Difference between revisions of "2014 AIME I Problems/Problem 15"
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<cmath>r = \frac {25 \cdot \sqrt{2}}{28}\implies ED = \frac {25 \cdot \sqrt{2}}{14}\implies \boxed{\textbf{041}}.</cmath> | <cmath>r = \frac {25 \cdot \sqrt{2}}{28}\implies ED = \frac {25 \cdot \sqrt{2}}{14}\implies \boxed{\textbf{041}}.</cmath> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 6== | ||
+ | Let <math>(BEFGD) = \omega</math>. | ||
+ | By Incenter-Excenter(Fact <math>5</math>), <math>F</math> is the angle bisector of <math>\angle B</math>. | ||
+ | Then by Ratio Lemma we have | ||
+ | <cmath>\frac{AG}{CG} = \frac{\sin(ABG)}{\sin(CBG)} \cdot \frac{AB}{BC} = \frac{\sin(GDE)}{\sin(DEG)} \cdot \frac{3}{4} = 1</cmath> | ||
+ | Thus, <math>G</math> is the midpoint of <math>AC</math>. | ||
+ | |||
+ | We can calculate <math>AF</math> and <math>CF</math> to be <math>\frac{15}{7}</math> and <math>\frac{20}{7}</math> respectively. | ||
+ | And then by Power of a Point, we have | ||
+ | <math>\newline</math> | ||
+ | <cmath>\operatorname{Pow}_{\omega}(A) = AE \cdot AB = AF \cdot AG \implies AE = \frac{25}{14}</cmath> | ||
+ | And then similarly, we have <math>CD = AE = \frac{25}{14}</math>. | ||
+ | <math>\newline</math> | ||
+ | |||
+ | Then <math>EB = \frac{17}{14}</math> and <math>DB = \frac{31}{14}</math> and by Pythagorean we have <math>DE = \\frac{25\sqrt{2}}{14}</math>, so our answer is <math>\boxed{\textbf{041}}.</math> | ||
==Video Solution by mop 2024== | ==Video Solution by mop 2024== |
Revision as of 22:28, 28 January 2024
Contents
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 . ~First
Solution 4
See inside the , we can find that since if , we can see that Ptolemy Theorem inside cyclic quadrilateral doesn't work. Now let's see when , since , we can assume that , since we know so is isosceles right triangle. We can denote .Applying Ptolemy Theorem inside the cyclic quadrilateral we can get the length of can be represented as . After observing, we can see , whereas so we can see is isosceles triangle. Since is a triangle so we can directly know that the length of AF can be written in the form of . Denoting a point on side with that is perpendicular to side . Now with the same reason, we can see that is a isosceles right triangle, so we can get while the segment is since its 3-4-5 again. Now adding all those segments together we can find that and and the desired which our answer is ~bluesoul
Solution 5
The main element of the solution is the proof that is bisector of
Let be the midpoint of
is the center of the circle BF is bisector of Let vladimir.shelomovskii@gmail.com, vvsss
Solution 6
The main element of the solution is the proof that is midpoint of
As in Solution 5 we get
is isosceles triangle with
Similarly
Let vladimir.shelomovskii@gmail.com, vvsss
Solution 6
Let . By Incenter-Excenter(Fact ), is the angle bisector of . Then by Ratio Lemma we have Thus, is the midpoint of .
We can calculate and to be and respectively. And then by Power of a Point, we have And then similarly, we have .
Then and and by Pythagorean we have , so our answer is
Video Solution by mop 2024
~r00tsOfUnity
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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