Quick and cute summer Sangaku #1

Geometry Level pending

The diagram shows a pink circle of radius 1 and a green circle of radius 4. The are tangent to each other and to the same black line. We inscribe a yellow square between the black line and the two circles.

  • Question : The side length of the square can be written as a a + a ( a c − b b ) b − a b b a b − a ( a b c − b a ⋅ a ) b − a b \boxed{a^a+\frac{a\left(a\sqrt{c}-b^b\right)^{\frac{b-a}{b}}}{b^{\frac{a}{b}}}-\frac{a}{\left(ab\sqrt{c}-b^{a\cdot a}\right)^{\frac{b-a}{b}}}}
  • Evaluate a + b + c \boxed{a+b+c}


The answer is 188.

Valentin Duringer by Valentin Duringer , 30, Belgium

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1 solution

  • We can draw a coordinate system and give the pink circle the equation : x 2 + ( y − 1 ) 2 − 1 = 0 \boxed{x^{2}+(y-1)^{2}-1=0} and the green circle this equation : ( x − 4 ) 2 + ( y − 4 ) 2 − 16 = 0 \boxed{(x-4)^{2}+(y-4)^{2}-16=0}
  • There must be a point M ( x ; y ) \boxed{M(x;y)} on the pink circle and a point N ( x ′ ; y ) \boxed{N(x';y)} on the green circle to construct the square.
  • Since point M ( x ; y ) \boxed{M(x;y)} is on the pink circle, we can express x \boxed{x} in terms of y \boxed{y} using the equation x 2 + ( y − 1 ) 2 − 1 = 0 \boxed{x^{2}+(y-1)^{2}-1=0} <=> x = 2 y − y 2 \boxed{x=\sqrt{2y-y^{2}}}
  • Since point N ( x ′ ; y ) \boxed{N(x';y)} is on the green circle, we can express x \boxed{x} in terms of y \boxed{y} using the equation ( x − 4 ) 2 + ( y − 4 ) 2 − 16 = 0 \boxed{(x-4)^{2}+(y-4)^{2}-16=0} <=> x ′ = 4 − 8 y − y 2 \boxed{x'=4-\sqrt{8y-y^{2}}}
  • In a square, the length and the width are the same, then we need to solve the equation : y = 4 − 8 y − y 2 − 2 y − y 2 \boxed{y=4-\sqrt{8y-y^2}-\sqrt{2y-y^2}} < = > <=> ( y − 4 + 8 y − y 2 ) 2 = ( − 2 y − y 2 ) 2 \boxed{\left(y-4+\sqrt{8y-y^2}\right)^2=\left(-\sqrt{2y-y^2}\right)^2} < = > <=> 2 8 y − y 2 y − 8 8 y − y 2 + 16 = 2 y − y 2 \boxed{2\sqrt{8y-y^2}y-8\sqrt{8y-y^2}+16=2y-y^2} < = > <=> ( 2 8 y − y 2 ( y − 4 ) ) 2 = ( 2 y − y 2 − 16 ) 2 \boxed{\left(2\sqrt{8y-y^2}\left(y-4\right)\right)^2=\left(2y-y^2-16\right)^2} < = > <=> − 4 y 4 + 64 y 3 − 320 y 2 + 512 y = − 4 y 3 + y 4 + 36 y 2 − 64 y + 256 \boxed{-4y^4+64y^3-320y^2+512y=-4y^3+y^4+36y^2-64y+256} < = > <=> − 5 y 4 + 68 y 3 − 356 y 2 + 576 y − 256 = 0 \boxed{-5y^4+68y^3-356y^2+576y-256=0} < = > <=> − ( 5 y − 8 ) ( y 3 − 12 y 2 + 52 y − 32 ) = 0 \boxed{-\left(5y-8\right)\left(y^3-12y^2+52y-32\right)=0}
  • The first solution is 8 5 \boxed{\frac {8}{5}} and this is unacceptable. We need to find the solution of the cubic factor.
  • We can use the cubic formula to find the value of y \boxed{y}
  • We get y = ( − 24 − ( 15616 27 ) 1 2 ) 1 3 + ( − 24 + ( 15616 27 ) 1 2 ) 1 3 + 4 \boxed{y=\left(-24-\left(\frac{15616}{27}\right)^{\frac{1}{2}}\right)^{\frac{1}{3}}+\left(-24+\left(\frac{15616}{27}\right)^{\frac{1}{2}}\right)^{\frac{1}{3}}+4} < = > <=> ( − 16 61 3 3 − 24 ) 1 3 + ( 16 61 3 3 − 24 ) 1 3 + 4 \boxed{\left(-\frac{16\sqrt{61}}{3\sqrt{3}}-24\right)^{\frac{1}{3}}+\left(\frac{16\sqrt{61}}{3\sqrt{3}}-24\right)^{\frac{1}{3}}+4} < = > <=> ( − 16 183 9 − 24 ) 1 3 + ( 16 183 9 − 24 ) 1 3 + 4 \boxed{\left(-\frac{16\sqrt{183}}{9}-24\right)^{\frac{1}{3}}+\left(\frac{16\sqrt{183}}{9}-24\right)^{\frac{1}{3}}+4}
  • ( 16 183 9 − 24 ) 1 3 = ( 16 183 − 216 3 2 ) 1 3 = 2 ( 2 183 − 27 3 2 ) 1 3 = 2 ( 2 183 − 27 ) 1 3 3 2 3 \boxed{\left(\frac{16\sqrt{183}}{9}-24\right)^{\frac{1}{3}}=\left(\frac{16\sqrt{183}-216}{3^2}\right)^{\frac{1}{3}}=2\left(\frac{2\sqrt{183}-27}{3^2}\right)^{\frac{1}{3}}=2\frac{\left(2\sqrt{183}-27\right)^{\frac{1}{3}}}{3^{\frac{2}{3}}}}
  • ( − 16 183 9 − 24 ) 1 3 = − 2 ( 6 183 − 81 ) 1 3 \boxed{\left(-\frac{16\sqrt{183}}{9}-24\right)^{\frac{1}{3}}=-\frac{2}{\left(6\sqrt{183}-81\right)^{\frac{1}{3}}}}
  • We find that the side length is equal to : 4 + 2 ( 2 183 − 27 ) 1 3 3 2 3 − 2 ( 6 183 − 81 ) 1 3 \boxed{4+\frac{2\left(2\sqrt{183}-27\right)^{\frac{1}{3}}}{3^{\frac{2}{3}}}-\frac{2}{\left(6\sqrt{183}-81\right)^{\frac{1}{3}}}}
  • a = 2 \boxed{a=2} b = 3 \boxed{b=3} c = 183 \boxed{c=183}
  • a + b + c = 188 \boxed{a+b+c=188}
1 Helpful 1 Interesting 2 Brilliant 2 Confused

I am confused as to how exactly did you find the numeric expression for the side length from the equation just before it. Can you elaborate on that ?

Hosam Hajjir - 11 months ago

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I did an editing, i hope you'r not "confused" anymore =)

Valentin Duringer - 11 months ago

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Thank you for explaining that.

Hosam Hajjir - 11 months ago

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@Hosam Hajjir – You are welcome

Valentin Duringer - 11 months ago

You mean how to solve the equation or how to get the required form? I'll try to make an edit in 10 hours anyway...

Valentin Duringer - 11 months ago

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