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| − | ==Problem==
| + | #redirect [[2022 AMC 12A Problems/Problem 20]] |
| − | 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>
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| − | ==Solution 1 (Reflections + Ptolemy's Theorem)==
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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 [[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>.
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| − | ==Solution 2 (Extensions + Stewart's Theorem)==
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| − | <asy>
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| − | size(7.5cm);
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| − | 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);
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| − | label("$Q$",(2.1,3),N);
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| − | draw((2.1,3)--(1,0.5));
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| − | </asy>
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| − | 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.
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| − | 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>.
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| − | ~kred9
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| − | ==Solution 3 (Coordinate Bashing)==
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| − | 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.
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| − | 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>.
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| − | 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>
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| − | 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>
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| − | Again, we can subtract these equations, yielding <cmath>-4bc = 5</cmath>
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| − | We can now divide the equations to eliminate <math>c</math>, yielding <cmath>\frac{b}{a} = \frac{5}{15} = \frac{1}{3}</cmath>
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| − | 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.
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| − | Therefore <math>\frac{BC}{AD} = \boxed{\textbf{(B) }\frac{1}{3}}</math>.
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| − | ~KingRavi
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| − | ==Solution 4 (Coordinate Bashing)==
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| − | 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:
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| − | <cmath>x_1^2 + y_1^2 = 1 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (1)</cmath>
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| − | <cmath>x_2^2 + y_2^2 = 4 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (2)</cmath>
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| − | <cmath>x_3^2 + y_2^2 = 9 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (3)</cmath>
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| − | <cmath>x_4^2 + y_1^2 = 16 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (4)</cmath>
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| − | 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
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| − | <cmath> x_1 + x_4 = x_2 + x_3 \;\;\;\;\;\;\;\;\;\;\;\;\; (5)</cmath>
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| − | Subtracting Equation <math>2</math> from Equation <math>3</math>, and Equation <math>1</math> from Equation <math>4</math>, we have
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| − | <cmath>x_3^2 - x_2^2 = 5 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (6)</cmath>
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| − | <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
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| − | <cmath>\frac{x_3^2-x_2^2}{x_4^2-x_1^2}=\frac{1}{3}</cmath>
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| − | <cmath>\frac{(x_3-x_2)(x_3+x_2)}{(x_4-x_1)(x_4+x_1)}=\frac{1}{3}.</cmath>
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| − | Cancelling <math>(x_3+x_2)</math> and <math>(x_4+x_1)</math> with Equation <math>5</math>, we get
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| − | <cmath>\frac{(x_3-x_2)}{(x_4-x_1)}=\frac{1}{3}.</cmath>
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| − | In other words,
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| − | <cmath>\frac{BC}{AD}=\frac{1}{3}=\boxed{\textbf{(B) }\frac{1}{3}}.</cmath>
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| − | ~G63566
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| − | ==Solution 5 (Cheese)==
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| − | 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>.
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| − | Now all we have to find is <math>\frac{BC}{AD} = \boxed{\textbf{(B) }\frac{1}{3}}</math>.
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| − | ~KingRavi
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| − | == Video Solution By ThePuzzlr ==
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| − | https://youtu.be/KqtpaHy-eoU
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| − | ~ MathIsChess
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| − | ==Video Solution by Punxsutawney Phil==
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| − | https://youtube.com/watch?v=uYXtEzX4fb0
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| − | ==Video Solution==
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| − | https://youtu.be/hvIOvjjQvIw
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| − | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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| − | == Video Solution by OmegaLearn using Pythagorean Theorem ==
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| − | https://youtu.be/jnm2alniaM4
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| − | ~ pi_is_3.14
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| − | ==See also==
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| − | {{AMC10 box|year=2022|ab=A|num-b=22|num-a=24}}
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| − | {{AMC12 box|year=2022|ab=A|num-b=19|num-a=21}}
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| − | [[Category:Intermediate Geometry Problems]]
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| − | {{MAA Notice}}
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