2022 AMC 10A Problems/Problem 23
- The following problem is from both the 2022 AMC 10A #23 and 2022 AMC 12A #20, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1 (Reflections + Ptolemy's Theorem)
- 3 Solution 2 (Extensions + Stewart's Theorem)
- 4 Solution 3 (Coordinate Bashing)
- 5 Solution 4 (Coordinate Bashing)
- 6 Solution 5 (Cheese)
- 7 Solution 6 (Pythagorean Theorem)
- 8 Video Solution by OmegaLearn
- 9 Video Solution By ThePuzzlr
- 10 Video Solution by Punxsutawney Phil
- 11 Video Solution by Steven Chen
- 12 Video Solution by MRENTHUSIASM (English & Chinese)
- 13 See also
Problem
Isosceles trapezoid has parallel sides and with and There is a point in the plane such that and What is
Solution 1 (Reflections + Ptolemy's Theorem)
Consider the reflection of over the perpendicular bisector of , creating two new isosceles trapezoids and . Under this reflection, , , , and .
Since and are isosceles trapezoids, they are cyclic. Using Ptolemy's theorem on , we get that , so Then, using Ptolemy's theorem again on , we get that , so Thus, and ; dividing these two equations and taking the reciprocal yields .
Solution 2 (Extensions + Stewart's Theorem)
Extend and to a point as shown, and let . Then let and . Notice that by similar triangles.
By Stewart's theorem on and , we have
Subtracting, , and so .
~kred9 (minor edit by gwang2008)
Solution 3 (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 4 (Coordinate Bashing)
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
Solution 5 (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 6 (Pythagorean Theorem)
Consider the reflection of over the perpendicular bisector of . Let Similarly, vladimir.shelomovskii@gmail.com, vvsss
Video Solution by OmegaLearn
~ pi_is_3.14
Video Solution By ThePuzzlr
~ MathIsChess
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=uYXtEzX4fb0
Video Solution by Steven Chen
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by MRENTHUSIASM (English & Chinese)
~MRENTHUSIASM
See also
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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.