Difference between revisions of "2009 AIME I Problems/Problem 4"
Mathisfun286 (talk | contribs) m (→Coordinate Geometry) |
m (→Solution 6(Coordinate Geometry)) |
||
(One intermediate revision by one other user not shown) | |||
Line 52: | Line 52: | ||
Then, since <math>\frac{AM}{AB} = \frac{17}{1000}</math> and <math>AB = 1000</math>, we can place <math>M</math> at <math>(17, 0).</math> Similarly, place <math>AN</math> at <math>(0, 17).</math> | Then, since <math>\frac{AM}{AB} = \frac{17}{1000}</math> and <math>AB = 1000</math>, we can place <math>M</math> at <math>(17, 0).</math> Similarly, place <math>AN</math> at <math>(0, 17).</math> | ||
− | Then, the equation of line <math>MN</math> is <math>y=x-17,</math> and the equation of <math>AC</math> is <math>y=\frac{-2009}{1000}x.</math> Solve to find point <math>P</math> at <math>\left( \frac{1000}{ | + | Then, the equation of line <math>MN</math> is <math>y=x-17,</math> and the equation of <math>AC</math> is <math>y=\frac{-2009}{1000}x.</math> Solve to find point <math>P</math> at <math>\left( \frac{1000}{177}, \frac{-2009}{177} \right)</math>. |
We can calculate vectors to represent the distances: <cmath>\overrightarrow{AC}= <1000, -2009></cmath> <cmath>\overrightarrow{AP}= \frac{1}{177}<1000, -2009>.</cmath> In this way, we can see that <cmath>AC:AP = 177:1,</cmath> and our answer is <math>\boxed{177}.</math> | We can calculate vectors to represent the distances: <cmath>\overrightarrow{AC}= <1000, -2009></cmath> <cmath>\overrightarrow{AP}= \frac{1}{177}<1000, -2009>.</cmath> In this way, we can see that <cmath>AC:AP = 177:1,</cmath> and our answer is <math>\boxed{177}.</math> | ||
~ [https://artofproblemsolving.com/community/user/357326 HappyHuman] | ~ [https://artofproblemsolving.com/community/user/357326 HappyHuman] | ||
+ | |||
+ | ===Note=== | ||
+ | It is possible to use [[coordinate geometry]] without setting <math>ABCD</math> as a rectangle, either by projecting the plane onto another (tilted) plane or removing the restriction that the [[axes]] have to be [[perpendicular]]. | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 00:23, 9 January 2023
Contents
Problem
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,
Solution 6(Coordinate Geometry)
Assign . Since there are no constraints in the problem against this, assume
to be a rectangle with dimensions
Now, we can assign
Then, since and
, we can place
at
Similarly, place
at
Then, the equation of line
is
and the equation of
is
Solve to find point
at
.
We can calculate vectors to represent the distances:
In this way, we can see that
and our answer is
Note
It is possible to use coordinate geometry without setting as a rectangle, either by projecting the plane onto another (tilted) plane or removing the restriction that the axes have to be perpendicular.
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.