Baltic Way 1998 · Problem 9

Solution: Let $f(t)=\sqrt{1+t^{2}}$. Since $f''(t)=\left(1+t^{2}\right)^{-3 / 2}>0$, the function $f(t)$ is strictly convex on $(0, \infty)$. Consequently,

$$\begin{aligned} \frac{1}{\cos \gamma} & =\sqrt{1+\tan ^{2} \gamma}=f(\tan \gamma)=f\left(\frac{\tan \alpha+\tan \beta}{2}\right)< \\ & <\frac{f(\tan \alpha)+f(\tan \beta)}{2}=\frac{1}{2}\left(\frac{1}{\cos \alpha}+\frac{1}{\cos \beta}\right)=\frac{1}{\cos \delta}, \end{aligned}$$

and hence $\gamma<\delta$.

Alternative solution. Draw a unit segment $O P$ in the plane and take points $A$ and $B$ on the same side of line $O P$ so that $\angle P O A=\angle P O B=90^{\circ}$, $\angle O P A=\alpha$ and $\angle O P B=\beta$ (see Figure 1). Then we have $|O A|=\tan \alpha$, $|O B|=\tan \beta,|P A|=\frac{1}{\cos \alpha}$ and $|P B|=\frac{1}{\cos \beta}$. Figure 1 Let $C$ be the midpoint of the segment $A B$. By hypothesis, we have $|O C|=\frac{\tan \alpha+\tan \beta}{2}=\tan \gamma$, hence $\angle O P C=\gamma$ and $|P C|=\frac{1}{\cos \gamma}$. Let $Q$ be the point symmetric to $P$ with respect to $C$. The quadrilateral $P A Q B$ is a parallelogram, and therefore $|A Q|=|P B|=\frac{1}{\cos \beta}$. Eventually,

$$\frac{2}{\cos \delta}=\frac{1}{\cos \alpha}+\frac{1}{\cos \beta}=|P A|+|A Q|>|P Q|=2 \cdot|P C|=\frac{2}{\cos \gamma},$$

and hence $\delta>\gamma$.

Another solution. Set $x=\frac{\alpha+\beta}{2}$ and $y=\frac{\alpha-\beta}{2}$, then $\alpha=x+y, \beta=x-y$ and

$$\begin{aligned} \cos \alpha \cos \beta & =\frac{1}{2}(\cos 2 x+\cos 2 y)= \\ & =\frac{1}{2}\left(1-2 \sin ^{2} x\right)+\frac{1}{2}\left(2 \cos ^{2} y-1\right)=\cos ^{2} y-\sin ^{2} x . \end{aligned}$$

By the conditions of the problem,

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

and

$$\frac{1}{\cos \delta}=\frac{1}{2}\left(\frac{1}{\cos \alpha}+\frac{1}{\cos \beta}\right)=\frac{1}{2} \cdot \frac{\cos \alpha+\cos \beta}{\cos \alpha \cos \beta}=\frac{\cos x \cos y}{\cos \alpha \cos \beta} .$$

Using (3) we hence obtain

$$\begin{aligned} \tan ^{2} \delta-\tan ^{2} \gamma & =\frac{1}{\cos ^{2} \delta}-1-\tan ^{2} \gamma=\frac{\cos ^{2} x \cos ^{2} y-\sin ^{2} x \cos ^{2} x}{\cos ^{2} \alpha \cos ^{2} \beta}-1= \\ & =\frac{\cos ^{2} x\left(\cos ^{2} y-\sin ^{2} x\right)}{\left(\cos ^{2} y-\sin ^{2} x\right)^{2}}-1=\frac{\cos ^{2} x}{\cos ^{2} y-\sin ^{2} x}-1= \\ & =\frac{\cos ^{2} x-\cos ^{2} y+\sin ^{2} x}{\cos ^{2} y-\sin ^{2} x}=\frac{\sin ^{2} y}{\cos \alpha \cos \beta}>0, \end{aligned}$$

showing that $\delta>\gamma$.