Johnny was given 16 coins by his older, somewhat meaner brother, Mark. He told him that he could keep them all if he could place all 16 on the table in such a way that they formed 15 rows with 4 coins in each row.
After 10 minutes, Johnny walked away with the coins and Mark, after complaining futilely to his mother, left with nothing.
How did Johnny place the coins?
Look below for a hint.
Hint: Stars and pentagons
I'm posting one puzzle, riddle, math, or statistical problem a day. Try to answer each one and post your answers in the comments section. I'll post the answer the next day. Even if you have the same answer as someone else, feel free to put up your answer, too!
Friday, September 29, 2006
You think you can figure this one out?
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First, draw a large pentagram, and put dots at all the points of the star and all the intersections between lines. Then, use the five bisection points on the hexagon that is formed in the center of that pentagram to make a new, smaller pentagram with each of the five points being the end of one point on the star. Make dots at the bisection points as usual. Then, put a dot in the center of the whole thing. Coins go where dots are, 15 rows of 4 are formed.ReplyDelete
I am having trouble with Abe's solution. To make things clearer, I put the pentagon on the outside of the star, then placed dots at the mid-point of each side.ReplyDelete
My issue is that I still only come up with 10 rows (see the 10-tree solution: http://statspuzzles.blogspot.com/2006/09/government-subsidies.html)
Where do the third set of five rows come from?
Never mind, I got it. The smaller, internal star is an inversion of the orginal star. (Duh!)ReplyDelete
Good answer... I didn't even think of one i am so braindead today.ReplyDelete
That's it... I liked this extension of the previous problem (see mr dons link).ReplyDelete
If you draw a 5-pointed star with all sides of equal length, you will create a pentagon in the middle with all 5 sides of equal length. Then draw another 5 pointed star, upside-down, using the 5 points of the interior pentagon as the points of the inner star, This will give you another 5-sided pentagon in the interior of the second, smaller star.
Now, take your 16 coins and place 5 on the outside points of the outside star. Then place 5 more on the points of each of the two pentagons you have created. Finally, place the last coin in the dead center of this drawing.
The lines are as follows:
5 lines for the actual drawing of the outside star
5 lines for the actual drawing of the inside star
5 lines that start from any outside point of the outside star and go to the opposite point of the inside star, passing through 1 point of the inner pentagon, the dead center coin on the way.
Total lines: 15