如图所示，15 个美分排成一个正三角形， 有一些是正面，有些是反面。证明无论如何放，必有一个正三角形的顶点上都是正面（或反面）。 |
Still no solution at WXC yet. It is not very difficult, and I give it a difficulty of ++. |
use the symmetry of the problem and reduce it. let F, T denotes the two sides of the coin. We use the following system: (x,y) means x row, yth coin counting from the left. WLOG, consider T(3,2), F(4,2), F(4,3), so T(5,3). Now we must have T(4,1) and T(4,4). So we have F(1,1). One of (3,1), (3,3) is F. Let's have F(3,1), so T(3,3). then F(2,2), F(5,5), T(5,1), and F(5,2). Now, what to put on (2,2)? |
I believe the original problem requires the equilateral triangle has sides parallel to the original triangle. If so, your argument needs to go a little further. |
Start with F(3,2), T(4,2), T(4,3), then F(4,1), F(4,4). So we have T(1,1). One of (3,1), (3,3) is F. Let's have F(3,1), then T(3,3). So T(2,1), F(2,2). Now only the last row is undetermined. If T(5,1), then F(5,5). How to put (5,4)? If F(5,1), then T(5,2), How to put (5,3)? |
Prove it by contradiction Assume there is no such triangle. WLOG, we may assume T(3,2), F(4,2), F(4,3) (Same notation as sean9991's). Then T(5,3) => F(4,4) => T(2,2) => F(2,1) => T(4,1) Thus T(3,2), T(4,1), T(5,3) is a such triangle, contradiction.
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Both your proof and Sean's proof still rely on a sideway equilateral triangle ((3,2),(5,3),(4,4)). This is not needed, ie we will always have an equilateral triangle of the same face with its sides parallel to the original triangle's. The proof can go this way: Because T(3,2), we will have either F(2,1) & F(2,2) or F(2,1) & T(2,2). Then F(2,1) & F(2,2) => T(1,1) & T(4,1) & T(4,4) F(2,1) & T(2,2) => T(4,1) & F(3,3) => T(4,4) => F(1,1) => T(3,1) => F(5,1) => T(5,4) & T(5,5) If sideway can be used, we do not need 15 pennies. 10, or even the 7 in the middle, can produce a contradiction. |
But the question doesn't require that , does it? |
in your standard, right? Thanks for your good proof. (I forgot to say, ) |
because you need to think about side way. The reason I think it should have the extra condition is that otherwise it does not need 15 pennies. |
It may look lengthy, but I guess it did not take that guy much time, because it looks so smooth. You guys have given very nice proofs. Many people can do it, but only a few can do it beautifully. |
Since I did not notice fzy's remark after Sean9991's 1st proof (Actually I opened, but I thought it just need be an equilateral triangle,) I even did not read his 2nd proof carefully, I found mine was a little shorter than his, so I posted. Sorry for wasting some people's time. |
There are 15 pennies forming an equilateral triangle. Prove that it exists an equilateral triangle whose vertices have the same face. The latter triangle may have 15 coins or fewer whose sides are parallel to the big triangle. The 15 coins is arranged like the picture below: x x x x x x x x x x x x x x x Proof: We will prove it by contradiction. Lets try to arrange the 15 coins to avoid forming a triangle with same face vertices. There is a pattern we need to avoid. We cannot have hhh or ttt. If we had it, we would get h t t h h h For the big triangle, the 3 vertices cannot be the same. WLOG (without loss of generality), we can assume they are one h and 2 ts: h x x x x x x x x x t x x x t Let’s look at xx below h. They can only be tt, ht or th. Because of symmetry, ht and th are of the same situation. So, we only need to analyze tt and th. 1. tt h t t x x x => x x x x t x x x t h t t x h x x x x x t h x h t and we have a triangle with same faced vertices (3 hs). Therefore, it cannot be tt. 2. th h t h (no hhh) x x x => x x x x t x x x t h t h x x t => x x x x t x h h t h t h x x t h x t h t t h h t Again we have a triangle with same faced vertices (3 hs). Therefore, no matter how to arrange the coins, there must be a triangle with the same faced vertices. |
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