5 circs in a hex - harder version

Geometry Level 5

The smallest regular hexagon that can hold five unit circles has a side length s s .

Enter the first 4 non-zero digits of s 3 |s-3| .

If s s is exactly 3, enter 0.

The answer is 6446.

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1 solution

Jeremy Galvagni
Nov 14, 2018

We seek the length A B = N K e t c . AB=NK etc. Call this length x x . The plan: find the lengths of the legs in terms of x x of right C G F \triangle CGF which has hypotenuse 2 2 . The Pythagorean Theorem will give an equation we can solve for x x .

There are many 30-60-90 right triangles. I can use the ratios to find many of the lengths in the picture.

D I = H J = E C = M F = 1 , I B = 3 , K J = K L = 3 / 3 , R C = P O = 2 3 / 3 , F D = 2 , P O = O M = x / 2 DI=HJ=EC=MF=1, IB=\sqrt{3}, KJ=KL=\sqrt{3}/3, RC=PO=2\sqrt{3}/3, FD=2, PO=OM=x/2 .

G M = 3 x / 2 GM=\sqrt{3}x/2 so G F = 3 x 2 1 GF=\frac{\sqrt{3}x}{2}-1

J I = H D = x 4 3 / 3 JI=HD=x-4\sqrt{3}/3 , F H D = 120 \angle FHD=120 , H F D = θ \angle HFD=\theta , F D H = 60 θ \angle FDH=60-\theta .

Solving F H D \triangle FHD as it is S s A SsA is a bear. I'll leave out most of the details. Using the law of sines to find θ \theta and then again will allow us to find an expression for F H FH : Simplified as much as possible F H = 2 3 x 2 16 + 3 x 2 x 2 + 2 3 3 FH=2\sqrt{\frac{-3x^{2}}{16}+\frac{\sqrt{3}x}{2}}-\frac{x}{2}+\frac{2\sqrt{3}}{3}

O N = x / 2 2 3 / 3 ON= x/2-2\sqrt{3}/3

O M = C G = 2 x 5 3 3 2 3 x 2 16 + 3 x 2 OM=CG=2x-\frac{5\sqrt{3}}{3}-2\sqrt{\frac{-3x^{2}}{16}+\frac{\sqrt{3}x}{2}}

Now for that Pythagorean theorem! I'll spare some more gory details. C G 2 + F G 2 = 2 2 CG^{2}+FG^{2}=2^{2} simplifies to

0 = 4 x 2 17 3 x 3 + 16 3 + ( 8 x + 20 3 3 ) 3 x 2 16 + 3 x 2 0=4x^2-\frac{17\sqrt{3}x}{3}+\frac{16}{3}+(-8x+\frac{20\sqrt{3}}{3})\sqrt{\frac{-3x^{2}}{16}+\frac{\sqrt{3}x}{2}} You can solve this numerically, but to try for an exact form let's get this to a polynomial.

Move the square root term to one side, square both sides, then simplify to the quartic polynomial equation:

63 x 4 219 3 x 3 + 729 x 2 286 3 x + 64 = 0 63x^{4}-219\sqrt{3}x^{3}+729x^{2}-286\sqrt{3}x+64=0

s s is the greatest of the four real solutions to this equation. Wolfram|Alpha can give the exact form, but they aren't pretty.

s 2.999355399 s \approx 2.999355399 so s 3 0.000644601 |s-3| \approx 0.000644601 and the first 4 non-zero digits are 6446 \boxed{6446}

awfully close to exactly 3

A Former Brilliant Member - 2 years, 6 months ago

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Indeed. I was surprised. Nice job solving it.

Jeremy Galvagni - 2 years, 6 months ago

For 6 circles the answer is exactly three, of course. It turns out removing a circle just doesn't save you much.

Jeremy Galvagni - 2 years, 6 months ago

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Well, right, that's the amazing thing about it....it shrinks the hexagon only just a tiny bit.

A Former Brilliant Member - 2 years, 6 months ago

For 6 circles the answer is NOT exactly three. I will write this as a new problem right now. It's around 3.15

Jeremy Galvagni - 2 years, 6 months ago

I guessed this configuration. But did not have patience to finish. What is the reason that this gives the smallest regular hexagon. Only guess is answer should be symmetrical.

Srikanth Tupurani - 2 years, 6 months ago

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It is very hard to prove that such packings are minimal. All I can tell you is many people have tried to improve this result.

Jeremy Galvagni - 2 years, 6 months ago

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It is frustrating that most of the questions of this type are difficult to prove. In most cases one common observation is the solution has some symmetry.

Srikanth Tupurani - 2 years, 6 months ago

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@Srikanth Tupurani In these kind of problems it is difficult to solve by using multivariable calculus. It becomes too complex. The hexagon with smallest possible length should be such that each circle touches one of the sides. Suppose we have a line segment AB which touches a circle of unit radius at X for what point X and what length AB we can draw another four circles of unit radius and five segments such these six segments form a hexagon and the five unit circles are inside it. and each circle touches one of the sides of the hexagon. If we answer this question there is a little hope. But it is not at all easy.

Srikanth Tupurani - 2 years, 6 months ago

Excellent problem and solution! I took the slightly different approach of fixing the side of the hexagon as 1 1 and using coordinate geometry (equations coming from the distances between centres of adjacent blue circles and from the tangency points between the circles and hexagons). That particular forest was also full of bears. Thanks for a fun problem.

Chris Lewis - 2 years, 6 months ago

Iliya Hristov - 6 months, 4 weeks ago

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