IMO 1991 · Problem 5

Let \(ABC\) be a triangle and \(P\) an interior point of \(ABC\). Show that at least one of the angles \(\angle PAB,\angle PBC,\angle PCA\) is less than or equal to \(30^\circ\).

Official text of the problem. Official problem paper (PDF)

Solution

Put \[ \alpha=\angle PAB,\qquad \beta=\angle PBC,\qquad \gamma=\angle PCA, \] and \[ x=\angle PAC,\qquad y=\angle PBA,\qquad z=\angle PCB. \] Since \(P\) is interior, all six angles are positive, and \[ \alpha+\beta+\gamma+x+y+z=180^\circ. \]

Applying the sine rule in triangles \(APB\), \(BPC\), and \(CPA\), respectively, gives \[ \frac{PB}{PA}=\frac{\sin\alpha}{\sin y},\qquad \frac{PC}{PB}=\frac{\sin\beta}{\sin z},\qquad \frac{PA}{PC}=\frac{\sin\gamma}{\sin x}. \] Multiplication yields \[ \sin\alpha\sin\beta\sin\gamma=\sin x\sin y\sin z. \tag{1} \]

Suppose, for a contradiction, that \(\alpha,\beta,\gamma>30^\circ\). Their sum is less than \(180^\circ\), so each is also less than \(120^\circ\). Thus each of their sines exceeds \(1/2\), and consequently \[ \sin\alpha\sin\beta\sin\gamma>\frac18. \tag{2} \]

On the other hand, \[ x+y+z=180^\circ-(\alpha+\beta+\gamma)<90^\circ. \] Therefore \(0<x+y<90^\circ-z<90^\circ\), so \[ \cos(x+y)>\cos(90^\circ-z)=\sin z. \] Using the product-to-sum identity and \(\cos(x-y)\le1\), we obtain \[ \begin{aligned} \sin x\sin y\sin z &=\frac{\sin z}{2}\bigl(\cos(x-y)-\cos(x+y)\bigr)\\ &<\frac{\sin z}{2}(1-\sin z)\\ &\le\frac18. \end{aligned} \] The last inequality follows from \[ t(1-t)=\frac14-\left(t-\frac12\right)^2\le\frac14 \] with \(t=\sin z\).

This contradicts (1) and (2). Hence at least one of \(\angle PAB,\angle PBC,\angle PCA\) is at most \(30^\circ\). \(\square\)

The solution was checked against John Scholes' IMO solution archive (Kalva).