Difference between revisions of "2000 AIME II Problems/Problem 11"
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<cmath>7 - 8m = \pm 1, \quad 7 + 6m = \pm 1, \quad 7 - 6m = \pm 5, 7 + 4m = \pm 5</cmath> | <cmath>7 - 8m = \pm 1, \quad 7 + 6m = \pm 1, \quad 7 - 6m = \pm 5, 7 + 4m = \pm 5</cmath> | ||
− | These yield <math>m = 1, \frac 34, -1, -\frac 43, 2, \frac 13, -3, - \frac 12</math>, and | + | These yield <math>m = 1, \frac 34, -1, -\frac 43, 2, \frac 13, -3, - \frac 12</math>. Therefore, the corresponding slopes of <math>\overline{AB}</math> are <math>-1, -\frac 43, 1, \frac 34, -\frac 12, -3, \frac 13</math>, and <math>2</math>. The sum of their absolute values is <math>\frac{119}{12}</math>. The answer is <math>m+n= \boxed{131}</math> |
<br /> | <br /> | ||
− | {{note|1}} < | + | {{note|1}} <small>In other words, since <math>ABCD'</math> is a parallelogram, the difference between the x-coordinates and the y-coordinates of <math>C</math> and <math>D'</math> are, respectively, the difference between the x-coordinates and the y-coordinates of <math>A</math> and <math>B</math>. But since the latter are integers, then the former are integers also, so <math>C</math> has integer coordinates [[iff]] <math>D'</math> has integer coordinates.</small> |
== Solution 2 == | == Solution 2 == | ||
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we plug in the necessary values for <math>Y</math> and <math>Z</math> after simplifying. | we plug in the necessary values for <math>Y</math> and <math>Z</math> after simplifying. | ||
Since the slopes of <math>AB</math> and <math>CD</math> must both be <math>k</math>, | Since the slopes of <math>AB</math> and <math>CD</math> must both be <math>k</math>, | ||
− | <math>\frac{7 - kx \pm Z}{1 - x \pm Y} = k \implies k = \frac{7 \pm Z}{1 \pm Y}</math>. Plugging in the possible values of <math>\pm 7, \pm 1, \pm 5</math> in | + | <math>\frac{7 - kx \pm Z}{1 - x \pm Y} = k \implies k = \frac{7 \pm Z}{1 \pm Y}</math>. Plugging in the possible values of <math>\pm 7, \pm 1, \pm 5</math> in their respective pairs and ruling out degenerate cases, we find the sum is <math>\frac{119}{12} \implies m + n = \boxed{131}</math> |
- whatRthose | - whatRthose | ||
+ | |||
+ | (Note: This Solution is a lot faster if you rule out <math>(Y, Z) = (1, 7)</math> due to degeneracy.) | ||
== See also == | == See also == |
Latest revision as of 13:41, 16 July 2024
Contents
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 . Therefore, the corresponding slopes of
are
, 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.
Solution 2
A very natural solution:
. Shift to the origin. Suppose point
was
. Note
is the slope we're looking for. Note that point
must be of the form:
or
or
. Note that we want the slope of the line connecting
and
so also be
, since
and
are parallel.
Instead of dealing with the 12 cases, we consider
point
of the form
where
we plug in the necessary values for
and
after simplifying.
Since the slopes of
and
must both be
,
. Plugging in the possible values of
in their respective pairs and ruling out degenerate cases, we find the sum is
- whatRthose
(Note: This Solution is a lot faster if you rule out due to degeneracy.)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.