Difference between revisions of "2022 AMC 12A Problems/Problem 20"
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==Solution 2 (Coordinate Bashing)== | ==Solution 2 (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. | ||
+ | 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> | ||
+ | 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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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> | ||
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+ | 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> | ||
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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. | ||
+ | |||
+ | Therefore <math>\frac{BC}{AD} = \frac{1}{3} = \boxed{B}</math> | ||
Solution in Progress | Solution in Progress |
Revision as of 14:04, 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)
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 , and let the coordinates of
and
be at
and
, respectively. Then let
and
be at
and
, respectively. This follows the rules that this is an isosceles trapezoid since the origin is centered on the middle of
. Finally, let
be located at point
.
The distance from to
is
, so by the distance formula:
The distance from
to
is
, so
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
Next, the distance from to
is
, so
The distance from
to
is
, so
Again, we can subtract these equations, yielding
We can now divide the equations to eliminate , yielding
We wanted to find . But since
is half of
and
is half of
, this ratio is equal to the ratio we want.
Therefore
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.