### Brain Teasers

# Coin Circle

Place any circular coin down flat on a table. Can you do a simple calculation to estimate to the nearest whole number of coins the maximum number of coins of the same size that will fit in a concentric circle around the center coin? Each surrounding coin must touch the center coin and also lie flat on the table

### Hint

Use the formula Circumference = pi x diameter.### Answer

SixSince the surrounding coins will touch approximately at their diameters (d) we are looking for how many diameters of this size coin there are at the circumference (C) where the coins touch. This circumference falls at approximately 2 x the radius (r) of the coins from the center of the circle. The diameter of this circumference would be approximately 4 x r or 2 d. The circumference would then be pi (3.14) x 2 d = 6.28 d. Since each diameter of a coin represents one coin there would be approximately 6.28 coins. The coins would only touch at their widest part or diameter if they were laid out in a straight line. Since they are in a circular pattern they actually touch slightly below their widest diameters so the actual diameter of the circumference is slightly less than 2d. Thus we know that there would be slightly fewer than 6.28 coins and the closest whole number estimate is six.

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## Comments

But how do you know that the closest whole number estimate is six? We know that the coins touch each other at a point below the true diameter of each individual coin. While this means that

that the circumfrence is less than 2d or 6.28, it does not follow that 6 coins is the best estimate. It may very well be the case that the circumrence is 5.99, in which case 5 coins would be the maximum.

that the circumfrence is less than 2d or 6.28, it does not follow that 6 coins is the best estimate. It may very well be the case that the circumrence is 5.99, in which case 5 coins would be the maximum.

yirro has a good idea but the method used to solve the problem is not the best proof. The best way to prove this is to ask what geometrical shape can i create that has nodes placed exactly the same length apart as they are from the center. The best configuration is one with 12 right triangles, the hypotenuse should be twice the shortest base length. The only shape that has this geometrical design is a hexagon. Six sides, six nodes and made up of 12, 30-60-90 triangles. QED

Actually, if all the coins are the same size, we could fit exactly six coins around the central coin without leaving any gap. To see this, consider three coins touching each other. Their centres form an equilateral triangle. Hence the angle at the central coin is 60 degrees. So in 360 degrees we could fit six coins exactly.

I got it, but did not use math. I recalled from other areas that the most economical form is a hexagon, which is why bees do what they do and then recalled that this is what happens when you take several straws and evenly mash them together. They become hexagonal. This is also why many pencils were originally hexagonal. They could get more pencils when the rounds were removed from the rounded pencil shape. All this pointed me to the answer. Good one, nonetheless.

I thought EVERONE would have placed several coins of the same value in a circle - I certainly did as a child! I don't know how you can't know 6 coins fit round the middle coin!

Agree with krishna. and would also say: In terms of r, the centers of two coins will be 2r apart. All encircling coins would have centers on a circle of radius 2r. Starman's circle of touching points has a radius of SQR(3)r, which does not "exactly" make for "easy calculation."

Yes I can!!! (He/she only ever asks if you "can" do this, I assumed there was a way, but i was just to lazy to figure it out)

I'm with boodler on this one - didn't need any complicated mathmatical explainations, just life experience!!

boodler's answer is the easiest if you have enough coins in your pocket right now. But if you want to show the math, I'll go with krishnan's answer.

I didn't find the answer given to be a simple calculation. I didn't need to estimate or calculate since a hexagon is the one with the most sides (and hence tightest packing of inscribed circles) of the three polygons that can be tiled using only a single shape. Each regular hexagon in the tiling circumscribes a circle.

Forced to prove this knowlege, I suppose I'd use similar math given in the answer.

Forced to prove this knowlege, I suppose I'd use similar math given in the answer.

The implication in the teaser (and the given answer) that it is an approximation is incorrect.

Krishnan has the solution perfectly correct. It is exactly 6 for the reasons given - well done Krishnan!

Krishnan has the solution perfectly correct. It is exactly 6 for the reasons given - well done Krishnan!

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