Difference between revisions of "2001 AIME II Problems/Problem 4"
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== Solution == | == Solution == | ||
| − | + | The coordinates of P can be written as <math>(a, 15a/8)</math> and the coordinates of point Q can be written as <math>(b,3b/10)</math>. By the midpoint formula, we have <math>(a+b)/2=8</math> and <math>15a/16+3b/20=6</math>. Solving for b gives <math>b=80/7</math>, so the point Q is <math>(80/7,24/7)</math>. The answer is twice the distance from Q to <math>(8,6)</math>, which by the distance formula is <math>60/7</math>. Thus, the answer is 67. | |
== See also == | == See also == | ||
{{AIME box|year=2001|n=II|num-b=3|num-a=5}} | {{AIME box|year=2001|n=II|num-b=3|num-a=5}} | ||
Revision as of 15:10, 15 January 2008
Problem
Let 
. The lines whose equations are 
and 
contain points 
and 
, respectively, such that 
is the midpoint of 
. The length of 
equals 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
The coordinates of P can be written as 
and the coordinates of point Q can be written as 
. By the midpoint formula, we have 
and 
. Solving for b gives 
, so the point Q is 
. The answer is twice the distance from Q to 
, which by the distance formula is 
. Thus, the answer is 67.
See also
| 2001 AIME II (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 | ||
