Square Formed From Intersecting Arcs

I don't have the heart to leave this without solution. So...

Solution:

Let $a$ be a length of the bigger square's edge. Now, let's concentrate on point $Q$ (or: $M$ , $N$ or $P$ , conclusion will be the same). I believe, that it can be spotted that $BQ$ and $AQ$ length is $a$ . Therefore, triangle $ABQ$ is equilateral. $E$ and $F$ shall be midpoints of $AB$ and $CD$ , respectively. Length of $EQ$ is $\frac { a\sqrt { 3 } }{ 2 }$ , $EQ + QF$ is $a$ , therefore $QF = a(1 - \frac { \sqrt { 3 } }{ 2 } )$ .

Finally, we can calculate $QN$ ( $NE = QR$ , $QN = a - QR - NE$ ), which is $a( \sqrt{3} - 1)$ . $QN$ and $MP$ both are equal and are $MNPQ$ 's diagonals. Using formula $\frac{ d^{2} }{ 2 }$ we will get that area of $MNPQ$ is $a^{2}(2 - \sqrt{3})$ . Since $a^{2}$ is $ABCD$ 's area, we get $3(2 + \sqrt{3})(2 - \sqrt{3})$ which is 3 .

I solved it a bit differently. To start, I knew the vesica was used to construct equilateral triangles. So, we know the angle QCP, for example, is 60 degrees. So: $90=60+2\theta$ . We quickly find theta is 15. We kind the adjacent side of a right triangle easily: $a=\ sqrt{(6+3\sqrt{3})}/2\approx 1.67$ . And we find the hypotenuse: $h= 1.67/cos(15)\approx 1.73$ . Squaring this gives us the answer: $\boxed{3}$ .

Equation of circle with center at A(0,0)

x^2 + y^2 = r^2 = 6 + 3*sqrt(3) [1]

Equation of circle with center at B(6 + 3*sqrt(3), 0)

(x - 6 - 3 sqrt(3))^2 +y^2 = 6 + 3 sqrt(3) [2]

Subtracting [1] from [2] :

(x - 6 - 3*sqrt(3))^2 - x^2 = 0

Expanding and simplifying:

x^2 - 2 x * sqrt(6 + 3 sqrt(3)) + 6 + 3sqrt(3) - x^2 = 0

    6 + 3 * sqrt(3) = 2 * sqrt( 6 + 3*sqrt(3)) * x

    x = ( 6 + 3 * sqrt(3))/ 2*sqrt(6 + 3*sqrt(3))   = app. 1.673

Substituting into [1]

   y = 2.8978

By reflection across y = x a second point is:

 x = 2.8978       y = 1.673

Distance formula yields length of side = 1.7321

Area = 1.7321^2 = 3

The way i did it was by forming an equilateral triangle, solving for the area of the sectors and then solving for the arrow shapes and so forth. I was left with a reuleauxish type square in the centre with an area of 3.52. I then let A denote the side of the square MNPQ, substituting the values of 3,2 and 1 seeing which one would equal (6+3*(sqrt3)), and three was the answer

Take a vertex as origin and side length a and find two of the points on the square by solving quarter circle double equations. Knowing the side or required square in terms of a, area can be easily calculated. The two points are (asin30,acos30) and (asin60,acos60).

Excellent problem! Thank you!

make a line between point Q&A,P&A,project a 90° line form p to AB let us call intersection point z&connect point A to median point of line PQ call it( u),at∆ pzA sin(mAp)=opposite \hypotenuse,opp=square side\2,hypotenuse=square side =Ap (radius ),so sin(mAP)=square side\2÷square side =.5,angle=30°,at∆cAB angle (cAB)=45°,so angle (pau)=45-30=15°,from symmetry angle (pAQ)=15×2=30°,from geometry of ∆PAQ,cos30=PA^2+QA^2- PQ^2 (2×PA×QA),pa=qa=square side ,cos30=2(PA)^2-(PQ^2)\2PA^2, √3\2×2×(6+3√3)=2×(6+3√3)-PQ^2 ,,,,,,,,,,,,pQ^2 =required area =12+6√3 -6√3-9=3