Equilateral nine-side-polygon
Consider the above regular 9-gon.
Which area is larger: the black or the white are?
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2 solutions
Let AN be the altitude of isosceles triangle AFE in 9-gon ABCDEF.....So AN is the line of symmetry.
So consider only half 9-gon.
WLOG let the sides of the 9-gon be unit. So AB=BC=CD=DE=EF=1, EN=1/2.
Let AM be the angle bisector of DAE, M on DE.
Construct AKC congruent to ABC, .....AHE congruent to AME.
So net white area not canceled is EHN +CKD, net black area not canceled is AMD.
S u m o f a n g l e s o f a t r i a n g l e i s 1 8 0 . E x t e r n a l a n g l e o f a t r i a n g l e i s e q u a l t o t h e s u m o f i n t e r n a l a n g l e s . ∴ w e g e t t h e a n g l e s s h o w n i n t h e f i g . A r e a o f r t . ∠ Δ E H N = 1 / 2 ∗ 1 / 2 ∗ 1 / 2 / t a n 7 0 = . 0 4 5 5 . I n i s o s c e l e s Δ C K D , v e r t e x ∠ D C K = 1 0 0 , a n d e q u a l s i d e s = 1 . ∴ a r e a C K D = 1 / 2 ∗ 1 ∗ 1 ∗ s i n 1 0 0 = 0 . 4 9 2 4 . I n A M D , u s i n g S i n L a w , A D / s i n 7 0 = M D / s i n 1 0 . A r e a A M D = 1 / 2 ∗ A D ∗ A M ∗ s i n D A M . A D = A K + K D = 1 + 2 ∗ c o s 4 0 . A r e a A M D = 1 / 2 ∗ ( 1 + 2 ∗ c o s 4 0 ) 2 ∗ s i n 1 0 / s i n 7 0 ∗ s i n 1 0 0 = . 5 8 3 4 . ∴ B l a c k a r e a = 2 ∗ . 5 8 3 4 > w h i t e a r e a = 2 ∗ ( 0 . 4 9 2 4 + . 0 4 5 5 ) .
If we filled two triangles with same way, then their areas are equal.