Difference between revisions of "2009 AIME II Problems/Problem 10"
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== Solution 1== | == Solution 1== | ||
− | Let <math>O</math> be the intersection of <math>BC</math> and <math>AD</math>. By the [[Angle Bisector Theorem]], <math>\frac {5}{BO}</math> = <math>\frac {13}{CO}</math>, so <math>BO</math> = <math>5x</math> and <math>CO</math> = <math>13x</math>, and <math>BO</math> + <math>OC</math> = <math>BC</math> = <math>12</math>, so <math>x</math> = <math>\frac {2}{3}</math>, and <math>OC</math> = <math>\frac {26}{3}</math>. Let <math>P</math> be the foot of the altitude from <math>D</math> to <math>OC</math>. It can be seen that triangle <math>DOP</math> is similar to triangle <math>AOB</math>, and triangle <math>DPC</math> is similar to triangle <math>ABC</math>. If <math>DP</math> = <math>15y</math>, then <math>CP</math> = <math>36y</math>, <math>OP</math> = <math>10y</math>, and <math>OD</math> = <math>5y\sqrt {13}</math>. Since <math>OP</math> + <math>CP</math> = <math>46y</math> = <math>\frac {26}{3}</math>, <math>y</math> = <math>\frac {13}{69}</math>, and <math>AD</math> = <math>\frac {60\sqrt{13}}{23}</math>. The answer is <math>60</math> + <math>13</math> + <math>23</math> = <math>\boxed{096}</math>. | + | Let <math>O</math> be the intersection of <math>BC</math> and <math>AD</math>. By the [[Angle Bisector Theorem]], <math>\frac {5}{BO}</math> = <math>\frac {13}{CO}</math>, so <math>BO</math> = <math>5x</math> and <math>CO</math> = <math>13x</math>, and <math>BO</math> + <math>OC</math> = <math>BC</math> = <math>12</math>, so <math>x</math> = <math>\frac {2}{3}</math>, and <math>OC</math> = <math>\frac {26}{3}</math>. Let <math>P</math> be the foot of the altitude from <math>D</math> to <math>OC</math>. It can be seen that triangle <math>DOP</math> is similar to triangle <math>AOB</math>, and triangle <math>DPC</math> is similar to triangle <math>ABC</math>. If <math>DP</math> = <math>15y</math>, then <math>CP</math> = <math>36y</math>, <math>OP</math> = <math>10y</math>, and <math>OD</math> = <math>5y\sqrt {13}</math>. Since <math>OP</math> + <math>CP</math> = <math>46y</math> = <math>\frac {26}{3}</math>, <math>y</math> = <math>\frac {13}{69}</math>, and <math>AD</math> = <math>\frac {60\sqrt{13}}{23}</math> (by the pythagorean theorem on triangle ABO). The answer is <math>60</math> + <math>13</math> + <math>23</math> = <math>\boxed{096}</math>. |
==Solution 2== | ==Solution 2== |
Revision as of 13:03, 14 August 2015
Four lighthouses are located at points , , , and . The lighthouse at is kilometers from the lighthouse at , the lighthouse at is kilometers from the lighthouse at , and the lighthouse at is kilometers from the lighthouse at . To an observer at , the angle determined by the lights at and and the angle determined by the lights at and are equal. To an observer at , the angle determined by the lights at and and the angle determined by the lights at and are equal. The number of kilometers from to is given by , where , , and are relatively prime positive integers, and is not divisible by the square of any prime. Find + + .
Solution 1
Let be the intersection of and . By the Angle Bisector Theorem, = , so = and = , and + = = , so = , and = . Let be the foot of the altitude from to . It can be seen that triangle is similar to triangle , and triangle is similar to triangle . If = , then = , = , and = . Since + = = , = , and = (by the pythagorean theorem on triangle ABO). The answer is + + = .
Solution 2
Extend and to intersect at . Note that since and by ASA congruency we have . Therefore .
By the angle bisector theorem, and . Now we apply Stewart's theorem to find :
and our final answer is .
See Also
2009 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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