Problem 9.5. Positive numbers $a, b$, and $c$ satisfy the equations

$$
a^{2}+a b+b^{2}=1, \quad b^{2}+b c+c^{2}=3, \quad c^{2}+c a+a^{2}=4
$$

Find $a+b+c$.
Answer: $\sqrt{7}$.

## Kurchatov Mathematics Olympiad for School Students - 2019 Final Stage.

Solution. From a single point $T$, we lay out segments $T A, T B$, and $T C$ with lengths $a, b$, and $c$ respectively, such that $\angle A T B = \angle B T C = \angle C T A = 120^{\circ}$. Then, by the cosine theorem, taking into account the relation $\cos 120^{\circ} = -1 / 2$, we obtain $A B = 1, B C = \sqrt{3}, C A = 2$. We see that by the Pythagorean theorem, triangle $A B C$ is a right triangle ($\angle B = 90^{\circ}$), and its leg $A B$ is half the length of the hypotenuse $A C$, from which the equalities $\angle B A C = 60^{\circ}, \angle B C A = 30^{\circ}$ follow.

We mark point $B_{1}$ as the midpoint of the hypotenuse $A C$ and such a point $C_{1}$ that $\triangle A B C = \triangle A B_{1} C_{1}$ and points $C_{1}$ and $B$ are on opposite sides of $A C$ (Fig. 2). By construction, triangles $A B C$ and $A B_{1} C_{1}$ differ by a $60^{\circ}$ rotation around point $A$. We mark point $T_{1}$ in triangle $A B_{1} C_{1}$, corresponding to point $T$ in triangle $A B C$. Then $B T = B_{1} T_{1}, C T = C_{1} T_{1}$, and $A T = A T_{1} = T T_{1}$. The last equality is due to the fact that triangle $A T T_{1}$ becomes equilateral, since points $T$ and $T_{1}$ differ by a $60^{\circ}$ rotation around point $A$.

It remains to note that points $B, T, T_{1}$, and $C_{1}$ lie on the same line, since $\angle A T B = \angle A T_{1} C_{1} = 120^{\circ}$ and $\angle A T T_{1} = \angle A T_{1} T = 60^{\circ}$. In the end, we get that

$$
a + b + c = A T + B T + C T = B T + T T_{1} + T_{1} C_{1} = B C_{1}
$$

and $B C_{1}$ can be calculated using the cosine theorem for triangle $B A C_{1}$:

$$
B C_{1}^{2} = A B^{2} + A C_{1}^{2} + A B \cdot A C_{1} = 1 + 4 + 1 \cdot 2 = 7
$$

Another solution. We will obtain the answer using algebraic methods. Subtract the second equation from the first. We get $(a - c)(a + c) + b(a - c) = -2$, i.e.,

$$
a + b + c = \frac{-2}{a - c}
$$

Similarly, by subtracting the third equation from the second and the first from the third, we get

$$
a + b + c = \frac{-2}{a - c} = \frac{-1}{b - a} = \frac{3}{c - b}
$$

If we denote $s = a + b + c$, then the previous relations can be rewritten as

$$
a - c = -2 s^{-1}, \quad b - a = -s^{-1}, \quad c - b = 3 s^{-1}
$$

## Kurchatov Mathematics Olympiad for School Students - 2019 Final Stage.

Now we add all the original equations:

$$
2 a^{2} + 2 b^{2} + 2 c^{2} + a b + b c + c a = 8
$$

It is not difficult to see that the left side can be expressed as follows:

$$
(a + b + c)^{2} + \frac{1}{2}\left((a - c)^{2} + (b - a)^{2} + (c - b)^{2}\right) = 8
$$

which means

$$
s^{2} + \frac{1}{2}\left(4 s^{-2} + s^{-2} + 9 s^{-2}\right) = 8
$$

Multiplying by $s^{2}$, we get a quadratic equation in terms of $s^{2}$

$$
s^{4} - 8 s^{2} + 7 = 0
$$

the roots of which are $s^{2} = 1$ and $s^{2} = 7$. However, the first value clearly contradicts the equality (1):

$$
s^{2} = (a + b + c)^{2} > a^{2} + b^{2} + c^{2} + \frac{1}{2} a b + \frac{1}{2} b c + \frac{1}{2} c a = 4
$$

Thus, we are left with $s^{2} = 7$, i.e., $a + b + c = s = \sqrt{7}$.

## Criteria

One of the largest suitable criteria is used:

7 6. Any correct solution to the problem.

6 6. A correct solution with an incorrect answer due to an arithmetic error.

6 6. Several answers are obtained and incorrect ones are not excluded.

3 6. The transition to a geometric problem is made: a triangle with sides 1, $\sqrt{3}$, 2 and a point inside it at distances $a, b, c$ from the vertices is considered.

2 6. Differences $a - b, b - c$, and $c - a$ are expressed in terms of $a + b + c$.

1 6. The correct answer is given.

0 6. Algebraic transformations that did not lead to a solution.