Geometry question on square $ABCD$ is a square and $AB$ = 1. Equilateral triangles $AYB$ and $CXD$ are drawn such that $X$ and $Y$ are inside the square.How can I find the length of $XY$ ?
 A: This is actually a very simple problem - in fact, it's possible to solve it mentally. Let me take you through how I did it in my head. You might find it helpful to draw it out, but once you get enough practice, you'll be able to toss problems like this off mentally too. :)
Let $AB$ and $CD$ be the vertical sides (without loss of generality). By visualisation (or sketching), you know this is a highly symmetrical figure. Because of the symmetries involved (opposite sides of a square forming the bases of two equilateral triangles), you should be able to immediately deduce that the apices of the equilateral triangles lie on the horizontal line that exactly divides the square in two (i.e. the line connecting the midpoints of $AB$ and $CD$).
Since I was working mentally, I scaled everything up by a factor of $2$ so I wouldn't have to deal with fractions. So I let the side of the square be $2$ (instead of $1$). Therefore the sides of the equilateral triangles would all be $2$ as well. 
I will denote all vertices and points of the scaled figure by the prime symbol (e.g. $A'$) to avoid confusion. In the upper half of the square, the side of the triangle forms a hypotenuse. Consider the right triangle formed by half the side $A'B'$, the horizontal side lying on the square's midline and the hypotenuse I just mentioned. By applying Pythagoras' Theorem, you can show that the horizontal side has length $\sqrt{2^2 - 1^2} = \sqrt 3$. So $Y'$ lies $\sqrt 3$ units away from side $A'B'$, which means it lies $2 - \sqrt 3$ units away from $C'D'$. By symmetry, $X'$ lies $2 - \sqrt 3$ units from $A'B'$.
Hence you should be able to see that the distance $X'Y'$ can be found by taking the sum of those two distances and then subtracting it from the total length of the midline (which is also the side length of the square). Therefore, $X'Y' = 2 - 2(2 - \sqrt 3) = 2(\sqrt 3 - 1)$.
Now reverse the scaling to get the answer to the original problem: $XY = \sqrt 3 - 1$.
I don't want my notation to seem unnecessarily complex, I just wanted to reflect exactly how I did it in case this is helpful to you as well. Scaling by integer factors to simplify the math is a very useful technique in geometry.
A: HINT: Use the co-ordinates axes and origin i.e. take x-axis along $DC$ and take y-axis along $DA$ and $D$ as the origin. 
So,if you calculate now using the properties of an equilateral triangle ( Length of altitude of an eq. triangle is $\frac{\sqrt{3}a}{2}$ and median and altitude are the same. ), you will get that the co-ordinates of $X$ are $(\frac{1}{2},\frac{\sqrt{3}}{2})$and $Y$ are $(\frac{1}{2},1-\frac{\sqrt{3}}{2})$.
Get the distance using distance formula.
A: It is best if you draw, and I don't know how to draw here. So I will just describe my solution in the form of a hint. 
The interior angle of a square is $90^\circ$. The angle that the diagonal makes to a side is $45^\circ$. Hence, to be able to make an equilateral triangle (which has an interior angle of $60^\circ$, you will have to draw $15^\circ$ more from the diagonal.)
You can construct a triangle with sides composed of (a) half the diagonal, (length is $\dfrac{\sqrt{2}}{2}$), (b) one side of the equilateral triangle ($BY$) (length is 1), and (c ) the side from point $Y$ to the center of square, $x$. 
Find the length $x$ by using the cosine law. That is just half the distance from $X$ to $Y$. then you are done!
A: 
You can use also this approach if you like,also if, as already said, you can really solve it by simmetry arguments.
Area of $\Delta DPR = \cfrac {DP ^2 \cdot \sin P \cdot \sin D}{2 \sin R}=\cfrac {1}{4} \cdot \cfrac{ \sin 90 \cdot \sin 30 }{2 \sin60} =\cfrac { \sqrt {3}}{24}$. 
( ----> Area of any triangle $\Delta ABC$ given side $a$ and two angles is : $\cfrac{a^2 \sin B \sin C}{2 \sin A}$)
So we have that the area of $DROY' =\cfrac {1}{4} - \cfrac { \sqrt {3}}{24}$.
This is the equivalent to say that $DROY'=\cfrac {(DY'+RO) \cdot OY'}{2}= \cfrac {1}{4} - \cfrac { \sqrt {3} }{24}$, now since $DY=OY'=\cfrac {1}{2}$ , we find that $RO= \cfrac {3- \sqrt{3}}{6} $.
Now note that $OX=\sqrt{3} \cdot OR$  (30-60-90 triangle properties) ,so we find that $OX=\cfrac {3- \sqrt{3}}{6} \cdot \sqrt{3} =\cfrac {\sqrt {3}}{2} -\cfrac {1}{2} $.
Now since $XY=2OX$ we find $XY=\sqrt{3} -1$
