Consecutive sides of a regular polygon with 9 vertices

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Consecutive sides of a regular polygon with 9 vertices

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Source: Romanian JBMO TST, Day 4, Problem 16

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Valentin Vornicu

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Valentin Vornicu

#1 h Mar 18, 2006, 10:30 PM • 1 Y

Y by Adventure10

Let $AB$ and $BC$ be two consecutive sides of a regular polygon with 9 vertices inscribed in a circle of center $O$ . Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of the radius perpendicular to $BC$ . Find the measure of the angle $\angle OMN$ .

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#2 h Apr 16, 2006, 11:51 PM • 2 Y

Y by Adventure10, Mango247

In the triangle $\triangle OMN$ we have $\measuredangle O = \frac{2 \pi}{9}$ , $OM = R \cos \frac{\pi}{9}$ , $ON = \frac12$ . Using the Sine Law in this triangle we get after some computations (sum-to-product formula twice) that $\tan \measuredangle M = \frac{\sqrt{3}}{3}$ , so $\measuredangle M = \frac{\pi}{6}$ .

But without trig, I don't see any solution. My "synthetic" geometry was/is/will be very bad.

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#3 h Apr 29, 2016, 9:53 AM • 1 Y

Y by Adventure10

Elementary solution:
Let K be intersection of ON and circle with center O of this polygon. <AOK = <AOB + <BOK = 40 + 20 = 60 thus triangle AOK is equilateral. N is midpoint of OK thus <ANO=90, and <AMO=90 (M is midpoint of AB) so AMNO is cyclic quadrilateral. From here <MNO=180 - <MAO= 180 - 70 = 110 and < MON = <MOB + <NOB = <MOB + 1/2* <BOC = 40 so <OMN= 180 - 110 - 40 = 30.

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#4 h Apr 29, 2016, 5:25 PM • 1 Y

Y by Adventure10

Remark: $AC=2MN$ !!

Best regards,
sunken rock

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