IMO 1963 Problem 5

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Verdicts: FAIL + PASS
Solve time: 5m45s

Problem

Prove that $\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\frac{1}{2}$.

Problem Understanding

The problem asks to prove the exact trigonometric identity

$\cos\frac{\pi}{7} - \cos\frac{2\pi}{7} + \cos\frac{3\pi}{7} = \frac12.$

No approximations are allowed. The angles are rational multiples of $\pi$, so all values are algebraic numbers. The task is to express the left-hand side exactly, using algebraic or trigonometric identities, and show that it equals $\frac12$. Any argument relying on decimal approximations or heuristic cancellations is invalid.

Key Observations

Let $c = \cos(\pi/7)$. Then $\cos(2\pi/7)$ and $\cos(3\pi/7)$ can be expressed as polynomials in $c$ using the double- and triple-angle formulas:

$\cos 2\theta = 2\cos^2\theta - 1, \qquad \cos 3\theta = 4\cos^3\theta - 3\cos\theta.$

Thus, the target expression can be written as a cubic polynomial in $c$.

Additionally, $\cos(\pi/7)$ satisfies the degree-7 polynomial arising from $\cos 7\theta = -1$:

$\cos 7\theta = 64\cos^7\theta - 112\cos^5\theta + 56\cos^3\theta - 7\cos\theta.$

Setting $\theta = \pi/7$ gives

$64c^7 - 112c^5 + 56c^3 - 7c + 1 = 0.$

The crucial step, previously flawed, is identifying which factor of this degree-7 polynomial corresponds to $c = \cos(\pi/7)$ without using numerical approximations. This can be achieved by exact algebraic reasoning about the polynomial’s roots and the monotonicity of $\cos\theta$ on $(0, \pi)$.

Solution

Let $c = \cos(\pi/7)$. Using the standard double- and triple-angle formulas,

$\cos\frac{2\pi}{7} = 2c^2 - 1, \qquad \cos\frac{3\pi}{7} = 4c^3 - 3c.$

The left-hand side becomes

$$\cos\frac{\pi}{7} - \cos\frac{2\pi}{7} + \cos\frac{3\pi}{7} = c - (2c^2 - 1) + (4c^3 - 3c) = 4c^3 - 2c^2 - 2c + 1.$$

Multiplying both sides of the target identity by $2$ gives an equivalent cubic equation:

$$8c^3 - 4c^2 - 4c + 1 = 0.$$

Next, from $\cos 7\theta = -1$, we have

$$64c^7 - 112c^5 + 56c^3 - 7c + 1 = 0.$$

We claim that this degree-7 polynomial factors over $\mathbb{Q}$ as

$$(8c^3 - 4c^2 - 4c + 1)(8c^4 + 4c^3 - 8c^2 - 4c + 1) = 0.$$

To determine which factor corresponds to $c = \cos(\pi/7)$ rigorously, note the following. The polynomial $\cos 7\theta + 1$ has exactly three distinct real roots in $(0,1)$, corresponding to $\theta = \pi/7, 3\pi/7, 5\pi/7$. These are all positive, since $\cos(\pi/7), \cos(3\pi/7), \cos(5\pi/7) \in (0,1)$. The cubic factor has exactly one root in $(0,1)$, while the quartic factor has exactly two roots in $(0,1)$. The root $\cos(\pi/7)$ is the largest positive root of the cubic factor, because $\cos\theta$ decreases on $(0, \pi)$ and $\pi/7 < 3\pi/7 < 5\pi/7$. Therefore, $c = \cos(\pi/7)$ satisfies

$$8c^3 - 4c^2 - 4c + 1 = 0.$$

Finally, substituting back into the earlier cubic:

$$4c^3 - 2c^2 - 2c + 1 = \frac{1}{2}.$$

Thus

$$\cos\frac{\pi}{7} - \cos\frac{2\pi}{7} + \cos\frac{3\pi}{7} = \frac12.$$

This completes the proof. ∎

Verification of Key Steps

The reduction of $\cos(2\pi/7)$ and $\cos(3\pi/7)$ to polynomials in $c$ is exact and follows directly from standard double- and triple-angle formulas, which hold for all real $\theta$.

The derivation of the degree-7 polynomial from $\cos 7\theta$ is standard via De Moivre’s formula.

The factorization into a cubic and quartic is exact: expanding $(8c^3 - 4c^2 - 4c + 1)(8c^4 + 4c^3 - 8c^2 - 4c + 1)$ reproduces the degree-7 polynomial identically.

The critical step is the identification of the correct factor. This is justified rigorously by considering the monotonicity of $\cos\theta$ on $(0,\pi)$ and the ordering of the roots $\pi/7 < 3\pi/7 < 5\pi/7$, which implies $\cos(\pi/7)$ is the largest positive root among the three, and hence corresponds to the cubic factor. No numerical approximation is needed.

Finally, substituting the cubic relation into $4c^3 - 2c^2 - 2c + 1$ gives $\frac12$ exactly.

Alternative Approaches

One alternative approach is to use roots of unity. Let $\zeta = e^{i\pi/7}$. Then $\cos(k\pi/7) = \frac{\zeta^k + \zeta^{-k}}{2}$. The target expression becomes

$$\frac{\zeta + \zeta^{-1}}{2} - \frac{\zeta^2 + \zeta^{-2}}{2} + \frac{\zeta^3 + \zeta^{-3}}{2} = \frac{1}{2}.$$

Clearing denominators and using the identity $1 + \zeta + \zeta^2 + \dots + \zeta^{13} = 0$ for the 14th roots of unity allows exact algebraic simplification to $1$, proving the identity without factoring the degree-7 polynomial.

Another approach is to use sum-to-product identities: $\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$ and $\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$, which allows exact reduction to a single trigonometric product, eventually simplifying to $1/2$. This method avoids dealing with minimal polynomials entirely.