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Iranian Mathematical Olympiad

Geometry Difficulty 6.2 National Olympiad Prove it Iran

By angle, we mean its vertex together with its two rays. Find the maximum value of $n$ such that we can put $n$ $60^\circ$ angles on the plane, in a way that each pair of which has 4 intersection points.

Solution

Assume that two $60^\circ$ angles with vertices $A$ and $B$ have four intersection points $M$ , $N$ , $P$ and $Q$ as shown in the figure.

We have
$\widehat{AKB} = \widehat{KAP} + \widehat{APB} + \widehat{PBK} = 30^\circ + \widehat{APB} + 30^\circ > 60^\circ$
And
$\widehat{AKB} = 180^\circ - \widehat{KAB} - \widehat{KBA} < 180^\circ - 30^\circ - 30^\circ = 120^\circ$

So the value of the angle between two bisectors is between $60^\circ$ and $120^\circ$ , meaning that the acute angle between two lines is greater than $60^\circ$ . If we have three $60^\circ$ angles with the above properties, then we should have three lines such that the value of the acute angle between each pair is greater than $60^\circ$ , which is impossible, due to a triangle with three angles greater than $60^\circ$ , or in the case of concurrency six angles greater than $60^\circ$ on a point. The example for only two $60^\circ$ angles is shown in the figure! ■

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