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Difference between revisions of "2022 AMC 12A Problems/Problem 20"

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#redirect [[2022 AMC 10A Problems/Problem 23]]
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{{duplicate|[[2022 AMC 12A Problems#Problem 20|2022 AMC 12A #20]] and [[2022 AMC 10A Problems#Problem 23|2022 AMC 10A #23]]}}
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==Problem==
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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>
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<math>\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}</math>
 +
 
 +
==Solution 1 (Reflections + Ptolemy's Theorem)==
 +
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 [[Ptolemy's theorem|Ptolemy'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 (Extensions + Stewart's Theorem)==
 +
<asy>
 +
size(7.5cm);
 +
draw((0,0)--(4.2,0));
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draw((0,0)--(1.4,2)--(2.8,2)--(4.2,0));
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draw((1.4,2)--(2.1,3)--(2.8,2));
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draw((0,0)--(1,0.5)--(1.4,2)--(1,0.5)--(2.8,2)--(1,0.5)--(4.2,0));
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label("$A$",(0,0),SW);
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label("$B$",(1.4,2),NW);
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label("$C$",(2.8,2),NE);
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label("$D$",(4.2,0),SE);
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label("$P$",(1,0.5),NW);
 +
label("$Q$",(2.1,3),N);
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draw((2.1,3)--(1,0.5));
 +
</asy>
 +
 
 +
Extend <math>AB</math> and <math>CD</math> to a point <math>Q</math> as shown, and let <math>PQ = s</math>. Then let <math>BQ=CQ = b</math> and <math>AQ=DQ = a</math>. Notice that <math>\frac{BC}{AD} = \frac{QC}{QD} = \frac{a}{a+b}</math> by similar triangles.
 +
 
 +
By [[Stewart's theorem|Stewart's theorem]] on <math>APQ</math> and <math>DPQ</math>, we have <cmath>\begin{align*} ab(a+b) + 9(a+b) &= 16a + s^2b \\ ab(a+b) + 4(a+b) &= a + s^2b \\ \end{align*}</cmath>
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Subtracting, <math>5(a+b) = 15a</math>, and so <math>\frac{BC}{AD} = \frac{a}{a+b} = \frac{5}{15} = \boxed{\textbf{(B) }\frac{1}{3}}</math>.
 +
 
 +
~kred9
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 +
==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 <math>h</math>, and let the coordinates of <math>A</math> and <math>D</math> be at <math>(-a,0)</math> and <math>(a,0)</math>, respectively. Then let <math>B</math> and <math>C</math> be at <math>(-b,h)</math> and <math>(b,h)</math>, respectively. This follows the rules that this is an isosceles trapezoid since the origin is centered on the middle of <math>AD</math>. Finally, let <math>P</math> be located at point <math>(c,d)</math>.
 +
 
 +
The distance from <math>P</math> to <math>A</math> is <math>1</math>, so by the distance formula: <cmath>\sqrt{(c+a)^2+(d-h)^2} = 1 \implies (c+a)^2+(d-h)^2 = 1</cmath>
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The distance from <math>P</math> to <math>D</math> is <math>4</math>, so <cmath>\sqrt{(c-a)^2+(d-h)^2} = 1 \implies (c-a)^2+(d-h)^2 = 16</cmath>
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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 <cmath>-4ac = 15</cmath>
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Next, the distance from <math>P</math> to <math>B</math> is <math>2</math>, so <cmath>\sqrt{(c+b)^2+(d-h)^2} = 2 \implies (c+b)^2+(d-h)^2 = 4</cmath>
 +
The distance from <math>P</math> to <math>C</math> is <math>3</math>, so <cmath>\sqrt{(c-b)^2+(d-h)^2} = 3 \implies (c-b)^2+(d-h)^2 = 9</cmath>
 +
 
 +
Again, we can subtract these equations, yielding <cmath>-4bc = 5</cmath>
 +
 
 +
We can now divide the equations to eliminate <math>c</math>, yielding <cmath>\frac{b}{a} = \frac{5}{15} = \frac{1}{3}</cmath>
 +
 
 +
We wanted to find <math>\frac{BC}{AD}</math>. But since <math>b</math> is half of <math>BC</math> and <math>a</math> is half of <math>AD</math>, this ratio is equal to the ratio we want.
 +
 
 +
Therefore <math>\frac{BC}{AD} = \boxed{\textbf{(B) }\frac{1}{3}}</math>.
 +
 
 +
~KingRavi
 +
 
 +
==Solution 4 (Coordinate Bashing)==
 +
 
 +
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:
 +
<cmath>x_1^2 + y_1^2 = 1 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (1)</cmath>
 +
<cmath>x_2^2 + y_2^2 = 4 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (2)</cmath>
 +
<cmath>x_3^2 + y_2^2 = 9 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (3)</cmath>
 +
<cmath>x_4^2 + y_1^2 = 16 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (4)</cmath>
 +
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
 +
<cmath> x_1 + x_4 = x_2 + x_3 \;\;\;\;\;\;\;\;\;\;\;\;\; (5)</cmath>
 +
Subtracting Equation <math>2</math> from Equation <math>3</math>, and Equation <math>1</math> from Equation <math>4</math>, we have
 +
<cmath>x_3^2 - x_2^2 = 5 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (6)</cmath>
 +
<cmath>x_4^2 - x_1^2 = 15 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (7)</cmath>
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Dividing Equation <math>6</math> by Equation <math>7</math>, we have
 +
<cmath>\frac{x_3^2-x_2^2}{x_4^2-x_1^2}=\frac{1}{3}</cmath>
 +
<cmath>\frac{(x_3-x_2)(x_3+x_2)}{(x_4-x_1)(x_4+x_1)}=\frac{1}{3}.</cmath>
 +
Cancelling <math>(x_3+x_2)</math> and <math>(x_4+x_1)</math> with Equation <math>5</math>, we get
 +
<cmath>\frac{(x_3-x_2)}{(x_4-x_1)}=\frac{1}{3}.</cmath>
 +
In other words,
 +
<cmath>\frac{BC}{AD}=\frac{1}{3}=\boxed{\textbf{(B) }\frac{1}{3}}.</cmath>
 +
~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 <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} = \boxed{\textbf{(B) }\frac{1}{3}}</math>.
 +
 
 +
~KingRavi
 +
 
 +
== Video Solution By ThePuzzlr ==
 +
https://youtu.be/KqtpaHy-eoU
 +
 
 +
~ MathIsChess
 +
 
 +
==Video Solution by Punxsutawney Phil==
 +
 
 +
https://youtube.com/watch?v=uYXtEzX4fb0
 +
 
 +
==Video Solution by Steven Chen==
 +
 
 +
https://youtu.be/hvIOvjjQvIw
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 
 +
==Video Solution by Sohil Rathi==
 +
https://youtu.be/jnm2alniaM4
 +
 
 +
~ pi_is_3.14
 +
 
 +
==See also==
 +
{{AMC10 box|year=2022|ab=A|num-b=22|num-a=24}}
 +
{{AMC12 box|year=2022|ab=A|num-b=19|num-a=21}}
 +
 
 +
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Revision as of 02:36, 19 November 2022

The following problem is from both the 2022 AMC 12A #20 and 2022 AMC 10A #23, so both problems redirect to this page.

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 (Reflections + Ptolemy's Theorem)

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 Ptolemy'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 (Extensions + Stewart's Theorem)

[asy] size(7.5cm); draw((0,0)--(4.2,0)); draw((0,0)--(1.4,2)--(2.8,2)--(4.2,0)); draw((1.4,2)--(2.1,3)--(2.8,2)); draw((0,0)--(1,0.5)--(1.4,2)--(1,0.5)--(2.8,2)--(1,0.5)--(4.2,0)); label("$A$",(0,0),SW); label("$B$",(1.4,2),NW); label("$C$",(2.8,2),NE); label("$D$",(4.2,0),SE); label("$P$",(1,0.5),NW); label("$Q$",(2.1,3),N); draw((2.1,3)--(1,0.5)); [/asy]

Extend $AB$ and $CD$ to a point $Q$ as shown, and let $PQ = s$. Then let $BQ=CQ = b$ and $AQ=DQ = a$. Notice that $\frac{BC}{AD} = \frac{QC}{QD} = \frac{a}{a+b}$ by similar triangles.

By Stewart's theorem on $APQ$ and $DPQ$, we have \begin{align*} ab(a+b) + 9(a+b) &= 16a + s^2b \\ ab(a+b) + 4(a+b) &= a + s^2b \\ \end{align*}

Subtracting, $5(a+b) = 15a$, and so $\frac{BC}{AD} = \frac{a}{a+b} = \frac{5}{15} = \boxed{\textbf{(B) }\frac{1}{3}}$.

~kred9

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 $h$, 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 $\frac{BC}{AD} = \boxed{\textbf{(B) }\frac{1}{3}}$.

~KingRavi

Solution 4 (Coordinate Bashing)

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: \[x_1^2 + y_1^2 = 1 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (1)\] \[x_2^2 + y_2^2 = 4 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (2)\] \[x_3^2 + y_2^2 = 9 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (3)\] \[x_4^2 + y_1^2 = 16 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (4)\] In addition, because the trapezoid is isosceles ($AB=CD$), the midpoints of the two bases would then have the same x-coordinate, giving us \[x_1 + x_4 = x_2 + x_3 \;\;\;\;\;\;\;\;\;\;\;\;\; (5)\] Subtracting Equation $2$ from Equation $3$, and Equation $1$ from Equation $4$, we have \[x_3^2 - x_2^2 = 5 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (6)\] \[x_4^2 - x_1^2 = 15 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (7)\] 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 get \[\frac{(x_3-x_2)}{(x_4-x_1)}=\frac{1}{3}.\] In other words, \[\frac{BC}{AD}=\frac{1}{3}=\boxed{\textbf{(B) }\frac{1}{3}}.\] ~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 $AC=BD$. 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 $\frac{BC}{AD} = \boxed{\textbf{(B) }\frac{1}{3}}$.

~KingRavi

Video Solution By ThePuzzlr

https://youtu.be/KqtpaHy-eoU

~ MathIsChess

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=uYXtEzX4fb0

Video Solution by Steven Chen

https://youtu.be/hvIOvjjQvIw

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

Video Solution by Sohil Rathi

https://youtu.be/jnm2alniaM4

~ pi_is_3.14

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

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