Descartes Sangaku : The Beast part 1

Geometry Level pending
  • The diagram shows a semi-circle of radius 1. We place a point on its diameter.
  • This point allows us to draw two semi-circles (green and cyan).
  • We inscribe a red circle so that this circle is tangent to the other three semi-circles.
  • The center of the semi-circle of radius 1 is the origin of the coordinate system.

The question: What is the area of the red circle when the sum of the coordinates of its center is maximum? The answer can be expressed as a − b ⋅ c d ⋅ π \boxed{\frac{a-b\cdot \sqrt{c}}{d}\cdot \pi } where a a , b b , c c , d d are positive integers. Evaluate d − a − b − c \boxed{\sqrt{d-a-b-c}}


The answer is 5.

Valentin Duringer by Valentin Duringer , 30, Belgium

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1 solution

Valentin Duringer
Aug 18, 2020
  • To solve this problem, we shall denote the radius of the cyan semi-circle as x \boxed{x} , the radius of the green semi-circle as 1 − x \boxed{1-x} and the radius of the blue circle as R ( x ) \boxed{R(x)} , l ( x ) \boxed{l(x)} is the x x coordinate of the red circle's center and h ( x ) \boxed{h(x)} is the y y coordinate of the red circle's center.

  • We shall use the Pythagorean theorem to express R ( x ) \boxed{R(x)} , l ( x ) \boxed{l(x)} and h ( x ) \boxed{h(x)} in terms of x \boxed{x}

  • We shall write three equations using the diagram:

  • ( 1 − R ( x ) ) 2 = l ( x ) 2 + h ( x ) 2 \boxed{(1-R(x))^{2}=l(x)^{2}+h(x)^{2}}
  • ( 1 − x + R ( x ) ) 2 = ( x − l ( x ) ) 2 + h ( x ) 2 \boxed{(1-x+R(x))^{2}=(x-l(x))^{2}+h(x)^{2}}
  • ( x + R ( x ) ) 2 = ( 1 − x + l ( x ) ) 2 + h ( x ) 2 \boxed{(x+R(x))^{2}=(1-x+l(x))^{2}+h(x)^{2}}
  • After combining the equation and using minor algebra skills we find:
  • R ( x ) = x − x 2 x 2 − x + 1 \boxed{R(x)=\frac{x-x^{2}}{x^{2}-x+1}}
  • l ( x ) = 2 x − 1 x 2 − x + 1 \boxed{l\left(x\right)=\frac{2x-1}{x^2-x+1}}
  • h ( x ) = 2 ⋅ ( x 2 − x ) − x 2 + x − 1 \boxed{h\left(x\right)=2\cdot \frac{\left(x^2-x\right)}{-x^2+x-1}}
  • Then the function S ( x ) = 2 ⋅ ( x 2 − x ) − x 2 + x − 1 + 2 x − 1 x 2 − x + 1 \boxed{S(x)=2\cdot \frac{\left(x^2-x\right)}{-x^2+x-1}+\frac{2x-1}{x^2-x+1}} is the sum of the coordinates.
  • S ′ ( x ) = ( 3 − 2 x − 2 x 2 ) ( − 1 + x − x 2 ) 2 \boxed{S'(x)=\frac{\left(3\:-\:2\:x\:-\:2\:x^2\right)}{\left(-1\:+\:x\:-\:x^2\right)^2}}
  • After solving S ′ ( x ) = 0 \boxed{S'(x)=0} we find x = 7 − 1 2 \boxed{x=\frac{\sqrt{7}-1}{\:2}}
  • We calculate the area of the red circle for x = 7 − 1 2 \boxed{x=\frac{\sqrt{7}-1}{\:2}} : π ⋅ R ( x ) 2 = π ⋅ 23 − 8 7 63 \boxed{\pi \cdot R(x)^{2}=\pi \cdot \frac{23-8\sqrt{7}}{63}}
  • d − a − b − c = 5 \boxed{\sqrt{d-a-b-c=5}}
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