Difference between revisions of "2009 AIME I Problems/Problem 4"
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===Solution 3=== | ===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>. | 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>AC</math> bisects <math>BD</math>. Therefore, we 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=2965</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{3009}{17}=177</math>. | ||
== 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:30, 6 July 2013
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
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, $\overrightarrow{AB}+\overrightarrow{AD}}=177\overrightarrow{AP}$ (Error compiling LaTeX. Unknown error_msg) 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
bisects
. Therefore, we 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
.
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.