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Difference between revisions of "1992 IMO Problems/Problem 3"

(created the problem and solution for the 1992 IMO Problem 3 because it wasn't created yet)
 
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Clearly <math>X</math> is not the vertex of a monochrome square, because if <math>XY</math> and <math>XZ</math> are
 
Clearly <math>X</math> is not the vertex of a monochrome square, because if <math>XY</math> and <math>XZ</math> are
 
the same color then, <math>YZ</math> is either uncolored or the opposite color. There is no triangle within the red square or the blue square, and hence no monochrome triangle. It remains to consider triangles of the form <math>R_iR_jB_k</math> and <math>B_iB_jR_k.</math> But if <math>i</math> and <math>j</math> have the same parity, then <math>R_iR_j</math> is uncolored (and similarly <math>B_iB_j</math>), whereas if
 
the same color then, <math>YZ</math> is either uncolored or the opposite color. There is no triangle within the red square or the blue square, and hence no monochrome triangle. It remains to consider triangles of the form <math>R_iR_jB_k</math> and <math>B_iB_jR_k.</math> But if <math>i</math> and <math>j</math> have the same parity, then <math>R_iR_j</math> is uncolored (and similarly <math>B_iB_j</math>), whereas if
−
they have opposite parity, then <math>R_iB_k</math> and <math>R_jB_k</math> have opposite colors (and similarly <math>B_iR_k and </math>B_jR_k<math>).
+
they have opposite parity, then <math>R_iB_k</math> and <math>R_jB_k</math> have opposite colors (and similarly <math>B_iR_k</math> and <math>B_jR_k</math>).
−
It remains to show that for </math>n = 33<math> we can always find a monochrome triangle.
+
It remains to show that for <math>n = 33</math> we can always find a monochrome triangle.
 
There are three uncolored edges. Take a point on each of the uncolored edges.
 
There are three uncolored edges. Take a point on each of the uncolored edges.
−
The edges between the remaining </math>6<math> points must all be colored. Take one of
+
The edges between the remaining <math>6</math> points must all be colored. Take one of
−
these, </math>X.<math> At least </math>3<math> of the </math>5<math> edges to </math>X<math>, say </math>XA<math>, </math>XB<math>, </math>XC<math> must be the same color
+
these, <math>X.</math> At least <math>3</math> of the <math>5</math> edges to <math>X</math>, say <math>XA</math>, <math>XB</math>, <math>XC</math> must be the same color
−
(say red). If </math>AB<math> is also red, then </math>XAB<math> is monochrome. Similarly, for </math>BC<math> and </math>CA.<math>
+
(say red). If <math>AB</math> is also red, then <math>XAB</math> is monochrome. Similarly, for <math>BC</math> and <math>CA.</math>
−
But if </math>AB<math>, </math>BC<math> and </math>CA<math> are all blue, then </math>ABC$ is monochrome.
+
But if <math>AB</math>, <math>BC</math> and <math>CA</math> are all blue, then <math>ABC</math> is monochrome.

Revision as of 04:57, 7 August 2023

3. Consider nine points in space, no four of which are coplanar. Each pair 

of points is joined by an edge (that is, a line segment) and each edge is
either colored blue or red or left uncolored. Find the smallest value of
$n$ such that whenever exactly n edges are colored, the set of colored
edges necessarily contains a triangle all of whose edges have the same
color.


Solution: We show that for $n = 32$ we can find a coloring without a monochrome triangle. Take two squares $R_1R_2R_3R_4$ and $B_1B_2B_3B_4$. Leave the diagonals of each square uncolored, color the remaining edges of $R$ red and the remaining edges of $B$ blue. Color blue all the edges from the ninth point $X$ to the red square, and red all the edges from $X$ to the blue square. Color $R_iB_j$ red if $i$ and $j$ have the same parity and blue otherwise. Clearly $X$ is not the vertex of a monochrome square, because if $XY$ and $XZ$ are the same color then, $YZ$ is either uncolored or the opposite color. There is no triangle within the red square or the blue square, and hence no monochrome triangle. It remains to consider triangles of the form $R_iR_jB_k$ and $B_iB_jR_k.$ But if $i$ and $j$ have the same parity, then $R_iR_j$ is uncolored (and similarly $B_iB_j$), whereas if they have opposite parity, then $R_iB_k$ and $R_jB_k$ have opposite colors (and similarly $B_iR_k$ and $B_jR_k$). It remains to show that for $n = 33$ we can always find a monochrome triangle. There are three uncolored edges. Take a point on each of the uncolored edges. The edges between the remaining $6$ points must all be colored. Take one of these, $X.$ At least $3$ of the $5$ edges to $X$, say $XA$, $XB$, $XC$ must be the same color (say red). If $AB$ is also red, then $XAB$ is monochrome. Similarly, for $BC$ and $CA.$ But if $AB$, $BC$ and $CA$ are all blue, then $ABC$ is monochrome.

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