Difference between revisions of "1997 IMO Problems/Problem 1"
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case: <math>m</math> and <math>n</math> which are both even | case: <math>m</math> and <math>n</math> which are both even | ||
− | Since <math>m</math> and <math>n</math> | + | Since <math>m</math> and <math>n</math> are both even, the total area of the rectangle <math>ABCD</math> is <math>m \times n</math> which is also even |
Since every row has an even number of squares there are equally as many white squares than black squares for each row. | Since every row has an even number of squares there are equally as many white squares than black squares for each row. | ||
Line 50: | Line 50: | ||
Therefore <math>f(m,n)=0</math> when both <math>m</math> and <math>n</math> are even. | Therefore <math>f(m,n)=0</math> when both <math>m</math> and <math>n</math> are even. | ||
+ | case: <math>m</math> and <math>n</math> which are both odd | ||
+ | |||
+ | Since <math>m</math> and <math>n</math> are both odd, the total area of the rectangle <math>ABCD</math> is <math>m \times n</math> which is also odd | ||
+ | |||
+ | Since the total area is odd, then <math>\frac{mn}{2}</math> is not an integer but <math>\frac{mn+1}{2}</math> and <math>\frac{mn-1}{2}</math> are. | ||
+ | |||
+ | This means that in the rectangle there are <math>\frac{mn+1}{2}</math> squares of one color and <math>\frac{mn+1}{2}</math> squares of the other color | ||
+ | |||
+ | <math>g(m,n)=|T_{1}-T_{2}|=\left| \frac{mn+1}{2}-\frac{mn-1}{2} \right|=1</math> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>, which gives <math>f(m,n)=\frac{g(m,n)}{2}=0</math> | ||
+ | |||
+ | Therefore <math>f(m,n)=0</math> when both <math>m</math> and <math>n</math> are even. | ||
Revision as of 11:50, 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 and
, consider a right-angled triangle whose vertices have integer coordinates and whose legs, of lengths
and
, lie along edges of the squares.
Let be the total area of the black part of the triangle and
be the total area of the white part.
Let
(a) Calculate for all positive integers
and
which are either both even or both odd.
(b) Prove that for all
and
.
(c) Show that there is no constant such that
for all
and
.
Solution
For any pair of positive integers and
, consider a rectangle
whose vertices have integer coordinates and whose legs, of lengths
and
, lie along edges of the squares.
Let ,
,
, and
, be the lower left vertex, lower right vertex, upper right vertex, and upper left vertex of rectangle
respectively.
Let be the total area of the black part of the rectangle and
be the total area of the white part.
Let
Part (a):
case: and
which are both even
Since and
are both even, the total area of the rectangle
is
which is also even
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 and
Let be the midpoint of line
. Them
is at coordinate
Since both
and
are even, then
has integer coordinates.
Starting with vertex , because the length of
is even, then the color for the square inside rectangle
closest to
is the opposite color of the square inside rectangle
closest to
, then starting with vertex
, because the length of
is even, then the color of the square inside rectangle
closest to
is the opposite color of the square inside rectangle
closest to
. this means that the color of the square inside rectangle
closest to
is the same as the color of the square inside rectangle
closest to
. Likewise, the color of the square inside rectangle
closest to
is the same as the color of the square inside rectangle
closest to
.
This color pattern and the fact that the midpoint has integer coordinates indicates that triangle
has the same color pattern as triangle
rotated 180 degrees.
Therefore, the white area in triangle is the same as the white area in triangle
and the black area in triangle
is the same as the black area in triangle
.
Thus and
, which gives
Therefore when both
and
are even.
case: and
which are both odd
Since and
are both odd, the total area of the rectangle
is
which is also odd
Since the total area is odd, then is not an integer but
and
are.
This means that in the rectangle there are squares of one color and
squares of the other color
Let be the midpoint of line
. Them
is at coordinate
Since both
and
are even, then
has integer coordinates.
Starting with vertex , because the length of
is even, then the color for the square inside rectangle
closest to
is the opposite color of the square inside rectangle
closest to
, then starting with vertex
, because the length of
is even, then the color of the square inside rectangle
closest to
is the opposite color of the square inside rectangle
closest to
. this means that the color of the square inside rectangle
closest to
is the same as the color of the square inside rectangle
closest to
. Likewise, the color of the square inside rectangle
closest to
is the same as the color of the square inside rectangle
closest to
.
This color pattern and the fact that the midpoint has integer coordinates indicates that triangle
has the same color pattern as triangle
rotated 180 degrees.
Therefore, the white area in triangle is the same as the white area in triangle
and the black area in triangle
is the same as the black area in triangle
.
Thus and
, which gives
Therefore when both
and
are even.
(a) Calculate for all positive integers
and
which are either both even or both odd.
(b) Prove that for all
and
.
(c) Show that there is no constant such that
for all
and
.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.