Difference between revisions of "2006 AIME II Problems/Problem 12"
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Solution by e_power_pi_times_i/edited by srisainandan6 | Solution by e_power_pi_times_i/edited by srisainandan6 | ||
− | Let the center of the circle be <math>O</math> and the origin. Then, <math>A (0,2)</math>, <math>B (-\sqrt{3}, -1)</math>, <math>C (\sqrt{3}, -1)</math>. <math>D</math> and <math>E</math> can be calculated easily knowing <math>AD</math> and <math>AE</math>, <math>D (-\dfrac{13}{2}, \dfrac{-13\sqrt{3}+4}{2})</math>, <math>E (\dfrac{11}{2}, \dfrac{-11\sqrt{3}+4}{2})</math>. As <math>DF</math> and <math>EF</math> are parallel to <math>AE</math> and <math>AD</math>, <math>F (-1, -12\sqrt{3}+2)</math>. <math>G</math> and <math>A</math> is the intersection between <math>AF</math> and circle <math>O</math>. Therefore <math>G (-\dfrac{48\sqrt{3}}{433}, -\dfrac{862}{433})</math>. Using the Shoelace Theorem, <math>[CBG] = \dfrac{429\sqrt{3}}{433}</math>, so the answer is <math>\boxed{865}</math> | + | Let the center of the circle be <math>O</math> and the origin. Then, <math>A (0,2)</math>, <math>B (-\sqrt{3}, -1)</math>, <math>C (\sqrt{3}, -1)</math>. <math>D</math> and <math>E</math> can be calculated easily knowing <math>AD</math> and <math>AE</math>, <math>D (-\dfrac{13}{2}, \dfrac{-13\sqrt{3}+4}{2})</math>, <math>E (\dfrac{11}{2}, \dfrac{-11\sqrt{3}+4}{2})</math>. As <math>DF</math> and <math>EF</math> are parallel to <math>AE</math> and <math>AD</math>, <math>F (-1, -12\sqrt{3}+2)</math>. <math>G</math> and <math>A</math> is the intersection between <math>AF</math> and circle <math>O</math>. Therefore <math>G (-\dfrac{48\sqrt{3}}{433}, -\dfrac{862}{433})</math>. Using the Shoelace Theorem, <math>[CBG] = \dfrac{429\sqrt{3}}{433}</math>, so the answer is <math>\boxed{865}</math>. Note that although the solution may appear short, actually getting all the coordinates take a while as there is plenty of computation |
== Solution 3: Trig == | == Solution 3: Trig == |
Revision as of 11:42, 2 April 2020
Contents
Problem
Equilateral is inscribed in a circle of radius . Extend through to point so that and extend through to point so that Through draw a line parallel to and through draw a line parallel to Let be the intersection of and Let be the point on the circle that is collinear with and and distinct from Given that the area of can be expressed in the form where and are positive integers, and are relatively prime, and is not divisible by the square of any prime, find
Solution
Notice that because . Also, because they both correspond to arc . So .
Because the ratio of the area of two similar figures is the square of the ratio of the corresponding sides, . Therefore, the answer is .
Solution 2: Analytic Geometry/Cord Bash
Solution by e_power_pi_times_i/edited by srisainandan6
Let the center of the circle be and the origin. Then, , , . and can be calculated easily knowing and , , . As and are parallel to and , . and is the intersection between and circle . Therefore . Using the Shoelace Theorem, , so the answer is . Note that although the solution may appear short, actually getting all the coordinates take a while as there is plenty of computation
Solution 3: Trig
Lines and are constructed such that is a parallelogram, hence . Since is equilateral with angle of , angle is . Use law of cosines to find . Then use law of sines to find angle and . Next we use Ptolemy's Theorem on to find that . Next we use law of cosine on triangles and , solving for BG and CG respectively. Subtract the two equations and divide out a to find the value of . Next, , where R is radius of circle and angle . We already know sine of the angle so find cosine, hence we have found . At this point it is system of equation yielding and . Given , and by triangle, we can evaluate to find , to give answer = . The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.