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

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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; | 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); | draw(P--Pp--Ppp--Pppp--cycle); | ||

− | </asy> | + | </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 <math>x</math> then <math>\dfrac{x}{x-1}=\dfrac{5-x}{1}</math>. | 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=-x^2+6x-5</cmath> |

## Revision as of 16:40, 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. 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