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 15:40, 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?
![[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]](http://latex.artofproblemsolving.com/9/f/2/9f2a6546513bf520224bbd20f8b0204324baaf6e.png)
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]](http://latex.artofproblemsolving.com/3/e/8/3e874a5e571ccf35b0e76aec24d330f7b5df7a59.png)
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 
.
![\[x=-x^2+6x-5\]](http://latex.artofproblemsolving.com/c/7/7/c77d96e62cdcd23bf81e073e0148887e1bf01d80.png)
![\[x^2-5x+5=0\]](http://latex.artofproblemsolving.com/0/f/2/0f26da441af2f91c38eb4fb47b5886d5da9775b0.png)
![\[x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}\]](http://latex.artofproblemsolving.com/9/4/6/946c0be301e9cbfa5e5802f3eaa1f29d4083b560.png)
![\[x=\dfrac{5\pm \sqrt{5}}{2}\]](http://latex.artofproblemsolving.com/6/1/3/61327043a82f4aa7c550aee56bbea0ba47e9a91e.png)
Which means 
This means the area of each triangle is 
This the area of the square is 
