Difference between revisions of "2009 AIME I Problems/Problem 4"
(→Solution) |
|||
| Line 4: | Line 4: | ||
[[2009 AIME I Problems/Problem 4|Solution]] | [[2009 AIME I Problems/Problem 4|Solution]] | ||
| − | ==Solution 1== | + | ==Solution== |
| + | |||
| + | ===Solution 1=== | ||
One of the ways to solve this problem is to make this parallelogram a straight line. | One of the ways to solve this problem is to make this parallelogram a straight line. | ||
| Line 13: | Line 15: | ||
So the answer is <math>3009x/17x = \boxed{177}</math> | So the answer is <math>3009x/17x = \boxed{177}</math> | ||
| − | ==Solution 2== | + | ===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>. | ||
Revision as of 14:37, 23 March 2009
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 
(
or 
), and 
is 
And 
(
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, 
.
A diagram would be appreciated (and whoever makes it can take out the descriptive stuff).
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 | ||
