Difference between revisions of "2000 AIME II Problems/Problem 11"
(solution) |
m (→Solution: typo fix) |
||
Line 11: | Line 11: | ||
</asy></center> | </asy></center> | ||
− | Let the slope of | + | Let the slope of the perpendicular be <math>m</math>. Then the [[midpoint]] of <math>\overline{DD'}</math> lies on the line <math>y=mx</math>, so <math>\frac{b+7}{2} = m \cdot \frac{a+1}{2}</math>. Also, <math>AD = AD'</math> implies that <math>a^2 + b^2 = 1^2 + 7^2 = 50</math>. Combining these two equations yields |
<cmath>a^2 + \left(7 - (a+1)m\right)^2 = 50</cmath> | <cmath>a^2 + \left(7 - (a+1)m\right)^2 = 50</cmath> |
Revision as of 18:33, 5 September 2008
Problem
The coordinates of the vertices of isosceles trapezoid are all integers, with
and
. The trapezoid has no horizontal or vertical sides, and
and
are the only parallel sides. The sum of the absolute values of all possible slopes for
is
, where
and
are relatively prime positive integers. Find
.
Solution
For simplicity, we translate the points so that is on the origin and
. Suppose
has integer coordinates; then
is a vector with integer parameters (vector knowledge is not necessary for this solution). We construct the perpendicular from
to
, and let
be the reflection of
across that perpendicular. Then
is a parallelogram, and
. Thus, for
to have integer coordinates, it suffices to let
have integer coordinates.[1]
![[asy] pathpen = linewidth(0.7); pair A=(0,0), D=(1,7), Da = MP("D'",D((-7,1)),N), B=(-8,-6), C=B+Da, F=foot(A,C,D); D(MP("A",A)--MP("B",B)--MP("C",C,N)--MP("D",D,N)--cycle); D(F--A--Da,linetype("4 4")); [/asy]](http://latex.artofproblemsolving.com/f/b/f/fbf4e8b84d218d7e34781471aeb31f8011858a42.png)
Let the slope of the perpendicular be . Then the midpoint of
lies on the line
, so
. Also,
implies that
. Combining these two equations yields
Since is an integer, then
must be an integer. There are
pairs of integers whose squares sum up to
, namely
. We exclude the cases
because they lead to degenerate trapezoids (rectangle, line segment, vertical and horizontal sides). Thus we have
These yield , and the sum of their absolute values is
. The answer is
^ In other words, since is a parallelogram, the difference between the x-coordinates and the y-coordinates of
and
are, respectively, the difference between the x-coordinates and the y-coordinates of
and
. But since the latter are integers, then the former are integers also, so
has integer coordinates iff
has integer coordinates.
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |