Difference between revisions of "2005 Indonesia MO Problems/Problem 1"
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The line <math>x+y>n</math> and <math>y \ge x</math> intersect at <math>(\frac{n}{2},\frac{n}{2})</math>, but that value is not an integer. By symmetry, for each of the four line segments from the diagonal, there are <math>\frac{n+1}{2}</math> lattice points. Since there are a total of <math>(n+1)^2</math> lattice points within <math>0 \le x,y \le n</math>, by symmetry, each section formed from the diagonals has <math>\frac{(n+1)^2 - 2n - 2}{4} = \frac{n^2 - 1}{4}</math> lattice points. We want the points on lines <math>y = n</math>, <math>y = x</math> and not <math>x+y=n</math>, so there are <math>\frac{n^2 + 2n + 1}{4}</math> points that satisfy the conditions if <math>n</math> is odd. | The line <math>x+y>n</math> and <math>y \ge x</math> intersect at <math>(\frac{n}{2},\frac{n}{2})</math>, but that value is not an integer. By symmetry, for each of the four line segments from the diagonal, there are <math>\frac{n+1}{2}</math> lattice points. Since there are a total of <math>(n+1)^2</math> lattice points within <math>0 \le x,y \le n</math>, by symmetry, each section formed from the diagonals has <math>\frac{(n+1)^2 - 2n - 2}{4} = \frac{n^2 - 1}{4}</math> lattice points. We want the points on lines <math>y = n</math>, <math>y = x</math> and not <math>x+y=n</math>, so there are <math>\frac{n^2 + 2n + 1}{4}</math> points that satisfy the conditions if <math>n</math> is odd. | ||
− | In summary, the number of triangles that satisfy the conditions are <math>\boxed{\frac{n^2 + 2n}{4} (n \text{ is even)}}</math> and <math>\boxed{\frac{n^2 + 2n + 1}{4} (n \text{ is odd)}}</math>, or <math> | + | In summary, the number of triangles that satisfy the conditions are <math>\boxed{\frac{n^2 + 2n}{4} (n \text{ is even)}}</math> and <math>\boxed{\frac{n^2 + 2n + 1}{4} (n \text{ is odd)}}</math>, or <math>\boxed{\lceil \frac{n^2 + 2n}{4} \rceil}</math> |
==See Also== | ==See Also== |
Latest revision as of 01:17, 3 February 2023
Problem
Let be a positive integer. Determine the number of triangles (non congruent) with integral side lengths and the longest side length is
.
Solution
WLOG, let . The original problem is essentially asking for the number of lattice points that lie within this bound as well as
.
By experimenting with smaller graphs, we can split into two cases.
Case 1: is even
Below is the case where .
The line and
intersect at
. By symmetry, for each of the four line segments from the diagonal, there are
lattice points. Since there are a total of
lattice points within
, by symmetry, each section formed from the diagonals has
lattice points. We want the points on lines
,
and not
, so there are
points that satisfy the conditions if
is even.
Case 1: is odd
Below is the case where .
The line and
intersect at
, but that value is not an integer. By symmetry, for each of the four line segments from the diagonal, there are
lattice points. Since there are a total of
lattice points within
, by symmetry, each section formed from the diagonals has
lattice points. We want the points on lines
,
and not
, so there are
points that satisfy the conditions if
is odd.
In summary, the number of triangles that satisfy the conditions are and
, or
See Also
2005 Indonesia MO (Problems) | ||
Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 | Followed by Problem 2 |
All Indonesia MO Problems and Solutions |