Shiny Sternenhimmel

Geometry Level 4

As shown above, the yellow polygon is surrounded by six unit regular nonagons. If the area of the yellow polygon can be expressed as

A sin ⁡ π 9 + B sin ⁡ π 3 A\sin\dfrac{\pi}{9} + B\sin\dfrac{\pi}{3}

where A A and B B are integers, input the product A B AB as your answer.


The answer is 36.

Michael Huang by Michael Huang , 29, USA

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2 solutions

David Vreken
Jan 27, 2021

The yellow polygon can be divided into 12 12 unit equilateral triangles and 6 6 unit rhombi (with angles of π 9 \cfrac{\pi}{9} and 8 π 9 \cfrac{8\pi}{9} ):

Since each equilateral triangle has an area of 1 2 sin ⁡ π 3 \cfrac{1}{2}\sin \cfrac{\pi}{3} and each rhombi has an area of 2 ⋅ 1 2 ⋅ 1 ⋅ 1 ⋅ sin ⁡ π 9 = sin ⁡ π 9 2 \cdot \cfrac{1}{2} \cdot 1 \cdot 1 \cdot \sin \cfrac{\pi}{9} = \sin \cfrac{\pi}{9} , the total area is 12 ⋅ 1 2 sin ⁡ π 9 + 6 sin ⁡ π 3 12 \cdot \cfrac{1}{2}\sin \cfrac{\pi}{9} + 6 \sin \cfrac{\pi}{3} or:

6 sin ⁡ π 9 + 6 sin ⁡ π 3 6 \sin \cfrac{\pi}{9} + 6 \sin \cfrac{\pi}{3}

Therefore, A = B = 6 A = B = 6 and A B = 36 AB = \boxed{36} .

4 Helpful 3 Interesting 7 Brilliant 0 Confused

This is indeed a brilliant solution! Can't stop looking at it! Congratulations, sir!

P.S.: Rhombi sounds funny.

Veselin Dimov - 4 months, 2 weeks ago

But is there only one solution of ( A , B ) (A,B) ?

Pi Han Goh - 4 months, 2 weeks ago

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Yep :-) 

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import math
G = 6.0 * (math.sin(math.pi / 9) + math.sin(math.pi / 3))
for A in range(1, int(1.0 * G / math.sin(math.pi / 9)) + 1):
    for B in range(1, int(1.0 * G / math.sin(math.pi / 3)) + 1):
        V = A * math.sin(math.pi / 9) + B * math.sin(math.pi / 3)
        if math.isclose(V, G):
            print(G, V, A, B)

David Vreken - 4 months, 2 weeks ago

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Oh, I guess this only proves one solution for positive integers...

David Vreken - 4 months, 2 weeks ago
Chew-Seong Cheong
Jan 27, 2021

The yellow polygon is a 12-sided polygon with unit side length symmetrical about its center O O . The area of the yellow polygon is 12 12 times that of △ O A B \triangle OAB . The central angle ∠ A O B = 2 π 12 = π 6 \angle AOB = \dfrac {2\pi}{12} = \dfrac \pi 6 . A B AB and B C BC are sides of a unit regular nonagon. Therefore ∠ A B C = 9 − 2 9 π = 7 π 9 \angle ABC = \dfrac {9-2}9 \pi = \dfrac {7\pi}9 and the supplementary angle of ∠ A B O = ∠ A B C 2 = 7 π 18 \angle ABO = \dfrac {\angle ABC}2 = \dfrac {7\pi}{18} and ∠ O A B = 7 π 18 − π 6 = 2 π 9 \angle OAB = \dfrac {7\pi}{18}-\dfrac \pi 6 = \dfrac {2\pi}9 . Let O A = r OA = r . By sine rule ,

O A A B = sin ⁡ ∠ A B O sin ⁡ ∠ A O B r 1 = sin ⁡ ( π − 7 π 18 ) sin ⁡ π 6 Note that sin ⁡ ( π − θ ) = sin ⁡ θ ⟹ r = 2 sin ⁡ 7 π 18 \begin{aligned} \frac {OA}{AB} & = \frac {\sin \angle ABO}{\sin \angle AOB} \\ \frac r1 & = \frac {\sin \left(\pi - \frac {7\pi}{18}\right)}{\sin \frac \pi 6} & \small \blue{\text{Note that }\sin (\pi -\theta) = \sin \theta} \\ \implies r & = 2 \sin \frac {7\pi}{18} \end{aligned}

The area of the yellow polygon,

A = 12 ⋅ O A ⋅ A B ⋅ sin ⁡ ∠ O A B 2 = 12 ⋅ 2 sin ⁡ 7 π 18 ⋅ 1 ⋅ sin ⁡ 2 π 9 2 = 12 sin ⁡ 7 π 18 sin ⁡ 2 π 9 Note that cos ⁡ ( A − B ) − cos ⁡ ( A + B ) = 2 sin ⁡ A sin ⁡ B = 6 ( cos ⁡ π 6 − cos ⁡ 11 π 18 ) and cos ⁡ ( π − θ ) = − cos ⁡ θ = 6 ( cos ⁡ π 6 + cos ⁡ 7 π 18 ) also cos ⁡ θ = sin ⁡ ( π 2 − θ ) = 6 ( sin ⁡ π 3 + sin ⁡ π 9 ) \begin{aligned} A & = 12 \cdot \frac {OA \cdot AB \cdot \sin \angle OAB}2 \\ & = 12 \cdot \frac {2 \sin \frac {7\pi}{18} \cdot 1 \cdot \sin \frac {2\pi}9}2 \\ & = 12 \sin \frac {7\pi}{18} \sin \frac {2\pi}9 & \small \blue{\text{Note that }\cos(A-B)-\cos (A+B) = 2\sin A \sin B} \\ & = 6 \left(\cos \frac \pi 6 - \blue{\cos \frac {11\pi}{18}}\right) & \small \blue{\text{and }\cos (\pi - \theta) = - \cos \theta} \\ & = 6 \left(\cos \frac \pi 6 + \blue{\cos \frac {7\pi}{18}}\right) & \small \blue{\text{also }\cos \theta = \sin \left(\frac \pi 2 - \theta\right)} \\ & = 6 \left(\sin \frac \pi 3 + \sin \frac \pi 9 \right) \end{aligned}

Therefore A B = 6 ⋅ 6 = 36 AB = 6 \cdot 6 = \boxed{36} .

3 Helpful 2 Interesting 3 Brilliant 0 Confused

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