Difference between revisions of "2022 AMC 12A Problems/Problem 20"
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Consider the reflection <math>P^{\prime}</math> of <math>P</math> over the perpendicular bisector of <math>\overline{BC}</math>, creating two new isosceles trapezoids <math>DAPP^{\prime}</math> and <math>CBPP^{\prime}</math>. Under this reflection, <math>P^{\prime}A=PD=4</math>, <math>P^{\prime}D=PA=1</math>, <math>P^{\prime}C=PB=2</math>, and <math>P^{\prime}B=PC=3</math>. By Ptolmey's theorem <cmath>\begin{align*} PP^{\prime}\cdot AD+1=16 \\ PP^{\prime}\cdot BC+4=9\end{align*}</cmath> Thus <math>PP^{\prime}\cdot AD=15</math> and <math>PP^{\prime}\cdot BC=5</math>; dividing these two equations and taking the reciprocal yields <math>\frac{BC}{AD}=\boxed{\textbf{(B)}~\frac{1}{3}}</math>. | Consider the reflection <math>P^{\prime}</math> of <math>P</math> over the perpendicular bisector of <math>\overline{BC}</math>, creating two new isosceles trapezoids <math>DAPP^{\prime}</math> and <math>CBPP^{\prime}</math>. Under this reflection, <math>P^{\prime}A=PD=4</math>, <math>P^{\prime}D=PA=1</math>, <math>P^{\prime}C=PB=2</math>, and <math>P^{\prime}B=PC=3</math>. By Ptolmey's theorem <cmath>\begin{align*} PP^{\prime}\cdot AD+1=16 \\ PP^{\prime}\cdot BC+4=9\end{align*}</cmath> Thus <math>PP^{\prime}\cdot AD=15</math> and <math>PP^{\prime}\cdot BC=5</math>; dividing these two equations and taking the reciprocal yields <math>\frac{BC}{AD}=\boxed{\textbf{(B)}~\frac{1}{3}}</math>. | ||
| − | ==Solution 2 (Cheese)== | + | ==Solution 2 (Coordinate Bashing)== |
| + | |||
| + | Solution in Progress | ||
| + | |||
| + | ~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 <math>AC=BD</math>. Therefore, we can say WLOG that the height of the trapezoid is <math>0</math> and all <math>5</math> points, including <math>P</math>, lie on the same line with <math>PA=AB=BC=CD=1</math>. Notice that this satisfies the problem requirements because <math>PA=1, PB=2, PC=3,PD=4</math>, and <math>AC=BD=2</math>. | Notice that the question never says what the height of the trapezoid is; the only property we know about it is that <math>AC=BD</math>. Therefore, we can say WLOG that the height of the trapezoid is <math>0</math> and all <math>5</math> points, including <math>P</math>, lie on the same line with <math>PA=AB=BC=CD=1</math>. Notice that this satisfies the problem requirements because <math>PA=1, PB=2, PC=3,PD=4</math>, and <math>AC=BD=2</math>. | ||
Now all we have to find is <math>\frac{BC}{AD} = \frac{1}{3}= \boxed{B}</math> | Now all we have to find is <math>\frac{BC}{AD} = \frac{1}{3}= \boxed{B}</math> | ||
Revision as of 13:54, 12 November 2022
Contents
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 (Coordinate Bashing)
Solution in Progress
~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
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. 
