Quick and cute summer Sangaku #13

Geometry Level pending

The diagram shows a circular segment of radius 4 and height 2. We inscribe a red circle inside the circular segment, it's the largest circle possible so its radius will be 1 since the height is 2. We inscribe circles tangent to each other inside circular segment and this sequence of circles goes infinitely. The green circle (not the red) is considered the first circle of the sequence, the orange is the second, the purple is the third etc...

  • Question : The area of the nth circle can be expressed as : π ⋅ a ⋅ b n ( c n + d ) e \boxed{\pi \cdot \frac{a\cdot b^n}{\left(c^n+d\right)^e}}
  • Calculate : ( a − b ) ( c − d ) − e \boxed{\left(a-b\right)^{\left(c-d\right)}-e}


The answer is 45.

Valentin Duringer by Valentin Duringer , 30, Belgium

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1 solution

Valentin Duringer
Jul 19, 2020
  • We can find the radii of the circles one after the other using the Pythagorean theorem:
  • We solve this system of equations, R 1 \boxed{R_1} being the radius of the green circle:
  • ( 1 + R 1 ) 2 = ( 1 − R 1 ) 2 + y 2 \boxed{(1+R_1)^{2}=(1-R_1)^{2}+y^{2}}
  • ( 4 − R 1 ) 2 = ( 2 + R 1 ) 2 + y 2 \boxed{(4-R_1)^{2}=(2+R_1)^{2}+y^{2}}
  • We find R 1 = 3 4 \boxed{R_1=\frac{3}{4}}
  • In general we shall use a similar system of equation to find the next radii and we get the list of radii :
  • R 1 = 3 4 \boxed{R_1=\frac{3}{4}}
  • R 2 = 9 25 \boxed{R_2=\frac{9}{25}}
  • R 3 = 27 196 \boxed{R_3=\frac{27}{196}}
  • R 4 = 81 1681 \boxed{R_4=\frac{81}{1681}}
  • R 5 = 243 14884 \boxed{R_5=\frac{243}{14884}}
  • We need to express R n \boxed{R_n} in terms of n \boxed{n}
  • We easily see that the numerator of the radius is 3 n \boxed{3^{n}}
  • The denominator is trickier. But we see that each time, the denominator is a fourth of a perfect square :
  • 4 ⋅ 4 = 4 2 \boxed{4\cdot 4=4^2}
  • 4 ⋅ 25 = 1 0 2 \boxed{4\cdot 25=10^2}
  • 4 ⋅ 196 = 2 8 2 \boxed{4\cdot 196=28^2}
  • 4 ⋅ 1681 = 8 2 2 \boxed{4\cdot 1681=82^2}
  • 4 ⋅ 14884 = 24 4 2 \boxed{4\cdot 14884=244^2}
  • We see that to obtain 4 , 10 , 28 , 82 , 244 4, 10, 28, 82, 244 we actually add 1 to a power of 3 :
  • 4 = 3 1 + 1 \boxed{4=3^{1}+1}
  • 10 = 3 2 + 1 \boxed{10=3^{2}+1}
  • 28 = 3 3 + 1 \boxed{28=3^{3}+1}
  • 82 = 3 4 + 1 \boxed{82=3^{4}+1}
  • 244 = 3 5 + 1 \boxed{244=3^{5}+1}
  • We can conclude :
  • R 1 = 3 1 ( 3 1 + 1 ) 2 4 = 3 4 \boxed{R_1=\frac{3^1}{\frac{\left(3^1+1\right)^2}{4}}=\frac{3}{4}}
  • R 2 = 3 2 ( 3 2 + 1 ) 2 4 = 9 25 \boxed{R_2=\frac{3^2}{\frac{\left(3^2+1\right)^2}{4}}=\frac{9}{25}}
  • R 3 = 3 3 ( 3 3 + 1 ) 2 4 = 27 196 \boxed{R_3=\frac{3^3}{\frac{\left(3^3+1\right)^2}{4}}=\frac{27}{196}}
  • R 4 = 3 4 ( 3 4 + 1 ) 2 4 = 81 1681 \boxed{R_4=\frac{3^4}{\frac{\left(3^4+1\right)^2}{4}}=\frac{81}{1681}}
  • R 5 = 3 5 ( 3 5 + 1 ) 2 4 = 243 14884 \boxed{R_5=\frac{3^5}{\frac{\left(3^5+1\right)^2}{4}}=\frac{243}{14884}}
  • R n = 3 n ( 3 n + 1 ) 2 4 \boxed{R_n=\frac{3^n}{\frac{\left(3^n+1\right)^2}{4}}}
  • We simplify the formula and get : R n = 4 ⋅ 3 n ( 3 n + 1 ) 2 \boxed{R_n=\frac{4\cdot \:3^n}{\left(3^n+1\right)^2}}
  • Then A ( n ) = 16 ⋅ 9 n ( 3 n + 1 ) 4 ⋅ π \boxed{A(n)=\frac{16\cdot 9^n}{\left(3^n+1\right)^4}\cdot \pi }
  • ( 16 − 9 ) ( 3 − 1 ) − 4 = 45 \boxed{\left(16-9\right)^{\left(3-1\right)}-4=45}
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