Descartes Sangaku : The End part 1

Geometry Level pending
  • The diagram shows a black circle of radius 25 \boxed{25} .
  • We drawn an horizontal black line to divide this circle into two circular segments. This line a length 48 \boxed{48} .
  • Now we pack the top circulair segment with 3 types of circles : Y e l l o w \color{#CEBB00}Yellow , C y a n \color{cyan}Cyan and P u r p l e \color{#69047E}Purple .
  • Then we pack the bottom circular segment with 3 types of circles : G r e e n \color{#20A900}Green , R e d \color{#D61F06}Red and P i n k \color{#E81990}Pink
  • Note : The two biggest yellow circles are congruent.
  • The bottom circular segment is larger than the top circular segment
  • We count the circles starting from the largest of its sequence to the right (or left) towards the infinitely small circles
  • We focus on the Y e l l o w \color{#CEBB00}Yellow and G r e e n \color{#20A900}Green circles in this problem. The largest G r e e n \color{#20A900}Green circle is the 0 t h 0^{th} circle of its sequence. The largest Y e l l o w \color{#CEBB00}Yellow circle is the 0 t h 0^{th} circle of its sequence.

The question: The radius of the n t h n^{th} Y e l l o w \color{#CEBB00}Yellow circle (for n ≥ 0 n≥0 ) can be expressed as ( a + b ) b ⋅ b ( b ⋅ n + a + b ) ( b b ⋅ n + a + a ) b \boxed{\frac{\left(a+b\right)^b\cdot b^{\left(b\cdot \:n+a+b\right)}}{\left(b^{b\cdot n+a}+a\right)^b}} and the radius of the n t h n^{th} G r e e n \color{#20A900}Green circle (for n ≥ 0 n≥0 ) can be expressed as c d ⋅ d ( n ⋅ c ) ( d ( n ⋅ c ) + c − d ) c \boxed{\frac{c^d\cdot d^{\left(^{n\cdot \:\sqrt{c}}\right)}}{\left(d^{\left(^{n\cdot \:\sqrt{c}}\right)}+c-d\right)^{\sqrt{c}}}} where a a , b b , c c , d d are positive integers. Calculate a + b + c + d \boxed{a+b+c+d}


The answer is 10.

Valentin Duringer by Valentin Duringer , 30, Belgium

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1 solution

Valentin Duringer
Aug 20, 2020
  • First we need to know the heights of both circular segments. We use the product of chords theorem in a circle :
  • ( 48 2 ) 2 = h 1 ⋅ h 2 \boxed{\left(\frac{48}{2}\right)^2=h_1\cdot \:h_2} < = > <=> h 1 = 18 \boxed{h_1=18} and h 2 = 32 \boxed{h_2=32} since h 1 = 50 − h 2 \boxed{h_1=50-h_2}
  • Then the largest green circle has radius 32 2 = 16 \boxed{\frac{32}{2}=16}
  • To calculate the radii of consecutive green circles we use the pythagorean theorem:
  • We use those two right triangles to write two equations: { ( R 0 + R 1 ) 2 = ( R 0 − R 1 ) 2 + x 2 ( 25 − R 1 ) 2 = ( R 0 − R 1 − 9 ) 2 + x 2 \begin{cases}\left(R_0+R_1\right)^2=\left(R_0-R_1\right)^2+x^2\\\left(25-R_1\right)^2=\left(R_0-R_1-9\right)^2+x^2\end{cases}
  • R 0 = 16 \boxed{R_0=16} and 9 \boxed{9} is the distance bewteen the center of the largest circle containing all the others and the center of the green circle.
  • After solving the system of equations we find R 1 = \color{#20A900}R_1= 144 25 \boxed{\frac{144}{25}}
  • We can repeat this process to find the first 5 terms of the sequence:
  • R 0 = \color{#20A900}R_0= 16 1 \boxed{\frac{16}{1}}
  • R 1 = \color{#20A900}R_1= 144 25 \boxed{\frac{144}{25}}
  • R 2 = \color{#20A900}R_2= 1296 1681 \boxed{\frac{1296}{1681}}
  • R 3 = \color{#20A900}R_3= 11664 133225 \boxed{\frac{11664}{133225}}
  • R 4 = \color{#20A900}R_4= 104976 10764961 \boxed{\frac{104976}{10764961}}
  • Let's look at the numerator. It's difficult to identify a pattern, but with some effort and research, we can rewrite them :
  • 16 = ( 4 ) 2 = ( 4 ⋅ 3 0 ) 2 \boxed{16=\left(4\right)^2=\left(4\cdot 3^0\right)^2}
  • 144 = ( 12 ) 2 = ( 4 ⋅ 3 1 ) 2 \boxed{144=\left(12\right)^2=\left(4\cdot 3^1\right)^2}
  • 1296 = ( 36 ) 2 = ( 4 ⋅ 3 2 ) 2 \boxed{1296=\left(36\right)^2=\left(4\cdot 3^2\right)^2}
  • 11664 = ( 108 ) 2 = ( 4 ⋅ 3 3 ) 2 \boxed{11664=\left(108\right)^2=\left(4\cdot 3^3\right)^2}
  • 104976 = ( 324 ) 2 = ( 4 ⋅ 3 4 ) 2 \boxed{104976=\left(324\right)^2=\left(4\cdot 3^4\right)^2}
  • The numerator can be written as ( 4 ⋅ 3 n ) 2 \boxed{\left(4\cdot 3^n\right)^2}
  • Let's look at the denominator. It's difficult to identify a pattern, but with some effort and research, we can rewrite them :
  • 1 = 4 4 = 2 2 4 = ( 3 2 ⋅ 0 + 1 ) 2 4 \boxed{1=\frac{4}{4}=\frac{2^2}{4}=\frac{\left(3^{2\cdot 0}+1\right)^2}{4}}
  • 25 = 100 4 = 1 0 2 4 = ( 3 2 ⋅ 1 + 1 ) 2 4 \boxed{25=\frac{100}{4}=\frac{10^2}{4}=\frac{\left(3^{2\cdot 1}+1\right)^2}{4}}
  • 1681 = 6724 4 = 8 2 2 4 = ( 3 2 ⋅ 2 + 1 ) 2 4 \boxed{1681=\frac{6724}{4}=\frac{82^2}{4}=\frac{\left(3^{2\cdot 2}+1\right)^2}{4}}
  • 133225 = 532900 4 = 73 0 2 4 = ( 3 2 ⋅ 3 + 1 ) 2 4 \boxed{133225=\frac{532900}{4}=\frac{730^2}{4}=\frac{\left(3^{2\cdot 3}+1\right)^2}{4}}
  • 10764961 = 43059844 4 = 656 2 2 4 = ( 3 2 ⋅ 4 + 1 ) 2 4 \boxed{10764961=\frac{43059844}{4}=\frac{6562^2}{4}=\frac{\left(3^{2\cdot 4}+1\right)^2}{4}}
  • The denominator can be written as ( 3 2 ⋅ n + 1 ) 2 4 \boxed{\frac{\left(3^{2\cdot n}+1\right)^2}{4}}
  • Then R n = \color{#20A900}R_n= 64 ⋅ 3 2 ⋅ n ( 3 2 ⋅ n + 1 ) 2 = 4 3 ⋅ 3 n ⋅ 4 ( 3 n ⋅ 4 + 4 − 3 ) 4 \boxed{\frac{64\cdot \:\:3^{2\cdot n}}{\left(3^{2\cdot \:n}+1\right)^2}=\frac{4^3\cdot \:\:\:3^{n\cdot \sqrt{4}}}{\left(3^{n\cdot \sqrt{4}}+4-3\right)^{\sqrt{4}}}}
  • Then c = 4 \boxed{c=4} and d = 3 \boxed{d=3}
  • We can use a similar method to get the radii of the yellow circles, the consecutive Pythagorean equations give the sequence:
  • R 0 = \color{#CEBB00}R_0= 8 1 \boxed{\frac{8}{1}}
  • R 1 = \color{#CEBB00}R_1= 32 9 \boxed{\frac{32}{9}}
  • R 2 = \color{#CEBB00}R_2= 128 121 \boxed{\frac{128}{121}}
  • R 3 = \color{#CEBB00}R_3= 512 1849 \boxed{\frac{512}{1849}}
  • R 4 = \color{#CEBB00}R_4= 2048 29241 \boxed{\frac{2048}{29241}}
  • Let's look at the numerator. We can rewrite them :
  • 8 = 2 2 ⋅ 0 + 3 \boxed{8=2^{2\cdot 0+3}}
  • 32 = 2 2 ⋅ 1 + 3 \boxed{32=2^{2\cdot 1+3}}
  • 128 = 2 2 ⋅ 2 + 3 \boxed{128=2^{2\cdot 2+3}}
  • 512 = 2 2 ⋅ 3 + 3 \boxed{512=2^{2\cdot 3+3}}
  • 2048 = 2 2 ⋅ 4 + 3 \boxed{2048=2^{2\cdot 4+3}}
  • The numerator can be written as 2 2 ⋅ n + 3 \boxed{2^{2\cdot n+3}}
  • Let's look at the denominator. It's difficult to identify a pattern, but with some effort and research, we can rewrite them :
  • 1 = 1 2 = ( 3 3 ) 2 = ( 2 2 ⋅ 0 + 1 + 1 3 ) 2 \boxed{1=1^2=\left(\frac{3}{3}\right)^2=\left(\frac{2^{2\cdot 0+1}+1}{3}\right)^2}
  • 9 = 3 2 = ( 9 3 ) 2 = ( 2 2 ⋅ 1 + 1 + 1 3 ) 2 \boxed{9=3^2=\left(\frac{9}{3}\right)^2=\left(\frac{2^{2\cdot 1+1}+1}{3}\right)^2}
  • 121 = 1 1 2 = ( 33 3 ) 2 = ( 2 2 ⋅ 2 + 1 + 1 3 ) 2 \boxed{121=11^2=\left(\frac{33}{3}\right)^2=\left(\frac{2^{2\cdot 2+1}+1}{3}\right)^2}
  • 1849 = 4 3 2 = ( 129 3 ) 2 = ( 2 2 ⋅ 3 + 1 + 1 3 ) 2 \boxed{1849=43^2=\left(\frac{129}{3}\right)^2=\left(\frac{2^{2\cdot 3+1}+1}{3}\right)^2}
  • 29241 = 17 1 2 = ( 513 3 ) 2 = ( 2 2 ⋅ 4 + 1 + 1 3 ) 2 \boxed{29241=171^2=\left(\frac{513}{3}\right)^2=\left(\frac{2^{2\cdot 4+1}+1}{3}\right)^2}
  • We can then write the denominator as ( 2 2 n + 1 + 1 3 ) 2 \boxed{\left(\frac{2^{2n+1}+1}{3}\right)^2}
  • Then R n = \color{#CEBB00}R_n= 9 ⋅ 2 2 n + 3 ( 2 2 n + 1 + 1 ) 2 = ( 1 + 2 ) 2 ⋅ 2 2 n + 1 + 2 ( 2 2 n + 1 + 1 ) 2 \boxed{\frac{9\cdot \:\:\:2^{2n+3}}{\left(2^{2n+1}+1\right)^2}=\frac{\left(1+2\right)^2\cdot \:2^{2n+1+2}}{\left(2^{2n+1}+1\right)^2}}
  • Then a = 1 \boxed{a=1} and b = 2 \boxed{b=2}
  • Finally a + b + c + d = 1 + 2 + 3 + 4 = 10 \boxed{a+b+c+d=1+2+3+4=10}
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