Can the constants $a, b, c$ be determined such that the equality

$$
(x+a)^{2}+(2 x+b)^{2}+(2 x+c)^{2}=(3 x+1)^{2}
$$

holds for all $x$?
Both sides have the set of non-negative numbers as their value range, their smallest value being 0. The right side only takes this value at $x=-\frac{1}{3}$, while the left side does so precisely when simultaneously

$$
a=-x=\frac{1}{3}, \quad b=-2 x=\frac{2}{3}, \quad \text { and } \quad c=-2 x=\frac{2}{3}
$$

This value assignment for the triplet $(a, b, c)$ is thus a necessary condition for the equality to be an identity.

The condition is also sufficient, for after substitution and factoring, the equality can be transformed into

$$
\left(x+\frac{1}{3}\right)^{2}+2^{2}\left(x+\frac{1}{3}\right)^{2}+2^{2}\left(x+\frac{1}{3}\right)^{2}=3^{2}\left(x+\frac{1}{3}\right)^{2}
$$

which holds for all $x$, even when $x+\frac{1}{3} \neq 0$, because $1+2^{2}+2^{2}=3^{2}$.

Remarks. 1. Most of the submissions expanded the parentheses, thus getting into a lengthy "neck-breaking" work. Some of them, fortunately, arrived at interesting conclusions in exchange.

Those who started the problem this way, after rearranging the desired identity, could arrive at a first-degree polynomial on the left side:

$$
2(a+2 b+2 c-3) x+\left(a^{2}+b^{2}+c^{2}-1\right)=0
$$

This holds for all $x$ if and only if the values in the parentheses are 0. This leads to a system of equations for the unknowns $a, b, c$. It might seem insufficient that we only have two equations for three unknowns, but this "prejudice" is based on our knowledge of first-degree systems. We will get a unique solution, and again, $c=b=\frac{2}{3}, a=\frac{1}{3}$ will be the only solution.

2. Several created a third equation for themselves by assuming $b=c$, reasoning that the roles of $b$ and $c$ are "equal." Fortunately, the calculated values also satisfy this.
3. Many noticed that the disappearance of the coefficients in (2) can be nicely interpreted in the language of coordinate geometry in a three-dimensional orthogonal coordinate system spanned by the $a, b, c$ axes. The second requirement in parentheses

$$
a^{2}+b^{2}+c^{2}=1
$$

is the equation of a unit sphere centered at the origin, and the first, when rearranged,

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

is the equation of the plane that intersects the positive parts of the axes at distances of $3, 3/2, 3/2$ units from the origin. This plane is tangent to the sphere at the point $\left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right)$.

4. Many of the unsuccessful submissions' authors - unjustifiably - only searched for solutions among the integers.
5. The above considerations are, of course, only valid if we stay within the real number system, as usual. However, it is worth noting what happens if we step into the complex number system. In the obtained

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

system of equations, consider $a$ as a parameter. Then, with not too complicated transformations, we get that

$$
(b-c)^{2}=b^{2}-2 b c+c^{2}=2\left(1-a^{2}\right)-\left(\frac{3-a}{2}\right)^{2}=-\left(\frac{3 a-1}{2}\right)^{2}
$$

Thus, the following two parametric solution systems are obtained:

![](https://cdn.mathpix.com/cropped/2024_05_02_4015417d55f2560d1609g-1.jpg?height=105&width=1289&top_left_y=2121&top_left_x=407)

This is of the form of the parametric equations of two lines in space. We usually interpret this as the plane and sphere intersecting in these two (intersecting) lines. This is not an incomprehensible thing, since a plane can also intersect a surface of revolution in intersecting lines. Indeed, in the complex space, there is not much difference between a sphere and a cone.

Well, as we said, these two lines intersect at one point (since they are in the same plane!); interestingly, this intersection point falls into the real space. This point is the point of tangency of the plane on the sphere.

If the plane intersects or "avoids" the sphere, the situation is different. For example, consider the sphere $a^{2}+b^{2}+c^{2}=2$. This is intersected by the plane given by the equation $a=1$ in a circle $\left(b^{2}+c^{2}=1\right)$. Although this also has complex points, it remains a circle. If the plane is defined by the equation $a=2$, then the "intersecting circle" has the equation $b^{2}+c^{2}=-2$. This is also a circle, but it only has complex points. The fact that in our original example the intersection was a pair of lines is due to the fact that the tangential behavior is not "regular" (regular, normal), but "singular" (irregular, exceptional).