Size of a square in a square

Question

Given a unit square (blue in the picture), pick a point on one edge and label it A. Label the distance from the nearest corner to A as x. Pick one of the corners opposite A and label it B. Call the remaining edge C. There is a unique square with one corner at A, one corner on edge C and with the remaining two corners forming an edge passing through B. What is the area of the new square in terms of x?

The blue square is a unit square. What is the area of the orange square?

(I haven’t been able to work out a complete answer to this yet.)

(I haven't been able to work out a complete answer to this yet.)

Oray · Accepted Answer

Here is the answer:

More is coming!
Here is our original diagram completed with

for sure later;

the length $|DE|$ is our $x$ and let's put some specific angle that we are going to work with in our main square as below;

I call the length of side of the other square as $y$ and as you can see

from the $alpha$ values, right triangles, and hypotenuse as $y$ all four right triangles in the biggest square is the same triangles. I do not want to get into much detail since it is kinda obvious. ($Delta {GEI}$,$Delta {EFC}$,$Delta {FHK}$ and $Delta {HGJ}$)

so we now know that;

And Let's zoom where we want to focus and put our known equations;

Method 1
We know that from the figure above;

from here we solve $z$ as;

then using the equation below

changing z value in terms of x later;

simply we find $y^2$ which is the area of the square we are looking for;
Method 2

and we  know something else from sinus;

and we also know that;
5.

if we combine these without using $1$ and notated that as $cos{alpha }$ alone we get;

as a result;
6.

and using 6. and 1. we are going to figure out what is $z$ in terms of $y$ and $x$ as below;
7.

so let's find our y value using these knowns;

then put our new z value in terms of x and y as in 7;

and solve for $y^2$ which is the area of the new square we are trying to find;

and z in terms of only x becomes;

Weckar E. · Answer

I tried this problem for fun, but got quite a different answer than the others. Posting it for commentary, and educational purposes.
Definitions:

Points A, B and C and length x as described in the problem
Point D, diagonally across from B on the original square
Point E, top left of the original square
Length y (also: AC), side length of the created square

First, I bisected the quadrangle ABCD along the line AC.
The area of the triangle ABC is simple, as both the base and height are y. Therefore, the area is y^2 / 2
The area of ACD is the same base (y), but the height is sqr(2)-y. The diagonal of the original square, minus the height of the other triangle. (y*sqrt(2) - y^2)/2
This means that the area of ABCD is y*sqrt(2)/2
The area of ABD is half the square, minus ADE. In other words, ABD = (1-x)/2, or AD/2
By the same logic, we can say that the area of BCD is CD/2, we just don't quite know what CD is, yet.
Knowing the area of ABCD and ABD, however, BCD = y*sqrt(2)/2 - (1-x)/2. This means that CD = y*sqrt(2) - (1-x)
The Pythagorean triangle ACD is of course y^2 = (1-x)^2+(y*sqrt(2) - (1-x))^2
That's a bit of a garbled mess, but if we work it out (it's just calculation by this point) we find that y = sqrt(2)(1-x)
Finally, we go from a length to a square, so the area is 2(1-x)^2

Mind you, this is clearly wrong, as it implies AD = CD... which in many cases can't possibly be true... I fear I may have made a mistake in calculating the area of ACD, that's the only place where the error would be like this.

Uli Egg · Answer

humn · Answer

As this tessellation of the tilted new square
neatly matches an overlapped tiling of the unit square,
the $w{small,times,}x$ overlap of unit squares equals
the $1 ! - r^2$ difference in the areas of the two types of squares.

(The “$ smallpm $” was deduced to be
“$ small +  $”
by testing the formula on the case where 
$    x   = large{1 over 3}       $, 
$ theta = 45^circ                $, 
$    r   = large{2sqrt2 over 3} $  and 
$   r^2  = large{8 over 9}       $.)
Here are some tiling experiments,
beginning with the easiest-by-hand 45° case,
that led to selecting the straightforward version presented.
The 45° case’s symmetry naturally created some fun red herrings.

hexomino · Answer

I think this matches up with the other two solutions but uses coordinate geometry which is quicker here.

Rand al'Thor · Answer

Firstly let's label everything:

We have five similar right-angled triangles, which must be enough to get a lot of information about the interrelationships between the quantities $x,y,d,e,f,p,q,r$.

Simplifying this expression, the final answer is

Size of a square in a square

6 Answers

Method 1

Method 2

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