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Difference between revisions of "1979 IMO Problems/Problem 2"

(Created page with "==Problem== We consider a prism which has the upper and inferior basis the pentagons: <math>A_{1}A_{2}A_{3}A_{4}A_{5}</math> and <math>B_{1}B_{2}B_{3}B_{4}B_{5}</math>. Each o...")
 
 
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Let us prove first that the edges <math>A_1A_2,A_2A_3,\cdots ,A_5A_1</math> are of the same color. Assume the contrary, and let w.l.o.g. <math>A_1A_2</math> be red and <math>A_2A_3</math> be green. Three of the segments <math>A_2B_l, (l = 1, 2, 3, 4, 5)</math>, say <math>A_2B_i,A_2B_j,A_2B_k</math>,have to be of the same color, let it w.l.o.g. be red. Then <math>A_1B_i,A_1B_j,A_1B_k</math> must be green. At least one of the sides of triangle <math>B_iB_jB_k</math>, say <math>B_iB_j</math> ,must be an edge of the prism. Then looking at the triangles <math>A_1B_iB_j</math> and <math>A_2B_iB_j</math> we deduce that <math>B_iB_j</math> can be neither green nor red, which is acontradiction. Hence all five edges of the pentagon <math>A_1A_2A_3A_4A_5</math> have thesame color. Similarly, all five edges of <math>B_1B_2B_3B_4B_5</math> have the same color.We now show that the two colors are the same. Assume otherwise, i.e.,that w.l.o.g. the <math>A</math> edges are painted red and the <math>B</math> edges green. Let us call segments of the form <math>A_iB_j</math> diagonal (<math>i</math> and <math>j</math> may be equal). We now count the diagonal segments by grouping the red segments based on their <math>A</math> point, and the green segments based on their <math>B</math> point. As above, the assumption that three of <math>A_iB_j</math> for fixed <math>i</math> are red leads to a contradiction. Hence at most two diagonal segments out of each <math>A_i</math> may be red, which counts up to at most <math>10</math> red segments. Similarly, at most <math>10</math> diagonal segments can be green. But then we can paint at most <math>20</math> diagonal segments out of <math>25</math>, which is a contradiction. Hence all edges in the pentagons <math>A_1A_2A_3A_4A_5</math> and <math>B_1B_2B_3B_4B_5</math> have the same color.
 
Let us prove first that the edges <math>A_1A_2,A_2A_3,\cdots ,A_5A_1</math> are of the same color. Assume the contrary, and let w.l.o.g. <math>A_1A_2</math> be red and <math>A_2A_3</math> be green. Three of the segments <math>A_2B_l, (l = 1, 2, 3, 4, 5)</math>, say <math>A_2B_i,A_2B_j,A_2B_k</math>,have to be of the same color, let it w.l.o.g. be red. Then <math>A_1B_i,A_1B_j,A_1B_k</math> must be green. At least one of the sides of triangle <math>B_iB_jB_k</math>, say <math>B_iB_j</math> ,must be an edge of the prism. Then looking at the triangles <math>A_1B_iB_j</math> and <math>A_2B_iB_j</math> we deduce that <math>B_iB_j</math> can be neither green nor red, which is acontradiction. Hence all five edges of the pentagon <math>A_1A_2A_3A_4A_5</math> have thesame color. Similarly, all five edges of <math>B_1B_2B_3B_4B_5</math> have the same color.We now show that the two colors are the same. Assume otherwise, i.e.,that w.l.o.g. the <math>A</math> edges are painted red and the <math>B</math> edges green. Let us call segments of the form <math>A_iB_j</math> diagonal (<math>i</math> and <math>j</math> may be equal). We now count the diagonal segments by grouping the red segments based on their <math>A</math> point, and the green segments based on their <math>B</math> point. As above, the assumption that three of <math>A_iB_j</math> for fixed <math>i</math> are red leads to a contradiction. Hence at most two diagonal segments out of each <math>A_i</math> may be red, which counts up to at most <math>10</math> red segments. Similarly, at most <math>10</math> diagonal segments can be green. But then we can paint at most <math>20</math> diagonal segments out of <math>25</math>, which is a contradiction. Hence all edges in the pentagons <math>A_1A_2A_3A_4A_5</math> and <math>B_1B_2B_3B_4B_5</math> have the same color.
  
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This solution was posted and copyrighted by [user]. The original thread for this problem can be found here: [link]
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This solution was posted and copyrighted by Learner94. The original thread for this problem can be found here: [https://aops.com/community/p2448979]
  
 
== See Also == {{IMO box|year=1979|num-b=1|num-a=3}}
 
== See Also == {{IMO box|year=1979|num-b=1|num-a=3}}

Latest revision as of 21:50, 29 January 2021

Problem

We consider a prism which has the upper and inferior basis the pentagons: $A_{1}A_{2}A_{3}A_{4}A_{5}$ and $B_{1}B_{2}B_{3}B_{4}B_{5}$. Each of the sides of the two pentagons and the segments $A_{i}B_{j}$ with $i,j=1,\ldots$,5 is colored in red or blue. In every triangle which has all sides colored there exists one red side and one blue side. Prove that all the 10 sides of the two basis are colored in the same color.

Solution

Let us prove first that the edges $A_1A_2,A_2A_3,\cdots ,A_5A_1$ are of the same color. Assume the contrary, and let w.l.o.g. $A_1A_2$ be red and $A_2A_3$ be green. Three of the segments $A_2B_l, (l = 1, 2, 3, 4, 5)$, say $A_2B_i,A_2B_j,A_2B_k$,have to be of the same color, let it w.l.o.g. be red. Then $A_1B_i,A_1B_j,A_1B_k$ must be green. At least one of the sides of triangle $B_iB_jB_k$, say $B_iB_j$ ,must be an edge of the prism. Then looking at the triangles $A_1B_iB_j$ and $A_2B_iB_j$ we deduce that $B_iB_j$ can be neither green nor red, which is acontradiction. Hence all five edges of the pentagon $A_1A_2A_3A_4A_5$ have thesame color. Similarly, all five edges of $B_1B_2B_3B_4B_5$ have the same color.We now show that the two colors are the same. Assume otherwise, i.e.,that w.l.o.g. the $A$ edges are painted red and the $B$ edges green. Let us call segments of the form $A_iB_j$ diagonal ($i$ and $j$ may be equal). We now count the diagonal segments by grouping the red segments based on their $A$ point, and the green segments based on their $B$ point. As above, the assumption that three of $A_iB_j$ for fixed $i$ are red leads to a contradiction. Hence at most two diagonal segments out of each $A_i$ may be red, which counts up to at most $10$ red segments. Similarly, at most $10$ diagonal segments can be green. But then we can paint at most $20$ diagonal segments out of $25$, which is a contradiction. Hence all edges in the pentagons $A_1A_2A_3A_4A_5$ and $B_1B_2B_3B_4B_5$ have the same color.

This solution was posted and copyrighted by Learner94. The original thread for this problem can be found here: [1]

See Also

1979 IMO (Problems) • Resources
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 3
All IMO Problems and Solutions
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