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There are n (bigger or equal to 3) given points in the plane such that any three of them formed a right triangle.Find the largest possible n.

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You mean "a right angled triangle"? It seems too easy for an olympiad problem. – Alex B. Jul 20 '11 at 16:55
Uh... not that easy man. How about you give us an answer if it's so easy? =) (Not saying its impossible though.) – Patrick Da Silva Jul 20 '11 at 16:58
it is from 23rd moscow olympiad – Victor Jul 20 '11 at 16:59
@Patrick You are right, I was too fast, it's not as easy as I thought. – Alex B. Jul 20 '11 at 17:02
Can someone use combinatoric way to do this,since it is from a combinatoric book – Victor Jul 20 '11 at 17:16
up vote 5 down vote accepted

Clearly, you can arrange 4 points in such a way, namely in a rectangle. I claim that 4 is the highest possible.

Among your $n$ points, let $A$ and $B$ be two points with the smallest distance between them. That means that in any triangle with vertices $A$ and $B$, the angle opposite the side $\overline{AB}$ will always be the smallest. Thus, in any triangle $ABC$, the right angle must be at the vertex $A$ or at the vertex $B$. Since no three points can be on a line, there is at most one point $C_1$ with the property that $ABC_1$ has a right angle at $A$, and similarly there is at most one point $C_2$ such that $ABC_2$ has a right angle at $B$.

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We basically shoot the same ideas, but your proof is more rigorous and mine more intuitive. +1 to yours. – Patrick Da Silva Jul 20 '11 at 17:13
i prefer combinatorics way,is it possible? – Victor Jul 20 '11 at 17:18
How the hell do you wanna use combinatorics to prove a problem about geometry? Do you have any suggestion or idea in mind? I don't see how this could be possible, enlighten me. – Patrick Da Silva Jul 20 '11 at 19:15
Nice. Alternative: Use the largest distance. Then all other points are on the circle with diameter AB, and there can be no more than one on each side. – Max Jul 20 '11 at 23:21

How about this. Let $n$ be the maximum number of points you can place on the plane having this property. Therefore you can obtain a particular solution of this by letting three points on the plane at first, and then adding the next points to reach n points. This is because if the set has $n$ points, then any arbitrary subset of $n-1$ points out of this set has also the property that any three points from it form a right angle, so we can "build" the set of $n$ points out of the one with $n-1$ (inductive argument).

Place three points on the plane that form a right angle. Say they are placed so that $\{a,b,c\}$ has a right angle at $b$. If I am trying to place $d$ on the plane, $ad$ must be perpendicular to $ab$ and $cd$ must be perpendicular to $bc$, which leaves the possibility that $\{a,b,c,d\}$ form a rectangle.

Let's try to place a fifth point on the plane. We already know that a set of four points who satisfy this property, call it $\{a,b,c,d\}$, MUST form a rectangle. Suppose we had a fifth point possibly placed on the plane, call it $e$, so that we now have $\{a,b,c,d,e\}$ satisfying this property. Consider two sets of four points out of this set, say $\{a,b,c,d\}$ and $\{a,b,c,e\}$. They both form a rectangle because they satisfy your property, but they have three vertices in common, so that $d = e$, contradicting the fact that the set contains $5$ distinct points. Since any set with $n$ points and $n \ge 5$ would contain a set with $5$ points with this property, $n \ge 5$ also leads to a contradiction, thus the maximum is $n = 4$.

Hope that helps,

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i prefer combinatorics way,is it possible? – Victor Jul 20 '11 at 17:17

A combinatorial proof can be given by using the Fubini Principle. Please refer to "principles and techniques in combinatorics" by Chen chuang-chong and khee-meng.pp 68

Anyway, it is almost the same as counting the maximum number of points that can be placed on an integer lattice such that no three are collinear and the answer is logically 4.


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