Trapped in a Pentagon
The figure shows a square trapped in a regular pentagon of side length 1, symmetrically touching the pentagon at 3 points.
What is the distance between the lowest point of the square and the base of the pentagon?
The answer is of the form .
Submit your answer as the value of .
The answer is 25.
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1 solution
A B = 1 . A C , s i d e o f t h e s q u a r e , = s . D i s t h e l o w e s t p o i n t . A D , d i a g o n a l o f t h e s q u a r e , = d . H e i g h t o f t h e p e n t a g o n = h . D E , t h e r e q u i r e d d i s t a n c e , = h − d = x . . . . . . . . . . . ( 1 ) ∠ E A B = 5 4 o . ∠ D A C = 4 5 o . r ∴ ∠ C A B = 9 o . I n A S A Δ C A B , u s i n g S i n L a w , s = 1 / S i n 6 3 ∗ S i n 1 0 8 S o d = s ∗ 2 . . . . . . . . . . . ( 2 ) h = 1 / 2 ∗ 1 ∗ ( C o t 3 6 + C s c 3 6 ) . . . . . . . . . . ( 3 ) F r o m ( 1 ) x = h − d = 1 / 2 ∗ A − 2 5 − 5 . a s g i v e n . ∴ A = ( ( h − d ) + 2 ∗ 5 ) 2 + 2 ∗ 5 . S u b s t i t u t i n g v a l u e s f r o m ( 3 ) a n d ( 2 ) , A = { C o t 3 6 + C s c 3 6 − 2 ∗ S i n 6 3 2 ∗ S i n 1 0 8 + 2 ∗ 5 } 2 + 2 ∗ 5 = 2 5 .