Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AMC 12A Problems/Problem 20"
Line 1: Line 1:
−
\usepackage{amsmath}
 
−
\usepackage{amssymb}
 
−
\usepackage{amsthm}
 
−
 
 
==Problem==
 
==Problem==
 
Isosceles trapezoid <math>ABCD</math> has parallel sides <math>\overline{AD}</math> and <math>\overline{BC},</math> with <math>BC < AD</math> and <math>AB = CD.</math> There is a point <math>P</math> in the plane such that <math>PA=1, PB=2, PC=3,</math> and <math>PD=4.</math> What is <math>\tfrac{BC}{AD}?</math>
 
Isosceles trapezoid <math>ABCD</math> has parallel sides <math>\overline{AD}</math> and <math>\overline{BC},</math> with <math>BC < AD</math> and <math>AB = CD.</math> There is a point <math>P</math> in the plane such that <math>PA=1, PB=2, PC=3,</math> and <math>PD=4.</math> What is <math>\tfrac{BC}{AD}?</math>
Line 44: Line 40:
  
 
Let the point <math>P</math> be at the origin, and draw four concentric circles around <math>P</math> each with radius <math>1</math>, <math>2</math>, <math>3</math>, and <math>4</math>, respectively. The vertices of the trapezoid would be then on each of the four concentric circles. WLOG, let <math>BC</math> and <math>AD</math> be parallel to the x-axis. Assigning coordinates to each point, we have <cmath>A=(x_1,y_1)</cmath> <cmath>B=(x_2,y_2)</cmath> <cmath>C=(x_3,y_2)</cmath> <cmath>D=(x_4,y_1)</cmath> which satisfy the following:  
 
Let the point <math>P</math> be at the origin, and draw four concentric circles around <math>P</math> each with radius <math>1</math>, <math>2</math>, <math>3</math>, and <math>4</math>, respectively. The vertices of the trapezoid would be then on each of the four concentric circles. WLOG, let <math>BC</math> and <math>AD</math> be parallel to the x-axis. Assigning coordinates to each point, we have <cmath>A=(x_1,y_1)</cmath> <cmath>B=(x_2,y_2)</cmath> <cmath>C=(x_3,y_2)</cmath> <cmath>D=(x_4,y_1)</cmath> which satisfy the following:  
−
\begin{align*}
+
\begin{equation}
−
x_1^2 + y_1^2 &= 1 \text{1} \\
+
x_1^2 + y_1^2 = 1
−
x_2^2 + y_2^2 &= 4 \text{2} \\
+
x_2^2 + y_2^2 = 4  
−
x_3^2 + y_2^2 &= 9 \text{3} \\
+
x_3^2 + y_2^2 = 9
−
x_4^2 + y_1^2 &= 16 \text{4}
+
x_4^2 + y_1^2 = 16
−
\end{align*}
+
\end{equation}
 
In addition, because the trapezoid is isosceles (<math>AB=CD</math>), the midpoints of the two bases would then have the same x-coordinate, giving us  
 
In addition, because the trapezoid is isosceles (<math>AB=CD</math>), the midpoints of the two bases would then have the same x-coordinate, giving us  
−
\begin{align*}
+
\begin{equation}
−
x_1 + x_4 &= x_2 + x_3 \text{5}
+
x_1 + x_4 = x_2 + x_3
−
\end{align*}
+
\end{equation}
 
Subtracting Equation <math>2</math> from Equation <math>3</math>, and Equation <math>1</math> from Equation <math>4</math>, we have
 
Subtracting Equation <math>2</math> from Equation <math>3</math>, and Equation <math>1</math> from Equation <math>4</math>, we have
−
\begin{align*}
+
\begin{equation}
−
x_3^2 - x_2^2 &= 5 \text{6} \\
+
x_3^2 - x_2^2 = 5
−
x_4^2 - x_1^2 &= 15 \text{7}
+
x_4^2 - x_1^2 = 15
−
\end{align*}
+
\end{equation}
  
 
Dividing Equation <math>6</math> by Equation <math>7</math>, we have
 
Dividing Equation <math>6</math> by Equation <math>7</math>, we have

Revision as of 13:41, 12 November 2022

Problem

Isosceles trapezoid $ABCD$ has parallel sides $\overline{AD}$ and $\overline{BC},$ with $BC < AD$ and $AB = CD.$ There is a point $P$ in the plane such that $PA=1, PB=2, PC=3,$ and $PD=4.$ What is $\tfrac{BC}{AD}?$

$\textbf{(A) }\frac{1}{4}\qquad\textbf{(B) }\frac{1}{3}\qquad\textbf{(C) }\frac{1}{2}\qquad\textbf{(D) }\frac{2}{3}\qquad\textbf{(E) }\frac{3}{4}$

Solution 1

Consider the reflection $P^{\prime}$ of $P$ over the perpendicular bisector of $\overline{BC}$, creating two new isosceles trapezoids $DAPP^{\prime}$ and $CBPP^{\prime}$. Under this reflection, $P^{\prime}A=PD=4$, $P^{\prime}D=PA=1$, $P^{\prime}C=PB=2$, and $P^{\prime}B=PC=3$. By Ptolmey's theorem \begin{align*} PP^{\prime}\cdot AD+1=16 \\ PP^{\prime}\cdot BC+4=9\end{align*} Thus $PP^{\prime}\cdot AD=15$ and $PP^{\prime}\cdot BC=5$; dividing these two equations and taking the reciprocal yields $\frac{BC}{AD}=\boxed{\textbf{(B)}~\frac{1}{3}}$.

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 $h$, and let the coordinates of $A$ and $D$ be at $(-a,0)$ and $(a,0)$, respectively. Then let $B$ and $C$ be at $(-b,h)$ and $(b,h)$, respectively. This follows the rules that this is an isosceles trapezoid since the origin is centered on the middle of $AD$. Finally, let $P$ be located at point $(c,d)$.

The distance from $P$ to $A$ is $1$, so by the distance formula: \[\sqrt{(c+a)^2+(d-h)^2} = 1 \implies (c+a)^2+(d-h)^2 = 1\] The distance from $P$ to $D$ is $4$, so \[\sqrt{(c-a)^2+(d-h)^2} = 1 \implies (c-a)^2+(d-h)^2 = 16\]

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 \[-4ac = 15\]

Next, the distance from $P$ to $B$ is $2$, so \[\sqrt{(c+b)^2+(d-h)^2} = 2 \implies (c+b)^2+(d-h)^2 = 4\] The distance from $P$ to $C$ is $3$, so \[\sqrt{(c-b)^2+(d-h)^2} = 3 \implies (c-b)^2+(d-h)^2 = 9\]

Again, we can subtract these equations, yielding \[-4bc = 5\]

We can now divide the equations to eliminate $c$, yielding \[\frac{b}{a} = \frac{5}{15} = \frac{1}{3}\]

We wanted to find $\frac{BC}{AD}$. But since $b$ is half of $BC$ and $a$ is half of $AD$, this ratio is equal to the ratio we want.

Therefore $\frac{BC}{AD} = \frac{1}{3} = \boxed{B}$


~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 $AC=BD$. Therefore, we can say WLOG that the height of the trapezoid is $0$ and all $5$ points, including $P$, lie on the same line with $PA=AB=BC=CD=1$. Notice that this satisfies the problem requirements because $PA=1, PB=2, PC=3,PD=4$, and $AC=BD=2$. Now all we have to find is $\frac{BC}{AD} = \frac{1}{3}= \boxed{B}$

~KingRavi

Solution 4 (Coordinate Bashing 2)

Let the point $P$ be at the origin, and draw four concentric circles around $P$ each with radius $1$, $2$, $3$, and $4$, respectively. The vertices of the trapezoid would be then on each of the four concentric circles. WLOG, let $BC$ and $AD$ be parallel to the x-axis. Assigning coordinates to each point, we have \[A=(x_1,y_1)\] \[B=(x_2,y_2)\] \[C=(x_3,y_2)\] \[D=(x_4,y_1)\] which satisfy the following: \begin{equation} x_1^2 + y_1^2 = 1 x_2^2 + y_2^2 = 4 x_3^2 + y_2^2 = 9 x_4^2 + y_1^2 = 16 \end{equation} In addition, because the trapezoid is isosceles ($AB=CD$), the midpoints of the two bases would then have the same x-coordinate, giving us \begin{equation} x_1 + x_4 = x_2 + x_3 \end{equation} Subtracting Equation $2$ from Equation $3$, and Equation $1$ from Equation $4$, we have \begin{equation} x_3^2 - x_2^2 = 5 x_4^2 - x_1^2 = 15 \end{equation}

Dividing Equation $6$ by Equation $7$, we have \[\frac{x_3^2-x_2^2}{x_4^2-x_1^2}=\frac{1}{3}\] \[\frac{(x_3-x_2)(x_3+x_2)}{(x_4-x_1)(x_4+x_1)}=\frac{1}{3}\]

Cancelling $(x_3+x_2)$ and $(x_4+x_1)$ with Equation $5$, we then arrive at

\[\frac{(x_3-x_2)}{(x_4-x_1)}=\frac{1}{3}\]

i.e.

\[\frac{BC}{AD}=\frac{1}{3}=\boxed{B}\]

Video Solution By ThePuzzlr

https://youtu.be/KqtpaHy-eoU

~ MathIsChess


Video Solution

https://youtu.be/hvIOvjjQvIw

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by OmegaLearn using Pythagorean Theorem

https://youtu.be/jnm2alniaM4

~ 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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems/Problem_20&oldid=181092"