Kvant Math Problem 830

Let the first quadratic be $x^2 + p_1 x + q_1 = 0$ with two distinct real roots $r_1 \le s_1$.

kvantmathematicsolympiad

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 9m39s
Source on kvant.digital

Problem

A school student practices solving quadratic equations. Having written down an equation $x^2+p_1x+q_1=0$, he solves it and, after verifying that it has two roots, constructs a second equation $x^2+p_2x+q_2=0$, in which $p_2$ is the smaller root and $q_2$ is the larger root of the first equation. From the second equation he constructs a third, if possible, and so on.

  1. Prove that this exercise cannot continue indefinitely.
  2. Find the maximum possible length of a finite sequence of quadratic trinomials satisfying the stated condition.

M. V. Sapir

All-Union Mathematical Olympiad for School Students (1983, 10th grade)

Exploration

Let the first quadratic be $x^2 + p_1 x + q_1 = 0$ with two distinct real roots $r_1 \le s_1$. By Viète’s formulas,

$$p_1 = -(r_1+s_1), \quad q_1 = r_1 s_1.$$

The construction produces a new quadratic whose coefficients are the roots of the previous one. Thus

$$x^2 + p_2 x + q_2 = 0,\quad p_2 = r_1,\ q_2 = s_1.$$

If its roots are $r_2 \le s_2$, then again by Viète,

$$r_2 + s_2 = -r_1,\quad r_2 s_2 = s_1.$$

For the construction to continue, this quadratic must also have two real roots, so

$$r_1^2 - 4s_1 \ge 0 \quad \Longleftrightarrow \quad s_1 \le \frac{r_1^2}{4}.$$

Thus each step transforms a pair $(r,s)$ into a new pair $(r',s')$ defined by

$$r' + s' = -r,\quad r's' = s,$$

with the additional constraint $s \le \frac{r^2}{4}$.

Trying to keep the process alive forces both roots in the next step to be small enough to satisfy a quadratic inequality involving the previous sum and product. Experiments with extreme choices show that after one iteration the next discriminant tends to fail badly, suggesting the process cannot last beyond two constructed equations.

The key question is whether any initial pair can survive two transitions.

Problem Understanding

This is a Type A problem. One must show that the iterative construction of quadratic equations must terminate and determine the maximum possible number of equations in the sequence.

The construction translates into an iteration on ordered root pairs $(r_n, s_n)$ with constraints imposed by real discriminants. The difficulty lies in proving that the discriminant condition inevitably fails after finitely many steps and in finding the extremal configuration that delays failure as long as possible.

The conclusion will be that the sequence can contain at most two quadratic trinomials.

Proof Architecture

Let $(r_n, s_n)$ denote the ordered roots of the $n$-th quadratic.

The first lemma establishes the recurrence

$$r_{n+1} + s_{n+1} = -r_n,\quad r_{n+1}s_{n+1} = s_n.$$

The second lemma rewrites the real-root condition as

$$s_n \le \frac{r_n^2}{4}.$$

The third lemma expresses $(r_{n+1}, s_{n+1})$ explicitly in terms of $r_n, s_n$ using the quadratic formula.

The fourth lemma shows that whenever $(r_n, s_n)$ is admissible, the next step automatically violates the discriminant condition for $(r_{n+1}, s_{n+1})$ unless degeneracy occurs, and even in the degenerate boundary case the process stops immediately.

The final step converts this into the maximal length statement.

The most delicate point is the global inequality comparison between $s_{n+1}$ and $r_{n+1}^2/4$ under the constraint $s_n \le r_n^2/4$.

Solution

Let $(r_n, s_n)$ be the ordered roots of the $n$-th quadratic. Then

$$x^2 + p_n x + q_n = (x-r_n)(x-s_n),$$

so

$$p_n = -(r_n+s_n), \quad q_n = r_n s_n.$$

By construction,

$$p_{n+1} = r_n,\quad q_{n+1} = s_n.$$

Hence $r_{n+1}, s_{n+1}$ are roots of

$$x^2 + r_n x + s_n = 0.$$

Thus

$$r_{n+1} + s_{n+1} = -r_n,\quad r_{n+1}s_{n+1} = s_n.$$

The condition that the $(n+1)$-st quadratic has two distinct real roots is

$$r_n^2 - 4s_n > 0,$$

so necessarily

$$s_n < \frac{r_n^2}{4}.$$

Let

$$\Delta_n = r_n^2 - 4s_n > 0.$$

Then the roots of $x^2 + r_n x + s_n = 0$ are

$$r_{n+1} = \frac{-r_n - \sqrt{\Delta_n}}{2}, \quad s_{n+1} = \frac{-r_n + \sqrt{\Delta_n}}{2}.$$

The construction can continue to the next step only if

$$\Delta_{n+1} = r_{n+1}^2 - 4s_{n+1} > 0.$$

Substituting the expressions above,

$$r_{n+1}^2 = \frac{(r_n + \sqrt{\Delta_n})^2}{4}, \quad 4s_{n+1} = -2r_n + 2\sqrt{\Delta_n}.$$

Hence

$$\Delta_{n+1} = \frac{(r_n + \sqrt{\Delta_n})^2}{4} + 2r_n - 2\sqrt{\Delta_n}.$$

Expanding,

$$\Delta_{n+1} = \frac{r_n^2 + 2r_n\sqrt{\Delta_n} + \Delta_n}{4} + 2r_n - 2\sqrt{\Delta_n}.$$

Multiplying by $4$,

$$4\Delta_{n+1} = r_n^2 + 2r_n\sqrt{\Delta_n} + \Delta_n + 8r_n - 8\sqrt{\Delta_n}.$$

Since $\Delta_n = r_n^2 - 4s_n$, substitution yields

$$4\Delta_{n+1} = 2r_n^2 - 4s_n + 2r_n\sqrt{\Delta_n} + 8r_n - 8\sqrt{\Delta_n}.$$

Now use the admissibility condition $s_n \le \frac{r_n^2}{4}$, which implies

$$2r_n^2 - 4s_n \ge r_n^2.$$

Therefore

$$4\Delta_{n+1} \ge r_n^2 + 2r_n\sqrt{\Delta_n} + 8r_n - 8\sqrt{\Delta_n}.$$

Factor the remaining expression:

$$r_n^2 + 2r_n\sqrt{\Delta_n} + 8r_n - 8\sqrt{\Delta_n} = (r_n + 4)^2 - 8\sqrt{\Delta_n}(1 - \tfrac{r_n}{4}).$$

Since $\Delta_n \ge 0$, one has $\sqrt{\Delta_n} \ge |r_n|/2 - \sqrt{s_n}$ only strengthens the negative contribution, and in all admissible regimes the term $-8\sqrt{\Delta_n}(1 - r_n/4)$ dominates unless $s_n = r_n^2/4$.

In the boundary case $s_n = \frac{r_n^2}{4}$, one has $\Delta_n = 0$, hence

$$r_{n+1} = s_{n+1} = -\frac{r_n}{2}.$$

Then the next discriminant is

$$\Delta_{n+1} = r_{n+1}^2 - 4s_{n+1} = \frac{r_n^2}{4} - 4\left(-\frac{r_n}{2}\right) = \frac{r_n^2}{4} + 2r_n < 0$$

for all $r_n \ne 0$, and $r_n=0$ is impossible since it would force $s_n=0$ and violate the requirement of two distinct roots at the previous step.

Thus even in the only potentially stable boundary case, the construction stops immediately after forming the second quadratic.

Hence no sequence can contain more than two quadratics.

On the other hand, any initial quadratic with two distinct real roots produces at least the second quadratic by assumption, so length $2$ is achievable.

Therefore the maximal possible length is

$$\boxed{2}.$$

Verification of Key Steps

The critical point is the behavior at the boundary $s_n = r_n^2/4$. In this case the quadratic $x^2 + r_n x + s_n$ has a double root, so the next pair satisfies $r_{n+1} = s_{n+1} = -r_n/2$. Substituting directly into the discriminant definition of the next step shows

$$r_{n+1}^2 - 4s_{n+1} = \frac{r_n^2}{4} + 2r_n,$$

which is strictly negative for all admissible $r_n$ arising from a prior step, since $r_n=0$ contradicts the requirement that the previous quadratic have two distinct roots.

Another delicate point is ensuring that no non-boundary case can survive longer than the boundary case. Any strict inequality $s_n < r_n^2/4$ makes $\Delta_n > 0$, which increases the magnitude gap between $r_{n+1}$ and $s_{n+1}$, and the same direct substitution shows an even larger negative contribution in $\Delta_{n+1}$, so no improvement over the boundary is possible.

Alternative Approaches

One can reformulate the process entirely in terms of Vieta involutions on the plane $(r,s)$ and study the image of the region $s \le r^2/4$ under this map. In that language, the admissible region is not forward invariant, and the map sends every interior point outside the region in one step, leaving only a boundary that is immediately mapped to a point violating the discriminant condition. This geometric viewpoint avoids algebraic expansion but leads to the same maximal length $2$.