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Difference between revisions of "1997 IMO Problems/Problem 1"

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Therefore <math>T_{1}=T_{2}=\frac{mn}{2}</math> and <math>g(m,n)=|T_{1}-T_{2}|=0</math>
 
Therefore <math>T_{1}=T_{2}=\frac{mn}{2}</math> and <math>g(m,n)=|T_{1}-T_{2}|=0</math>
  
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Let <math>M</math> be the midpoint of line <math>AC</math>.  Them <math>M</math> is at coordinate <math>(A_{x}+\frac{m}{2},A_{y}+\frac{n}{2})</math>
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Let <math>M</math> be the midpoint of line <math>AC</math>.  Them <math>M</math> is at coordinate <math>(A_{x}+\frac{m}{2},A_{y}+\frac{n}{2})</math> Since both <math>m</math> and <math>n</math> are even, then <math>M</math> has integer coordinates.
  
 
Starting with vertex <math>A</math>, because the length of <math>AB</math> is even, then the color for the square inside rectangle <math>ABCD</math> closest to <math>B</math> is the opposite color of the square inside rectangle <math>ABCD</math> closest to <math>A</math>, then starting with vertex <math>B</math>, because the length of <math>BC</math> is even, then the color of the square inside rectangle <math>ABCD</math> closest to <math>C</math> is the opposite color of the square inside rectangle <math>ABCD</math> closest to <math>B</math>.  this means that the color of the square inside rectangle <math>ABCD</math> closest to <math>A</math> is the same as the color of the square inside rectangle <math>ABCD</math> closest to <math>C</math>.  Likewise, the color of the square inside rectangle <math>ABCD</math> closest to <math>B</math> is the same as the color of the square inside rectangle <math>ABCD</math> closest to <math>D</math>.
 
Starting with vertex <math>A</math>, because the length of <math>AB</math> is even, then the color for the square inside rectangle <math>ABCD</math> closest to <math>B</math> is the opposite color of the square inside rectangle <math>ABCD</math> closest to <math>A</math>, then starting with vertex <math>B</math>, because the length of <math>BC</math> is even, then the color of the square inside rectangle <math>ABCD</math> closest to <math>C</math> is the opposite color of the square inside rectangle <math>ABCD</math> closest to <math>B</math>.  this means that the color of the square inside rectangle <math>ABCD</math> closest to <math>A</math> is the same as the color of the square inside rectangle <math>ABCD</math> closest to <math>C</math>.  Likewise, the color of the square inside rectangle <math>ABCD</math> closest to <math>B</math> is the same as the color of the square inside rectangle <math>ABCD</math> closest to <math>D</math>.
 +
 +
This color pattern and the fact that the midpoint <math>M</math> has integer coordinates indicates that triangle <math>ABC</math> has the same color pattern as triangle <math>CDA</math> rotated 180 degrees.
 +
 +
Therefore, the white area in triangle <math>ABC</math> is the same as the white area in triangle <math>CDA</math> and the black area in triangle <math>ABC</math> is the same as the black area in triangle <math>CDA</math>.
 +
 +
Thus <math>S_{1}=\frac{T_{1}}{2}</math> and <math>S_{2}=\frac{T_{2}}{2}</math>
  
  

Revision as of 10:40, 16 November 2023

Problem

In the plane the points with integer coordinates are the vertices of unit squares. The squares are colored alternatively black and white (as on a chessboard).

For any pair of positive integers $m$ and $n$, consider a right-angled triangle whose vertices have integer coordinates and whose legs, of lengths $m$ and $n$, lie along edges of the squares.

Let $S_{1}$ be the total area of the black part of the triangle and $S_{2}$ be the total area of the white part.

Let $f(m,n)=|S_{1}-S_{2}|$

(a) Calculate $f(m,n)$ for all positive integers $m$ and $n$ which are either both even or both odd.

(b) Prove that $f(m,n) \le \frac{1}{2} max\left\{ m,n \right\}$ for all $m$ and $n$.

(c) Show that there is no constant $C$ such that $f(m,n)<C$ for all $m$ and $n$.

Solution

For any pair of positive integers $m$ and $n$, consider a rectangle $ABCD$ whose vertices have integer coordinates and whose legs, of lengths $m$ and $n$, lie along edges of the squares.

Let $A$, $B$, $C$, and $D$, be the lower left vertex, lower right vertex, upper right vertex, and upper left vertex of rectangle $ABCD$ respectively.

Let $T_{1}$ be the total area of the black part of the rectangle and $T_{2}$ be the total area of the white part.

Let $g(m,n)=|T_{1}-T_{2}|$

Now we do part (a) case: $m$ and $n$ which are both even

Since $m$ and $n$ which are both even, the total area of the rectangle $ABCD$ is $m \times n$

Since every row has an even number of squares there are equally as many white squares than black squares for each row.

Since every column has an even number of squares there are equally as many white squares than black squares for each column.

This means that in the rectangle there are equal number of white squares and black squares.

Therefore $T_{1}=T_{2}=\frac{mn}{2}$ and $g(m,n)=|T_{1}-T_{2}|=0$

Let $M$ be the midpoint of line $AC$. Them $M$ is at coordinate $(A_{x}+\frac{m}{2},A_{y}+\frac{n}{2})$ Since both $m$ and $n$ are even, then $M$ has integer coordinates.

Starting with vertex $A$, because the length of $AB$ is even, then the color for the square inside rectangle $ABCD$ closest to $B$ is the opposite color of the square inside rectangle $ABCD$ closest to $A$, then starting with vertex $B$, because the length of $BC$ is even, then the color of the square inside rectangle $ABCD$ closest to $C$ is the opposite color of the square inside rectangle $ABCD$ closest to $B$. this means that the color of the square inside rectangle $ABCD$ closest to $A$ is the same as the color of the square inside rectangle $ABCD$ closest to $C$. Likewise, the color of the square inside rectangle $ABCD$ closest to $B$ is the same as the color of the square inside rectangle $ABCD$ closest to $D$.

This color pattern and the fact that the midpoint $M$ has integer coordinates indicates that triangle $ABC$ has the same color pattern as triangle $CDA$ rotated 180 degrees.

Therefore, the white area in triangle $ABC$ is the same as the white area in triangle $CDA$ and the black area in triangle $ABC$ is the same as the black area in triangle $CDA$.

Thus $S_{1}=\frac{T_{1}}{2}$ and $S_{2}=\frac{T_{2}}{2}$




(a) Calculate $f(m,n)$ for all positive integers $m$ and $n$ which are either both even or both odd.

(b) Prove that $f(m,n) \le \frac{1}{2} max\left\{ m,n \right\}$ for all $m$ and $n$.

(c) Show that there is no constant $C$ such that $f(m,n)<C$ for all $m$ and $n$.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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