On Friday I posted this puzzle.

Find 5 identical coins. Can you arrange them so that every coin is touching every other coin? I will warn you now, it’s tricky!

If you have not tried to solve it, have a go now. For everyone else the answer is after the break.

OK, it is an old puzzle, and here is the answer….

Did you solve it? Any other answers?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called

**PUZZLED** and is available for the

**Kindle **(UK

here and USA

here) and on the

**iBookstore** (UK

here in the USA

here). You can try 101 of the puzzles for free

here.

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Give me a break! I’ve not had a wink of sleep since Friday wrestling with this and it was that simple.

I’d pretty much got there, although mine wasn’t as elegant, having to keep hold of the coins so that contact was retained.

I struggled for hours. Then the penny dropped.

I did it by taking arranging four coins vertically as if they were the four wheels of a car, with all pressed together so that there was one point at which all four met. Then I lowered the fifth coin vertically into the gap between the front and rear pairs so that it was touched by all of the others.

Colin

I would have thought that if the rear left “wheel” touches the front right then front left can not touch the rear right or vice versa.

The Masked Twit.

My solution assumed that the coins could be any disc so I was working with the presumption that each disc was perfect with equally perfectly sharp corners. With my theoretical precision laser-cut discs there would be one point at which they all touch.

I don’t think it is reasonable to assume the coins have zero thickness, so this doesn’t work.

But the top two coins aren’t touching the bottom two!,

Sorry- should have said the top two coins aren’t touching the bottom one.

But the bottoms of the top two coins do indeed touch the bottom two coins! You just have to angle them just right. Just try it with five coins.

you are correct. the gap is very tiny, but it is still a gap.

i put four in a circle and dropped one on top.

Doesn’t work. Each of the “four in a circle” cannot be touching all of the other three. Just try to do it.

confused….their all touching!

Each coin needs to be touching ALL the coins. Name them A, B, C, D and E. In a ‘circle’, before you place coin D on top, coins A and B touch, as do B and C, C and D, and D and A. The top coin, E, touches A, B, C and D. However, coins A and C, and B and D actually make no contact.

Thus, ALL of the coins are not touching.

That is supposed to say ‘before you place coin E on top’.

If the first four weren’t in a circle, per se, but arranged closer together so that each of those first four were resembling corners of a square, then add the fifth coin in the middle on top of the first four. All touch, so…? (Sorry, couldn’t do those arrangements with multiple coins on one or with angular placements.)

yes, miss chilli, four in a circle all touching. then one on top. this is the simplest arrangement.

With 4 in a circle and 1 on top, not every coin is touching every other coin. I think you misunderstand the question.

Dharmaruci, Miss Chili, No, it does not work. They four in the circle/square cannot all touch each other.

This solution was the nearest I could get, but couldn’t get the top two to touch the bottom one, no matter what angle I tried.

My solution (drawn but not actually verified with real coins) was different but from the solution given, I cannot see from the picture that the top two coins will be touching the bottom coin. Has anyone verified this by actually using coins ? I’ll have a go and with my ‘solution’ too.

The top two coins rest on the bottom coin and touch the edges of the two side coins – tried, tested and proven. What was your solution, out of curiosity?

I thought this was very tricky, which is why I was surprised by the various responses in the original posting declaring it was easy and solved within a matter of seconds.

They wither knew the puzzle already, or were showing off and actually hadn’t tried it at all 😉

either*

I also think quite a few folks didn’t realise that not all coins were touching each other in their solution, as illustrated in the discussion above.

My solution was elegant, simple, kinda genius and very very wrong. I can’t justify it no matter how hard I try. Grrrr. Won’t be so cocky next time.

I had three touching each other in a triangle, with one on top and one underneath.

Sorry; I’ve just looked at it and it doesn’t work. Silly me.

here we go….in the old days…

My Grandad kept us quiet at the tea table with this puzzle and the promise of winning the 5 coins (yup, old pennies but maybe 100 years younger than those in your pic) if we could solve it. I remember the delight of winning them, then the dissapointment of dividing up the prize between my brothers, and giving my Dad a penny for the tip-off.

In those days Dad was Google.

and all this used to be fields, this coffee’s gone cold, have you seen my keys?

My solution with 5x1p coins matched the real one but I was wondering whether the size and/or thickness of the coins made it easier or harder, comparing say old pennies with pound coins. I would imagine the thickness might be extremely significant (i don’t have the spare cash to check this)

answer is right,just have to glue coins to each other.

I got the answer Richard gives but I was hoping there would be a better one. I can’t make it work in real life using 2p or 5p coins. I’d love to see video of someone actually doing it.

I’ve discovered another solution. Hard to draw so I’ll try to describe it. Sandwich two coins touching at one edge between 2 more, off-centered enough so the top and bottom coins can form a sidways narrow V and thus touch each other as well as the first two coins. Now take the 5th coin and align it perpendicularly to the first four and carefully place it touching the other four coins at the top of the V. The only question is, does the 5th coin force the first two apart? If it does, the V can be widened slightly to allow the first two to be pushed together and touch again. If you’re careful, this can actually be done by holding all 5 coins in your hand (no table required).

I meant to say “in your hands”. I don’t think one hand would be sufficent. 😉

OK, I understand. That seems to work. Well done.

I think you’ve got it! It seems to work better than the official answer. Because, according to the diagram, there is no way that the top two coins touch the bottom one. And I can’t get it to work with real coins, either.

This seems to work with 2 EURO coins, I failed in reproducing the original answer with those

http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln005.html

You can do this with six coins!

Mathematically, two coins, considered to be perfect cylinders, can touch (intersect) at a single point, a line segment, or at a planar surface. Richard’s solution contains combinations of points and planar surfaces. Create an equilateral triangle on a table with three coins touching each other. They mutually intersect at a line segment. Place another identical equilateral triangle of three coins directly on top of the first. The coins on the top layer touch (intersect) the bottom layer coins either at a planar surface or at a single point. This is very easy to show mathematically if you express equations for intersecting cylinders.

You can always take one coin off the top if you only want five.

For the true physicists among you, it can be argued that no two objects can actually truly “touch” each other due to electron repulsion at the surfaces. See MinutePhysics: http://www.youtube.com/watch?v=BksyMWSygnc&list=PLED25F943F8D6081C&index=19

Jerry:

This depends on the the definition of “touch” as you point out. I think it’s reasonable to assume, for purposes of this puzzle, that if four coins meet at a single point (and only there), then *either* of the 2 diagonally opposite pairs coins can be considered to touch, but not *both*. With that assumption, your solution wouldn’t work.

For the true Buddhists among you, you can claim that all is Maya and that the coins don’t exist anyway.

You can certainly make 6 cylinders all touch each other in a colloqual sense if they are pencil-shaped cylinders rather than coin-shaped cylinders. Place two cylinders on the table in a V-shape and put a third in between flat on the table so it touches both others. Then repeat the pattern but place it on top and rotated by a right angle. however I can’t see how to make this pattern work with discs rather than rods.

Ken:

In my scenario of three coins (perfect cylinders) each of the three top (or bottom) coins touches (intersects) two coins in the opposite layer. This allows three coins to touch (intersect) at a point.

If you had eight perfect, identical thin square “coins” stacked as two sets of four squares, all eight would touch (intersect) at a single point in the center.

Jerry:

I look at this like the 4-color mapping problem. Two regions that share only a single point of contact are not considered to be “touching”.

In this case, consider the coins to be slightly soft, so that “touching” means if you pushed them together even more, the area of contact would increase. This can’t happen with your six-coin solution, but it can with the 5-coin answers that Richard and others have submitted.

Ivan:

Interesting. I wonder what the minimum cylinder height (with respect to the diameter) would need to be for your answer to work.

My solution was to melt them all down together. A bit tricky though with a melting point of ~1400 degrees C.

very close to my initial thought: stack the 5 coins and apply sufficient pressure

i can not solve it

I think the picture shown is incorrect because it seems to clearly show that the top two coins do not go through the middle two coins to touch the bottom coin. In other words, the top two coins leaning against each other do not touch the bottom coin as shown in the picture.

But, with a slight variation, I think I have a solution.

Remove the top two coins and replace them as follows: Notice the small area of the bottom coin where it is visible from above on the left and right side of it. Take the top two coins and turn them sideways/vertical like a stop sign so that one of the sides is facing you and put one at 9 o’clock on top of the left bottom coin that is viewable and put the other coin at 3 o’clock on top of the right bottom coin that is viewable so that both of these coins are touching in the middle above where they are touch the middle two coins. So, in my configuration, the bottom three coins will be exactly as shown in the

given diagram (lying flat like a pancake) but the top two coins will be vertical like a stop sign coming in from opposite sides (at 3 o’clock and 9 o’clock) and each and every coin will touch the other four.

Steven Prizant

Anyone know how we can get some pictures on this discussion ? I can’t visualize from this description unfortunately.

perfect for me!

The answer drawing is wrong as shown. Plain and simple, if you put the two top coins on top of the middle two, they can not touch the bottom coin unless someone hypnotizes to believe such Hhaa.. And that is what is shown. But, it looks like it would work if you place the top two coins the same way angled against each other but the bottom of the top left coin must be placed so it lies on the bottom coin on the left exposed side and the top right coin must be placed so it lies on the bottom coin on the right exposed side angled as shown in the picture so the top two coins touch each other as show. In other words, place the top two coins as shown but spread out the bottom of the two coins so they are placed on the bottom coin.

Steven Prizant

Have you tried this?! I have, and it’s impossible to get the two top coins into that position without them falling over. Perhaps with Superglue?

Anita, try using wide enough and thin enough coins that attract to a magnet.

Good idea about the magnet! But is that cheating…?

Reply to Anita Dunne:

I couldn’t get the top pennies in my post at 5:41 pm to stay in position correctly without holding them, but the problem

implies or leaves open that it is searching for the position of the 5 coins that would work if they were positioned as such. So it works theoretically if you hold them or glue them even though the top two don’t balance easily by themselves.

In other words, if there was a problem that asked

how high would 1000 Honda Accords be stacked one on top of the other and so on.

It doesn’t mean you have to actually do it – it just has to be do-able. And it is do-able if you hold the top two pennies in position. I think it is possible to balance the top two pennies in my position/solution per my post at 5:32 pm although I held them also.

OK, fair enough!

How Do The 2 On Top Touch Every Other Coin? Especially The Bottom Coin?

If You Were To Put The Bottom Coin On Top Of The 2 That It Is Under And Hold The Top Two In The Angled Position, Then Every Coin Touches Every Other Coin.

I can’t tell what most of you are on about.

A picture is worth a thousand words.

Richard’s answer works fine for me.

Something doesn’t seem right here.

Each one of the angled coins is touching all other four coins in the same plane (same side of the angled coin). but you MUST have three points to define a plane, not more & not less. so unless these coins are designed in a really specific geometry – the above solution fails.

Lots of great comments and challenges about Richard’s solution.

I’d suggest that the coins would have to have a suitable diameter to thickness ratio to enable the too two to be able to touch the bottom coin the edges of the next two and each other.

That is to say £2 coins may not work, owing to their thickness while five new pence may not because of their diameter.

I was just about to leave a comment about this too. The key to whether it works or not is not if they stack (like a house of cards) and don’t fall over but the thickness of the coins. If the coins were ‘ideal’ discs with no thickness (as in a line drawing) then no problem but real coins aren’t. There must be maximum thickness (compared to the coin diameter) that allows contact with all coins and anything above that thickness doesn’t work. I’m trying to work this out at the moment.

The solution given by Richard is virtually identical to the solution which I thought of on Friday (after about 20 minutes of thinking), except that I had the two top coins at a more shallow angle – i.e. with the “feet” of the top two coins at the edges of the bottom coin, rather than closer together.

On Friday I tried doing it with five £1 coins (because they were the only coins of which i had 5 of the same) but they were too fiddly to work with, and I wasn’t able to hold them still enough to check whether all the connections were correct.

Then, on Sunday, I remembered that i had a stash of old pennies in a drawer, so i used them, and I quickly found that the solution worked as shown in the diagram.

In other words, it’s much easier with large thin coins than small fat ones.

By the way, the other solution which I thought of was based on using five identical chocolate coins, and letting them melt so that the top one drooped down to touch the bottom one as well as the three middle ones. I wasn’t sure if that was going to count as cheating because the shape of the coin was being altered.

Yeah i did it differently – one central coin, upright in the centre, and then the other four forming a diamond formation around it. I’m actually surprised it wasn’t THE answer!

The four coins in the diamond don’t all touch each other. At least one pair of opposite-side coins will not be touching.

That solution doesn’t seem right. Got a picture as proof ?

Nice one! By the way, I don’t know if you’ve seen this, but it has been buzzing around the internet for a while. It’s quite clever.

http://www.theatlanticcities.com/neighborhoods/2013/06/every-library-and-museum-america-mapped/5826/

Put mine into a blast furnace and melted them into one larger coin.

Well, I was quite smug when, after working quite a bit, I was sure I had it. When I showed my husband a few hours later, though, he informed me that two were NOT touching, and he was SO right. We both messed with it, but came up dry. Now I can show him the answer. 🙂 this one was quite tough!

I think I have another solution but I’m not sure if anyone has stated it already. Take the solution as given with the picture at the start of this discussion. Remove the top 2 coins that are lent against each other. Stand each coin on its end but touching each other at a point (or line as the coins have thickness) i.e. end to end – much like the bottom 2 coins but in a vertical plane instead. Now place these 2 (joined) coins on top of the bottom coin to complete the arrangement. The top 2 coins touch each other, the bottom coin and the horizontal (flat) 2 coins. The bottom coin and middle 2 coins touch surfaces but the top 2 coins (vertical) touch at points (or line as the coins have thickness).

That’s a much better solution than Richard Wiseman’s!

Hat dies auf Pfeffermatz rebloggt und kommentierte:

Wie kann man fünf gleiche Münzen so arrangieren, dass jede Münze jede andere berührt? Dies ist eine Aufgabe, die in der Regel als Trockenübung (also ohne Münzen) sehr schwer zu lösen ist (ich habe zwei Tage erfolglos darüber nachgedacht). Aber durch praktisches Ausprobieren finden die meisten Leute innerhalb weniger Sekunden eine Lösung (galt auch für mich :)). Wie ich den Kommentaren zum Rätsel (hier habe ich gleich die Lösung gepostet) geht es vielen Knoblern bei diesem Rätsel genau so.

Hier ist also eine Lösung , wobei es auch weitere gibt.

Can someone translate please ?

What’s the problem Steve?

I found Pfeffermatz’ post as easy to understand as a lot of the other posts on this site elaborating alternative solutions – i.e. not understandable at all.

I reblogged this post on my German blog site. Obviously, WordPress automatically posts my German explanation as a comment here.

Answer to the Friday Puzzle! « Richard Wiseman, me ha parecido muy ameno, me hubiera gustado que fuese más extenso pero ya saeis si lo bueno es breve es dos veces bueno. Enhorabuena por vuestra web. Besotes.

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