**From New Scientist #3241, 27th July 2019** [link] [link]

Seven coins have been placed in the “H” shape above. Altogether there are five lines of three, including the diagonals.

Your challenge is to place two more coins so that you can make 10 straight lines of three. No stacking of coins or other sneaky trick is required.

If you find a way to do this, give yourself a silver medal. If you find a second way to do it that isn’t a mirror image of the first, award yourself a gold.

[puzzle#14]

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My first thought was draw the lines that pass through exactly two of the coins.

Any coin placed on one of these lines will make a set of three collinear coins that we didn’t have before, and a coin placed at the intersection of two of the lines will create two new sets.

So if we place our two extra coins at intersections we will end up with 5 + 2 + 2 = 9 sets of collinear coins. And if we can place the two extra coins so that they form a collinear set with one of the existing coins we will have 10 sets in total.

This gives us two obvious scenarios:

(We can count the number of lines by counting the ends and dividing by 2 (each line having two ends)).

In the above I am looking for 10 different lines, where each passes through 3 (or more) coins. But if instead we think about looking for 10 different sets of 3 collinear coins, then we could just place 5 coins in a straight line. Any of the

C(5, 3) = 10subsets of 3 coins would be collinear, so we could just place the 2 extra coins on any of the existing lines in the initial diagram to give us more than 10 sets of 3 collinear coins. But maybe that is a sneaky trick.