Problem 11.8. Real numbers $a, b, c$ are such that

$$
\left\{\begin{array}{l}
a^{2}+a b+b^{2}=11 \\
b^{2}+b c+c^{2}=11
\end{array}\right.
$$

(a) (1 point) What is the smallest value that the expression $c^{2}+c a+a^{2}$ can take?

(b) (3 points) What is the largest value that the expression $c^{2}+c a+a^{2}$ can take?
Answer: (a) 0; (b) 44.

Solution. (a) Note that if $a=c=0$, and $b=\sqrt{11}$, then the given system of equalities is satisfied, and $c^{2}+c a+a^{2}$ equals 0. On the other hand,

$$
c^{2}+c a+a^{2}=\frac{c^{2}}{2}+\frac{a^{2}}{2}+\frac{(c+a)^{2}}{2} \geqslant 0
$$

Thus, the smallest value of the expression $c^{2}+c a+a^{2}$ is 0.

(b) We will use the formula $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$. Multiply the first equation by $a-b$, and the second by $b-c$, we get

$$
\left\{\begin{array}{l}
a^{3}-b^{3}=11(a-b) \\
b^{3}-c^{3}=11(b-c)
\end{array}\right.
$$

Adding these two equations, we get $a^{3}-c^{3}=11(a-c)$, from which it follows that $a^{2}+a c+c^{2}=11$ if $a \neq c$. It remains to determine what happens if $a=c$.

In the case $a=c$, both equalities in the system are the same, and we need to maximize the expression $3 a^{2}$.

The case $a=c=0$ has already been considered in the previous part, so we will consider $a \neq 0$ from now on. Let $b=k a$ for some real $k$. Then from the condition we have

$$
a^{2}+k a^{2}+k^{2} a^{2}=11 \quad \text { and } \quad a^{2}=\frac{11}{1+k+k^{2}}
$$

Note that

$$
1+k+k^{2}=\frac{3}{4}+\left(\frac{1}{2}+k\right)^{2} \geqslant \frac{3}{4}
$$

Thus,

$$
a^{2} \leqslant \frac{11}{3 / 4}=\frac{44}{3} \quad \text { and } \quad 3 a^{2} \leqslant \frac{44}{3} \cdot 3=44
$$

It is clear that this value is achieved if $a=c=\sqrt{44 / 3}, b=-\sqrt{11 / 3}$.

Another solution for part (b). We complete the squares in the left-hand sides of the given equations:

$$
\left\{\begin{array}{l}
\frac{3}{4} a^{2}+\left(\frac{1}{2} a+b\right)^{2}=11 \\
\frac{3}{4} c^{2}+\left(\frac{1}{2} c+b\right)^{2}=11
\end{array}\right.
$$

From the fact that $\left(\frac{1}{2} a+b\right)^{2}$ and $\left(\frac{1}{2} c+b\right)^{2}$ are non-negative, it follows that $\frac{3}{4} a^{2} \leqslant 11$ and $\frac{3}{4} c^{2} \leqslant 11$. Thus, $|a| \leqslant \sqrt{44 / 3}$ and $|c| \leqslant \sqrt{44 / 3}$.

We estimate the desired expression:

$$
c^{2}+a c+a^{2} \leqslant c^{2}+|a| \cdot|c|+a^{2} \leqslant \frac{44}{3}+\sqrt{\frac{44}{3}} \cdot \sqrt{\frac{44}{3}}+\frac{44}{3}=44
$$

Thus, the desired expression does not exceed 44, and this value is achieved when $a=c=\sqrt{44 / 3}$ and $b=-a / 2=-\sqrt{11 / 3}$.

Another solution. Mark a point $O$ on the plane and draw three coordinate lines $O a, O b, O c$ through it at angles of $120^{\circ}$ to each other (Fig. 12a). Considering $O$ as the origin on each of the lines, mark points $A, B, C$ on them with coordinates equal to the given numbers $a, b, c$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9703912434d201fc4f33g-08.jpg?height=418&width=460&top_left_y=294&top_left_x=172)

(a)

![](https://cdn.mathpix.com/cropped/2024_05_06_9703912434d201fc4f33g-08.jpg?height=418&width=470&top_left_y=292&top_left_x=758)

(b)

Fig. 12: to the solution of problem 11.8

Note that $A B^{2}=a^{2}+a b+b^{2}$. Indeed, in the case $a, b>0$ this is simply the cosine rule for triangle $A O B$ with sides $a$ and $b$ and an angle of $120^{\circ}$ between them. But this equality holds in all other cases as well. For example, if $a>0$ and $b<0$ (Fig. 12b), we have $O B=-b$ and $\angle A O B=60^{\circ}$, from which

$$
A B^{2}=A O^{2}+B O^{2}-2 \cos \left(60^{\circ}\right) \cdot A O \cdot B O=a^{2}+b^{2}+a b
$$

Similarly, we get $B C^{2}=b^{2}+b c+c^{2}$ and $C A^{2}=c^{2}+c a+a^{2}$. Now it is not difficult to rephrase the original problem:

On three lines intersecting at one point at angles of $60^{\circ}$ to each other, points $A, B, C$ are marked respectively (possibly coinciding). Given $A B=B C=\sqrt{11}$; we need to find the smallest and largest possible value of $A C^{2}$.

(a) It is clear that the smallest value is 0, which is achieved when $A=C=O$ (Fig. 13a).

(b) On the other hand, from the triangle inequality we have $A C \leqslant A B+B C=2 \sqrt{11}$, which is achieved if the points $A, B, C$ lie on the same line (Fig. 13b).

![](https://cdn.mathpix.com/cropped/2024_05_06_9703912434d201fc4f33g-09.jpg?height=402&width=460&top_left_y=744&top_left_x=174)

(a)

![](https://cdn.mathpix.com/cropped/2024_05_06_9703912434d201fc4f33g-09.jpg?height=401&width=470&top_left_y=740&top_left_x=758)

(b)

Fig. 13: to the solution of problem 11.8

## 11th Grade