My ultimate Sangaku part 2

Geometry Level pending
  • The diagram shows a black semi-circle with radius 1 \boxed{1} .
  • The blue semi-circles are equal.
  • The green circles interestects such that they pass through each others center.
  • The blue and red semi-circles and the green circles are tangent to each other. The green circles and black semi-circle are tangent to each other.
  • We use the centers of the green circles and their 2 intersection points to drawn a quadrilateral.

Question : The area of the orange quadrilateral can be expressed as a b − c d e \boxed{\frac{a\sqrt{b}-c\sqrt{d}}{e}} . Evaluate ( e − b ) ( a − c + d ) \boxed{\sqrt{\left(e-b\right)\left(a-c+d\right)}}


The answer is 5.

Valentin Duringer by Valentin Duringer , 30, Belgium

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1 solution

Valentin Duringer
Aug 24, 2020
  • We use the pythagorean theorem to solve this problem:
  • { ( x + R ) 2 = ( 1 − R 2 − x ) 2 + h 2 ( R + 1 − 2 x ) 2 = ( R 2 ) 2 + h 2 R + 1 − 2 x = 1 − R \begin{cases}\left(x+R\right)^2=\left(1-\frac{R}{2}-x\right)^2+h^2\\\left(R+1-2x\right)^2=\left(\frac{R}{2}\right)^2+h^2\\R+1-2x=1-R\end{cases}
  • We find x = 33 − 5 2 \boxed{x=\frac{\sqrt{33}-5}{2}} and R = 33 − 5 2 \boxed{R=\frac{\sqrt{33}-5}{2}}
  • Since the sides of the orange quadrilateral are equal to R R , then we have a rhombus
  • Then its area is 2 ⋅ 3 4 ( − 5 2 + 33 2 ) 2 = 29 3 − 15 11 4 \boxed{2\cdot \:\frac{\sqrt{3}}{4}\left(\frac{-5}{2}+\frac{\sqrt{33}}{2}\right)^2=\frac{29\sqrt{3}-15\sqrt{11}}{4}}
  • Finally ( 4 − 3 ) ( 29 − 15 + 11 ) = 5 \boxed{\sqrt{\left(4-3\right)\left(29-15+11\right)}=5}
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