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# 3 for 1

## Question

The father told his son to make as many non-intersecting triangles with his eight coins.
And for every such triangle (3 centers) counted on the table he will pay 1 coin but take
all coins used to make those triangles . How many coins shall the boy win or lose?

## 10 answers to this question

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I believe if the boy makes a wheel with 7 nodes and a center point, the boy can claim 7 coins and lose three....

unless we claim that after selecting coins the remaining coins either placed by the father or the originals placed by the boy are reexamined for new triangles?/

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I believe if the boy makes a wheel with 7 nodes and a center point, the boy can claim 7 coins and lose three....

unless we claim that after selecting coins the remaining coins either placed by the father or the originals placed by the boy are reexamined for new triangles?/

No, the boy only lose a coin for making 7 and only the boy places all his coins

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k-man    26

The boy can make 13 triangles with 8 coins, so he can win 5 coins.

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The boy can make 13 triangles with 8 coins, so he can win 5 coins.

he can do much better than ten

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bonanova    76

The boy can make 13 triangles with 8 coins, so he can win 5 coins.

This calls for a distinction (is there one?) between non-intersecting and non-overlapping.

Hmm, yeah, I guess there is.

TSLF, care to comment?

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i was thinking that triangle intersection is overlapping of area.. line intersection is overlapping of point

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k-man    26

Well, if overlapping is not allowed then...

11 is the max

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Well, if overlapping is not allowed then...

11 is the max

Bingo! Flatlander solution gives 11.

o

o o o o o o

o

But he can do much better.

Remember the centers of identical coins are "above" the table.

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k-man    26

Well, if overlapping is not allowed then...

11 is the max

Bingo! Flatlander solution gives 11.

o

o o o o o o

o

But he can do much better.

Remember the centers of identical coins are "above" the table.

So, we're solving for flat triangles formed by coins in 3d space?

Lay 4 coins on the table forming a triangle with the 4th coin in the center. Stack the remaining 4 coins on top of the center coin. If I didn't miss anything then this produces 27 non-overlapping triangles.

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Good one!

A roof more but still can be improved because 3 for 1 means atleast winning 24 coin pays for 8 of his coins.
32 triangles is just what he made .

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