My ultimate Sangaku part 3

Geometry Level pending
  • The diagram shows a black semi-circle of radius 1 \boxed{1} .
  • We place a point P \boxed{P} and a point P ′ \boxed{P'} on its diameter.
  • This allows us to draw 3 semi-circles.
  • Since point P \boxed{P} and point P ′ \boxed{P'} can move freely on the diameter as long as there are always 3 semi-circles drawn on the figure, for instance, P ′ \boxed{P'} can not be place to the left side of P \boxed{P}

  • Finally we draw two circles so they are tangent to the black circle, to the middle semi-circle and to one of the 2 other semi-circles.

  • The raduis of the orange semi-circle is called x x

  • Depending on the position of P \boxed{P} and P ′ \boxed{P'} , the red and the purple circles can be identical and form a unique pink circle.

Question : When this happens, the area of this pink circle can be written as : a ⋅ x 4 − b ⋅ x 3 + a ⋅ x 2 c ⋅ x 4 + d ⋅ x 2 + c ⋅ π \boxed{\frac{a\cdot x^4-b\cdot x^3+a\cdot x^2}{c\cdot x^4+d\cdot x^2+c}\cdot \pi } where a a , b b , c c and d d are positive integers. Evaluate a + b − c − d \boxed{\sqrt{a+b-c-d}}


The answer is 3.

Valentin Duringer by Valentin Duringer , 30, Belgium

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1 solution

Valentin Duringer
Aug 24, 2020
  • Let us call the radius of the green semi-circle y y , then the radius of the middle blue semi-circle is 1 − x − y 1-x-y

  • Now we use this diagram and the pythagorean theorem to write some equalities:

  • { ( x + R ) 2 = l 2 + h 2 ( 1 − x − y + R ) 2 = ( 1 − l − y ) 2 + h 2 ( 1 − R ) 2 = ( 1 − x − l ) 2 + h 2 \begin{cases}\left(x+R\right)^2=l^2+h^2\\\left(1-x-y+R\right)^2=\left(1-l-y\right)^2+h^2\\\left(1-R\right)^2=\left(1-x-l\right)^2+h^2\end{cases}

  • We need to do some algebra to express R ( x ; y ) R(x;y) in terms of x x and y y
  • After some work we find R ( x ; y ) = ( − x + x 2 ) ( − 1 + y ) 1 − x + x 2 − y \boxed{R(x;y)=\frac{\left(-x+x^2\right)\left(-1+y\right)}{1-x+x^2-y}}
  • We can call the radius of the other circle (on the right side) : r ( x ; y ) r(x;y) and its expression in terms of x x and y y is :
  • r ( x ; y ) = ( − y + y 2 ) ( − 1 + x ) 1 − y + y 2 − x \boxed{r(x;y)=\frac{\left(-y+y^2\right)\left(-1+x\right)}{1-y+y^2-x}}
  • If both circles are equal, then R ( x ; y ) = r ( x ; y ) \boxed{R(x;y)=r(x;y)} and after some algebra we find:
  • y = x − 1 x + 1 \boxed{y=\frac{x-1}{x+1}}
  • Now we can express the area of the unique pink circle in terms of x x :
  • R ( x ) = 4 x 4 − 8 x 3 + 4 x 2 x 4 + 2 x 2 + 1 ⋅ π \boxed{R(x)=\frac{4x^4-8x^3+4x^2}{x^4+2x^2+1}\cdot \pi }
  • Finally 4 + 8 − 1 − 2 = 3 \boxed{\sqrt{4+8-1-2}=3}
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