Difference between revisions of "2022 AMC 12A Problems/Problem 20"
Pi is 3.14 (talk | contribs) (→Video Solution by OmegaLearn) |
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==Video Solution== | ==Video Solution== |
Revision as of 09:45, 12 November 2022
Contents
[hide]Problem
Isosceles trapezoid has parallel sides and with and There is a point in the plane such that and What is
Solution 1
Consider the reflection of over the perpendicular bisector of , creating two new isosceles trapezoids and . Under this reflection, , , , and . By Ptolmey's theorem Thus and ; dividing these two equations and taking the reciprocal yields .
Solution 2 (Cheese)
Notice that the question never says what the height of the trapezoid is; the only property we know about it is that . Therefore, we can say WLOG that the height of the trapezoid is and all points, including , lie on the same line with . Notice that this satisfies the problem requirements because , and . Now all we have to find is
~KingRavi
Video Solution By ThePuzzlr
~ MathIsChess
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by OmegaLearn using Pythagorean Theorem
~ pi_is_3.14
See also
2022 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 |
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.