Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2014 AMC 10A Problems/Problem 18"

m (→‎Solution)
(→‎Solution 3)
 
(11 intermediate revisions by 4 users not shown)
Line 4: Line 4:
  
 
<math> \textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 27 </math>
 
<math> \textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 27 </math>
 +
[[Category: Introductory Geometry Problems]]
  
−
==Solution==
+
==Solution 1==
 
Let the points be <math>A=(x_1,0)</math>, <math>B=(x_2,1)</math>, <math>C=(x_3,5)</math>, and <math>D=(x_4,4)</math>
 
Let the points be <math>A=(x_1,0)</math>, <math>B=(x_2,1)</math>, <math>C=(x_3,5)</math>, and <math>D=(x_4,4)</math>
  
−
Note that the difference in <math>y</math> value of <math>B</math> and <math>C</math> is <math>4</math>. By rotational symmetry of the square, the difference in <math>x</math> value of <math>A</math> and <math>B</math> is also <math>4</math>. Note that the difference in <math>y</math> value of <math>A</math> and <math>B</math> is <math>1</math>. We now know that <math>AB</math>, the side length of the square, is equal to <math>\sqrt{1^2+4^2}=\sqrt{17}</math>, so the area is <math>\textbf{(B) }17</math>.
+
Note that the difference in <math>y</math> value of <math>B</math> and <math>C</math> is <math>4</math>. By rotational symmetry of the square, the difference in <math>x</math> value of <math>A</math> and <math>B</math> is also <math>4</math>. Note that the difference in <math>y</math> value of <math>A</math> and <math>B</math> is <math>1</math>. We now know that <math>AB</math>, the side length of the square, is equal to <math>\sqrt{1^2+4^2}=\sqrt{17}</math>, so the area is <math>\boxed{\textbf{(B) }17}</math>.
 +
 
 +
==Solution 2==
 +
By translation, we can move the square with point <math>A</math> at the origin. Then, <math>A=(0,0), B=(x_1,1), C=(x_2,5), D=(x_3,4)</math>. We will use the relationship among the 4 sides of being perpendicular and equal.
 +
 
 +
The slope of <math>AB</math> is <math>\frac{1-0}{x_1-0}=\frac{1}{x_1}</math>.
 +
 
 +
Because <math>BC</math> is perpendicular to <math>AB</math>, the slope of <math>BC=-x_1</math>.
 +
From the information above we could have the equation:
 +
 
 +
<math>\frac{5-1}{x_2-x_1}=-x_1</math>
 +
<math>-x_1 \cdot x_2+x_1^2=4</math>
 +
<math>x_1 \cdot x_2=x_1^2-4</math>
 +
<math>x_2=\frac{x_1^2-4}{x_1}</math>
 +
 
 +
Because <math>CD</math> is perpendicular to <math>BC</math>, the slope of <math>CD=\frac{1}{x_1}</math>.
 +
From the information above we could have the equation:
 +
 
 +
<math>\frac{5-4}{x_2-x_3}=\frac{1}{x_1}</math>
 +
<math>x_2-x_3=x_1</math>
 +
<math>\frac{x_1^2-4}{x_1}-x_3=x_1</math>
 +
<math>x_1^2-4-x_1 \cdot x_3 = x_1^2</math>
 +
<math>x_1 \cdot x_3=-4</math>
 +
<math>x_3=- \frac{4}{x_1}</math>
 +
 
 +
Because <math>AD=AB,</math>
 +
 
 +
<math>\sqrt{x_3^2 +4^2} = \sqrt{x_1^2+1^2}</math>
 +
<math>\sqrt{(- \frac{4}{x_1})^2 +4^2} = \sqrt{x_1^2+1^2}</math>
 +
<math>\frac{16}{x_1^2}+16=x_1^2+1</math>
 +
<math>\frac{16}{x_1^2}+15=x_1^2</math>
 +
<math>16+15x_1^2=x_1^4</math>
 +
<math>Let</math> <math>y=x_1^2</math>
 +
<math>16+15y=y^2</math>
 +
<math>y^2-15y-16=0</math>
 +
<math>(y-16)(y+1)=0</math>
 +
<math>y=16</math>
 +
<math>x_1=\pm4</math>
 +
 
 +
Note that the square with <math>x_1=-4</math> is just the reflection of square with <math>x_1=4</math> over the origin. I will use <math>x_1=4</math>.
 +
<math>B=(4,1), AB=\sqrt{17}, [ABCD]=\boxed{\textbf{(B) }17}</math>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 +
 
 +
==Solution 3==
 +
 
 +
In this solution, we will use the fact that the diagonals of a square bisect each other, they are perpendicular to each other, and they are equal in length.
 +
 
 +
Using the fact that the diagonals bisect each other, we get the equation:
 +
<math>\frac{x_2}{2}=\frac{x_1+x_3}{2}</math>
 +
<math>x_2=x_1+x_3</math>
 +
 
 +
Now we use the fact that the diagonals are perpendicular to each other:
 +
<math>Slope</math> <math>of</math> <math>AC=\frac{5-0}{x_2}=\frac{5}{x_2}</math>
 +
<math>Slope</math> <math>of</math> <math>BD=\frac{4-1}{x_3-x_1}=\frac{3}{x_3-x_1}</math>
 +
<math>\frac{5}{x_2} \cdot \frac{3}{x_3-x_1} = -1</math>
 +
<math>x_2(x_1-x_3)=15</math>
 +
 
 +
Using the fact that the diagonals are equal in length, we get the equation:
 +
<math>BD=AC</math>
 +
<math>(4-1)^2+(x_3-x_1)^2=5^2+x_2^2</math>
 +
<math>9+(x_3-x_1)^2=25+x_2^2</math>
 +
<math>(x_3-x_1)^2-x_2^2=16</math>
 +
 
 +
Now we have 3 equations with 3 variables:
 +
 
 +
<math>\begin{cases} x_2=x_1+x_3 \\ x_2(x_1-x_3)=15 \\ (x_3-x_1)^2-x_2^2=16 \end{cases}</math>
 +
 
 +
We substitute <math>x_2</math> into the 2 other equations:
 +
<math>(x_1+x_3)(x_1-x_3)=15</math>
 +
<math>x_1^2-x_3^2=15</math>
 +
 
 +
<math>(x_3-x_1)^2-(x_3+x_1)^2=16</math>
 +
<math>x_3^2-2x_1x_3+x_3^2-x_1^2-2x_1x_3-x_3^2=16</math>
 +
<math>-4x_1x_3=16</math>
 +
<math>x_1x_3=-4</math>
 +
 
 +
Now we have 2 equations of <math>x_1</math> and <math>x_3</math>:
 +
 
 +
<math>\begin{cases} x_1^2-x_3^2=15 \\ x_1x_3=-4 \end{cases}</math>
 +
 
 +
<math>x_3=- \frac{4}{x_1}</math>
 +
<math>x_1^2-(- \frac{4}{x_1})^2=15</math>
 +
<math>x_1^2- \frac{16}{x_1^2}=15</math>
 +
 
 +
This is the same equation as solution <math>2</math>. So <math>x_1= \pm 4, AB=\sqrt{17}, [ABCD]=\boxed{\textbf{(B) }17}</math>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 +
 
 +
==Solution 4 (not rigorous)==
 +
Draw it out, by inspection the coordinates are <math>(-1, 4)</math>, <math>(0, 0)</math>, <math>(4, 1)</math>, and <math>(3, 5)</math>. The side length is <math>\sqrt{17}, [ABCD]=\boxed{\textbf{(B) }17}</math>
 +
~JH. L
 +
 
 +
== Video Solution==
 +
 
 +
https://www.youtube.com/watch?v=iPPQUrNE4RE
 +
 
 +
~ naren_pr
  
 
==See Also==
 
==See Also==

Latest revision as of 02:18, 20 June 2022

Problem

A square in the coordinate plane has vertices whose $y$-coordinates are $0$, $1$, $4$, and $5$. What is the area of the square?

$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 27$

Solution 1

Let the points be $A=(x_1,0)$, $B=(x_2,1)$, $C=(x_3,5)$, and $D=(x_4,4)$

Note that the difference in $y$ value of $B$ and $C$ is $4$. By rotational symmetry of the square, the difference in $x$ value of $A$ and $B$ is also $4$. Note that the difference in $y$ value of $A$ and $B$ is $1$. We now know that $AB$, the side length of the square, is equal to $\sqrt{1^2+4^2}=\sqrt{17}$, so the area is $\boxed{\textbf{(B) }17}$.

Solution 2

By translation, we can move the square with point $A$ at the origin. Then, $A=(0,0), B=(x_1,1), C=(x_2,5), D=(x_3,4)$. We will use the relationship among the 4 sides of being perpendicular and equal.

The slope of $AB$ is $\frac{1-0}{x_1-0}=\frac{1}{x_1}$.

Because $BC$ is perpendicular to $AB$, the slope of $BC=-x_1$. From the information above we could have the equation: 

$\frac{5-1}{x_2-x_1}=-x_1$
$-x_1 \cdot x_2+x_1^2=4$
$x_1 \cdot x_2=x_1^2-4$
$x_2=\frac{x_1^2-4}{x_1}$

Because $CD$ is perpendicular to $BC$, the slope of $CD=\frac{1}{x_1}$. From the information above we could have the equation: 

$\frac{5-4}{x_2-x_3}=\frac{1}{x_1}$
$x_2-x_3=x_1$
$\frac{x_1^2-4}{x_1}-x_3=x_1$
$x_1^2-4-x_1 \cdot x_3 = x_1^2$
$x_1 \cdot x_3=-4$
$x_3=- \frac{4}{x_1}$

Because $AD=AB,$ 

$\sqrt{x_3^2 +4^2} = \sqrt{x_1^2+1^2}$
$\sqrt{(- \frac{4}{x_1})^2 +4^2} = \sqrt{x_1^2+1^2}$
$\frac{16}{x_1^2}+16=x_1^2+1$
$\frac{16}{x_1^2}+15=x_1^2$
$16+15x_1^2=x_1^4$
$Let$ $y=x_1^2$
$16+15y=y^2$
$y^2-15y-16=0$
$(y-16)(y+1)=0$
$y=16$
$x_1=\pm4$

Note that the square with $x_1=-4$ is just the reflection of square with $x_1=4$ over the origin. I will use $x_1=4$. $B=(4,1), AB=\sqrt{17}, [ABCD]=\boxed{\textbf{(B) }17}$

~isabelchen

Solution 3

In this solution, we will use the fact that the diagonals of a square bisect each other, they are perpendicular to each other, and they are equal in length.

Using the fact that the diagonals bisect each other, we get the equation: 

$\frac{x_2}{2}=\frac{x_1+x_3}{2}$
$x_2=x_1+x_3$

Now we use the fact that the diagonals are perpendicular to each other: 

$Slope$ $of$ $AC=\frac{5-0}{x_2}=\frac{5}{x_2}$
$Slope$ $of$ $BD=\frac{4-1}{x_3-x_1}=\frac{3}{x_3-x_1}$
$\frac{5}{x_2} \cdot \frac{3}{x_3-x_1} = -1$
$x_2(x_1-x_3)=15$

Using the fact that the diagonals are equal in length, we get the equation: 

$BD=AC$
$(4-1)^2+(x_3-x_1)^2=5^2+x_2^2$
$9+(x_3-x_1)^2=25+x_2^2$
$(x_3-x_1)^2-x_2^2=16$

Now we have 3 equations with 3 variables:

$\begin{cases} x_2=x_1+x_3 \\ x_2(x_1-x_3)=15 \\ (x_3-x_1)^2-x_2^2=16 \end{cases}$

We substitute $x_2$ into the 2 other equations: 

$(x_1+x_3)(x_1-x_3)=15$
$x_1^2-x_3^2=15$

$(x_3-x_1)^2-(x_3+x_1)^2=16$
$x_3^2-2x_1x_3+x_3^2-x_1^2-2x_1x_3-x_3^2=16$
$-4x_1x_3=16$
$x_1x_3=-4$

Now we have 2 equations of $x_1$ and $x_3$:

$\begin{cases} x_1^2-x_3^2=15 \\ x_1x_3=-4 \end{cases}$ 

$x_3=- \frac{4}{x_1}$
$x_1^2-(- \frac{4}{x_1})^2=15$
$x_1^2- \frac{16}{x_1^2}=15$

This is the same equation as solution $2$. So $x_1= \pm 4, AB=\sqrt{17}, [ABCD]=\boxed{\textbf{(B) }17}$

~isabelchen

Solution 4 (not rigorous)

Draw it out, by inspection the coordinates are $(-1, 4)$, $(0, 0)$, $(4, 1)$, and $(3, 5)$. The side length is $\sqrt{17}, [ABCD]=\boxed{\textbf{(B) }17}$ ~JH. L

Video Solution

https://www.youtube.com/watch?v=iPPQUrNE4RE

~ naren_pr

See Also

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_18&oldid=175078"