The Lemon Fool

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Arrange ten touching pennies in the familiar snooker formation shown. What is the smallest number of coins you must remove so that no equilateral triangle, of any size, will have its three corners marked by the centres of the pennies that remain? What is the remaining formation?

Cinelli

Far too easy for one of yours? Or have I completely misunderstood?
By simple inspection, you can do it by removing four coins: the top, the centre, and the two middle coins from the bottom row.

Can you remove fewer?
1. There are three non-overlapping triangles of the smallest size, comprising each vertex of the big triangle together with its neighbours. So you must remove at least one coin from each of those. That's three coins excluding the central one.
2. There are six smallest triangles including the central coin. To eliminate those, either the central coin or three of its neighbours must go.
3. At least one corner coin must go to kill the big triangle. So to satisfy (2) while removing fewer than 4 coins, we must remove the central coin.
4. But (1) already compels us to remove 3 coins excluding the centre coin.
5. So there is no solution removing fewer than four coins, and our simple solution is also optimal.

[edit] Hmmm, I don't draw, but maybe this time I can draw the result. The spurious dot is needed to stop the lemonfool editor collapsing whitespace and putting the top line there.

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UncleEbenezer wrote:Far too easy for one of yours? Or have I completely misunderstood?

Your solution is quite right, UncleEbenezer. I like to mix these puzzles up for all tastes and talents.

Cinelli