Difference between revisions of "2009 AIME I Problems/Problem 4"
Geowhiz1116 (talk | contribs) m (→Solution 2) |
Aeryde.xin (talk | contribs) (→Solution) |
||
Line 6: | Line 6: | ||
===Solution 1=== | ===Solution 1=== | ||
− | One of the ways to solve this problem is to make this parallelogram a straight | + | One of the ways to solve this problem is to make this parallelogram a straight lin |
− | |||
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> | 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> | ||
Line 16: | Line 15: | ||
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>. | ||
== 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}} |
Revision as of 17:31, 24 August 2012
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 lin 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, .
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. ! Extra }, or forgotten $.) and the desired quantity 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 |