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Difference between revisions of "2015 AMC 8 Problems/Problem 25"

(→‎See Also)
m (→‎Solution 2)
Line 38: Line 38:
 
We draw a square as shown:
 
We draw a square as shown:
 
<asy>
 
<asy>
−
pair Q,R,S,T;
 
−
Q=(1.381966,0);
 
−
R=(5,1.381966);
 
−
S=(3.618034,5);
 
−
T=(0,3.618034);
 
−
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);
 
−
draw(Q--R--S--T--cycle);
 
−
draw((1,1)--(4,1)--(4,4)--(1,4)--cycle,dashed);
 
−
</asy>
 
  
 
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 <math>3</math> and height <math>1</math>, so the combined area of the four triangles is <math>4 \cdot \frac 32=6</math>.  The area of the smaller square is <math>9</math>.  We add these to see that the area of the large square is <math>9+6=\boxed{{\textbf{(C)}}~15}</math>.
 
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 <math>3</math> and height <math>1</math>, so the combined area of the four triangles is <math>4 \cdot \frac 32=6</math>.  The area of the smaller square is <math>9</math>.  We add these to see that the area of the large square is <math>9+6=\boxed{{\textbf{(C)}}~15}</math>.

Revision as of 11:14, 1 October 2016

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 <math>3</math> and height <math>1</math>, so the combined area of the four triangles is <math>4 \cdot \frac 32=6</math>.  The area of the smaller square is <math>9</math>.  We add these to see that the area of the large square is <math>9+6=\boxed{{\textbf{(C)}}~15}</math>.

===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 making remote request. Unknown error_msg)

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 $25- (4+6) = \boxed{\textbf{(C)}~15}$

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
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