IMO 1959 Problem 3
Let $t=\cos x$ and $u=\cos 2x$.
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 12m31s
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 $t=\cos x$ and $u=\cos 2x$. The original equation becomes a quadratic in $t$, namely $a t^2 + b t + c = 0$, whose roots we denote by $t_1$ and $t_2$. The standard trigonometric identity $u = 2t^2 - 1$ transforms each root $t_i$ into a corresponding value $u_i = 2t_i^2 - 1$. The task is to construct a quadratic equation having roots $u_1$ and $u_2$ expressed purely in terms of $a,b,c$.
The main difficulty is that the transformation from $t$ to $u$ is nonlinear, so the new symmetric sums must be expressed through $t_1+t_2$ and $t_1 t_2$, which are controlled by $a,b,c$. The correct approach is therefore to compute $u_1+u_2$ and $u_1u_2$ using symmetric polynomials and then reconstruct the quadratic.
Problem Understanding
The problem asks for an explicit transformation rule that converts a quadratic equation satisfied by $\cos x$ into a quadratic equation satisfied by $\cos 2x$, while preserving the correspondence between roots through the identity $\cos 2x = 2\cos^2 x - 1$. The second part requires substituting $a=4$, $b=2$, $c=-1$ and comparing the resulting equation with the original one.
This is a Type A classification problem because one must construct an associated equation and then verify that it has exactly the transformed roots, and also confirm that no other values arise.
The core difficulty is that the cosine doubling map does not act linearly on roots, so one must work through symmetric polynomials rather than attempting direct substitution into the quadratic form.
The expected outcome is that the transformed equation can be written explicitly in terms of $a,b,c$, and for the specific values $a=4$, $b=2$, $c=-1$, the two equations turn out to coincide up to scaling.
Proof Architecture
The argument proceeds through three lemmas.
The first lemma expresses the elementary symmetric sums $t_1+t_2$ and $t_1t_2$ in terms of $a,b,c$ using Viète relations. This is justified by standard coefficient-root relationships for quadratics.
The second lemma expresses $u_1+u_2$ in terms of $t_1+t_2$ and $t_1t_2$ using the identity $u_i = 2t_i^2 - 1$, followed by expansion and regrouping.
The third lemma expresses $u_1u_2$ in terms of $t_1t_2$ and $t_1^2+t_2^2$, and then rewrites everything in terms of $t_1+t_2$ and $t_1t_2$.
After these lemmas, the quadratic with roots $u_1,u_2$ is constructed via $u^2 - (u_1+u_2)u + u_1u_2 = 0$, followed by substitution of $t_1,t_2$ expressions in terms of $a,b,c$.
The final comparison step substitutes $a=4$, $b=2$, $c=-1$ into both the original and transformed equations.
The most delicate step is the correct elimination of $t_1^2+t_2^2$ and $t_1^2 t_2^2$ without introducing extraneous expressions.
Solution
Let $t_1$ and $t_2$ be the roots of the equation $a t^2 + b t + c = 0$. By Viète relations, the sum and product of the roots satisfy
$t_1 + t_2 = -\frac{b}{a}, \qquad t_1 t_2 = \frac{c}{a}.$
This establishes the fundamental symmetric relations connecting the roots to the coefficients, and any further expression in $t_1,t_2$ must be reducible to these two quantities.
Define $u_i = \cos 2x_i$ corresponding to $t_i = \cos x_i$, so that
$u_i = 2t_i^2 - 1.$
Then the sum of the transformed roots satisfies
$u_1 + u_2 = 2t_1^2 - 1 + 2t_2^2 - 1 = 2(t_1^2 + t_2^2) - 2.$
Using the identity $t_1^2 + t_2^2 = (t_1+t_2)^2 - 2t_1 t_2$, substitution yields
$u_1 + u_2 = 2\bigl((t_1+t_2)^2 - 2t_1 t_2\bigr) - 2,$
which simplifies to
$u_1 + u_2 = 2(t_1+t_2)^2 - 4t_1 t_2 - 2.$
Substituting the Viète relations gives
$u_1 + u_2 = 2\frac{b^2}{a^2} - 4\frac{c}{a} - 2.$
Next, the product of the transformed roots satisfies
$u_1 u_2 = (2t_1^2 - 1)(2t_2^2 - 1) = 4t_1^2 t_2^2 - 2(t_1^2 + t_2^2) + 1.$
Using $t_1^2 t_2^2 = (t_1 t_2)^2$ and again substituting $t_1^2 + t_2^2 = (t_1+t_2)^2 - 2t_1 t_2$, one obtains
$u_1 u_2 = 4(t_1 t_2)^2 - 2\bigl((t_1+t_2)^2 - 2t_1 t_2\bigr) + 1.$
Substitution of Viète relations yields
$u_1 u_2 = 4\frac{c^2}{a^2} - 2\frac{b^2}{a^2} + 4\frac{c}{a} + 1.$
The quadratic equation with roots $u_1,u_2$ is therefore
$u^2 - (u_1+u_2)u + u_1u_2 = 0.$
Multiplying through by $a^2$ eliminates denominators and gives
$a^2 u^2 - a^2(u_1+u_2)u + a^2 u_1u_2 = 0.$
Substituting the expressions already obtained yields
$a^2 u^2 - \bigl(2b^2 - 4ac - 2a^2\bigr)u + \bigl(4c^2 - 2b^2 + 4ac + a^2\bigr) = 0.$
Rewriting the coefficient of $u$ produces the equivalent form
$a^2 u^2 + (2a^2 + 4ac - 2b^2)u + (4c^2 - 2b^2 + 4ac + a^2) = 0.$
This is the required quadratic in $\cos 2x$.
For $a=4$, $b=2$, $c=-1$, the original equation becomes
$4\cos^2 x + 2\cos x - 1 = 0.$
The transformed equation has coefficients
$a^2 = 16,$
$2a^2 + 4ac - 2b^2 = 32 + 4\cdot 4 \cdot (-1) - 8 = 8,$
$4c^2 - 2b^2 + 4ac + a^2 = 4 - 8 - 16 + 16 = -4.$
Thus the transformed equation is
$16\cos^2 2x + 8\cos 2x - 4 = 0.$
Dividing by $4$ yields
$4\cos^2 2x + 2\cos 2x - 1 = 0,$
which coincides with the original equation in the same algebraic form.
This completes the construction and comparison.
$\boxed{a^2 u^2 + (2a^2 + 4ac - 2b^2)u + (4c^2 - 2b^2 + 4ac + a^2) = 0}$
Verification of Key Steps
The expression for $u_1+u_2$ depends on correctly converting $t_1^2+t_2^2$ into $(t_1+t_2)^2 - 2t_1t_2$, and any omission of the mixed term $2t_1t_2$ produces an incorrect linear coefficient in $u$. Recomputing directly from $u_i = 2t_i^2 - 1$ confirms that both constant shifts contribute $-2$ in total.
The product $u_1u_2$ is sensitive to the identity $t_1^2 t_2^2 = (t_1 t_2)^2$, and failure to square the product term correctly leads to a missing quartic contribution in $t_i$ that manifests as an incorrect $c^2$ term in the final coefficient.
The final comparison for $a=4$, $b=2$, $c=-1$ hinges on consistent scaling after substitution, since the transformed equation naturally appears multiplied by $a^2$, and incorrect normalization would falsely suggest a different polynomial.
Alternative Approaches
A different approach proceeds by setting $t=\cos x$ and directly expressing the original quadratic as a relation in $t$ and then eliminating $t$ from the system $u=2t^2-1$ and $a t^2 + b t + c = 0$. This leads to a resultant computation in $t$ that produces the same quadratic in $u$ after elimination.
Another approach interprets the problem as determining how the map $t \mapsto 2t^2-1$ acts on the space of quadratic polynomials via symmetric functions, which yields a transformation rule on coefficients. The direct symmetric polynomial method used above is preferable because it avoids introducing auxiliary elimination steps and keeps all expressions explicitly tied to Viète relations.