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Difference between revisions of "2014 AMC 10A Problems/Problem 18"
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[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
  
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==Solution==
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==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>\boxed{\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>.
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 +
==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:
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 +
<math>\frac{5-1}{x_2-x_1}=-x_1</math>
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<math>-x_1 \cdot x_2+x_1^2=4</math>
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<math>x_1 \cdot x_2=x_1^2-4</math>
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<math>x_2=\frac{x_1^2-4}{x_1}</math>
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 +
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:
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 +
<math>\frac{5-4}{x_2-x_3}=\frac{1}{x_1}</math>
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<math>x_2-x_3=x_1</math>
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<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>
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<math>x_1 \cdot x_3=-4</math>
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<math>x_3=- \frac{4}{x_1}</math>
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 +
Because <math>AD=AB,</math>
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<math>\sqrt{(- \frac{4}{x_1})^2 +4^2} = \sqrt{(x_1)^2+1^2}</math>
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<math>\frac{16}{x_1^2}+16=x_1^2+1</math>
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<math>16+16x_1^2=x_1^4+x_1^2</math>
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<math>Let</math> <math>y=x_1^2</math>
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<math>16+16y=y^2+y</math>
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<math>y^2-15y-16=0</math>
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<math>(y-16)(y+1)=0</math>
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<math>y=16</math>
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<math>x_1=\pm4</math>
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 +
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>.
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<math>B=(4,1), AB=\sqrt{17}, [ABCD]=\boxed{\textbf{(B) }17}</math>
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 +
~isabelchen
  
 
=== Video Solution===
 
=== Video Solution===

Revision as of 05:23, 1 October 2021

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 $AD=AB,$ 

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

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

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

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