Difference between revisions of "2021 AMC 12B Problems/Problem 24"
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==Solution 2(Trig) == | ==Solution 2(Trig) == | ||
− | Let <math>X</math> denote the intersection point of the diagonals <math>AC</math> and <math>BD</math> | + | Let <math>X</math> denote the intersection point of the diagonals <math>AC</math> and <math>BD,</math> and let <math>\theta = \angle{COB}</math>. Then, by the given conditions, <math>XR = 4,</math> <math>XQ = 3,</math> <math>[XCB] = \frac{15}{4}</math>. So, |
<cmath> XC = \frac{3}{\cos \theta}</cmath> | <cmath> XC = \frac{3}{\cos \theta}</cmath> | ||
<cmath> XB \cos \theta = 4 </cmath> | <cmath> XB \cos \theta = 4 </cmath> | ||
Line 26: | Line 26: | ||
Combining the above 3 equations, we get | Combining the above 3 equations, we get | ||
<cmath>\frac{\sin \theta }{\cos^2 \theta} = \frac{5}{8}.</cmath> | <cmath>\frac{\sin \theta }{\cos^2 \theta} = \frac{5}{8}.</cmath> | ||
− | Since we want to find <math>d^2 = 4XB^2 = \frac{64}{\cos^2 \theta}</math> | + | Since we want to find <math>d^2 = 4XB^2 = \frac{64}{\cos^2 \theta},</math> we let <math>x = \frac{1}{\cos^2 \theta}.</math> Then |
<cmath> \frac{\sin^2 \theta }{\cos^4 \theta} = \frac{1-\cos ^2 \theta}{\cos^4 \theta} = x^2 - x = \frac{25}{64}.</cmath> | <cmath> \frac{\sin^2 \theta }{\cos^4 \theta} = \frac{1-\cos ^2 \theta}{\cos^4 \theta} = x^2 - x = \frac{25}{64}.</cmath> | ||
− | Solving this, we get <math>x = \frac{4 + \sqrt{41}}{8}</math> | + | Solving this, we get <math>x = \frac{4 + \sqrt{41}}{8},</math> so <math>d^2 = 64x = 32 + 8\sqrt{41}</math>. <math>\boxed{81}</math> |
~sequoia | ~sequoia |
Revision as of 02:08, 17 February 2021
Contents
Problem
Let be a parallelogram with area . Points and are the projections of and respectively, onto the line and points and are the projections of and respectively, onto the line See the figure, which also shows the relative locations of these points.
Suppose and and let denote the length of the longer diagonal of Then can be written in the form where and are positive integers and is not divisible by the square of any prime. What is
simple Video Solution Using trigonometry and Equations
https://youtu.be/ZB-VN02H6mU ~hippopotamus1
Solution
Let denote the intersection point of the diagonals and . Remark that by symmetry is the midpoint of both and , so and . Now note that since , quadrilateral is cyclic, and so which implies .
Thus let be such that and . Then Pythagorean Theorem on yields , and soSolving this for yields , and soThe requested answer is .
Solution 2(Trig)
Let denote the intersection point of the diagonals and and let . Then, by the given conditions, . So, Combining the above 3 equations, we get Since we want to find we let Then Solving this, we get so .
~sequoia
Video Solution by OmegaLearn (Cyclic Quadrilateral and Power of a Point)
~ pi_is_3.14
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.