Difference between revisions of "2014 AIME I Problems/Problem 15"
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<cmath>d\cdot FG+\frac{3d^2}{5\sqrt{2}}=\frac{4d^2}{5\sqrt{2}}\implies FG=\frac{d}{5\sqrt{2}}</cmath> | <cmath>d\cdot FG+\frac{3d^2}{5\sqrt{2}}=\frac{4d^2}{5\sqrt{2}}\implies FG=\frac{d}{5\sqrt{2}}</cmath> | ||
Thus <math>\frac{d}{5\sqrt{2}}=\frac{5}{14}\rightarrow d=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{d}{5\sqrt{2}}=\frac{5}{14}\rightarrow d=5\sqrt{2}\cdot\frac{5}{14}=\frac{25\sqrt{2}}{14}</math>. <math>a+b+c=25+2+14= \boxed{041}</math> | ||
+ | |||
+ | ===Solution 3=== | ||
+ | Call <math>DE=x</math> and as a result <math>DG=EF=\frac{x\sqrt{2}}{2}, EF=\frac{3x}{5}, FD=\frac{4x}{5}</math>. Since <math>EFGD</math> is cyclic we just need to get <math>DG</math> and using LoS(for more detail see the <math>2</math>nd paragraph of Solution <math>2</math>) we get <math>AG=\frac{5}{2}</math> and using a similar argument(use LoS again) and subtracting you get <math>FG=\frac{5}{14}</math> so you can use Ptolemy to get <math>x=\frac{25\sqrt{2}}{14} \implies \boxed{041}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2014|n=I|num-b=14|after=Last Question}} | {{AIME box|year=2014|n=I|num-b=14|after=Last Question}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:08, 25 June 2017
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 1
Since , is the diameter of . Then . But , so is a 45-45-90 triangle. Letting , we have that , , and .
Note that by SAS similarity, so and . Since is a cyclic quadrilateral, and , implying that and are isosceles. As a result, , so and .
Finally, using the Pythagorean Theorem on , Solving for , we get that , so . Thus, the answer is .
Solution 2
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 .
Solution 3
Call and as a result . Since is cyclic we just need to get and using LoS(for more detail see the nd paragraph of Solution ) we get and using a similar argument(use LoS again) and subtracting you get so you can use Ptolemy to get .
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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