Descartes Sangaku : The Beast part 2

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 coordinates of its center have the same absolute value? 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 8.

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 green semi-circle as x \boxed{x} , the radius of the cyan semi-circle as 1 − x \boxed{1-x} and the radius of the red 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}}
  • We can solve the equation l ( x ) = h ( x ) \boxed{l(x)=h(x)} and find x = 1 2 \boxed{x=\frac{1}{\sqrt{2}}}
  • Then we calculate the area of the red circle for x = 1 2 \boxed{x=\frac{1}{\sqrt{2}}} : π ⋅ R ( x ) 2 = π ⋅ 9 − 4 2 49 \boxed{\pi \cdot R(x)^{2}=\pi \cdot \frac{9-4\sqrt{2}}{49}}
  • d + a + b + c = 8 \boxed{\sqrt{d+a+b+c}=8}
1 Helpful 0 Interesting 0 Brilliant 0 Confused

I didn't understand how did you use pythagoras theorem, I made a graph to solve it, here , but I couldn't find a thread between the red circle and other two semicircles. But I did manage to guess x = 1 2 x=\frac{1}{\sqrt{2}}

A Former Brilliant Member - 9 months, 1 week ago

hi, the diagram in the solution does not help you?

Valentin Duringer - 9 months, 1 week ago

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I understand it now, I'm not very familiar with functions, I then realised you wrote every other length as the function of x

A Former Brilliant Member - 9 months, 1 week ago

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