IMO 1959 Problem 3

Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 4m43s

Problem

Let $a,b,c$ be real numbers. Consider the quadratic equation in $\cos{x}$ :

$a\cos ^{2}x + b\cos{x} + c = 0.$

Using the numbers $a,b,c$, form a quadratic equation in $\cos{2x}$, whose roots are the same as those of the original equation. Compare the equations in $\cos{x}$ and $\cos{2x}$ for $a=4, b=2, c=-1$.

Exploration

Let $t=\cos x$. The given condition is that $t$ satisfies a quadratic equation

$a t^{2}+bt+c=0.$

The task is to construct another quadratic equation whose variable is $u=\cos 2x$, and whose set of roots corresponds to the same admissible values of $x$.

Since $\cos 2x=2\cos^{2}x-1$, one has $u=2t^{2}-1$, so the passage from $t$ to $u$ depends only on $t^{2}$. The obstruction is that a quadratic equation in $t$ determines $t^{2}$ only through symmetric combinations of roots, so the correct approach is to express the symmetric polynomials in $u_1,u_2$ in terms of those of $t_1,t_2$.

The key quantities are $t_1+t_2$ and $t_1 t_2$, which come directly from Vieta’s formulas. From these, $t_1^{2}+t_2^{2}$ and $t_1^{2}t_2^{2}$ can be reconstructed, and these determine the quadratic equation in $u$.

For the specific case $a=4$, $b=2$, $c=-1$, the structure may simplify because the original equation already has a simple discriminant structure.

Problem Understanding

The problem asks to start with a quadratic equation satisfied by $\cos x$, then rewrite the same condition in terms of $\cos 2x$, producing a new quadratic equation whose roots correspond to the same underlying trigonometric solutions.

This is a Type A classification problem in the sense that we must explicitly construct the transformed equation and then compare it with the original in a special case.

The main difficulty is that the transformation $u=2t^{2}-1$ is nonlinear and does not preserve roots in a direct way. One must convert root data through symmetric polynomials rather than attempting substitution into the polynomial itself.

For the given coefficients $a=4$, $b=2$, $c=-1$, the structure suggests potential invariance under the transformation, so one expects the resulting equation in $\cos 2x$ to coincide with the original form after simplification.

Proof Architecture

The proof proceeds through the following claims.

First, if $t_1,t_2$ are roots of $a t^{2}+bt+c=0$, then $t_1+t_2=-\frac{b}{a}$ and $t_1 t_2=\frac{c}{a}$ by Vieta’s formulas. This provides the starting point for all symmetric expressions.

Second, one can express $t_1^{2}+t_2^{2}$ as $\left(t_1+t_2\right)^{2}-2t_1 t_2$. This follows from direct expansion of squares.

Third, one can express $t_1^{2}t_2^{2}$ as $\left(t_1 t_2\right)^{2}$, which follows from basic properties of multiplication.

Fourth, defining $u_i=2t_i^{2}-1$, one can compute $u_1+u_2$ and $u_1 u_2$ in terms of $t_1,t_2$, which determines the quadratic equation satisfied by $u$.

The most delicate step is the reconstruction of the quadratic from symmetric expressions in $u$, since a sign or scaling error there changes the final polynomial.

Solution

Let $t=\cos x$. Then $t$ satisfies

$a t^{2}+bt+c=0.$

Let its roots be $t_1,t_2$. By Vieta’s formulas,

$t_1+t_2=-\frac{b}{a}, \qquad t_1 t_2=\frac{c}{a}.$

From algebraic identities,

$t_1^{2}+t_2^{2}=(t_1+t_2)^{2}-2t_1 t_2,$

so substituting the Vieta relations gives

$t_1^{2}+t_2^{2}=\frac{b^{2}}{a^{2}}-\frac{2c}{a}.$

Also,

$t_1^{2}t_2^{2}=\left(t_1 t_2\right)^{2}=\frac{c^{2}}{a^{2}}.$

Define $u_i=2t_i^{2}-1$. Then

$u_1+u_2=2(t_1^{2}+t_2^{2})-2,$

hence

= \frac{2b^{2}}{a^{2}}-\frac{4c}{a}-2.$$Next,$$u_1 u_2=(2t_1^{2}-1)(2t_2^{2}-1)=4t_1^{2}t_2^{2}-2(t_1^{2}+t_2^{2})+1.$$Substituting the previously obtained expressions,$$u_1 u_2=\frac{4c^{2}}{a^{2}}-2\left(\frac{b^{2}}{a^{2}}-\frac{2c}{a}\right)+1,$$so$$u_1 u_2=\frac{4c^{2}}{a^{2}}-\frac{2b^{2}}{a^{2}}+\frac{4c}{a}+1.$$Thus $u_1,u_2$ are roots of the quadratic equation$$u^{2}-(u_1+u_2)u+u_1 u_2=0,$$which becomes$$u^{2}-\left(\frac{2b^{2}}{a^{2}}-\frac{4c}{a}-2\right)u+\left(\frac{4c^{2}}{a^{2}}-\frac{2b^{2}}{a^{2}}+\frac{4c}{a}+1\right)=0.$$Multiplying by $a^{2}$ yields an equivalent equation with integer coefficients:$$a^{2}u^{2}-\left(2b^{2}-4ac-2a^{2}\right)u+\left(4c^{2}-2b^{2}+4ac+a^{2}\right)=0.$$Now substitute $a=4$, $b=2$, $c=-1$. First compute the coefficients:$$t_1+t_2=-\frac{1}{2}, \qquad t_1 t_2=-\frac{1}{4}.$$Then$$t_1^{2}+t_2^{2}=\left(-\frac{1}{2}\right)^{2}-2\left(-\frac{1}{4}\right)=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}.$$Hence$$u_1+u_2=2\cdot\frac{3}{4}-2=\frac{3}{2}-2=-\frac{1}{2}.$$Also$$t_1^{2}t_2^{2}=\frac{1}{16},$$so$$u_1 u_2=4\cdot\frac{1}{16}-2\cdot\frac{3}{4}+1=\frac{1}{4}-\frac{3}{2}+1=-\frac{1}{4}.$$Therefore $u$ satisfies$$u^{2}-\left(-\frac{1}{2}\right)u-\frac{1}{4}=0,$$that is,$$u^{2}+\frac{1}{2}u-\frac{1}{4}=0.$$Multiplying by $4$ gives$$4u^{2}+2u-1=0.$$The original equation in $t=\cos x$ is$$4t^{2}+2t-1=0,$$while the transformed equation in $u=\cos 2x$ is$$4u^{2}+2u-1=0.$$Thus the two equations coincide in form.$$\boxed{4x^{2}+2x-1=0 \text{ (same form for } \cos x \text{ and } \cos 2x)}$$--- ## Verification of Key Steps The first delicate point is the computation of $t_1^{2}+t_2^{2}$. Recomputing directly from $t_1+t_2=-\frac{1}{2}$ and $t_1 t_2=-\frac{1}{4}$ gives$$t_1^{2}+t_2^{2}=\frac{1}{4}-2\left(-\frac{1}{4}\right)=\frac{3}{4},$$which confirms the intermediate value used to build $u_1+u_2$. The second sensitive point is $u_1 u_2$. Expanding from scratch,$$(2t_1^{2}-1)(2t_2^{2}-1)=4t_1^{2}t_2^{2}-2t_1^{2}-2t_2^{2}+1,$$and substituting $t_1^{2}t_2^{2}=\frac{1}{16}$ and $t_1^{2}+t_2^{2}=\frac{3}{4}$ yields$$\frac{1}{4}-\frac{3}{2}+1=-\frac{1}{4},$$confirming consistency. The third point is the final reconstruction of the quadratic from roots. Using $u^{2}-(u_1+u_2)u+u_1 u_2$ reproduces the correct sign pattern; any sign error in symmetric sums would flip linear or constant terms and produce a different equation. --- ## Alternative Approaches One alternative begins by substituting $\cos 2x=2\cos^{2}x-1$ directly into the original equation and rewriting everything in terms of $t=\cos x$, producing a polynomial identity in $t^{2}$. This approach leads to the same symmetric reconstruction but hides the role of Vieta’s formulas. Another approach interprets the transformation $t\mapsto 2t^{2}-1$ as a Chebyshev polynomial relation, since $T_{2}(t)=2t^{2}-1$. From this viewpoint, the problem becomes a statement about how quadratic polynomials behave under composition with $T_{2}$. This perspective explains why the specific coefficients $4,2,-1$ are preserved under the transformation.