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

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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>.
 
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 ===
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=== 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 <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>.
 
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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Edit by ngourise: if u dont think this is the best solution then smh
  
 
=== Solution 3 (Area) ===  
 
=== Solution 3 (Area) ===  
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Solution by Ilikeapos
 
Solution by Ilikeapos
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=== Solution 4 (centroid) (fastest way) ===
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<math>(\Sigma x_i /4, \Sigma y_i /4)</math> is the centroid, which generates <math>(19,99)</math>, so the answer is <math>\boxed{118}</math>. This is the fastest way because you do not need to find the opposite vertices by drawing.
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Solution by maxamc
  
 
== See also ==
 
== See also ==

Latest revision as of 10:06, 27 April 2023

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 $\boxed{118}$.

Edit by ngourise: if u dont think this is the best solution then smh

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

Solution 4 (centroid) (fastest way)

$(\Sigma x_i /4, \Sigma y_i /4)$ is the centroid, which generates $(19,99)$, so the answer is $\boxed{118}$. This is the fastest way because you do not need to find the opposite vertices by drawing.

Solution by maxamc

See also

1999 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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All AIME Problems and Solutions

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