Quick and cute summer Sangaku #4

Geometry Level 4

The diagram shows a square. Along the square's diagonal, we inscribe 3 circles so their radius are in arithmetic progression.

  • α \boxed{α} is the side of the square
  • k \boxed{k} is the common difference of successive terms with 0 < k < β \boxed{0<k<β}
  • Question : evaluate β α \boxed{\frac{β}{α}}
  • The answer can be express has a + b a \boxed{\frac{\sqrt{a}+b}{a}} where a \boxed{a} and b \boxed{b} are integers
  • Find a − b \boxed{a-b}


The answer is 3.

Valentin Duringer by Valentin Duringer , 30, Belgium

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2 solutions

Let the radius of the circle in the middle, which is described by the "Arithmetic Mean" be r r , that of the largest circle be r + b r+b and of the smallest circle be r − b r-b .

Then 2 ( r + b ) + r + b + 2 r + r − b + 2 ( r − b ) = α 2 \sqrt 2 (r+b)+r+b+2r+r-b+\sqrt 2 (r-b) =α\sqrt 2

⟹ β α = r α = 1 2 ( 2 + 1 ) \implies \dfrac {β}{α}=\dfrac {r}{α}=\dfrac {1}{2(\sqrt 2 +1)}

= 2 − 1 2 ⟹ a = 2 , b = − 1 =\dfrac {\sqrt 2 -1}{2}\implies a=2,b=-1

Hence a − b = 3 a-b=\boxed 3 .

3 Helpful 4 Interesting 2 Brilliant 0 Confused

Hello there ! If you are motivated you can try this problem and post your solution : https://brilliant.org/problems/revenge-of-the-deceiving-sangaku/?ref_id=1594858

Valentin Duringer - 11 months ago

I had the same solution, somewhere midway I got this wrong

A Former Brilliant Member - 9 months, 3 weeks ago
Valentin Duringer
Jul 11, 2020

Since the the three circles are in arithmetic progression we can a very useful equation using this diagram, x \boxed{x} is the radius of the smallest circle - The sum of the following distances are equal to α 2 \boxed{α\sqrt{2}} , we can write the equation :

  • ( x 2 + x ) + ( 2 x + 2 k ) + ( x + 2 k + 2 ( x + 2 k ) ) = α 2 \boxed{(x\sqrt{2}+x)+(2x+2k)+(x+2k+\sqrt{2}(x+2k))=α\sqrt{2}}
  • We do some algebra and express x \boxed{x} in terms of k \boxed{k}
  • x = α 2 − k ( 4 + 2 2 ) 4 + 2 2 \boxed{x=\frac{α\sqrt{2}-k(4+2\sqrt{2})}{4+2\sqrt{2}}}
  • Since x > 0 \boxed{x>0} we can solve the inequation : α 2 − k ( 4 + 2 2 > 0 \boxed{α\sqrt{2}-k(4+2\sqrt{2}>0}
  • k < α 2 2 + 2 \boxed{k<\frac{α}{2\sqrt{2}+2}}
  • β α = 1 2 2 + 2 = 2 − 1 2 \boxed{\frac{β}{α}=\frac{1}{2\sqrt{2}+2}=\frac{\sqrt{2}-1}{2}}
  • a − b = 2 + 1 = 3 \boxed{a-b=2+1=3}
3 Helpful 1 Interesting 3 Brilliant 1 Confused

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