2015 AMC 8 Problems/Problem 25

One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?

$\textbf{(A) } 9\qquad \textbf{(B) } 12\frac{1}{2}\qquad \textbf{(C) } 15\qquad \textbf{(D) } 15\frac{1}{2}\qquad \textbf{(E) } 17$

$[asy] draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); [/asy]$

Solution 1

We can draw a diagram as shown. $[asy] draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); path arc = arc((2.5,4),1.5,0,90); pair P = intersectionpoint(arc,(0,5)--(5,5)); pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; draw(P--Pp--Ppp--Pppp--cycle); [/asy]$ Let us focus on the big triangles taking up the rest of the space. The triangles on top of the unit square between the inscribed square, are similiar to the $4$ big triangles by $AA.$ Let the height of a big triangle be $x$ then $\tfrac{x}{x-1}=\tfrac{5-x}{1}$. $$x=-x^2+6x-5$$ $$x^2-5x+5=0$$ $$x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}$$ $$x=\dfrac{5\pm \sqrt{5}}{2}$$ Thus $x=\dfrac{5-\sqrt{5}}{2}$, because by symmetry, $x < \dfrac52$.

This means the area of each triangle is $\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}$ This the area of the square is $25-(4*\dfrac{5}{2})=\boxed{\textbf{(C)}~15}$

Solution 2

We draw a square as shown:


We wish to find the area of the square.  The area of the larger square is composed of the smaller square and the four triangles.  The triangles have base $3$ and height $1$, so the combined area of the four triangles is $4 \cdot \frac 32=6$.  The area of the smaller square is $9$.  We add these to see that the area of the large square is $9+6=\boxed{{\textbf{(C)}}~15}$.

===Solution 3===
<asy>
draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);
filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);
filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);
filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);
filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);
path arc = arc((2.5,4),1.5,0,90);
pair P = intersectionpoint(arc,(0,5)--(5,5));
pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;
draw(P--Pp--Ppp--Pppp--cycle);
(Error compiling LaTeX. We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four triangles. The triangles have base $3$ and height $1$, so the combined area of the four triangles is $4 \cdot \frac 32=6$. The area of the smaller square is $9$. We add these to see that the area of the large square is $9+6=\boxed{{\textbf{(C)}}~15}$.
^
3a2b058d72a4ff6f6b7c371c8005551cd008c13e.asy: 5.9: syntax error
error: could not load module '3a2b058d72a4ff6f6b7c371c8005551cd008c13e.asy')

Let us find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length $1$, and let's label the other legs $x$ for one of the triangles and $y$ for the other. Note that $x + y = 3$. The area of each of the triangles is $\frac{x}{2}$ and $\frac{y}{2}$, and there are $4$ of each. So now we need to find $4\left(\frac{x}{2}\right) + 4\left(\frac{y}{2}\right)$.

$(4)\frac{x}{2} + (4)\frac{y}{2}$ $\Rightarrow~~4\left(\frac{x}{2}+ \frac{y}{2}\right)$ $\Rightarrow~~4\left(\frac{x+y}{2}\right)$ Remember that $x+y=3$, so substituting this in we find that the area of all of the triangles is $4\left(\frac{3}{2}\right) = 6$. The area of the $4$ unit squares is $4$, so the area of the square we need is $25- (4+6) = \boxed{\textbf{(C)}~15}$