $x=15$.

By the Law of Cosines from triangle $ACM$, we find that

$$
CM^{2}=AC^{2}+AM^{2}-2 AC \cdot AM \cos 120^{\circ}=16^{2}+1+16=273
$$

Therefore, $CM=\sqrt{273}$.