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

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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 | ||

+ | 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 <math>x</math> then <math>\dfrac{x}{x-1}=\dfrac{5-x}{1}</math>. | ||

+ | <cmath>x=-x^2+6x-5</cmath> | ||

+ | <cmath>x^2-5x+5=0</cmath> | ||

+ | <cmath>x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}</cmath> | ||

+ | <cmath>x=\dfrac{5\pm \sqrt{5}}{2}</cmath> | ||

+ | Which means <math>x=\dfrac{5-\sqrt{5}}{2}</math> | ||

+ | 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 16:39, 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?

SOLUTION 1 Lets draw a diagram. (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 then . Which means This means the area of each triangle is This the area of the square is