2022 AMC 12A Problems/Problem 20: Difference between revisions
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==Solution== | ==Solution== | ||
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 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>. | ||
==See also== | ==See also== | ||
Revision as of 20:02, 11 November 2022
Problem
Isosceles trapezoid 
has parallel sides 
and 
with 
and 
There is a point 
in the plane such that 
and 
What is 
Solution
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 
.
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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