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− | # | + | {{duplicate|[[2022 AMC 10A Problems/Problem 23|2022 AMC 10A #23]] and [[2022 AMC 12A Problems/Problem 20|2022 AMC 12A #20]]}} |
+ | |||
+ | ==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> | ||
+ | |||
+ | <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>. | ||
+ | |||
+ | Since <math>DAPP'</math> and <math>CBPP'</math> are isosceles trapezoids, they are cyclic. Using [[Ptolemy's theorem|Ptolemy's theorem]] on <math>DAPP'</math>, we get that <math>(PP')(AD) + (PA)(P'D) = (AP')(PD)</math>, so | ||
+ | <cmath>PP' \cdot AD + 1 \cdot 1 = 4 \cdot 4.</cmath> Then, using Ptolemy's theorem again on <math>CBPP'</math>, we get that <math>(BC)(PP') + (BP)(CP') = (BP')(CP)</math>, so | ||
+ | <cmath>PP' \cdot BC + 2 \cdot 2 = 3 \cdot 3.</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>. | ||
+ | <asy> | ||
+ | size(300); | ||
+ | pair A = (0,0); | ||
+ | pair B = (1, 2); | ||
+ | pair C = (2,2); | ||
+ | pair D = (3,0); | ||
+ | label("$A$", A, SW); | ||
+ | label("$B$", B, NW); | ||
+ | label("$C$", C, NE); | ||
+ | label("$D$", D, SE); | ||
+ | draw(A--B--C--D--cycle, blue); | ||
+ | pair P = (0.8, 0.6); | ||
+ | dot("$P$", P, NW); | ||
+ | draw(P--A, magenta); | ||
+ | draw(P--B, magenta); | ||
+ | draw(P--C); | ||
+ | draw(P--D); | ||
+ | label("$1$", P--A, NW); | ||
+ | label("$2$", P--B, E); | ||
+ | label("$3$", P--C, NW); | ||
+ | label("$4$", P--D, S); | ||
+ | pair P1 = (2.2, 0.6); | ||
+ | dot("$P'$", P1, NE); | ||
+ | draw(P1--D, magenta); | ||
+ | draw(P1--C, magenta); | ||
+ | draw(P1--A); | ||
+ | draw(P1--B); | ||
+ | label("$1$", P1--D, NE); | ||
+ | label("$2$", P1--C, E); | ||
+ | label("$3$", P1--B, NE); | ||
+ | label("$4$", P1--A, SE); | ||
+ | draw(P--P1, dashed+magenta); | ||
+ | </asy> | ||
+ | (diagram by cinnamon_e) | ||
+ | |||
+ | ==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 <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 = a</math> and <math>AB=DC = b</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> | ||
+ | |||
+ | 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 (minor edit by gwang2008) | ||
+ | |||
+ | ==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> | ||
+ | The distance from <math>P</math> to <math>D</math> is <math>4</math>, so <cmath>\sqrt{(c-a)^2+(d-h)^2} = 4 \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> | ||
+ | |||
+ | 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 <math>x</math>-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 <math>x</math>-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> | ||
+ | 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 (Polar Coordinates)== | ||
+ | |||
+ | As <math>ABCD</math> is an isosceles trapezoid, it is cyclic. We will use polar coordinates with origin <math>O</math> as in the following figure. | ||
+ | |||
+ | [[File:2022AMC12AP20.png|center|500px]] | ||
+ | |||
+ | The <math>x</math>-coordinate of <math>C</math> is <math>r \cos \beta</math>, <math>BC = 2r \cos \beta</math>. The <math>x</math>-coordinate of <math>D</math> is <math>r \cos \alpha</math>, <math>AD = 2r \cos \alpha</math>, <math>\frac{BC}{AD} = \frac{ \cos \beta }{ \cos \alpha }</math> | ||
+ | |||
+ | By the law of cosines, | ||
+ | |||
+ | <cmath>OD^2 + OP^2 - 2 \cdot OD \cdot OP \cdot \cos (\theta - \alpha) = PD^2</cmath> | ||
+ | <cmath>OC^2 + OP^2 - 2 \cdot OC \cdot OP \cdot \cos (\theta - \beta) = PC^2</cmath> | ||
+ | <cmath>OB^2 + OP^2 - 2 \cdot OB \cdot OP \cdot \cos (\theta - \pi + \beta) = PB^2</cmath> | ||
+ | <cmath>OA^2 + OP^2 - 2 \cdot OA \cdot OP \cdot \cos (\theta - \pi + \alpha) = PA^2</cmath> | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | r^2 + d^2 - 2rd \cos (\theta - \alpha) = 16 \quad \quad (1) \\ | ||
+ | r^2 + d^2 - 2rd \cos (\theta - \beta) = 9 \quad \quad (2) \\ | ||
+ | r^2 + d^2 - 2rd \cos (\theta - \pi + \beta) = 4 \quad \quad (3) \\ | ||
+ | r^2 + d^2 - 2rd \cos (\theta - \pi + \alpha) = 1 \quad \quad (4) | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | (3) - (2), \quad 2rd \cos (\theta - \beta) - 2rd \cos (\theta - \pi + \beta) = 4-9, \quad 2rd(\cos (\theta - \beta) + \cos (\theta + \beta)) = -5 \quad \quad (5) \\ | ||
+ | (4) - (1), \quad 2rd \cos (\theta - \alpha) - 2rd \cos (\theta - \pi + \alpha) = 1-16, \quad 2rd(\cos (\theta - \alpha) + \cos (\theta + \alpha)) = -15 \quad \quad (6) | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | <cmath>(5)/(6), \quad \frac{ 2rd(\cos (\theta - \beta) + \cos (\theta + \beta)) }{ 2rd(\cos (\theta - \alpha) + \cos (\theta + \alpha)) } = \frac{-5}{ -15 }, \quad \frac{ \cos (\theta - \beta) + \cos (\theta + \beta) }{ \cos (\theta - \alpha) + \cos (\theta + \alpha) } = \frac13</cmath> | ||
+ | |||
+ | By the sum to product identity | ||
+ | |||
+ | <cmath> \frac{ \cos (\theta - \beta) + \cos (\theta + \beta) }{ \cos (\theta - \alpha) + \cos (\theta + \alpha) } = \frac{ 2 \cos ( \frac{\theta - \beta + \theta + \beta}{2}) \cos ( \frac{\theta - \beta - \theta - \beta}{2}) }{ 2 \cos ( \frac{\theta - \alpha + \theta + \alpha}{2}) \cos ( \frac{\theta - \alpha - \theta - \alpha}{2}) } = \frac{\cos \theta \cos \beta}{\cos \theta \cos \alpha} = \frac{\cos \beta}{\cos \alpha} = \boxed{\textbf{(B) }\frac{1}{3}}</cmath> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 6 (Cheese)== | ||
+ | [[File:AMC_12A_2022_Q20-Cheese_Diagram.png|500px|right]] | ||
+ | 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 assume 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 | ||
+ | |||
+ | ~ShawnX (Diagram) | ||
+ | |||
+ | ==Solution 7 (Pythagorean Theorem)== | ||
+ | [[File:2022 AMC 12A 24.png|360px|right]] | ||
+ | Consider the reflection <math>P^{\prime}</math> of <math>P</math> over the perpendicular bisector of <math>\overline{BC}</math>. | ||
+ | <cmath>PB = P^{\prime}C = a, PC = P^{\prime}B = b.</cmath> | ||
+ | Let <math>PE \perp BC, P^{\prime}E^{\prime} \perp BC.</math> | ||
+ | <cmath>BE = CE^{\prime}, PP^{\prime} = EE^{\prime} \implies </cmath> | ||
+ | <cmath>b^2 - a^2 = PC^2 - PB^2 = CE^2 - BE^2</cmath> | ||
+ | <cmath>=(CE - BE)(CE + BE) = EE^{\prime} \cdot BC = PP^{\prime} \cdot BC.</cmath> | ||
+ | Similarly, | ||
+ | <cmath>d^2 - c^2 = PP^{\prime} \cdot AD \implies \frac {BC}{AD} = \frac {b^2 - a^2}{d^2 - c^2}= \boxed{\textbf{(B) }\frac{1}{3}}.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/jnm2alniaM4 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | == 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 MRENTHUSIASM (English & Chinese) == | ||
+ | |||
+ | https://youtu.be/OXlyX-WzMfg | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Video Solution by Math-X (Smart and Simple)== | ||
+ | https://youtu.be/7yAh4MtJ8a8?si=DZJywiBKH7yK97Hw&t=6149 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==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 12:52, 26 March 2024
- The following problem is from both the 2022 AMC 10A #23 and 2022 AMC 12A #20, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1 (Reflections + Ptolemy's Theorem)
- 3 Solution 2 (Extensions + Stewart's Theorem)
- 4 Solution 3 (Coordinate Bashing)
- 5 Solution 4 (Coordinate Bashing)
- 6 Solution 5 (Polar Coordinates)
- 7 Solution 6 (Cheese)
- 8 Solution 7 (Pythagorean Theorem)
- 9 Video Solution by OmegaLearn
- 10 Video Solution By ThePuzzlr
- 11 Video Solution by Punxsutawney Phil
- 12 Video Solution by Steven Chen
- 13 Video Solution by MRENTHUSIASM (English & Chinese)
- 14 Video Solution by Math-X (Smart and Simple)
- 15 See also
Problem
Isosceles trapezoid has parallel sides
and
with
and
There is a point
in the plane such that
and
What is
Solution 1 (Reflections + Ptolemy's Theorem)
Consider the reflection of
over the perpendicular bisector of
, creating two new isosceles trapezoids
and
. Under this reflection,
,
,
, and
.
Since and
are isosceles trapezoids, they are cyclic. Using Ptolemy's theorem on
, we get that
, so
Then, using Ptolemy's theorem again on
, we get that
, so
Thus,
and
; dividing these two equations and taking the reciprocal yields
.
(diagram by cinnamon_e)
Solution 2 (Extensions + Stewart's Theorem)
Extend and
to a point
as shown, and let
. Then let
and
. Notice that
by similar triangles.
By Stewart's theorem on and
, we have
Subtracting, , and so
.
~kred9 (minor edit by gwang2008)
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 , 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 .
~KingRavi
Solution 4 (Coordinate Bashing)
Let the point be at the origin, and draw four concentric circles around
each with radius
,
,
, and
, respectively. The vertices of the trapezoid would be then on each of the four concentric circles. WLOG, let
and
be parallel to the
-axis. Assigning coordinates to each point, we have:
which satisfy the following:
In addition, because the trapezoid is isosceles (
), the midpoints of the two bases would then have the same
-coordinate, giving us
Subtracting Equation
from Equation
, and Equation
from Equation
, we have
Dividing Equation
by Equation
, we have
Cancelling
and
with Equation
, we get
In other words,
~G63566
Solution 5 (Polar Coordinates)
As is an isosceles trapezoid, it is cyclic. We will use polar coordinates with origin
as in the following figure.
The -coordinate of
is
,
. The
-coordinate of
is
,
,
By the law of cosines,
By the sum to product identity
Solution 6 (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 assume 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
~ShawnX (Diagram)
Solution 7 (Pythagorean Theorem)
Consider the reflection of
over the perpendicular bisector of
.
Let
Similarly,
vladimir.shelomovskii@gmail.com, vvsss
Video Solution by OmegaLearn
~ pi_is_3.14
Video Solution By ThePuzzlr
~ MathIsChess
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=uYXtEzX4fb0
Video Solution by Steven Chen
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by MRENTHUSIASM (English & Chinese)
~MRENTHUSIASM
Video Solution by Math-X (Smart and Simple)
https://youtu.be/7yAh4MtJ8a8?si=DZJywiBKH7yK97Hw&t=6149
~Math-X
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.