1966 IMO Problems/Problem 2

Problem

Let $a$ , $b$ , and $c$ be the lengths of the sides of a triangle, and $\alpha,\beta,\gamma$ respectively, the angles opposite these sides. Prove that if

$a+b=\tan{\frac{\gamma}{2}}(a\tan{\alpha}+b\tan{\beta}),$

the triangle is isosceles.

Solution

We'll prove that the triangle is isosceles with $a=b$ . We'll prove that $a=b$ . Assume by way of contradiction WLOG that $a>b$ . First notice that as $\gamma = \pi -\alpha-\beta$ then and the identity $\tan\left(\frac \pi 2 - x \right)=\cot x$ our equation becomes: $a+b=\cot \frac{\alpha +\beta}{2}\left(a\tan \alpha + b\tan \beta \right)$ $\iff a\tan\frac{\alpha +\beta}{2}+b\tan \frac{\alpha +\beta}{2}=a\tan \alpha + b\tan \beta$ $\iff a\left(\tan \alpha -\tan \frac{\alpha +\beta}{2}\right)+b\left(\tan \beta -\tan \frac{\alpha +\beta}{2} \right)=0$ Using the identity $\tan (A-B)=\frac {\tan A-\tan B}{1+\tan A\tan B}$ $\iff \tan A-\tan B=\tan(A-B)(1+\tan A\tan B)$ and inserting this into the above equation we get: $\iff a\tan \frac{\alpha -\beta}{2}\left(1+\tan \alpha \tan \frac{\alpha +\beta}{2}\right)+b\tan \frac{\beta -\alpha}{2}\left(1+\tan \beta \tan \frac{\alpha +\beta}{2} \right)=0$ $\underbrace{\iff}_{\tan -A=-\tan A}a\tan \frac{\alpha -\beta}{2}\left(1+\tan \alpha \tan \frac{\alpha +\beta}{2}\right)-b\tan \frac{\alpha -\beta}{2}\left(1+\tan \beta \tan \frac{\alpha +\beta}{2} \right)=0$ $\iff \tan \frac{\alpha -\beta}{2}\left(a-b+\tan \frac{\alpha +\beta}{2}(a\tan\alpha -b\tan \beta) \right)=0$ Now, since $a>b$ and the definitions of $a,b,\alpha,\beta$ being part of the definition of a triangle, $\alpha >\beta$ . Now, $\pi >\alpha -\beta >0$ (as $\alpha+\beta +\gamma = \pi$ and the angles are positive), $\tan \frac{\alpha -\beta}{2}\neq 0$ , and furthermore, $\tan \frac{\alpha+\beta}{2}>0$ . By all the above, $\left(a-b+\tan \frac{\alpha +\beta}{2}(a\tan\alpha -b\tan \beta) \right)>0$ Which contradicts our assumption, thus $a\leq b$ . By the symmetry of the condition, using the same arguments, $a\geq b$ . Hence $a=b$ .

Solution 2

First, we'll prove that both $\alpha$ and $\beta$ are acute. At least one of them has to be acute because these are angles of a triangle. We can assume that $\alpha$ is acute. We want to show that $\beta$ is acute as well. For a proof by contradiction, assume $\beta \ge \frac{\pi}{2}$ .

From the hypothesis, it follows that $(a + b) \tan \frac{\alpha + \beta}{2} = a \tan \alpha + b \tan \beta$ .

From $\alpha < \frac{\pi}{2} \le \beta$ it follows that $a < b$ . So,

$b \tan \beta = (a + b) \tan \frac{\alpha + \beta}{2} - a \tan \alpha > 2a \tan \frac{\alpha + \beta}{2} - a \tan \alpha \ge a (2 \tan \left( \frac{\alpha}{2} + \frac{\pi}{4} \right) - \tan \alpha) =$

$2a \left( \frac{\tan \frac{\alpha}{2} + 1}{1 - \tan \frac{\alpha}{2}} - \frac{\tan \frac{\alpha}{2}}{1 - \tan^2 \frac{\alpha}{2}} \right) = 2a \cdot \frac{\tan^2 \frac{\alpha}{2} + \tan \frac{\alpha}{2} + 1} {1 - \tan^2 \frac{\alpha}{2}} > 0$

because the numerator is $> 0$ (because $Y^2 + Y + 1 > 0$ for any real $Y$ ), and the denominator is also $> 0$ (because $\alpha < \frac{\pi}{2}$ , so $\tan \frac{\alpha}{2} < \tan \frac{\pi}{4} = 1$ ).

It follows that $\tan \beta > 0$ , so it can not be that $\beta \ge \frac{\pi}{2}$ .

Now, we will prove that $(a + b) \tan \frac{\alpha + \beta}{2} = a \tan \alpha + b \tan \beta$ implies $\alpha = \beta$ .

Replace $a = \sin \alpha \cdot 2R$ and $b = \sin \beta \cdot 2R$ (in fact, we don't care that $R$ is the radius of the circumscribed circle), and simplify by $2R$ . We get

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

This becomes

$\left( \sin \frac{\alpha + \beta}{2} \tan \frac{\alpha + \beta}{2} \right) \cdot \cos \frac{\alpha - \beta}{2} = \frac{1}{2}(\sin \alpha \tan \alpha + \sin \beta \tan \beta)$

We will show that the function $f(x) = \tan x \sin x$ is convex on the interval $\left( 0, \frac{\pi}{2} \right)$ . Indeed, the first derivative is $f'(x) = \frac{\sin x}{\cos^2 x} + \sin x$ , and the second derivative is $f''(x) = \frac{\cos^4 x - \cos ^2 x + 2}{\cos^3 x}$ .

We have $f''(x) > 0$ on $\left( 0, \frac{\pi}{2} \right)$ since the numerator is $> 0$ (because $Y^2 - Y + 1 > 0$ for any real $Y$ ), and the denominator is $> 0$ on the interval $\left( 0, \frac{\pi}{2} \right)$ . It follows that $f(x) = \tan x \sin x$ is convex on the interval $\left( 0, \frac{\pi}{2} \right)$ .

Using the convexity we have $f \left( \frac{x + y}{2} \right) \le \frac{1}{2} (f(x) + f(y))$ . In our case, we have

$\frac{1}{2}(\sin \alpha \tan \alpha + \sin \beta \tan \beta) = \left( \sin \frac{\alpha + \beta}{2} \tan \frac{\alpha + \beta}{2} \right) \cdot \cos \frac{\alpha - \beta}{2} \le \frac{1}{2}(\sin \alpha \tan \alpha + \sin \beta \tan \beta) \cdot \cos \frac{\alpha - \beta}{2}$ .

We can simplify by $\sin \alpha \tan \alpha + \sin \beta \tan \beta$ because it is positive (because both $\alpha, \beta$ are acute!), and we get

$1 \le \cos \frac{\alpha - \beta}{2}$ . This is possible only when $\cos \frac{\alpha - \beta}{2} = 1$ , i.e. $\alpha = \beta$ .

[Solution by pf02, September 2024]

1966 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
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1966 IMO Problems/Problem 2

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Solution 2

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