Difference between revisions of "1999 AIME Problems/Problem 2"

 
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== Problem ==
 
== Problem ==
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Consider the [[parallelogram]] with [[vertex|vertices]] <math>(10,45)</math>, <math>(10,114)</math>, <math>(28,153)</math>, and <math>(28,84)</math>.  A [[line]] through the [[origin]] cuts this figure into two [[congruent]] [[polygon]]s.  The [[slope]] of the line is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are [[relatively prime]] [[positive]] [[integer]]s.  Find <math>m+n</math>.
  
 
== Solution ==
 
== Solution ==
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=== Solution 1 ===
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Let the first point on the line <math>x=10</math> be <math>(10,45+a)</math> where a is the height above <math>(10,45)</math>.  Let the second point on the line <math>x=28</math> be <math>(28, 153-a)</math>. For two given points, the line will pass the origin if the coordinates are [[proportion]]al (such that <math>\frac{y_1}{x_1} = \frac{y_2}{x_2}</math>). Then, we can write that <math>\frac{45 + a}{10} = \frac{153 - a}{28}</math>. Solving for <math>a</math> yields that <math>1530 - 10a = 1260 + 28a</math>, so <math>a=\frac{270}{38}=\frac{135}{19}</math>. The slope of the line (since it passes through the origin) is <math>\frac{45 + \frac{135}{19}}{10} = \frac{99}{19}</math>, and the solution is <math>m + n = \boxed{118}</math>.
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=== Solution 2 (the best solution)===
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You can clearly see that a line that cuts a parallelogram into two congruent pieces must go through the center of the parallelogram. Taking the midpoint of <math>(10,45)</math>, and <math>(28,153)</math> gives <math>(19,99)</math>, which is the center of the parallelogram. Thus the slope of the line must be <math>\frac{99}{19}</math>, and the solution is <math>\boxed{118}</math>.
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=== Solution 3 (Area) ===
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Note that the area of the parallelogram is equivalent to <math>69 \cdot 18 = 1242,</math> so the area of each of the two trapezoids with congruent area is <math>621.</math> Therefore, since the height is <math>18,</math> the sum of the bases of each trapezoid must be <math>69.</math>
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The points where the line in question intersects the long side of the parallelogram can be denoted as <math>(10, \frac{10m}{n})</math> and <math>(28, \frac{28m}{n}),</math> respectively. We see that <math>\frac{10m}{n} - 45 + \frac{28m}{n} - 84 = 69,</math> so <math>\frac{38m}{n} = 198 \implies \frac{m}{n} = \frac{99}{19} \implies \boxed{118}.</math>
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Solution by Ilikeapos
  
 
== See also ==
 
== See also ==
* [[1999 AIME Problems]]
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{{AIME box|year=1999|num-b=1|num-a=3}}
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[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Revision as of 22:58, 6 April 2020

Problem

Consider the parallelogram with vertices $(10,45)$, $(10,114)$, $(28,153)$, and $(28,84)$. A line through the origin cuts this figure into two congruent polygons. The slope of the line is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n$.

Solution

Solution 1

Let the first point on the line $x=10$ be $(10,45+a)$ where a is the height above $(10,45)$. Let the second point on the line $x=28$ be $(28, 153-a)$. For two given points, the line will pass the origin if the coordinates are proportional (such that $\frac{y_1}{x_1} = \frac{y_2}{x_2}$). Then, we can write that $\frac{45 + a}{10} = \frac{153 - a}{28}$. Solving for $a$ yields that $1530 - 10a = 1260 + 28a$, so $a=\frac{270}{38}=\frac{135}{19}$. The slope of the line (since it passes through the origin) is $\frac{45 + \frac{135}{19}}{10} = \frac{99}{19}$, and the solution is $m + n = \boxed{118}$.

Solution 2 (the best solution)

You can clearly see that a line that cuts a parallelogram into two congruent pieces must go through the center of the parallelogram. Taking the midpoint of $(10,45)$, and $(28,153)$ gives $(19,99)$, which is the center of the parallelogram. Thus the slope of the line must be $\frac{99}{19}$, and the solution is $\boxed{118}$.

Solution 3 (Area)

Note that the area of the parallelogram is equivalent to $69 \cdot 18 = 1242,$ so the area of each of the two trapezoids with congruent area is $621.$ Therefore, since the height is $18,$ the sum of the bases of each trapezoid must be $69.$

The points where the line in question intersects the long side of the parallelogram can be denoted as $(10, \frac{10m}{n})$ and $(28, \frac{28m}{n}),$ respectively. We see that $\frac{10m}{n} - 45 + \frac{28m}{n} - 84 = 69,$ so $\frac{38m}{n} = 198 \implies \frac{m}{n} = \frac{99}{19} \implies \boxed{118}.$

Solution by Ilikeapos

See also

1999 AIME (ProblemsAnswer KeyResources)
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Problem 3
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