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| | + | == Problem == |
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| | 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? | | 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? |
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| − | <math>\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</math>
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| | <asy> | | <asy> |
| | + | size(75); |
| | draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); | | draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); |
| | filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); | | filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); |
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| | </asy> | | </asy> |
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| − | ===Solution 1===
| + | <math>\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</math> |
| − | We can draw a diagram as shown.
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| − | <asy> | |
| − | draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);
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| − | filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);
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| − | filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);
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| − | filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);
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| − | filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);
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| − | path arc = arc((2.5,4),1.5,0,90);
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| − | pair P = intersectionpoint(arc,(0,5)--(5,5));
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| − | 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;
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| − | draw(P--Pp--Ppp--Pppp--cycle);
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| − | </asy>
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| − | 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 <math>4</math> big triangles by <math>AA.</math> Let the height of a big triangle be <math>x</math> then <math>\tfrac{x}{x-1}=\tfrac{5-x}{1}</math>.
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| − | <cmath>x=-x^2+6x-5</cmath>
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| − | <cmath>x^2-5x+5=0</cmath>
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| − | <cmath>x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}</cmath>
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| − | <cmath>x=\dfrac{5\pm \sqrt{5}}{2}</cmath>
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| − | Thus <math>x=\dfrac{5-\sqrt{5}}{2}</math>, because by symmetry, <math>x < \dfrac52</math>.
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| − | This means the area of each triangle is <math>\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}</math>
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| − | Thus the area of the square is <math>25-(4*\dfrac{5}{2})=\boxed{\textbf{(C)}~15}</math>
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| − | ===Solution 2=== | + | ==Video Solutions== |
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| − | We draw a square as shown:
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| − | <asy>
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| − | draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);
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| − | filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, red);
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| − | filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);
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| − | filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);
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| − | filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);
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| − | path arc = arc((2.5,4),1.5,0,90);
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| − | pair P = intersectionpoint(arc,(0,5)--(5,5));
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| − | 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;
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| − | draw(P--Pp--Ppp--Pppp--cycle);
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| − | filldraw((1,4)--P--(4,4)--cycle,red);
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| − | filldraw((4,4)--Pppp--(4,1)--cycle,red);
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| − | filldraw((1,1)--Ppp--(4,1)--cycle,red);
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| − | filldraw((1,1)--Pp--(1,4)--cycle,red);
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| − | </asy>
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| − | We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four red triangles. The red 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>.
| + | https://youtu.be/rTljQV79PCY |
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| − | ===Solution 3===
| + | ~savannahsolver |
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| | + | https://youtu.be/51K3uCzntWs?t=3358 |
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| − | 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 <math>1</math>, and let's label the other legs <math>x</math> for one of the triangles and <math>y</math> for the other. Note that <math>x + y = 3</math>.
| + | ~ pi_is_3.14 |
| − | The area of each of the triangles is <math>\frac{x}{2}</math> and <math>\frac{y}{2}</math>, and there are <math>4</math> of each. So now we need to find <math>4\left(\frac{x}{2}\right) + 4\left(\frac{y}{2}\right)</math>.
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| − | <math>(4)\frac{x}{2} + (4)\frac{y}{2}</math>
| + | ==See Also== |
| − | <math>\Rightarrow~~4\left(\frac{x}{2}+ \frac{y}{2}\right)</math>
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| − | <math>\Rightarrow~~4\left(\frac{x+y}{2}\right)</math>
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| − | Remember that <math>x+y=3</math>, so substituting this in we find that the area of all of the triangles is <math>4\left(\frac{3}{2}\right) = 6</math>.
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| − | The area of the <math>4</math> unit squares is <math>4</math>, so the area of the square we need is <math>25- (4+6) = \boxed{\textbf{(C)}~15}</math>
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| − | ==See Also==
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| | {{AMC8 box|year=2015|num-b=24|after=Last Problem}} | | {{AMC8 box|year=2015|num-b=24|after=Last Problem}} |
| | {{MAA Notice}} | | {{MAA Notice}} |
![[asy] size(75); 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]](http://latex.artofproblemsolving.com/1/6/9/16905c0611e83380a7ff4045538ffe74a49c7cb7.png)
