1959 IMO Problems/Problem 3

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 $a=4, b=2, c=-1$ .

Solution

Let the original equation be satisfied only for $\cos{x}=m, \cos{x}=n$ . Then we wish to construct a quadratic with roots $2m^2 -1, 2n^2 -1$ .

Clearly, the sum of the roots of this quadratic must be

$2 (m^2 + n^2) - 2 = 2 \left(\frac{b^2-2ac}{a^2}\right) - 2 = \frac{2b^2 - 4ac - 2a^2}{a^2},$

and the product of its roots must be

$4m^2 n^2 - 2(m^2 + n^2) + 1 = \frac{4c^2}{a^2}+\frac{4ac - 2b^2}{a^2} + \frac{a^2}{a^2} = \frac{(a+2)^2 - 2b^2}{a^2}$

Thus the following quadratic fulfils the conditions:

$a^2 \cos ^2 {2x} + (2a^2 + 4ac - 2b^2)\cos{2x} + (a+2c)^2 - 2b^2 = 0$

Now, when we let $a=4, b=2, c= -1$ , our equations are

$4 \cos^2 {x} + 2 \cos {x} - 1 = 0$

and

$16 \cos^2 {2x} + 8 \cos {2x} - 4 = 0,$

i.e., they are multiples of each other. The reason behind this is that the roots of the first equation are $\cos {x} = \frac{-1 \pm \sqrt{5}}{4}$ , which implies that $x$ is one of two certain multiples of $\frac{\pi}{5}$ , and when $5 \nmid n$ , $\left|\cos\left(\frac{n\pi}{5}\right)\right|$ can only assume two distinct values. Q.E.D.

Solution 2.

Note that $cos^2 {2x} = /frac(1/2) (1 + cos{2x})$

After two lines of rearrangement, the solution,

$a^2 \cos ^2 {2x} + (2a^2 + 4ac - 2b^2)\cos{2x} + (a+2c)^2 - 2b^2 = 0$

is obtained.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1959 IMO (Problems) • Resources
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1959 IMO Problems/Problem 3

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