perimeter of 5-gon wanted, by vertices of a regular 12-gon

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perimeter of 5-gon wanted, by vertices of a regular 12-gon

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Source: 2021 Yasinsky Geometry Olympiad X-XI p1 , Ukraine

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#1 h Mar 31, 2021, 5:17 AM • 2 Y

Y by Mango247, Mango247

A regular dodecagon $A_1A_2...A_{12}$ is inscribed in a circle with a diameter of $20$ cm . Calculate the perimeter of the pentagon $A_1A_3A_6A_8A_{11}$ .

(Alexey Panasenko)

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EmilXM

#2 h Mar 31, 2021, 3:12 PM • 1 Y

Y by Mango247

parmenides51 wrote:

A regular dodecagon $A_1A_2...A_{12}$ is inscribed in a circle with a diameter of $20$ cm . Calculate the perimeter of the pentagon $A_1A_3A_6A_8A_{11}$ .

(Alexey Panasenko)

Note that, Since $A_1A_2...A_{12}$ is a regular dodecagon, we have $|A_1A_3| = |A_6A_8| = |A_{11}A_1|$ , $|A_3A_6| = |A_8A_{11}|$ and $A_iA_{i+1}A_{i+2} = \frac{180^o \cdot (12-2)}{12} = 150^o$ . Now let $O$ be circumcenter of $A_1A_2...A_{12}$ , Then:
$\angle A_1OA_3 = 360^o - 2\cdot \angle A_1A_2A_3 = 60^o$ Also since $A_1OA_3$ is isosceles we get $A_1OA_3$ is also equilateral triangle. So $A_1A_3 = OA_1 = OA_3 = \frac{20cm}{2} = 10cm$ .
It sufficies to calculate $A_3A_6$ , which is:
$\angle A_3OA_6 = 3\cdot \angle A_1OA_3 = 3 \cdot 30^o = 90^o$ So $A_3OA_6$ is isosceles right triangle, which means $A_3A_6 = OA_3 \cdot \sqrt{2} = 10\sqrt{2} cm$ .
So answer is $3 \cdot 10 cm + 2 \cdot 10 \sqrt{2} cm = \boxed{30+20\sqrt{2} cm}$

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