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

Revision as of 09:56, 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

First let's start by defining some functions:

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

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 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 $g(m,n)=|T_{1}-T_{2}|=0$

(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.

@@ Line 15: / Line 15: @@
 (c) Show that there is no constant <math>C</math> such that <math>f(m,n)<C</math> for all <math>m</math> and <math>n</math>.
 ==Solution==
-{{solution}}
+First let's start by defining some functions:
+For any pair of positive integers <math>m</math> and <math>n</math>, consider a RECTANGLE whose vertices have integer coordinates and whose legs, of lengths <math>m</math> and <math>n</math>, lie along edges of the squares.
+Let <math>T_{1}</math> be the total area of the black part of the rectangle and <math>T_{2}</math> be the total area of the white part.
+Let <math>g(m,n)=|T_{1}-T_{2}|</math>
+Now we do part (a) case: <math>m</math> and <math>n</math> which are both even
+Since <math>m</math> and <math>n</math> which are both even, the total area of the rectangle is <math>m \times n</math>
+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 <math>T_{1}=T_{2}=\frac{mn}{2}</math> and <math>g(m,n)=|T_{1}-T_{2}|=0</math>
+(a) Calculate <math>f(m,n)</math> for all positive integers <math>m</math> and <math>n</math> which are either both even or both odd.
+(b) Prove that <math>f(m,n) \le \frac{1}{2} max\left\{ m,n \right\}</math> for all <math>m</math> and <math>n</math>.
+(c) Show that there is no constant <math>C</math> such that <math>f(m,n)<C</math> for all <math>m</math> and <math>n</math>.
+{{alternate solutions}}

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

Revision as of 09:56, 16 November 2023

Problem

Solution