The sides of an acute-angled triangle $A B C$ are respectively equal to $a, b$ and $c$. Point $M$ is located inside the triangle. The angles $A M B, B M C$ and $C M A$ are equal to each other. Find the sum of the segments $A M, B M$ and $C M$.
If $x, y$ and $z$ are the lengths of segments $A M, B M$ and $C M$, then

$$
(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(x y+y z+x z)
$$

To find the sum $x^{2}+y^{2}+z^{2}$, use the Law of Cosines for triangles $A M B, B M C$ and $C M A$, and for the sum $x y+y z+x z$, add the areas of these triangles.

## Solution

Each of the specified angles is $120^{\circ}$. Let $A M=x, B M=y, C M=z, S_{\Delta \mathrm{ABC}}=S$. Then

$$
S=\frac{1}{2}(x y+x z+y z) \sin 120^{\circ}=\frac{(x y+x z+y z) \sqrt{3}}{4}
$$

From this, we find that $3(x y+y z+x z)=4 S \sqrt{3}$.

Express $a^{2}, b^{2}$ and $c^{2}$ using the Law of Cosines from triangles $BMC$, $AMC$, and $AMB$ respectively, and add the resulting equations term by term. Then

$$
2\left(x^{2}+y^{2}+z^{2}\right)+(x y+x z+y z)=a^{2}+b^{2}+c^{2} .
$$

Adding this equation term by term with the equation

$$
3(x y+x z+y z)=4 S \sqrt{3}
$$

we get

$$
2\left(x^{2}+y^{2}+z^{2}+2 x y+2 x z+2 y z\right)=a^{2}+b^{2}+c^{2}+4 S \sqrt{3}
$$

or

$$
2(x+y+z)^{2}=a^{2}+b^{2}+c^{2}+4 S \sqrt{3}
$$

Therefore,

$$
A M+B M+C M=x+y+z=\sqrt{\frac{1}{2}\left(a^{2}+b^{2}+c^{2}\right)+2 S \sqrt{3}}
$$

( $S$ can be found using Heron's formula).

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-58.jpg?height=454&width=640&top_left_y=619&top_left_x=709)

Answer

$\sqrt{\frac{1}{2}\left(a^{2}+b^{2}+c^{2}\right)+2 S \sqrt{3}}$, where $S=\sqrt{p(p-a)(p-b)(p-c)}, p=\frac{a+b+c}{2}$.