Difference between revisions of "1983 IMO Problems/Problem 4"

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==Solution==
 
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The answer is positive; there always a class of the partition will contain the three vertices of a right-angled triangle.
  
==Solution==
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First notice that if there are at least two points of the same color on a side, then all the points in <math>\mathcal{E}</math> which project orthogonally onto points of that color on the respective side must have the opposite color if we are to have no monochromatic right-angled triangles.
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Now assume we can find a side on which there is at most one point bearing one of the colors (blue, say; we take the two colors to be red and blue). Then, by the observation above, it's obvious that the other two sides contain three blue vertices of a right-angled triangle, and we're done (all the points of those two sides are blue, except maybe for the endpoints which they have in common with the other side and the point which projects orthogonally onto the unique blue point on the third side). We can now assume all three sides contain at least two points of each color. Take <math>U\in BC</math> s.t. <math>BC=3BU</math>. <math>U</math> projects orthogonally onto <math>V\in CA</math>, and <math>V</math> projects orthogonally onto <math>T\in AB</math>. We can easily see that <math>T</math> projects orthogonally onto <math>U</math>. Again, by the observation in the previous paragraph, the points <math>U,V,T</math> must have different colors if we are to have no monochromatic right-angled triangle, and this is impossible (we only have two colors).
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This solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [https://aops.com/community/p367087]
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== See Also == {{IMO box|year=1983|num-b=3|num-a=5}}

Latest revision as of 22:40, 29 January 2021

Problem

Let $ABC$ be an equilateral triangle and $\mathcal{E}$ the set of all points contained in the three segments $AB$, $BC$ and $CA$ (including $A$, $B$ and $C$). Determine whether, for every partition of $\mathcal{E}$ into two disjoint subsets, at least one of the two subsets contains the vertices of a right-angled triangle. Justify your answer.

Solution

The answer is positive; there always a class of the partition will contain the three vertices of a right-angled triangle.

First notice that if there are at least two points of the same color on a side, then all the points in $\mathcal{E}$ which project orthogonally onto points of that color on the respective side must have the opposite color if we are to have no monochromatic right-angled triangles.

Now assume we can find a side on which there is at most one point bearing one of the colors (blue, say; we take the two colors to be red and blue). Then, by the observation above, it's obvious that the other two sides contain three blue vertices of a right-angled triangle, and we're done (all the points of those two sides are blue, except maybe for the endpoints which they have in common with the other side and the point which projects orthogonally onto the unique blue point on the third side). We can now assume all three sides contain at least two points of each color. Take $U\in BC$ s.t. $BC=3BU$. $U$ projects orthogonally onto $V\in CA$, and $V$ projects orthogonally onto $T\in AB$. We can easily see that $T$ projects orthogonally onto $U$. Again, by the observation in the previous paragraph, the points $U,V,T$ must have different colors if we are to have no monochromatic right-angled triangle, and this is impossible (we only have two colors).

This solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]


See Also

1983 IMO (Problems) • Resources
Preceded by
Problem 3
1 2 3 4 5 6 Followed by
Problem 5
All IMO Problems and Solutions