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

(Created page with "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? <math> ...")
 
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filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);
 
filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);
 
</asy>
 
</asy>
 +
 +
SOLUTION 1
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Lets draw a diagram.
 +
<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> (Credits to djmathman )
 +
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 <math>x</math> then <math>\dfrac{x}{x-1}=\dfrac{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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Which means <math>x=\dfrac{5-\sqrt{5}}{2}</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>
 +
This the area of the square is <math>25-(4*\dfrac{5}{2})=\boxed{C,~15}</math>

Revision as of 15:39, 25 November 2015

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?

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

[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 Lets draw a diagram. [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] (Credits to djmathman ) 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 $\dfrac{x}{x-1}=\dfrac{5-x}{1}$. \[x=-x^2+6x-5\] \[x^2-5x+5=0\] \[x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}\] \[x=\dfrac{5\pm \sqrt{5}}{2}\] Which means $x=\dfrac{5-\sqrt{5}}{2}$ 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 $25-(4*\dfrac{5}{2})=\boxed{C,~15}$

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