Quick and cute summer Sangaku #5

Geometry Level pending

The diagram shows two circles of radius 2 passing through each other's center. We inscribe the biggest circle (orange) possible in the overlapping area. Then we inscribe a circle (green) tangent to the orange circle and tangent to the circles of radius 2. Then we inscribe a circle (blue) tangent to the green circle and tangent to the circles of radius 2. We repeat this process infinitely.

  • Question : The curvature of the colored tangent circles (starting from the orange circle) are in an integer sequence.
  • We denote k n \boxed{k_n} the curvature of the nth circle.
  • There exists a fromula : k n = α k n − 1 + β k n − 2 + γ k n − 3 \boxed{k_n=αk_{n-1}+βk_{n-2}+γk_{n-3}} where α \boxed{α} , β \boxed{β} and γ \boxed{γ} are integers

  • Evaluate α + β + γ \boxed{α+β+γ}

  • BONUS : Find a formula giving k n \boxed{k_n} in terms of n \boxed{n}


The answer is 1.

Valentin Duringer by Valentin Duringer , 30, Belgium

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1 solution

Valentin Duringer
Jul 12, 2020
  • We can solve for the radius of the green circle using the pythagorean theorem since the orange circle's radius is 1 :
  • ( 1 + R 1 ) ² + 1 ² = ( 2 − R 1 ) ² \boxed{(1+R_1)²+1²=(2-R_1)²} < = > <=> R 1 = 1 3 \boxed{R_1=\frac{1}{3}} < = > <=> k 1 = 1 R 1 \boxed{k_1=\frac{1}{R_1}} < = > <=> k 1 = 3 \boxed{k_1=3}
  • Actually, we can repeat this process to calculate the radius and the curvature of the next circles using this pythagorean equality : 1 ² + ( R n + 1 + 2 ⋅ ∑ a = 1 n − 1 R a ) ² + = ( 2 − R n ) ² \boxed{1²+(R_n+1+2\cdot \sum _{a=1}^{n-1}R_a\:)²+=(2-R_n)²}
  • We get the following curvatures :
  • k 0 = 1 \boxed{k_0=1}
  • k 1 = 3 \boxed{k_1=3}
  • k 2 = 33 \boxed{k_2=33}
  • k 3 = 451 \boxed{k_3=451}
  • k 4 = 6273 \boxed{k_4=6273}
  • k 5 = 87363 \boxed{k_5=87363}
  • Using these results, we can write three equations :
  • k 3 = α k 2 + β k 1 + γ k 0 \boxed{k_3=αk_2+βk_1+γk_0} < = > <=> 451 = 33 α + 3 β + γ \boxed{451=33α+3β+γ}
  • k 4 = α k 3 + β k 2 + γ k 1 \boxed{k_4=αk_3+βk_2+γk_1} < = > <=> 6273 = 451 α + 33 β + 3 γ \boxed{6273=451α+33β+3γ}
  • k 5 = α k 4 + β k 3 + γ k 2 \boxed{k_5=αk_4+βk_3+γk_2} < = > <=> 87363 = 6273 α + 451 β + 33 γ \boxed{87363=6273α+451β+33γ}
  • Solving the system of equation gives the solutions : α = 15 \boxed{α=15} , β = − 15 \boxed{β=-15} , γ = 1 \boxed{γ=1}
  • Finally : α + β + γ = 15 − 15 + 1 = 1 \boxed{α+β+γ=15-15+1=1}
  • Bonus : I created this problem but couldn't find the general formula, after doing some research i found : Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences (very interesting!)
  • k n = ( 7 + 4 3 ) n + ( 7 − 4 3 ) n + 4 6 \boxed{k_n=\frac{\left(7+4\sqrt{3}\right)^n+\left(7-4\sqrt{3}\right)^n+4}{6}}
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