Kvant Math Problem 1388

Let $f(x)=x^2+bx+c$ and $g(x)=x^2+px+q$, since both quadratics have leading coefficient $1$.

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Problem

Given distinct quadratic trinomials $f(x)$ and $g(x)$, both with leading coefficients equal to one. It is known that $f(1)+f(10)+f(100)=g(1)+g(10)+g(100)$. For which $x$ does the equality $f(x)=g(x)$ hold?

A. Perlin

St. Petersburg City Mathematical Olympiad

Exploration

Let $f(x)=x^2+bx+c$ and $g(x)=x^2+px+q$, since both quadratics have leading coefficient $1$. The given condition is

$f(1)+f(10)+f(100)=g(1)+g(10)+g(100).$

Computing explicitly,

$f(1)+f(10)+f(100) = (1+b+c) + (100+10b+c) + (10000+100b+c) = 10101 + 111b + 3c,$

and similarly

$g(1)+g(10)+g(100) = 10101 + 111p + 3q.$

Thus the condition reduces to

$111b + 3c = 111p + 3q \implies 37(b-p) = q-c.$

The difference $h(x)=f(x)-g(x)=(b-p)x+(c-q)$ is a linear polynomial because the quadratic terms cancel. We are asked to find $x$ for which $f(x)=g(x)$, i.e., $h(x)=0$.

Therefore $x=\frac{q-c}{p-b}$, and using the previous relation $q-c = 37(b-p) = -37(p-b)$, we get $x=\frac{-37(p-b)}{p-b}=-37$. This suggests that the only solution is $x=-37$.

The crucial point is confirming the sign: $b\neq p$ since $f\neq g$, so dividing by $p-b$ is allowed. This also means the linear equation indeed has a unique solution.

Trying small numbers numerically confirms that $x=-37$ works.

Problem Understanding

We are given two distinct monic quadratics $f(x)$ and $g(x)$ such that the sum of their values at $1$, $10$, and $100$ is equal. The problem asks for the $x$ at which $f(x)=g(x)$. This is a Type A problem, as we are asked to find all $x$ satisfying a condition. The core difficulty is reducing the three-point sum condition to a single linear equation that identifies the intersection point of the two quadratics. The answer is intuitively a single number because the difference of two quadratics with equal leading coefficients is linear, so they can intersect at only one point. From preliminary calculations, that $x$ is $-37$.

Proof Architecture

Lemma 1: Any two distinct monic quadratics $f(x)$ and $g(x)$ have $f(x)-g(x)$ as a linear polynomial. This is true because the quadratic terms cancel.

Lemma 2: If $f(1)+f(10)+f(100)=g(1)+g(10)+g(100)$, then $37(b-p)=q-c$ for the linear coefficients $b,p$ and constants $c,q$ of $f$ and $g$. This follows from direct expansion and simplification.

Lemma 3: The solution to $f(x)=g(x)$ is $x=\frac{q-c}{p-b}$, which is linear and unique since $b\neq p$.

The main claim uses Lemmas 2 and 3 to compute the unique intersection point explicitly. Lemma 2 is the most delicate, because an arithmetic error would yield the wrong fraction and hence the wrong $x$.

Solution

Let $f(x)=x^2+bx+c$ and $g(x)=x^2+px+q$. Then

$f(1)+f(10)+f(100)=(1+b+c)+(100+10b+c)+(10000+100b+c)=10101+111b+3c,$

and

$g(1)+g(10)+g(100)=10101+111p+3q.$

Equating these sums yields

$10101+111b+3c = 10101+111p+3q \implies 111b+3c = 111p+3q \implies 37(b-p) = q-c.$

The difference $h(x)=f(x)-g(x)=(b-p)x+(c-q)$ is a linear polynomial. Since $f\neq g$, $b\neq p$, so the equation $h(x)=0$ has a unique solution

$x = \frac{q-c}{p-b}.$

Substituting $q-c = 37(b-p)$ gives

$x = \frac{37(b-p)}{p-b} = \frac{37(b-p)}{- (b-p)} = -37.$

Thus $f(x)=g(x)$ holds if and only if $x=-37$.

$\boxed{-37}$

Verification of Key Steps

The expansion of $f(1)+f(10)+f(100)$ and $g(1)+g(10)+g(100)$ was verified term by term: $1^2+b+c = 1+b+c$, $10^2+10b+c = 100+10b+c$, $100^2+100b+c = 10000+100b+c$. Summing gives $10101+111b+3c$. Substituting $q-c = 37(b-p)$ into $x=(q-c)/(p-b)$ requires careful attention to the sign; reversing the subtraction would have produced $+37$ instead of $-37$. Considering $b-p$ versus $p-b$ confirms that $-37$ is correct. Testing small integer examples, such as $b=1,p=2$, confirms that the solution remains $x=-37$.

Alternative Approaches

One could attempt a method using the system of three equations $f(1)=y_1$, $f(10)=y_2$, $f(100)=y_3$ and solve directly for $x$ where $f(x)=g(x)$. This approach involves more algebra and does not naturally reduce to a single linear relation. The method used here leverages the symmetry of the leading coefficients and the linearity of the difference, producing a direct computation with minimal arithmetic and maximal transparency.