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

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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> \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</math>
 
  
 
<asy>
 
<asy>
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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
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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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Lets draw a diagram.
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<asy>
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draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);
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==Video Solutions==
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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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https://youtu.be/rTljQV79PCY
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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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~savannahsolver
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draw(P--Pp--Ppp--Pppp--cycle);
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</asy>
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https://youtu.be/51K3uCzntWs?t=3358
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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 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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~ pi_is_3.14
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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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==See Also==
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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>
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{{AMC8 box|year=2015|num-b=24|after=Last Problem}}
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This the area of the square is <math>25-(4*\dfrac{5}{2})=\boxed{C,~15}</math>
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{{MAA Notice}}

Latest revision as of 10:02, 23 July 2024

Problem

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?

[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]

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


Video Solutions

https://youtu.be/rTljQV79PCY

~savannahsolver

https://youtu.be/51K3uCzntWs?t=3358

~ pi_is_3.14

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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