Difference between revisions of "2014 AIME I Problems/Problem 15"
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Lastly, we apply power of a point from points <math>A</math> and <math>C</math> with respect to <math>\omega</math> and have <math>AE \times AB=AF \times AG</math> and <math>CD \times CB=CG \times CF</math>, so we can compute that <math>EB = \frac{17}{14}</math> and <math>DB = \frac{31}{14}</math>. From the Pythagorean Theorem, we result in <math>DE = \frac{25 \sqrt{2}}{14}</math>, so <math>a+b+c=25+2+14= \boxed{041}</math> | Lastly, we apply power of a point from points <math>A</math> and <math>C</math> with respect to <math>\omega</math> and have <math>AE \times AB=AF \times AG</math> and <math>CD \times CB=CG \times CF</math>, so we can compute that <math>EB = \frac{17}{14}</math> and <math>DB = \frac{31}{14}</math>. From the Pythagorean Theorem, we result in <math>DE = \frac{25 \sqrt{2}}{14}</math>, so <math>a+b+c=25+2+14= \boxed{041}</math> | ||
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Also: <math>FG=\frac{20}{7}-\frac{5}{2}=\frac{5}{2}-\frac{15}{7}=\frac{5}{14}</math>. We can also use Ptolemy's Theorem on quadrilateral <math>DEFG</math> to figure what <math>FG</math> is in terms of <math>\omega</math>: | Also: <math>FG=\frac{20}{7}-\frac{5}{2}=\frac{5}{2}-\frac{15}{7}=\frac{5}{14}</math>. We can also use Ptolemy's Theorem on quadrilateral <math>DEFG</math> to figure what <math>FG</math> is in terms of <math>\omega</math>: | ||
<cmath>DG\cdot EF+EF\cdot DG=DF\cdot EG</cmath> | <cmath>DG\cdot EF+EF\cdot DG=DF\cdot EG</cmath> | ||
<cmath>\frac{3\omega}{5}\cdot \frac{\omega\sqrt{2}}{2}+\omega\cdot FG=\frac{4\omega}{5}\cdot \frac{\omega\sqrt{2}}{2}</cmath> | <cmath>\frac{3\omega}{5}\cdot \frac{\omega\sqrt{2}}{2}+\omega\cdot FG=\frac{4\omega}{5}\cdot \frac{\omega\sqrt{2}}{2}</cmath> | ||
− | <cmath>\frac{3\omega\sqrt{2}}{10}+EF | + | <cmath>\frac{3\omega\sqrt{2}}{10}+\omega\cdot EF=(\frac{4\omega\sqrt{2}}{10}\implies FG=\frac{\omega\sqrt{2}}{10}</cmath> |
Thus <math>\frac{\omega\sqrt{2}}{10}=\frac{5}{14}\rightarrow \omega=5\sqrt{2}\cdot\frac{5}{14}=\frac{25\sqrt{2}}{14}</math>. <math>a+b+c=25+2+14= \boxed{041}</math> | Thus <math>\frac{\omega\sqrt{2}}{10}=\frac{5}{14}\rightarrow \omega=5\sqrt{2}\cdot\frac{5}{14}=\frac{25\sqrt{2}}{14}</math>. <math>a+b+c=25+2+14= \boxed{041}</math> | ||
Revision as of 15:01, 14 February 2016
Problem 15
In , , , and . Circle intersects at and , at and , and at and . Given that and , length , where and are relatively prime positive integers, and is a positive integer not divisible by the square of any prime. Find .
Solution
First we note that is an isosceles right triangle with hypotenuse the same as the diameter of . We also note that since is a right angle and the ratios of the sides are .
From congruent arc intersections, we know that , and that from similar triangles is also congruent to . Thus, is an isosceles triangle with , so is the midpoint of and . Similarly, we can find from angle chasing that . Therefore, is the angle bisector of . From the angle bisector theorem, we have , so and .
Lastly, we apply power of a point from points and with respect to and have and , so we can compute that and . From the Pythagorean Theorem, we result in , so
Also: . We can also use Ptolemy's Theorem on quadrilateral to figure what is in terms of :
Thus .
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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