2022 AMC 12A Problems/Problem 20
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[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 Ptolemy's theorem Thus and ; dividing these two equations and taking the reciprocal yields .
Solution 2 (Coordinate Bashing)
Since we're given distances and nothing else, we can represent each point as a coordinate and use the distance formula to set up a series of systems and equations. Let the height of the trapezoid be , and let the coordinates of and be at and , respectively. Then let and be at and , respectively. This follows the rules that this is an isosceles trapezoid since the origin is centered on the middle of . Finally, let be located at point .
The distance from to is , so by the distance formula: The distance from to is , so
Looking at these two equations alone, notice that the second term is the same for both equations, so we can subtract the equations. This yields
Next, the distance from to is , so The distance from to is , so
Again, we can subtract these equations, yielding
We can now divide the equations to eliminate , yielding
We wanted to find . But since is half of and is half of , this ratio is equal to the ratio we want.
Therefore
~KingRavi
Solution 3 (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
Solution 4 (Coordinate Bashing 2)
Let the point be at the origin, and draw four concentric circles around each with radius , , , and , respectively. The vertices of the trapezoid would be then on each of the four concentric circles. WLOG, let and be parallel to the x-axis. Assigning coordinates to each point, we have: which satisfy the following: In addition, because the trapezoid is isosceles (), the midpoints of the two bases would then have the same x-coordinate, giving us Subtracting Equation from Equation , and Equation from Equation , we have
Dividing Equation by Equation , we have
Cancelling and with Equation , we get
In other words,
~G63566
Video Solution By ThePuzzlr
~ MathIsChess
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=uYXtEzX4fb0
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 |
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