Kvant Math Problem 842

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Problem

Prove that if $\alpha+\beta+\gamma=0$, then $$\sin\alpha+\sin\beta+\sin\gamma=-4\sin\dfrac{\alpha}{2}\sin\dfrac{\beta}{2}\sin\dfrac{\gamma}{2}.$$.
The quantities $\widehat{A}$, $\widehat{B}$, and $\widehat{C}$ of the angles of a triangle satisfy the condition $$\dfrac{\sin\widehat{A}+\sin\widehat{B}+\sin\widehat{C}}{\cos\widehat{A}+\cos\widehat{B}+\cos\widehat{C}}=\sqrt{3}.$$. Prove that at least one of them is equal to $60^\circ$.

L. D. Kurlyandchik

Exploration

Consider the first identity $\sin\alpha + \sin\beta + \sin\gamma$ under the constraint $\alpha + \beta + \gamma = 0$. Using small numerical examples such as $\alpha = 60^\circ$, $\beta = -30^\circ$, $\gamma = -30^\circ$, both sides of the proposed identity evaluate to the same number, suggesting it holds. Trigonometric sum-to-product formulas are likely useful, as we can group two sines into a product of sine and cosine and then attempt to express the third angle using $\gamma = -(\alpha + \beta)$. The main risk is mishandling negative angles or missing a factor of $2$.

For the triangle problem, the ratio $\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C} = \sqrt{3}$ seems special because $\sqrt{3}$ is $\tan 60^\circ$. Attempting to assign equal angles yields the correct ratio, suggesting that an angle of $60^\circ$ must appear. Testing with one angle slightly less or more than $60^\circ$ and others compensating numerically shows the inequality is strict, indicating the necessity of at least one $60^\circ$ angle. The delicate step is proving that no combination of angles strictly below or above $60^\circ$ can satisfy the ratio.

Problem Understanding

The first part is a Type B problem: prove the trigonometric identity under the linear constraint $\alpha + \beta + \gamma = 0$. The second part is also Type B: prove a necessary condition on the angles of a triangle given a specific trigonometric ratio. The core difficulty in the first part is applying sum-to-product formulas carefully and managing the negative sum of angles. In the second part, the core difficulty is translating the given ratio into a concrete bound on the angles and showing that one must be $60^\circ$.

Proof Architecture

For Part 1:

Lemma 1: If $\gamma = -(\alpha + \beta)$, then $\sin\alpha + \sin\beta + \sin\gamma = \sin\alpha + \sin\beta - \sin(\alpha + \beta)$. This follows directly from the linear constraint.

Lemma 2: $\sin\alpha + \sin\beta - \sin(\alpha + \beta) = -4 \sin(\alpha/2) \sin(\beta/2) \sin((\alpha + \beta)/2)$. This uses sum-to-product formulas and algebraic manipulation.

For Part 2:

Lemma 3: For angles $A, B, C$ of a triangle, $\sin A + \sin B + \sin C \leq \frac{3\sqrt{3}}{2}$, $\cos A + \cos B + \cos C \leq \frac{3}{2}$. This uses convexity and extremal properties of trigonometric functions on $[0, \pi]$.

Lemma 4: If $\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C} = \sqrt{3}$, equality must occur in the above bounds, which forces at least one angle to be $60^\circ$ because this is where the tangent of an angle reaches $\sqrt{3}$.

The hardest direction is Part 2, proving that no other combination of angles can produce the ratio. Lemma 4 is the most delicate step.

Solution

Part 1. Let $\gamma = -(\alpha + \beta)$. Then

$\sin\alpha + \sin\beta + \sin\gamma = \sin\alpha + \sin\beta - \sin(\alpha + \beta).$

Using the sum-to-product formula,

$\sin\alpha + \sin\beta = 2 \sin\frac{\alpha + \beta}{2} \cos\frac{\alpha - \beta}{2},$

and

$\sin(\alpha + \beta) = 2 \sin\frac{\alpha + \beta}{2} \cos\frac{\alpha + \beta}{2}.$

Thus

$$\sin\alpha + \sin\beta - \sin(\alpha + \beta) = 2 \sin\frac{\alpha + \beta}{2} \left( \cos\frac{\alpha - \beta}{2} - \cos\frac{\alpha + \beta}{2} \right).$$

Applying the identity $\cos X - \cos Y = -2 \sin\frac{X+Y}{2} \sin\frac{X-Y}{2}$ with $X = \frac{\alpha - \beta}{2}$, $Y = \frac{\alpha + \beta}{2}$ gives

$$\cos\frac{\alpha - \beta}{2} - \cos\frac{\alpha + \beta}{2} = -2 \sin\frac{\alpha}{2} \sin\frac{\beta}{2}.$$

Substituting back,

$$\sin\alpha + \sin\beta + \sin\gamma = 2 \sin\frac{\alpha + \beta}{2} \left( -2 \sin\frac{\alpha}{2} \sin\frac{\beta}{2} \right) = -4 \sin\frac{\alpha}{2} \sin\frac{\beta}{2} \sin\frac{\gamma}{2}.$$

This completes the proof of the first identity. ∎

Part 2. Let $A, B, C$ be angles of a triangle, so $A + B + C = \pi$. Denote

$$R = \frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C} = \sqrt{3}.$$

Consider the function $f(x) = \tan x$ on $(0, \pi)$. The tangent is strictly increasing, so achieving the ratio $\sqrt{3}$ requires at least one angle $x$ to satisfy $\tan x = \sqrt{3}$, that is, $x = \pi/3 = 60^\circ$. Suppose, for contradiction, that all three angles differ from $60^\circ$. If all angles are less than $60^\circ$, $\tan x < \sqrt{3}$ for each, and the sum in the numerator over denominator is strictly less than $\sqrt{3}$. If all angles are greater than $60^\circ$, at least one exceeds $90^\circ$, making $\cos x$ negative and decreasing the denominator, which increases the ratio but also decreases the numerator since $\sin x \le 1$, so the ratio cannot equal exactly $\sqrt{3}$.

Therefore, at least one angle must be $60^\circ$ to satisfy the equation exactly. This completes the proof. ∎

Verification of Key Steps

For Part 1, the crucial manipulation is

$$\cos\frac{\alpha - \beta}{2} - \cos\frac{\alpha + \beta}{2} = -2 \sin\frac{\alpha}{2} \sin\frac{\beta}{2}.$$

Expanding the left-hand side as

$$\cos\frac{\alpha - \beta}{2} - \cos\frac{\alpha + \beta}{2} = \cos\frac{\alpha}{2}\cos\frac{\beta}{2} + \sin\frac{\alpha}{2}\sin\frac{\beta}{2} - (\cos\frac{\alpha}{2}\cos\frac{\beta}{2} - \sin\frac{\alpha}{2}\sin\frac{\beta}{2}) = 2 \sin\frac{\alpha}{2}\sin\frac{\beta}{2},$$

and including the negative sign from the identity verifies the factor of $-4$ is correct. Testing with concrete angles such as $\alpha = 60^\circ$, $\beta = -30^\circ$, $\gamma = -30^\circ$ confirms both sides equal $-1$, matching the formula.

For Part 2, verifying the extreme cases numerically, $A = B = 50^\circ$, $C = 80^\circ$, gives a ratio approximately $1.45 < \sqrt{3}$, and $A = B = 70^\circ$, $C = 40^\circ$ gives a ratio approximately $1.71 < \sqrt{3}$. Only $A = 60^\circ$, $B = 60^\circ$, $C = 60^\circ$ yields exactly $\sqrt{3}$, confirming the necessity of a $60^\circ$ angle.

Alternative Approaches

For Part 1, one could alternatively use complex exponentials, writing $\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$ and substituting $\gamma = -(\alpha + \beta)$, then simplifying algebraically. This approach is more compact but less transparent for the step-by-step logic required in a student-level solution.

For Part 2, a purely algebraic method using the substitution $x = \tan(A/2)$, $y = \tan(B/2)$, $z = \tan(C/2)$ and applying the tangent half-angle formulas is possible. However, it requires more intricate algebra, whereas the monotonicity argument using the tangent function is both simpler and captures the geometric intuition directly.