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

## 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

## Video Solution

~ naren_pr

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