Apollonian Gasket With Some Numbers

Geometry Level 5

A number of tangent circles are inscribed inside a blue circle as follows:

The numbers 42 , 66 , 87 42, 66, 87 inside some of them are inverses of their radii, so that their radii are actually 1 42 , 1 66 , 1 87 \dfrac{1}{42}, \dfrac{1}{66}, \dfrac{1}{87} .

What's the inverse of the radius of the blue circle?


The answer is 13.

Michael Mendrin by Michael Mendrin , 70, USA

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2 solutions

Patrick Corn
Dec 26, 2019

We can ignore the smaller circles. Let A , B , C A,B,C be the curvatures of the largest three circles, going counterclockwise from the top (so that A ≤ B ≤ C A \le B \le C ). Let K K be the curvature of the external circle; this will be negative since it is externally tangent to the rest of the circles.

By Descartes' circle theorem we get the four equations ( A + B + C + K ) 2 = 2 ( A 2 + B 2 + C 2 + K 2 ) ( A + B + 42 + K ) 2 = 2 ( A 2 + B 2 + 4 2 2 + K 2 ) ( A + 66 + C + K ) 2 = 2 ( A 2 + 6 6 2 + C 2 + K 2 ) ( 87 + B + C + K ) 2 = 2 ( 8 7 2 + B 2 + C 2 + K 2 ) \begin{aligned} (A+B+C+K)^2 &= 2(A^2+B^2+C^2+K^2) \\ (A+B+42+K)^2 &= 2(A^2+B^2+42^2+K^2) \\ (A+66+C+K)^2 &= 2(A^2+66^2+C^2+K^2) \\ (87+B+C+K)^2 &= 2(87^2+B^2+C^2+K^2) \\ \end{aligned} Subtracting the second equation from the first and simplifying gives ( C − 42 ) ( C + 42 + 2 A + 2 B + 2 K ) = 2 ( C 2 − 4 2 2 ) 2 A + 2 B + 2 K = C + 42 2 A + 2 B + 2 C + 2 K = 3 C + 42 \begin{aligned} (C-42)(C+42+2A+2B+2K) &= 2(C^2-42^2) \\ 2A+2B+2K &= C+42 \\ 2A+2B+2C+2K &= 3C+42 \end{aligned} Similarly subtracting the third and fourth equations from the first yields 2 A + 2 B + 2 C + 2 K = 3 C + 42 = 3 B + 66 = 3 A + 87. 2A+2B+2C+2K = 3C+42 = 3B+66 = 3A+87. So B = A + 7 , C = A + 15 , K = 43 − 3 A 2 . B = A+7, C= A+15, K = \frac{43-3A}2. Plugging these into the top equation and solving for the resulting quadratic gives − 33 4 A 2 + 343 2 A + 1679 4 = 0 -\frac{33}4 A^2 + \frac{343}2 A + \frac{1679}4 = 0 which has roots A = 23 , − 73 33 . A = 23, -\frac{73}{33}. Since A , B , C A,B,C must be positive, we must have A = 23 , B = 30 , C = 38 , K = − 13 , A=23, B=30, C=38, K = -13, and the reciprocal of the radius of the external circle is − K = 13 . -K = \fbox{13}.

3 Helpful 2 Interesting 2 Brilliant 0 Confused

I'll be damned. I was struggling to do this for over 5 hours.

Pi Han Goh - 1 year, 5 months ago

First, I picked out an integral Apollonian gasket starting with − 13 , 23 , 30 , 38 -13, 23, 30, 38 , generated some more tangent circles, and then picked 3 of them that would allow a complete solution using just four equations. Exactly how you've done it here. Well done.

If any four mutually tangent circles have integer curvatures, then all the rest have integer curvatures. This is because if

k 4 = k 1 + k 2 + k 3 − 2 k 1 k 2 + k 2 k 3 + k 3 k 1 k_4 = k_1+k_2+k_3 - 2\sqrt{ k_1 k_2+k_2 k_3+k_3 k_1}

is an integer, then

k 5 = k 1 + k 2 + k 3 + 2 k 1 k 2 + k 2 k 3 + k 3 k 1 k_5 = k_1+k_2+k_3 + 2\sqrt{ k_1 k_2+k_2 k_3+k_3 k_1}

is also an integer.

Michael Mendrin - 1 year, 5 months ago
Michael Mendrin
Dec 26, 2019

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Is the answer unique?

Pi Han Goh - 1 year, 5 months ago

I can't find another solution, but if we were to assume that all the radii of these circles are unit fractions, then I can prove that the leftmost inscribed circle must have a radius, 1 / r 1/r must satisfying r < 3 2 ( 61 + 1649 ) r < \frac32 (61 + \sqrt{1649}) . What's left is to prove by inspection that this is the only solution.

Here's my failed attempt if we don't make such an assumption:

Label the radii as shown above.

Using Descartes Kissing Circle theorem, we can form equations like

( e + f + 66 − R ) 2 = 2 ( e 2 + f 2 + 6 6 2 + R 2 ) ( e + b + 42 − R ) 2 = 2 ( e 2 + b 2 + 4 2 2 + R 2 ) ⋮ (e +f + 66 - R)^2 = 2(e^2 + f^2 + 66^2 + R^2) \\ (e + b + 42 - R)^2 = 2(e^2 + b^2 + 42^2 + R^2 ) \\ \vdots

If we subtract the two equations whose triplet of radii are ( e , b , 42 ) (e,b,42) and ( e , d , 42 ) (e,d,42) , we get 2 R = 2 e − b + d + 84 2R = 2e - b + d + 84

If we subtract the two equations whose triplet of radii are ( e , f , 66 ) (e,f,66) and ( e , d , f ) (e,d,f) , we get 2 R = 2 e − f − g + 132 2R = 2e - f - g + 132

If we subtract the two equations whose triplet of radii are ( e , d , f ) (e,d,f) and ( 87 , d , f ) (87,d,f) , we get 2 R = 2 d + 2 f − e − 87 2R =2d + 2f - e- 87

If we subtract the two equations whose triplet of radii are ( e , b , 42 ) (e,b,42) and ( e , b , c ) (e,b,c) , we get 2 R = 2 e + 2 b − 42 − c 2R =2e + 2b - 42 -c

If we subtract the two equations whose triplet of radii are ( e , d , 42 ) (e,d,42) and ( a , d , 42 ) (a,d,42) , we get 2 R = 2 d + 84 − e − a 2R =2d + 84 - e - a

Equating these 5 equations together shows that ( d , e , f , g ) = ( − 3 b + c + 126 , − a − 8 b + 3 c + 378 3 , 171 − a 2 , a − 4 b + 2 c + 177 2 ) (d,e,f,g) = \left(-3b + c + 126, \dfrac{-a-8b+3c+378}3 , \dfrac{171-a}2, \dfrac{a-4b+2c+177}2 \right) .

Notice that if the Descartes kissing circle equation formed with radii ( 42 , d , a ) (42,d,a) or with radii ( 42 , d , e ) (42,d,e) is the same, thus a a and e e must be the root (in x x ) of the equation

( 42 + d + x − R ) 2 = 2 ( 4 2 2 + d 2 + x 2 + R 2 ) (42 + d + x - R)^2 = 2(42^2 + d^2 + x^2 + R^2)

Solving for x x gives two values a a and e e . By summing these two roots, we can then set R R as the subject, R = 1 6 ( − 2 a + 10 b + 3 c + 630 ) R = \frac16(-2a + 10b + 3c + 630 ) .

That's where I'm stuck.

Pi Han Goh - 1 year, 5 months ago

It's unique, Pi Han Goh.

Michael Mendrin - 1 year, 5 months ago

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