be real numbers so that 
and 
Show that
Y by Adventure10, Mango247
A small square is constructed inside a square of area 1 by dividing each side of the unit square into 
equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of if the the area of the small square is exactly 1/1985.
![[asy]
size(200);
pair A=(0,1), B=(1,1), C=(1,0), D=origin;
draw(A--B--C--D--A--(1,1/6));
draw(C--(0,5/6)^^B--(1/6,0)^^D--(5/6,1));
pair point=( 0.5 , 0.5 );
//label("$A$", A, dir(point--A));
//label("$B$", B, dir(point--B));
//label("$C$", C, dir(point--C));
//label("$D$", D, dir(point--D));
label("$1/n$", (11/12,1), N, fontsize(9));[/asy]](//latex.artofproblemsolving.com/8/6/9/869771a29889569e7f650bf31a1241605514e6bd.png)
equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of if the the area of the small square is exactly 1/1985.![[asy]
size(200);
pair A=(0,1), B=(1,1), C=(1,0), D=origin;
draw(A--B--C--D--A--(1,1/6));
draw(C--(0,5/6)^^B--(1/6,0)^^D--(5/6,1));
pair point=( 0.5 , 0.5 );
//label("$A$", A, dir(point--A));
//label("$B$", B, dir(point--B));
//label("$C$", C, dir(point--C));
//label("$D$", D, dir(point--D));
label("$1/n$", (11/12,1), N, fontsize(9));[/asy]](http://latex.artofproblemsolving.com/8/6/9/869771a29889569e7f650bf31a1241605514e6bd.png)
This post has been edited 3 times. Last edited by djmathman, Dec 24, 2015, 4:29 AM
Reason: corners of square not labeled in original diagram
Reason: corners of square not labeled in original diagram
be the foot of the perpendicular from 
to 
.
(perpendicular sides), so 
,
,
and 
,
by the Pythagorean Theorem. Additionally, 
because 
is the altitude to the hypotenuse of right triangle 
.
.
.
is equal to the length of a side of the interior square, 
.
,
.
is equal to the length of a side of the interior square, 
.
![[asy]defaultpen(fontsize(8));
pair A=(0,1), C=(31/32,1), P, D=(0,0), B=(0,31/32), Q;
P=foot(A,D,C);Q=foot(B,D,C);
draw(A--C--D--A--P);
draw(B--Q);
label("$A$",A,(-1,1));label("$D$",D,(-1,-1));label("$D'$",C,(1,1));label("$P$",P,(0,1));label("$C'$",B,(-1,-1));label("$Q$",Q,(-1,0));[/asy]](http://latex.artofproblemsolving.com/7/0/5/705da5e2d58e053673b59c3a389183306b71c0c2.png)
.
.
,
.
.
?
and 
but they are not because that would imply 
which is clearly false.
by dividing each side of the unit square into 
.![[asy]
pair A=(0,1), B=(1,1), C=(1,0), D=origin;
draw(A--B--C--D--A--(1,1/6));
draw(C--(0,5/6)^^B--(1/6,0)^^D--(5/6,1));
pair point=( 0.5 , 0.5 );
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$1/n$", (11/12,1), N, fontsize(9));[/asy]](http://latex.artofproblemsolving.com/e/5/4/e54f0ee920b3de7707f7b6cec28d7fe8facbc581.png)
and 
so that 
and the sidelength 
of the small square.
. Therefore, 
.
.
.
.
.
and factoring, 
,
.
be the length of the big square and 
be the length of small square, then![$[BLDK]=\frac{1}{n} \times l$](http://latex.artofproblemsolving.com/2/e/8/2e886d53056ebd14f7a506d6c2f4f825f8ad4a5f.png)
![$[BLDK]=k \times \sqrt{l^2+(\frac{n-1}{n})^2}$](http://latex.artofproblemsolving.com/2/5/a/25a1a0b6143e3e2f1830ddcfe758e759449267cf.png)
,
