Difference between revisions of "2009 AIME I Problems/Problem 4"
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In parallelogram <math>ABCD</math>, point <math>M</math> is on <math>\overline{AB}</math> so that <math>\frac {AM}{AB} = \frac {17}{1000}</math> and point <math>N</math> is on <math>\overline{AD}</math> so that <math>\frac {AN}{AD} = \frac {17}{2009}</math>. Let <math>P</math> be the point of intersection of <math>\overline{AC}</math> and <math>\overline{MN}</math>. Find <math>\frac {AC}{AP}</math>. | In parallelogram <math>ABCD</math>, point <math>M</math> is on <math>\overline{AB}</math> so that <math>\frac {AM}{AB} = \frac {17}{1000}</math> and point <math>N</math> is on <math>\overline{AD}</math> so that <math>\frac {AN}{AD} = \frac {17}{2009}</math>. Let <math>P</math> be the point of intersection of <math>\overline{AC}</math> and <math>\overline{MN}</math>. Find <math>\frac {AC}{AP}</math>. | ||
− | ==Solution== | + | ==Solution 1== |
− | = | + | One of the ways to solve this problem is to make this parallelogram a straight line. |
+ | So the whole length of the line is <math>APC</math>(<math>AMC</math> or <math>ANC</math>), and <math>ABC</math> is <math>1000x+2009x=3009x.</math> | ||
− | + | <math>AP</math>(<math>AM</math> or <math>AN</math>) is <math>17x.</math> | |
− | So the | + | So the answer is <math>3009x/17x = \boxed{177}</math> |
− | + | ==Solution 2== | |
− | |||
− | |||
Draw a diagram with all the given points and lines involved. Construct parallel lines <math>\overline{DF_2F_1}</math> and <math>\overline{BB_1B_2}</math> to <math>\overline{MN}</math>, where for the lines the endpoints are on <math>\overline{AM}</math> and <math>\overline{AN}</math>, respectively, and each point refers to an intersection. Also, draw the median of quadrilateral <math>BB_2DF_1</math> <math>\overline{E_1E_2E_3}</math> where the points are in order from top to bottom. Clearly, by similar triangles, <math>BB_2 = \frac {1000}{17}MN</math> and <math>DF_1 = \frac {2009}{17}MN</math>. It is not difficult to see that <math>E_2</math> is the center of quadrilateral <math>ABCD</math> and thus the midpoint of <math>\overline{AC}</math> as well as the midpoint of <math>\overline{B_1}{F_2}</math> (all of this is easily proven with symmetry). From more triangle similarity, <math>E_1E_3 = \frac12\cdot\frac {3009}{17}MN\implies AE_2 = \frac12\cdot\frac {3009}{17}AP\implies AC = 2\cdot\frac12\cdot\frac {3009}{17}AP</math> | Draw a diagram with all the given points and lines involved. Construct parallel lines <math>\overline{DF_2F_1}</math> and <math>\overline{BB_1B_2}</math> to <math>\overline{MN}</math>, where for the lines the endpoints are on <math>\overline{AM}</math> and <math>\overline{AN}</math>, respectively, and each point refers to an intersection. Also, draw the median of quadrilateral <math>BB_2DF_1</math> <math>\overline{E_1E_2E_3}</math> where the points are in order from top to bottom. Clearly, by similar triangles, <math>BB_2 = \frac {1000}{17}MN</math> and <math>DF_1 = \frac {2009}{17}MN</math>. It is not difficult to see that <math>E_2</math> is the center of quadrilateral <math>ABCD</math> and thus the midpoint of <math>\overline{AC}</math> as well as the midpoint of <math>\overline{B_1}{F_2}</math> (all of this is easily proven with symmetry). From more triangle similarity, <math>E_1E_3 = \frac12\cdot\frac {3009}{17}MN\implies AE_2 = \frac12\cdot\frac {3009}{17}AP\implies AC = 2\cdot\frac12\cdot\frac {3009}{17}AP</math> | ||
<math>= \boxed{177}AP</math>. | <math>= \boxed{177}AP</math>. | ||
− | + | ==Solution 3== | |
+ | |||
+ | Using vectors, note that <math>\overrightarrow{AM}=\frac{17}{1000}\overrightarrow{AB}</math> and <math>\overrightarrow{AN}=\frac{17}{2009}\overrightarrow{AD}</math>. Note that <math>\overrightarrow{AP}=\frac{x\overrightarrow{AM}+y\overrightarrow{AN}}{x+y}</math> for some positive x and y, but at the same time is a scalar multiple of <math>\overrightarrow{AB}+\overrightarrow{AD}</math>. So, writing the equation <math>\overrightarrow{AP}=\frac{x\overrightarrow{AM}+y\overrightarrow{AN}}{x+y}</math> in terms of <math>\overrightarrow{AB}</math> and <math>\overrightarrow{AD}</math>, we have <math>\overrightarrow{AP}=\frac{\frac{17x}{1000}\overrightarrow{AB}+\frac{17y}{2009}\overrightarrow{AD}}{x+y}</math>. But the coefficients of the two vectors must be equal because, as already stated, <math>\overrightarrow{AP}</math> is a scalar multiple of <math>\overrightarrow{AB}+\overrightarrow{AD}</math>. We then see that <math>\frac{x}{x+y}=\frac{1000}{3009}</math> and <math>\frac{y}{x+y}=\frac{2009}{3009}</math>. Finally, we have <math>\overrightarrow{AP}=\frac{17}{3009}(\overrightarrow{AB}+\overrightarrow{AD})</math> and, simplifying, <math>\overrightarrow{AB}+\overrightarrow{AD}=177\overrightarrow{AP}</math> and the desired quantity is <math>177</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | We approach the problem using mass points on triangle <math>ABD</math> as displayed below. | ||
+ | <asy> | ||
+ | |||
+ | pair A=(0,0),B=(50,0),D=(10,40),C=B+D,M=(8,0),NN=(2,8); | ||
+ | draw(A--B--C--D--cycle); | ||
+ | draw(B--D^^A--C^^M--NN); | ||
+ | pair O=extension(A,C,B,D); | ||
+ | pair P=extension(A,C,M,NN); | ||
+ | dot(A);dot(B);dot(C);dot(D);dot(O);dot(M);dot(NN);dot(P); | ||
+ | label("$A$",A,SW); | ||
+ | label("$B$",B,SE); | ||
+ | label("$C$",C,NE); | ||
+ | label("$D$",D,NW); | ||
+ | label("$M$",M,S); | ||
+ | label("$N$",NN,NW); | ||
+ | label("$P$",P,NNE); | ||
+ | label("$O$",O,N); | ||
+ | </asy> | ||
+ | |||
+ | But as <math>MN</math> does not protrude from a vertex, we will have to "split the mass" at point <math>A</math>. First, we know that <math>DO</math> is congruent to <math>BO</math> because diagonals of parallelograms bisect each other. Therefore, we can assign equal masses to points <math>B</math> and <math>D</math>. In this case, we assign <math>B</math> and <math>D</math> a mass of 17 each. Now we split the mass at <math>A</math>, so we balance segments <math>AB</math> and <math>AD</math> separately, and then the mass of <math>A</math> is the sum of those masses. A mass of 983 is required to balance segment <math>AB</math>, while a mass of 1992 is required to balance segment <math>AD</math>. Therefore, <math>A</math> has a mass of <math>1992+983=2975</math>. Also, <math>O</math> has a mass of 34. Therefore, <math>\frac{AO}{AP}=\frac{2975+34}{34}=\frac{3009}{34}</math>, so <math>\frac{AC}{AP}=\frac{2 (3009)}{34}=177</math>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | Assume, for the ease of computation, that <math>AM=AN=17</math>, <math>AB=1000</math>, and <math>AD=2009</math>. Now, let line <math>MN</math> intersect line <math>CD</math> at point <math>X</math> and let <math>Y</math> be a point such that <math>XY\parallel AD</math> and <math>AY\parallel DX</math>. As a result, <math>ADXY</math> is a parallelogram. By construction, <math>\triangle MAN\sim \triangle MYX</math> so <cmath>\frac{MY}{MA}=\frac{YX}{AN}=\frac{AD}{AN}=\frac{2009}{17}\implies MY=2009</cmath> and <math>AY=DX=2009-17</math>. Also, because <math>AM\parallel XC</math>, we have <math>\triangle PAM\sim \triangle PCX</math> so <cmath>\frac{PC}{PA}=\frac{CX}{AM}=\frac{DX+CD}{AM}=\frac{2009-17+1000}{17}=176.</cmath> Hence, <math>\frac{AC}{AP}=\frac{PC}{PA}+1=\boxed{177}.</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | Unique solution: https://youtu.be/2Xzjh6ae0MU | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/kALrIDMR0dg | ||
+ | |||
+ | ~Shreyas S | ||
== See also == | == See also == | ||
{{AIME box|year=2009|n=I|num-b=3|num-a=5}} | {{AIME box|year=2009|n=I|num-b=3|num-a=5}} | ||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 17:19, 18 June 2020
Contents
Problem 4
In parallelogram , point is on so that and point is on so that . Let be the point of intersection of and . Find .
Solution 1
One of the ways to solve this problem is to make this parallelogram a straight line. So the whole length of the line is ( or ), and is
( or ) is
So the answer is
Solution 2
Draw a diagram with all the given points and lines involved. Construct parallel lines and to , where for the lines the endpoints are on and , respectively, and each point refers to an intersection. Also, draw the median of quadrilateral where the points are in order from top to bottom. Clearly, by similar triangles, and . It is not difficult to see that is the center of quadrilateral and thus the midpoint of as well as the midpoint of (all of this is easily proven with symmetry). From more triangle similarity, .
Solution 3
Using vectors, note that and . Note that for some positive x and y, but at the same time is a scalar multiple of . So, writing the equation in terms of and , we have . But the coefficients of the two vectors must be equal because, as already stated, is a scalar multiple of . We then see that and . Finally, we have and, simplifying, and the desired quantity is .
Solution 4
We approach the problem using mass points on triangle as displayed below.
But as does not protrude from a vertex, we will have to "split the mass" at point . First, we know that is congruent to because diagonals of parallelograms bisect each other. Therefore, we can assign equal masses to points and . In this case, we assign and a mass of 17 each. Now we split the mass at , so we balance segments and separately, and then the mass of is the sum of those masses. A mass of 983 is required to balance segment , while a mass of 1992 is required to balance segment . Therefore, has a mass of . Also, has a mass of 34. Therefore, , so .
Solution 5
Assume, for the ease of computation, that , , and . Now, let line intersect line at point and let be a point such that and . As a result, is a parallelogram. By construction, so and . Also, because , we have so Hence,
Video Solution
Unique solution: https://youtu.be/2Xzjh6ae0MU
~IceMatrix
Video Solution
~Shreyas S
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.