IMO 1959 · Problem 3

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 \(a=4, b=2, c=-1\).

Official text of the problem. Official problem paper (PDF)

Answer

The usual double-angle construction gives \[ \boxed{a^2\cos^2 2x+2(a^2+2ac-b^2)\cos 2x+(a+2c)^2-2b^2=0.} \]

There is a qualification to the statement: this equation generally has additional real solutions \(x\). An equation solely in \(\cos 2x\) cannot always have exactly the original \(x\)-solutions.

For \(a=4,\ b=2,\ c=-1\), the constructed equation is \[ \boxed{4\cos^2 2x+2\cos 2x-1=0.} \] The respective solution sets are \[ \boxed{ \begin{aligned} \text{Original:}\quad &x=\frac{2k\pi}{5}, &&k\in\mathbb Z,\quad 5\nmid k,\\ \text{Constructed:}\quad &x=\frac{k\pi}{5}, &&k\in\mathbb Z,\quad 5\nmid k. \end{aligned}} \]

Solution

Put \(P(t)=at^2+bt+c\). Using \(2\cos^2x=1+\cos 2x\), the original equation gives \[ a\cos 2x+a+2c=-2b\cos x. \] Squaring yields \[ (a\cos 2x+a+2c)^2=2b^2(1+\cos 2x). \] Expansion gives the displayed equation in the Answer. Its leading coefficient is \(a^2\), so it is genuinely quadratic whenever the original equation is quadratic, namely when \(a\ne0\).

To determine precisely whether squaring introduces solutions, define \[ Q(y)=a^2y^2+2(a^2+2ac-b^2)y+(a+2c)^2-2b^2. \] The identity \[ \begin{aligned} Q(\cos 2x) &=4\bigl((a\cos^2x+c)^2-b^2\cos^2x\bigr)\\ &=4P(\cos x)P(-\cos x) \end{aligned} \] shows that \[ Q(\cos 2x)=0 \quad\Longleftrightarrow\quad P(\cos x)=0\ \text{or}\ P(\cos(x+\pi))=0. \] Thus the constructed equation has exactly the original solutions together with their translates by \(\pi\).

This also identifies the wording issue. Every equation solely in \(\cos 2x\) is unchanged by replacing \(x\) with \(x+\pi\). The original equation need not have this property. Indeed, if \(P(t)=0\), then \[ P(-t)=P(t)-2bt=-2bt. \] Consequently, when \(b\ne0\), any original solution with \(\cos x\ne0\) has a translate by \(\pi\) which is not an original solution. In that situation, no equation solely in \(\cos 2x\) can have exactly the original solution set.

For completeness, the construction is equivalent to the original equation precisely when \(b=0\), or when \(P\) has no nonzero root in \([-1,1]\). The factorization proves this even for repeated roots, endpoint roots, and empty solution sets.

Although “quadratic” presumes \(a\ne0\), the identity also handles the degenerate coefficients. If \(a=0\) and \(b\ne0\), the constructed equation becomes \[ 4c^2-2b^2(1+\cos 2x)=0, \] equivalently \(\cos^2x=c^2/b^2\), whereas the original requires \(\cos x=-c/b\). If \(a=b=0\), both equations hold for every real \(x\) when \(c=0\), and neither has a solution when \(c\ne0\).

Now take \(a=4,\ b=2,\ c=-1\). Substitution gives \[ Q(y)=16y^2+8y-4=4(4y^2+2y-1), \] so the constructed equation is \[ 4\cos^2 2x+2\cos 2x-1=0. \] The two equations therefore have identical coefficients in their respective cosine variables. Their algebraic roots are \[ \frac{-1+\sqrt5}{4},\qquad \frac{-1-\sqrt5}{4}. \]

To find and compare all their real \(x\)-solutions, the sine addition formulas give \[ \begin{aligned} (4\cos^2t+2\cos t-1)\sin\frac t2 &=(1+2\cos t+2\cos 2t)\sin\frac t2\\ &=\sin\frac{5t}{2}. \end{aligned} \] If \(t\in2\pi\mathbb Z\), the quadratic expression equals \(5\), so it is nonzero. Otherwise, division by \(\sin(t/2)\) is valid, and \[ 4\cos^2t+2\cos t-1=0 \quad\Longleftrightarrow\quad t=\frac{2k\pi}{5},\qquad 5\nmid k. \] Taking \(t=x\) and then \(t=2x\) gives exactly the two solution sets stated above.

In particular, \(2\pi/5\) is an original solution but \(7\pi/5\) is not. Since these arguments differ by \(\pi\), even the specified example rules out a literal interpretation requiring identical real \(x\)-solutions.

The solution was checked against John Scholes' IMO solution archive (Kalva).