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Difference between revisions of "2000 AIME II Problems/Problem 11"
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== Problem ==
 
== Problem ==
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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>.
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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 ==
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The answer is <math>\boxed{131}</math>
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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}} 
  
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{{incomplete|solution}}
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<center><asy>
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pathpen = linewidth(0.7);
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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);
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D(MP("A",A)--MP("B",B)--MP("C",C,N)--MP("D",D,N)--cycle); D(F--A--Da,linetype("4 4"));
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</asy></center>
  
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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
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<cmath>a^2 + \left(7 - (a+1)m\right)^2 = 50</cmath>
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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
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<cmath>7 - 8m = \pm 1, \quad 7 + 6m = \pm 1, \quad 7 - 6m = \pm 5, 7 + 4m = \pm 5</cmath>
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 +
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>
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<br />
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{{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>
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== Solution 2 ==
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A very natural solution:
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. 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:
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<math>(x \pm 1, kx \pm 7)</math> or <math>(x \pm 7, kx \pm 1)</math> or
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<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.
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Instead of dealing with the 12 cases, we consider
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point <math>C</math> of the form <math>(x \pm Y, kx \pm Z)</math> where
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we plug in the necessary values for <math>Y</math> and <math>Z</math> after simplifying.
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Since the slopes of <math>AB</math> and <math>CD</math> must both be <math>k</math>,
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<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>
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- 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]]
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[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 12:41, 16 July 2024

Problem

The coordinates of the vertices of isosceles trapezoid $ABCD$ are all integers, with $A=(20,100)$ and $D=(21,107)$. The trapezoid has no horizontal or vertical sides, and $\overline{AB}$ and $\overline{CD}$ are the only parallel sides. The sum of the absolute values of all possible slopes for $\overline{AB}$ is $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

For simplicity, we translate the points so that $A$ is on the origin and $D = (1,7)$. Suppose $B$ has integer coordinates; then $\overrightarrow{AB}$ is a vector with integer parameters (vector knowledge is not necessary for this solution). We construct the perpendicular from $A$ to $\overline{CD}$, and let $D' = (a,b)$ be the reflection of $D$ across that perpendicular. Then $ABCD'$ is a parallelogram, and $\overrightarrow{AB} = \overrightarrow{D'C}$. Thus, for $C$ to have integer coordinates, it suffices to let $D'$ 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]

Let the slope of the perpendicular be $m$. Then the midpoint of $\overline{DD'}$ lies on the line $y=mx$, so $\frac{b+7}{2} = m \cdot \frac{a+1}{2}$. Also, $AD = AD'$ implies that $a^2 + b^2 = 1^2 + 7^2 = 50$. Combining these two equations yields

\[a^2 + \left(7 - (a+1)m\right)^2 = 50\]

Since $a$ is an integer, then $7-(a+1)m$ must be an integer. There are $12$ pairs of integers whose squares sum up to $50,$ namely $( \pm 1, \pm 7), (\pm 7, \pm 1), (\pm 5, \pm 5)$. We exclude the cases $(\pm 1, \pm 7)$ because they lead to degenerate trapezoids (rectangle, line segment, vertical and horizontal sides). Thus we have

\[7 - 8m = \pm 1, \quad 7 + 6m = \pm 1, \quad 7 - 6m = \pm 5, 7 + 4m = \pm 5\]

These yield $m = 1, \frac 34, -1, -\frac 43, 2, \frac 13, -3, - \frac 12$. Therefore, the corresponding slopes of $\overline{AB}$ are $-1, -\frac 43, 1, \frac 34, -\frac 12, -3, \frac 13$, and $2$. The sum of their absolute values is $\frac{119}{12}$. The answer is $m+n= \boxed{131}$



^ In other words, since $ABCD'$ is a parallelogram, the difference between the x-coordinates and the y-coordinates of $C$ and $D'$ are, respectively, the difference between the x-coordinates and the y-coordinates of $A$ and $B$. But since the latter are integers, then the former are integers also, so $C$ has integer coordinates iff $D'$ has integer coordinates.

Solution 2

A very natural solution: . Shift $A$ to the origin. Suppose point $B$ was $(x, kx)$. Note $k$ is the slope we're looking for. Note that point $C$ must be of the form: $(x \pm 1, kx \pm 7)$ or $(x \pm 7, kx \pm 1)$ or $(x \pm 5, kx \pm 5)$. Note that we want the slope of the line connecting $D$ and $C$ so also be $k$, since $AB$ and $CD$ are parallel. Instead of dealing with the 12 cases, we consider point $C$ of the form $(x \pm Y, kx \pm Z)$ where we plug in the necessary values for $Y$ and $Z$ after simplifying. Since the slopes of $AB$ and $CD$ must both be $k$, $\frac{7 - kx \pm Z}{1 - x \pm Y} = k \implies k = \frac{7 \pm Z}{1 \pm Y}$. Plugging in the possible values of $\pm 7, \pm 1, \pm 5$ in their respective pairs and ruling out degenerate cases, we find the sum is $\frac{119}{12} \implies m + n = \boxed{131}$ - whatRthose

(Note: This Solution is a lot faster if you rule out $(Y, Z) = (1, 7)$ due to degeneracy.)

See also

2000 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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All AIME Problems and Solutions

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