Kvant Math Problem 106

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Problem

Prove that if for the numbers $p_1$, $p_2$, $q_1$, $q_2$ the inequality $$(q_1-q_2)^2+(p_1-p_2)(p_1q_2-p_2q_1)\lt0,$$ holds, then the quadratic trinomials $$x^2+p_1x+q_1\quad\text{и}\quad x^2+p_2x+q_2$$ have real roots, and between the two roots of each of them lies a root of the other.

I. F. Sharygin

All-Union Mathematical Olympiad of School Students (V)

Exploration

Let $f_1(x)=x^2+p_1x+q_1$ and $f_2(x)=x^2+p_2x+q_2$. The given condition is

$$(q_1-q_2)^2+(p_1-p_2)(p_1q_2-p_2q_1)<0.$$

A natural object connecting two quadratics is the Wronskian-like expression

$$W(x)=f_1(x)f_2'(x)-f_2(x)f_1'(x).$$

For quadratics this simplifies to another quadratic, and its discriminant turns out to coincide exactly with the expression in the hypothesis. Thus the condition forces $W$ to have no real zeros, hence constant sign.

If $W$ has constant sign, then the ratio $f_2/f_1$ has derivative of constant sign wherever it is defined, suggesting strict monotonicity between roots. This is the mechanism behind interlacing.

The main difficulty is that the statement also requires existence of real roots for both quadratics. That must be derived from the same monotonicity idea by contradiction: if both were positive everywhere, the ratio would approach the same limit at both infinities yet remain strictly monotone, which is impossible.

The delicate point is then to convert constant sign of $W$ into alternating signs of $f_1$ and $f_2$ at the critical points of each other, forcing a root of one between consecutive roots of the other.

Problem Understanding

The problem is a Type B statement: one must prove that a single inequality between coefficients of two monic quadratic polynomials forces both polynomials to have real roots and forces their roots to interlace.

The core difficulty is that the condition does not mention roots directly. Instead it encodes a hidden comparison structure between the two quadratics. The goal is to extract this structure and show it forces Sturm-type interlacing behavior.

Proof Architecture

The Wronskian $W(x)=f_1f_2'-f_2f_1'$ is computed explicitly and shown to be a quadratic polynomial whose discriminant equals $4\big((q_1-q_2)^2+(p_1-p_2)(p_1q_2-p_2q_1)\big)$.

The hypothesis implies that this discriminant is negative, so $W(x)$ has constant sign on $\mathbb{R}$.

From constant sign of $W$, one proves that not both $f_1$ and $f_2$ can be positive everywhere, hence both must have real roots by symmetry.

The next step uses the identity

$$\left(\frac{f_2}{f_1}\right)'=\frac{W}{f_1^2},$$

which implies strict monotonicity of $f_2/f_1$ on intervals where $f_1\neq 0$, and similarly with roles reversed.

The hardest part is proving interlacing: constant sign of $W$ forces alternating signs of $f_1$ and $f_2$ at successive critical points, which forces a root of one polynomial between consecutive roots of the other.

The most delicate lemma is the deduction of real-root existence from monotonicity of the ratio.

Solution

Let

$$f_1(x)=x^2+p_1x+q_1,\qquad f_2(x)=x^2+p_2x+q_2.$$

Define

$$W(x)=f_1(x)f_2'(x)-f_2(x)f_1'(x).$$

Since $f_1'(x)=2x+p_1$ and $f_2'(x)=2x+p_2$, expansion yields

$$W(x)=(p_1-p_2)x^2+2(q_1-q_2)x+(p_2q_1-p_1q_2).$$

Its discriminant equals

$$\Delta_W=[2(q_1-q_2)]^2-4(p_1-p_2)(p_2q_1-p_1q_2),$$

which simplifies to

$$\Delta_W=4\big((q_1-q_2)^2+(p_1-p_2)(p_1q_2-p_2q_1)\big).$$

The hypothesis implies $\Delta_W<0$, hence $W(x)$ has no real roots and does not change sign on $\mathbb{R}$.

Assume first that $f_1$ has no real roots. Since its leading coefficient is positive, $f_1(x)>0$ for all real $x$, so the quotient $f_2(x)/f_1(x)$ is well defined and smooth on $\mathbb{R}$. Differentiation gives

$$\left(\frac{f_2}{f_1}\right)'=\frac{f_2'f_1-f_2f_1'}{f_1^2}=\frac{W(x)}{f_1(x)^2}.$$

The denominator is always positive, hence the derivative has constant nonzero sign. Therefore $f_2/f_1$ is strictly monotone on $\mathbb{R}$.

As $x\to\pm\infty$,

$$\frac{f_2(x)}{f_1(x)}\to 1.$$

A strictly monotone function on $\mathbb{R}$ cannot have identical finite limits at both infinities unless it is constant, which is impossible because $W(x)$ is not identically zero. This contradiction shows that $f_1$ has at least one real root.

By symmetry in $(p_1,q_1)$ and $(p_2,q_2)$, the same argument shows that $f_2$ also has a real root.

Each polynomial is monic quadratic, hence having at least one real root forces two real roots.

Let the roots of $f_1$ be $a_1<a_2$. Since $f_1(x)>0$ outside $[a_1,a_2]$ and $f_1(x)<0$ inside, consider the function $f_2/f_1$ on each interval of $\mathbb{R}\setminus{a_1,a_2}$. On each such interval the sign of $W$ implies that $f_2/f_1$ is strictly monotone, hence $f_2$ cannot change sign more than once on each interval.

Evaluating $W$ at a root $a$ of $f_1$ gives

$$W(a)= -f_2(a)f_1'(a),$$

so the sign of $f_2(a)$ is determined by the sign of $f_1'(a)$ together with the constant sign of $W$. Since $f_1'(a_1)<0$ and $f_1'(a_2)>0$, the values $f_2(a_1)$ and $f_2(a_2)$ have opposite signs. Hence $f_2$ has a root in $(a_1,a_2)$.

Repeating the argument with roles exchanged shows that between the two roots of $f_2$ lies a root of $f_1$.

Thus both quadratics have real roots and their roots strictly interlace.

This completes the proof.

∎

Verification of Key Steps

The computation of $W(x)$ was checked directly by expanding $f_1f_2'$ and $f_2f_1'$ and confirming cancellation of cubic terms and correct collection of coefficients.

The identification of $\Delta_W$ with $4((q_1-q_2)^2+(p_1-p_2)(p_1q_2-p_2q_1))$ was verified by direct algebraic rearrangement of the constant term in the discriminant expression.

The limit argument for $f_2/f_1$ uses only leading coefficient comparison; both quadratics are monic so the ratio tends to $1$ at both infinities.

The sign change argument at roots uses that $W(a)\neq 0$ and $f_1'(a)\neq 0$ at simple roots of a monic quadratic, ensuring opposite signs of $f_2$ at the two roots of $f_1$.

Alternative Approaches

One alternative approach interprets the pair $(p_i,q_i)$ as coefficients of a Sturm sequence and derives interlacing directly from Sturm comparison theory for second-order polynomials. Another approach uses projective geometry: the condition encodes negativity of a cross-ratio-type invariant ensuring separation of roots on the real projective line. The Wronskian method is preferable because it reduces the problem to a single explicit polynomial whose discriminant reproduces the hypothesis exactly, making the mechanism transparent and fully elementary.