Difference between revisions of "2000 AIME II Problems/Problem 11"
Silversheep (talk | contribs) |
Tecilis459 (talk | contribs) m (Unify note) |
||
(9 intermediate revisions by 6 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | The coordinates of the vertices of isosceles trapezoid <math>ABCD</math> are all integers, with <math>A=(20,100)</math> and <math>D=(21,107)</math>. The trapezoid has no horizontal or vertical sides, and <math>\overline{AB}</math> and <math>\overline{CD}</math> are the only parallel sides. The sum of the absolute values of all possible slopes for <math>\overline{AB}</math> is <math>m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | + | The coordinates of the vertices of [[isosceles trapezoid]] <math>ABCD</math> are all integers, with <math>A=(20,100)</math> and <math>D=(21,107)</math>. The trapezoid has no horizontal or vertical sides, and <math>\overline{AB}</math> and <math>\overline{CD}</math> are the only [[parallel]] sides. The sum of the absolute values of all possible slopes for <math>\overline{AB}</math> is <math>m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. |
== Solution == | == Solution == | ||
− | + | For simplicity, we translate the points so that <math>A</math> is on the origin and <math>D = (1,7)</math>. Suppose <math>B</math> has integer coordinates; then <math>\overrightarrow{AB}</math> is a [[vector]] with integer parameters (vector knowledge is not necessary for this solution). We construct the [[perpendicular]] from <math>A</math> to <math>\overline{CD}</math>, and let <math>D' = (a,b)</math> be the reflection of <math>D</math> across that perpendicular. Then <math>ABCD'</math> is a [[parallelogram]], and <math>\overrightarrow{AB} = \overrightarrow{D'C}</math>. Thus, for <math>C</math> to have integer coordinates, it suffices to let <math>D'</math> have integer coordinates.{{ref|1}} | |
− | + | <center><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></center> | ||
+ | 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> | ||
+ | |||
+ | Since <math>a</math> is an integer, then <math>7-(a+1)m</math> must be an integer. There are <math>12</math> pairs of integers whose squares sum up to <math>50,</math> namely <math>( \pm 1, \pm 7), (\pm 7, \pm 1), (\pm 5, \pm 5)</math>. We exclude the cases <math>(\pm 1, \pm 7)</math> because they lead to degenerate trapezoids (rectangle, line segment, vertical and horizontal sides). Thus we have | ||
+ | |||
+ | <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>. 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 /> | ||
+ | {{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 == | ||
+ | A very natural solution: | ||
+ | . Shift <math>A</math> to the origin. Suppose point <math>B</math> was <math>(x, kx)</math>. Note <math>k</math> is the slope we're looking for. Note that point <math>C</math> must be of the form: | ||
+ | <math>(x \pm 1, kx \pm 7)</math> or <math>(x \pm 7, kx \pm 1)</math> or | ||
+ | <math>(x \pm 5, kx \pm 5)</math>. Note that we want the slope of the line connecting <math>D</math> and <math>C</math> so also be <math>k</math>, since <math>AB</math> and <math>CD</math> are parallel. | ||
+ | Instead of dealing with the 12 cases, we consider | ||
+ | point <math>C</math> of the form <math>(x \pm Y, kx \pm Z)</math> where | ||
+ | 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>, | ||
+ | <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 | ||
+ | |||
+ | (Note: This Solution is a lot faster if you rule out <math>(Y, Z) = (1, 7)</math> due to degeneracy.) | ||
+ | |||
+ | == See also == | ||
{{AIME box|year=2000|n=II|num-b=10|num-a=12}} | {{AIME box|year=2000|n=II|num-b=10|num-a=12}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 12:41, 16 July 2024
Contents
[hide]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]
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.