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

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

~ naren_pr

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS