Descartes Sangaku : The End part II

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 C y a n \color{cyan}Cyan and R e d \color{#D61F06}Red 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} C y a n \color{cyan}Cyan circle (for n ≥ 0 n≥0 ) can be expressed as a a ⋅ n + b ( a a ⋅ n + b − a ) a \boxed{\frac{a^{a\cdot n+b}}{\left(a^{a\cdot \:n}+b-a\right)^a}} and the radius of the n t h n^{th} R e d \color{#D61F06}Red circle (for n ≥ 0 n≥0 ) can be expressed as c ⋅ d ( n + b − a ) ( d ( a ⋅ n + b − a ) + b − a ) a \boxed{\frac{c\cdot d^{\left(n+b-a\right)}}{\left(\sqrt{d}^{\left(a\cdot n+b-a\right)}+b-a\right)^a}} where a a , b b , c c , d d are positive integers. Calculate b ⋅ c ⋅ d a \boxed{b\cdot \sqrt[a]{c\cdot d}}


The answer is 18.

Valentin Duringer by Valentin Duringer , 30, Belgium

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1 solution

Valentin Duringer
Aug 20, 2020
  • To solve this question you'll need the height of the circular segments and the radii of the G r e e n \color{#20A900}Green and Y e l l o w \color{#CEBB00}Yellow circles. There is a solution for those steps in part 1 of this series.
  • Let's focus on the C y a n \color{cyan}Cyan circles. Using Descartes circle's theorem we find the 5 first C y a n \color{cyan}Cyan radii.
  • R 0 \color{cyan}R_0 = 8 4 \boxed{\frac{8}{4}}
  • R 0 \color{cyan}R_0 = 32 25 \boxed{\frac{32}{25}}
  • R 0 \color{cyan}R_0 = 128 289 \boxed{\frac{128}{289}}
  • R 0 \color{cyan}R_0 = 512 4225 \boxed{\frac{512}{4225}}
  • R 0 \color{cyan}R_0 = 2048 66049 \boxed{\frac{2048}{66049}}
  • Let's look at the numerator, we easily see that it's each time an odd power of 2, then it can be written as 2 2 n + 3 \boxed{2^{2n+3}}
  • The denominator is also an easy challenge, we can rewritte them and spot a pattern:
  • 4 = ( 2 ) 2 = ( 4 0 + 1 ) 2 \boxed{4=\left(2\right)^2=\left(4^0+1\right)^2}
  • 25 = ( 5 ) 2 = ( 4 1 + 1 ) 2 \boxed{25=\left(5\right)^2=\left(4^1+1\right)^2}
  • 289 = ( 17 ) 2 = ( 4 2 + 1 ) 2 \boxed{289=\left(17\right)^2=\left(4^2+1\right)^2}
  • 4225 = ( 65 ) 2 = ( 4 3 + 1 ) 2 \boxed{4225=\left(65\right)^2=\left(4^3+1\right)^2}
  • 66049 = ( 257 ) 2 = ( 4 4 + 1 ) 2 \boxed{66049=\left(257\right)^2=\left(4^4+1\right)^2}
  • We can conclude that the denominator can be written as ( 4 n + 1 ) 2 \boxed{\left(4^n+1\right)^2}
  • Finally R n \color{cyan}R_n = 2 2 n + 3 ( 4 n + 1 ) 2 = 2 2 ⋅ n + 3 ( 2 2 ⋅ n + 3 − 2 ) 2 \boxed{\frac{2^{2n+3}}{\left(4^n+1\right)^2}=\frac{2^{2\cdot n+3}}{\left(2^{2\cdot n}+3-2\right)^2}}
  • Then a = 2 \boxed{a=2} and b = 3 \boxed{b=3}
  • Let's focus on the r e d \color{#D61F06}red circles. Using Descartes circle's theorem we find the 5 first R e d \color{#D61F06}Red radii.
  • R 0 \color{#D61F06}R_0 = 9 4 \boxed{\frac{9}{4}}
  • R 1 \color{#D61F06}R_1 = 81 196 \boxed{\frac{81}{196}}
  • R 2 \color{#D61F06}R_2 = 729 14884 \boxed{\frac{729}{14884}}
  • R 3 \color{#D61F06}R_3 = 6561 1196836 \boxed{\frac{6561}{1196836}}
  • R 4 \color{#D61F06}R_4 = 59049 96864964 \boxed{\frac{59049}{96864964}}
  • We easily see that the numerators are odd power of 9 9 , then the denominator can be written as 9 n + 1 \boxed{9^{n+1}}
  • The denominator is also an easy challenge, we can rewritte them and spot a pattern:
  • 4 = 16 4 = ( 4 ) 2 4 = ( 3 2 ⋅ 0 + 1 + 1 ) 2 4 \boxed{4=\frac{16}{4}=\frac{\left(4\right)^2}{4}=\frac{\left(3^{2\cdot \:0+1}+1\right)^2}{4}}
  • 196 = 784 4 = ( 28 ) 2 4 = ( 3 2 ⋅ 1 + 1 + 1 ) 2 4 \boxed{196=\frac{784}{4}=\frac{\left(28\right)^2}{4}=\frac{\left(3^{2\cdot \:1+1}+1\right)^2}{4}}
  • 14884 = 59536 4 = ( 244 ) 2 4 = ( 3 2 ⋅ 2 + 1 + 1 ) 2 4 \boxed{14884=\frac{59536}{4}=\frac{\left(244\right)^2}{4}=\frac{\left(3^{2\cdot \:2+1}+1\right)^2}{4}}
  • 1196836 = 4787344 4 = ( 2188 ) 2 4 = ( 3 2 ⋅ 3 + 1 + 1 ) 2 4 \boxed{1196836=\frac{4787344}{4}=\frac{\left(2188\right)^2}{4}=\frac{\left(3^{2\cdot \:3+1}+1\right)^2}{4}}
  • 96864964 = 387459856 4 = ( 19684 ) 2 4 = ( 3 2 ⋅ 4 + 1 + 1 ) 2 4 \boxed{96864964=\frac{387459856}{4}=\frac{\left(19684\right)^2}{4}=\frac{\left(3^{2\cdot \:4+1}+1\right)^2}{4}}
  • We can conclude the denominator can be written as ( 3 2 n + 1 + 1 ) 2 4 \boxed{\frac{\left(3^{2n+1}+1\right)^2}{4}}
  • Thus, R n \color{#D61F06}R_n = 4 ⋅ 9 n + 1 ( 3 2 n + 1 + 1 ) 2 = 4 ⋅ 9 ( n + 3 − 2 ) ( 9 ( 2 ⋅ n + 3 − 2 ) + 3 − 2 ) 2 \boxed{\frac{4\cdot \:\:9^{n+1}}{\left(3^{2n+1}+1\right)^2}=\frac{4\cdot 9^{\left(n+3-2\right)}}{\left(\sqrt{9}^{\left(2\cdot n+3-2\right)}+3-2\right)^2}}
  • Then c = 4 \boxed{c=4} and d = 9 \boxed{d=9}
  • b ⋅ c ⋅ d a = 3 ⋅ 4 ⋅ 9 2 = 18 \boxed{b\cdot \sqrt[a]{c\cdot d}=3\cdot \sqrt[2]{4\cdot 9}=18}
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