Difference between revisions of "2021 AMC 12B Problems/Problem 24"
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<math>\textbf{(A) }81 \qquad \textbf{(B) }89 \qquad \textbf{(C) }97\qquad \textbf{(D) }105 \qquad \textbf{(E) }113</math> | <math>\textbf{(A) }81 \qquad \textbf{(B) }89 \qquad \textbf{(C) }97\qquad \textbf{(D) }105 \qquad \textbf{(E) }113</math> | ||
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==Solution 1== | ==Solution 1== | ||
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==simple Video Solution Using trigonometry and Equations== | ==simple Video Solution Using trigonometry and Equations== | ||
https://youtu.be/ZB-VN02H6mU | https://youtu.be/ZB-VN02H6mU |
Latest revision as of 00:03, 24 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
Solution 1
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 .
Solution 3 (Similar Triangles and Algebra)
Let be the intersection of diagonals and . By symmetry , and , so now we have reduced all of the conditions one quadrant. Let . , by similar triangles and using the area condition we get . Note that it suffices to find because we can double and square it to get . Solving for in the above equation, and then using .
Solution 4 (Similar Triangles)
Again, Let be the intersection of diagonals and . Note that triangles and are similar because they are right triangles and share . First, call the length of . By the definition of an area of a parallelogram, , so . Using similar triangles on and , . Therefore, finding , . Now, applying the Pythagorean theorem once, we find + = . Solving this equation for , we find .
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.