The Twelve-Coin Problem

Rounds of silver destined to be coins, at the Old Glory Mint in Utah.
Our challenges this week were suggested by Numberplay regular Stephan Peers, an investment banker from Lafayette, Calif. The first is a authoritative we ran in exchangeable imprint years ago, and the second is more strange and a lot of fun. I asked Mr. Peers where he found the two problems .

“ My Dad proposed the mint puzzle when I was, like, 10 years old. After working on it for weeks, I gave up and asked him for the answer. ‘ Sorry, ’ he said, ‘ I good know the problem, not the solution. ’ Thanks, Dad. I figured it out in my mid-20s. I hope it wasn ’ t the alone result of my graduate education. ”

And the second trouble ?

“ Same Dad publish : bang-up problem, no solution. I guess he got a kick out of things like that. He posed the problem and I could never figure it out. I think I was reading Mario Livio ’ s book The Golden Ratio. Somewhere else was a pass statement on the fact that the ratio of the long arms on a star to the base was the golden ratio. I put the two together and was thrilled. sol 40 years after, I told my Dad that I solved the trouble. His reaction : ‘ I have no idea what you are talking about. ‘ ”

Let ’ s give these a try. We ’ ll start by leaning into —
The Twelve-Coin Problem

You have 12 coins that all look precisely the same. One is counterfeit and is either heavier or lighter than the other 11. With a remainder glow scale, isolate the counterfeit coin in three moves .

Our second challenge involves a act of geometry. To share a movie of your solution, equitable upload to an image-hosting site ( like imgr ) and include the link in your remark. Gary Hewitt ’ south Enhancer ( see note at the buttocks of this mail ) will make the image appear in your post .
Let ’ s nowadays attempt :
The Perfect Pentagon

manufacture a arrant pentagon with a grok and a ruler .

That ’ s it for this week. As always, once you ’ re able to read comments for this post, use Gary Hewitt ’ s Enhancer to correctly view formulas and those aplomb pentagon diagrams. And send your favorite family puzzles to gary.antonick @ .
Let ’ s begin with the arrant pentagon. here ’ second Mr. Peers :

The Pentagon Problem gave me trouble for years. At some distributor point I learned that the ratio of the the long side to the shortstop slope of a star is in the Golden Ratio .


Of course it wasn ’ triiodothyronine for years that I found out what that intend. not til I read Mario Livio ’ s reserve The Golden Ratio .


Great stuff that quadratic equation solution.

See below for the structure. I assume you know how to drop perpendiculars and bisect lines .


There were some great answers, but they all seem a little harder than my solution, which requires only two pieces of cognition .
Jan ’ s carbon monoxide ( 72 ) equals extremist 5 minus 1 over 2 was great, but who knows that without looking something up ?
Thanks, Ravi, for knowing that the solidus of a pentagon is the Golden Ratio. That ’ s truly the same as mine .
I once saw a solution that not lone had the pentagon, but you started with a unit circle and the pentagon was inscribed in the circle, or is it the other room around ?
As for LAN and Gauss-Wanzel : I don ’ thymine think a square ( 4-gon ) is the product of a Fermat prime and a exponent of two. I ’ molarity sure I have a misinterpretation here .
Thanks for all the comments. fabulously concern .

And thank you, Mr. Peers. Our second challenge this workweek was the authoritative Twelve Coin Problem. Readers weighed in with a kind of solutions, including this by Andy, which identified the counterfeit coin — the purpose of the original problem — and besides determined whether the juke was heavy or light :

Number the coins 1 through 12 .
first weighing : 1,2,3,4 five 5,6,7,8 .
If peer, weigh 9,10,11 vanadium 1,2,3 ( not counterfeit ) .
If equal, 12 is the forge and weigh it against any early coin to determine if it ’ mho heavy or light. If not equal, the commission of 9,10,11 will determine heavy or light. The third deliberation is 9v10. If equal, 11 is counterfeit. If not, which always of 9 or 10 went the lapp direction as in the second weigh is counterfeit ( and you ’ ll know heavy or sparkle from the second weigh ) .
If 1,2,3,4 vanadium 5,6,7,8 is not equal, target which means each side moved. Let ’ s simulate 1,2,3,4 went toss off and 5,6,7,8 went up. The moment count is 1,2,5 five 3,4,6 where three of the coins change sides. If equal either 7 or 8 is counterfeit thus count them against each other and which evening one goes up is the counterfeit and it is unhorse .
If 1,2,5 v 3,4,6 is not peer then three of the coins changed direction ( since they ’ re on different sides of the symmetry scale for this weigh ). Those three are obviously not counterfeit since the counterfeit will always cause the scale to move the same way. Let ’ s assume that 1,2,5 went gloomy again. That means that either 1 or 2 is counterfeit and it is heavy. weigh 1 five 2 and which always one goes down again is the forge coin .
I first read this problem in a book of brusque stories by Ethan Canin called “ The Palace Thief. ” This was in the second floor.

Andy continued :

I have to add something to my answer. On the moment consider of 1,2,5 vanadium 3,4,6, if 1,2,5 goes down again, then either 1 or 2 is the counterfeit and it is heavy OR 6 is the counterfeit and it is light. so, if on the third weighing 1v2 is adequate, then 6 is the forge and it is light. This is now the complete answer to the 12 coin trouble .

Thank you, Andy, and giant thanks to Stephan Peers for this week ’ sulfur challenges. Thank you a well to everyone who participated in our discussion : Technic Ally, John W., Donald Quixote, Ramona D ’ Souza, D-Ferg, Jan, James, Tom, Ricardo Ech, Winston, Ravi, miami lawyer ma, Tim Lewis, LAN, Dave McRae, Allaisa, 2E, Pummy Kalsi, Jim, Golden Dragon, Sam, Hans, Andy, Dr W, Doc and mora nehama .

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