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==Problem== | ==Problem== | ||
− | The interior of a quadrilateral is bounded by the graphs of <math>(x+ay)^2 = 4a^2</math> and <math>(ax-y)^2 = a^2</math>, where <math>a</math> a positive real number. What is the area of this region in terms of <math>a</math>, valid for all <math>a > 0</math>? | + | The interior of a quadrilateral is bounded by the graphs of <math>(x+ay)^2 = 4a^2</math> and <math>(ax-y)^2 = a^2</math>, where <math>a</math> is a positive real number. What is the area of this region in terms of <math>a</math>, valid for all <math>a > 0</math>? |
<math>\textbf{(A)} ~\frac{8a^2}{(a+1)^2}\qquad\textbf{(B)} ~\frac{4a}{a+1}\qquad\textbf{(C)} ~\frac{8a}{a+1}\qquad\textbf{(D)} ~\frac{8a^2}{a^2+1}\qquad\textbf{(E)} ~\frac{8a}{a^2+1}</math> | <math>\textbf{(A)} ~\frac{8a^2}{(a+1)^2}\qquad\textbf{(B)} ~\frac{4a}{a+1}\qquad\textbf{(C)} ~\frac{8a}{a+1}\qquad\textbf{(D)} ~\frac{8a^2}{a^2+1}\qquad\textbf{(E)} ~\frac{8a}{a^2+1}</math> | ||
− | ==Solution 1== | + | ==Diagram== |
− | The | + | Graph in Desmos: https://www.desmos.com/calculator/satawguqsc |
− | + | ||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 1 (Generalized Value of a)== | ||
+ | The cases for <math>(x+ay)^2 = 4a^2</math> are <math>x+ay = \pm2a,</math> or two parallel lines. We rearrange each case and construct the table below: | ||
+ | <cmath>\begin{array}{c||c|c|c|c} | ||
+ | & & & & \\ [-2.5ex] | ||
+ | \textbf{Case} & \textbf{Line's Equation} & \boldsymbol{x}\textbf{-Intercept} & \boldsymbol{y}\textbf{-Intercept} & \textbf{Slope} \\ [0.5ex] | ||
+ | \hline | ||
+ | & & & & \\ [-1.5ex] | ||
+ | 1 & x+ay-2a=0 & 2a & 2 & -\frac1a \\ [2ex] | ||
+ | 2 & x+ay+2a=0 & -2a & -2 & -\frac1a \\ [0.75ex] | ||
+ | \end{array}</cmath> | ||
+ | The cases for <math>(ax-y)^2 = a^2</math> are <math>ax-y=\pm a,</math> or two parallel lines. We rearrange each case and construct the table below: | ||
+ | <cmath>\begin{array}{c||c|c|c|c} | ||
+ | & & & & \\ [-2.5ex] | ||
+ | \textbf{Case} & \textbf{Line's Equation} & \boldsymbol{x}\textbf{-Intercept} & \boldsymbol{y}\textbf{-Intercept} & \textbf{Slope} \\ [0.5ex] | ||
+ | \hline | ||
+ | & & & & \\ [-1.5ex] | ||
+ | 1* & ax-y-a=0 & 1 & -a & a \\ [2ex] | ||
+ | 2* & ax-y+a=0 & -1 & a & a \\ [0.75ex] | ||
+ | \end{array}</cmath> | ||
+ | Since the slopes of intersecting lines <math>(1)\cap(1*), (1)\cap(2*), (2)\cap(1*),</math> and <math>(2)\cap(2*)</math> are negative reciprocals, we get four right angles, from which the quadrilateral is a rectangle. | ||
+ | |||
+ | Two solutions follow from here: | ||
+ | |||
+ | ===Solution 1.1 (Distance Between Parallel Lines)=== | ||
+ | Recall that for constants <math>A,B,C_1</math> and <math>C_2,</math> the distance <math>d</math> between parallel lines | ||
+ | <math>\begin{cases} | ||
+ | Ax+By+C_1=0 \\ | ||
+ | Ax+By+C_2=0 | ||
+ | \end{cases}</math> is <cmath>d=\frac{\left|C_2-C_1\right|}{\sqrt{A^2+B^2}}.</cmath> | ||
+ | From this formula: | ||
+ | |||
+ | * The distance between lines <math>(1)</math> and <math>(2)</math> is <math>\frac{4a}{\sqrt{1+a^2}},</math> the length of this rectangle. | ||
− | + | * The distance between lines <math>(1*)</math> and <math>(2*)</math> is <math>\frac{2a}{\sqrt{a^2+1}},</math> the width of this rectangle. | |
− | |||
− | < | + | The area we seek is <cmath>\frac{4a}{\sqrt{1+a^2}}\cdot\frac{2a}{\sqrt{a^2+1}}=\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</cmath> |
+ | ~MRENTHUSIASM | ||
− | <math>(2) \ | + | ===Solution 1.2 (Distance Between Points)=== |
+ | The solutions to systems of equations <math>(1)\cap(1*), (1)\cap(2*), (2)\cap(2*), (2)\cap(1*)</math> are <cmath>(x,y)=\left(\frac{a(a+2)}{a^2+1},\frac{a(2a-1)}{a^2+1}\right), \left(-\frac{a(a-2)}{a^2+1},\frac{a(2a+1)}{a^2+1}\right), \left(-\frac{a(a+2)}{a^2+1},-\frac{a(2a-1)}{a^2+1}\right), \left(\frac{a(a-2)}{a^2+1},-\frac{a(2a+1)}{a^2+1}\right),</cmath> respectively, which are the consecutive vertices of this rectangle. | ||
− | + | By the Distance Formula, the length and width of this rectangle are <math>\frac{4a\sqrt{a^2+1}}{a^2+1}</math> and <math>\frac{2a\sqrt{a^2+1}}{a^2+1},</math> respectively. | |
− | < | + | The area we seek is <cmath>\frac{4a\sqrt{a^2+1}}{a^2+1}\cdot\frac{2a\sqrt{a^2+1}}{a^2+1}=\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</cmath> |
+ | ~MRENTHUSIASM | ||
− | + | ==Solution 2 (Specified Value of a)== | |
+ | In this solution, we will refer to equations <math>(1),(2),(1*),</math> and <math>(2*)</math> from Solution 1. | ||
− | + | Substituting <math>a=2</math> into the answer choices gives | |
<math>\textbf{(A)} ~\frac{32}{9}\qquad\textbf{(B)} ~\frac{8}{3}\qquad\textbf{(C)} ~\frac{16}{3}\qquad\textbf{(D)} ~\frac{32}{5}\qquad\textbf{(E)} ~\frac{16}{5}</math> | <math>\textbf{(A)} ~\frac{32}{9}\qquad\textbf{(B)} ~\frac{8}{3}\qquad\textbf{(C)} ~\frac{16}{3}\qquad\textbf{(D)} ~\frac{32}{5}\qquad\textbf{(E)} ~\frac{16}{5}</math> | ||
− | + | At <math>a=2,</math> the solutions to systems of equations <math>(1)\cap(1*), (1)\cap(2*), (2)\cap(2*), (2)\cap(1*)</math> are <cmath>(x,y)=\left(\frac 85, \frac 65\right), (0,2), \left(-\frac 85, -\frac 65\right), (0,-2),</cmath> respectively, which are the consecutive vertices of the quadrilateral. | |
− | ===Solution 2.1 (Rectangle)=== | + | |
− | + | Two solutions follow from here: | |
+ | |||
+ | ===Solution 2.1 (Area of a Rectangle)=== | ||
+ | From the tables in Solution 1, we conclude that the quadrilateral is a rectangle. | ||
+ | |||
+ | By the Distance Formula, the length and width of this rectangle are <math>\frac{8\sqrt5}{5}</math> and <math>\frac{4\sqrt5}{5},</math> respectively. | ||
− | The answer is <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math> | + | The area we seek is <cmath>\frac{8\sqrt5}{5}\cdot\frac{4\sqrt5}{5}=\frac{32}{5},</cmath> from which the answer is <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ===Solution 2.2 ( | + | ===Solution 2.2 (Area of a General Quadrilateral)=== |
− | Even if we do not recognize that the | + | Even if we do not recognize that the quadrilateral is a rectangle, we can apply the Shoelace Theorem to its <i><b>consecutive</b></i> vertices |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | (x_1,y_1) &= \left(\frac 85, \frac 65\right) \\ | + | (x_1,y_1) &= \left(\frac 85, \frac 65\right), \\ |
− | (x_2,y_2) &= (0,2) \\ | + | (x_2,y_2) &= (0,2), \\ |
− | (x_3,y_3) &= \left(-\frac 85, -\frac 65\right) \\ | + | (x_3,y_3) &= \left(-\frac 85, -\frac 65\right), \\ |
− | (x_4,y_4) &= (0,-2) | + | (x_4,y_4) &= (0,-2). |
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | The area we seek is <cmath>\frac{1}{2} \left|(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)\right| = \frac{32}{5}.</cmath> from which the answer is <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math> | ||
− | + | ~MRENTHUSIASM | |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | + | ==Solution 3 (Slopes and Intercepts)== | |
− | + | [[File:Diagram of Quadrilateral.png|600px|center]] | |
− | + | The quadrilateral is enclosed by four lines. Similar to Solution 1, we will use the equations from the four cases: | |
− | Similar to Solution | + | <ol style="margin-left: 1.5em;"> |
+ | <li><math>x+ay=2a.</math> This is a line with <math>x</math>-intercept <math>2a,</math> <math>y</math>-intercept <math>2,</math> and slope <math>-\frac 1a.</math></li><p> | ||
+ | <li><math>x+ay=-2a.</math> This is a line with <math>x</math>-intercept <math>-2a,</math> <math>y</math>-intercept <math>-2,</math> and slope <math>-\frac 1a.</math></li><p> | ||
+ | <li><math>ax-y=a.</math> This is a line with <math>x</math>-intercept <math>1,</math> <math>y</math>-intercept <math>-a,</math> and slope <math>a.</math></li><p> | ||
+ | <li><math>ax-y=-a.</math> This is a line with <math>x</math>-intercept <math>-1,</math> <math>y</math>-intercept <math>a,</math> and slope <math>a.</math></li><p> | ||
+ | </ol> | ||
+ | It follows that <math>DF = 4</math> and <math>DE = \sqrt{4^2 - s^2}</math>. | ||
− | + | Because the slope of line <math>y = -\frac{x}{a} + 2</math> is <math>-\frac{1}{a}</math>, <math>\frac{1}{a} = \frac{DE}{EF} = \frac{\sqrt{16-s^2}}{s}</math>, <math>s^2(a^2+1) = 16a^2</math>, <math>s = \frac{4a}{\sqrt{a^2+1}}</math>. | |
− | + | It follows that <math>AC = 2a</math> and <math>BC = \sqrt{(2a)^2 - w^2}</math>. | |
− | + | Because the slope of line <math>y = ax - a</math> is <math>a</math>, <math>a = \frac{BC}{AB} = \frac{\sqrt{4a^2-w^2}}{w}</math>, <math>w^2(a^2+1)=4a^2</math>, <math>w=\frac{2a}{\sqrt{a^2+1}}</math>. | |
− | + | Therefore, the answer is <cmath>\text{Area} = s \cdot w=\frac{4a}{\sqrt{a^2+1}} \cdot \frac{2a}{\sqrt{a^2+1}} = \boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</cmath> | |
− | + | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | |
− | = <math>8a\cos A \cdot \sin A</ | + | ==Solution 4 (Trigonometry)== |
+ | Similar to Solution 1, we will use the equations from the four cases: | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li><math>x+ay=2a.</math> This is a line with <math>x</math>-intercept <math>2a,</math> <math>y</math>-intercept <math>2,</math> and slope <math>-\frac 1a.</math></li><p> | ||
+ | <li><math>x+ay=-2a.</math> This is a line with <math>x</math>-intercept <math>-2a,</math> <math>y</math>-intercept <math>-2,</math> and slope <math>-\frac 1a.</math></li><p> | ||
+ | <li><math>ax-y=a.</math> This is a line with <math>x</math>-intercept <math>1,</math> <math>y</math>-intercept <math>-a,</math> and slope <math>a.</math></li><p> | ||
+ | <li><math>ax-y=-a.</math> This is a line with <math>x</math>-intercept <math>-1,</math> <math>y</math>-intercept <math>a,</math> and slope <math>a.</math></li><p> | ||
+ | </ol> | ||
+ | Let <math>\tan A=a.</math> The area of the rectangle created by the four equations can be written as | ||
+ | <cmath>\begin{align*} | ||
+ | 2a\cdot \cos A\cdot4\sin A &= 8a\cos A \cdot \sin A \\ | ||
+ | &= 8a\cdot~\frac{1}{\sqrt{a^2+1}}\cdot~\frac{a}{\sqrt{a^2+1}} \\ | ||
+ | &= \boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}. | ||
+ | \end{align*}</cmath> | ||
+ | ~fnothing4994 (Solution) | ||
− | + | ~MRENTHUSIASM (Code Adjustments) | |
− | = <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}} | + | ==Solution 5 (Observations)== |
+ | The conditions <math>(x+ay)^2 = 4a^2</math> and <math>(ax-y)^2 = a^2</math> give <math>|x+ay| = |2a|</math> and <math>|ax-y| = |a|</math> or <math>x+ay = \pm 2a</math> and <math>ax-y = \pm a</math>. The slopes here are perpendicular, so the quadrilateral is a rectangle. | ||
+ | Plug in <math>a=1</math> and graph it. We quickly see that the area is <math>2\sqrt{2} \cdot \sqrt{2} = 4</math>, so the answer can't be <math>\textbf{(A)}</math> or <math>\textbf{(B)}</math> by testing the values they give (test it!). Now plug in <math>a=2</math>. We see using a ruler that the sides of the rectangle are about <math>\frac74</math> and <math>\frac72</math>. So the area is about <math>\frac{49}8 = 6.125</math>. Testing <math>\textbf{(C)}</math>, we get <math>\frac{16}3</math> which is clearly less than <math>6</math>, so it is out. Testing <math>\textbf{(D)}</math>, we get <math>\frac{32}5</math> which is near our answer, so we leave it. Testing <math>\textbf{(E)}</math>, we get <math>\frac{16}5</math>, way less than <math>6</math>, so it is out. So, the only plausible answer is <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}</math>. | ||
− | + | ~firebolt360 | |
− | + | ==Solution 6 (Observations)== | |
+ | Trying <math>a = 1</math> narrows down the choices to options <math>\textbf{(C)}</math>, <math>\textbf{(D)}</math> and <math>\textbf{(E)}</math>. Trying <math>a = 2</math> and <math>a = 3</math> eliminates <math>\textbf{(C)}</math> and <math>\textbf{(E)}</math>, to obtain <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}</math> as our answer. Refer to Solution 2 for a detailed explanation. | ||
− | + | ~¢ | |
− | + | ==Solution 7 (Observations: Cheap)== | |
+ | Note that <math>a=2</math> yields different values for all answer choices. If we put in <math>a=2,</math> we find that the area of the quadrilateral is <math>\frac{32}{5}.</math> This means that the answer must be <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math> Refer to Solution 2 for a detailed explanation. | ||
== Video Solution by OmegaLearn (System of Equations and Shoelace Formula) == | == Video Solution by OmegaLearn (System of Equations and Shoelace Formula) == | ||
Line 87: | Line 146: | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
+ | |||
+ | == Video Solution by MRENTHUSIASM (English & Chinese) == | ||
+ | https://www.youtube.com/watch?v=oEY-kX4d87M | ||
+ | |||
+ | ~MRENTHUSIASM | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2021|ab=A|num-b=23|num-a=25}} | {{AMC10 box|year=2021|ab=A|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:31, 18 November 2022
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Generalized Value of a)
- 4 Solution 2 (Specified Value of a)
- 5 Solution 3 (Slopes and Intercepts)
- 6 Solution 4 (Trigonometry)
- 7 Solution 5 (Observations)
- 8 Solution 6 (Observations)
- 9 Solution 7 (Observations: Cheap)
- 10 Video Solution by OmegaLearn (System of Equations and Shoelace Formula)
- 11 Video Solution by MRENTHUSIASM (English & Chinese)
- 12 See also
Problem
The interior of a quadrilateral is bounded by the graphs of and
, where
is a positive real number. What is the area of this region in terms of
, valid for all
?
Diagram
Graph in Desmos: https://www.desmos.com/calculator/satawguqsc
~MRENTHUSIASM
Solution 1 (Generalized Value of a)
The cases for are
or two parallel lines. We rearrange each case and construct the table below:
The cases for
are
or two parallel lines. We rearrange each case and construct the table below:
Since the slopes of intersecting lines
and
are negative reciprocals, we get four right angles, from which the quadrilateral is a rectangle.
Two solutions follow from here:
Solution 1.1 (Distance Between Parallel Lines)
Recall that for constants and
the distance
between parallel lines
is
From this formula:
- The distance between lines
and
is
the length of this rectangle.
- The distance between lines
and
is
the width of this rectangle.
The area we seek is
~MRENTHUSIASM
Solution 1.2 (Distance Between Points)
The solutions to systems of equations are
respectively, which are the consecutive vertices of this rectangle.
By the Distance Formula, the length and width of this rectangle are and
respectively.
The area we seek is
~MRENTHUSIASM
Solution 2 (Specified Value of a)
In this solution, we will refer to equations and
from Solution 1.
Substituting into the answer choices gives
At the solutions to systems of equations
are
respectively, which are the consecutive vertices of the quadrilateral.
Two solutions follow from here:
Solution 2.1 (Area of a Rectangle)
From the tables in Solution 1, we conclude that the quadrilateral is a rectangle.
By the Distance Formula, the length and width of this rectangle are and
respectively.
The area we seek is from which the answer is
~MRENTHUSIASM
Solution 2.2 (Area of a General Quadrilateral)
Even if we do not recognize that the quadrilateral is a rectangle, we can apply the Shoelace Theorem to its consecutive vertices
The area we seek is
from which the answer is
~MRENTHUSIASM
Solution 3 (Slopes and Intercepts)
The quadrilateral is enclosed by four lines. Similar to Solution 1, we will use the equations from the four cases:
This is a line with
-intercept
-intercept
and slope
This is a line with
-intercept
-intercept
and slope
This is a line with
-intercept
-intercept
and slope
This is a line with
-intercept
-intercept
and slope
It follows that and
.
Because the slope of line is
,
,
,
.
It follows that and
.
Because the slope of line is
,
,
,
.
Therefore, the answer is
Solution 4 (Trigonometry)
Similar to Solution 1, we will use the equations from the four cases:
This is a line with
-intercept
-intercept
and slope
This is a line with
-intercept
-intercept
and slope
This is a line with
-intercept
-intercept
and slope
This is a line with
-intercept
-intercept
and slope
Let The area of the rectangle created by the four equations can be written as
~fnothing4994 (Solution)
~MRENTHUSIASM (Code Adjustments)
Solution 5 (Observations)
The conditions and
give
and
or
and
. The slopes here are perpendicular, so the quadrilateral is a rectangle.
Plug in
and graph it. We quickly see that the area is
, so the answer can't be
or
by testing the values they give (test it!). Now plug in
. We see using a ruler that the sides of the rectangle are about
and
. So the area is about
. Testing
, we get
which is clearly less than
, so it is out. Testing
, we get
which is near our answer, so we leave it. Testing
, we get
, way less than
, so it is out. So, the only plausible answer is
.
~firebolt360
Solution 6 (Observations)
Trying narrows down the choices to options
,
and
. Trying
and
eliminates
and
, to obtain
as our answer. Refer to Solution 2 for a detailed explanation.
~¢
Solution 7 (Observations: Cheap)
Note that yields different values for all answer choices. If we put in
we find that the area of the quadrilateral is
This means that the answer must be
Refer to Solution 2 for a detailed explanation.
Video Solution by OmegaLearn (System of Equations and Shoelace Formula)
~ pi_is_3.14
Video Solution by MRENTHUSIASM (English & Chinese)
https://www.youtube.com/watch?v=oEY-kX4d87M
~MRENTHUSIASM
See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
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