Difference between revisions of "2009 AIME I Problems/Problem 4"
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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. | ||
− | So the whole length of the line<math>APC(AMC or ANC), and ABC</math> is <math>1000x+2009x=3009x</math> | + | So the whole length of the line <math>APC</math>(<math>AMC</math> or <math>ANC</math>), and <math>ABC</math> is <math>1000x+2009x=3009x</math> |
− | And <math>AP(AM or AN | + | And <math>AP</math>(<math>AM</math> or <math>AN</math>) is <math>17x</math> |
− | So the answer is <math>3009x/17x = 177</math> | + | So the answer is <math>3009x/17x = \boxed{177}</math> |
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2009|n=I|num-b=3|num-a=5}} |
Revision as of 20:10, 20 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
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
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 |