Locate The D: Revised

The problem may seem short but this is a pretty confusing and long problem.

We can calculate the length of line segment $MN$ which is $1$ .

Point $P$ is the mid-point of line segment $CK$ .

We have:

$MP \parallel BE, MP=1/2EK=1/4EB, \angle MPN = \angle EBA = 60, \angle NMP = \angle EBA = 60 \Rightarrow MNP$ is an equilateral triangle. $\Rightarrow$ the side length of the square $ABCD$ is 2.

To find the location of D, we have to calculate the length of $DM$ and $DN$ .

We have:

$DN = DK/2$ so we need to calculate $DK$ .

Triangle $DEK$ with $EK = 2MP = 2$ and $\angle DEK = 30$ . $ED = EA - DA$ , $EA, DA$ can be determined $\Rightarrow$ $ED$ .

Apply the Cosine rule in a triangle, we can determine the length of $DK$ . Then $DN = DK/2$ (1)

$DM = EC/2$ . We can determine $EC$ , therefore $DM$ (2)

We already have the lengths of $DN,DM$ from (1), (2) and the coordinate of point $M,N$ so we can use a set of equations to solve for $D$ . We will find out that $x_d = 1/2 and 3/2$ and $y_d = \sqrt{3}/2$ in both cases. Given that we only take the value $1/2$ of $x_d$ from the requirement.

In conclusion, $x_d + y_d = 1/2 + \sqrt{3}/2 = \boxed{1.37}$ .

There are four unknowns: The two coordinates of point $A$ $(A_x,A_y)$ , the square side length $S$ , and the rotation angle for the square relative to the $x$ axis $\theta$ . I used a hill-climbing algorithm to find the set of values $(A_x,A_y, S, \theta)$ which makes point $M$ equal to $(1,2)$ and point $N$ equal to $(1,1)$ .

The answers come out to be:

$A = \Big(\frac{3}{2},-\frac{\sqrt{3}}{2} \Big) \\ S = 2 \\ \theta = 30^\circ \\ D = \Big(\frac{1}{2},\frac{\sqrt{3}}{2} \Big)$

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import math
import random

# Initial unknown parameters

Ax = -2.0 + 4.0 * random.random()   # Point A x coordinate
Ay = -2.0 + 4.0 * random.random()   # Point A y coordinate
S = 5.0 * random.random()           # Square side length
theta = math.pi * random.random()   # Rotation angle

delta = 0.001

minres = 99999999.0

########################################################################################

for j in range(0,2*10**6):

    #########

    # Mutate Parameters

    Axm = Ax + delta * (-1.0 + 2.0 * random.random())         
    Aym = Ay + delta * (-1.0 + 2.0 * random.random())          
    Sm = S + delta * (-1.0 + 2.0 * random.random())           
    thetam = theta + delta * (-1.0 + 2.0 * random.random())  

    #########

    # Create two orthogonal basis vectors

    b1x = math.cos(thetam)
    b1y = math.sin(thetam)

    b2x = math.cos(thetam + math.pi/2.0)
    b2y = math.sin(thetam + math.pi/2.0)

    #########

    # Coordinates of point C,B,D,E

    Cx = Axm + 1.0*Sm*b1x + 1.0*Sm*b2x
    Cy = Aym + 1.0*Sm*b1y + 1.0*Sm*b2y

    Bx = Axm + 1.0*Sm*b1x + 0.0*Sm*b2x
    By = Aym + 1.0*Sm*b1y + 0.0*Sm*b2y

    Dx = Axm + 0.0*Sm*b1x + 1.0*Sm*b2x
    Dy = Aym + 0.0*Sm*b1y + 1.0*Sm*b2y

    Ex = Axm + math.sqrt(3.0)*Sm*b2x
    Ey = Aym + math.sqrt(3.0)*Sm*b2y

    #########

    # Points M,K,N

    Mx = (Ex+Cx)/2.0
    My = (Ey+Cy)/2.0

    Kx = (Ex+Bx)/2.0
    Ky = (Ey+By)/2.0

    Nx = (Dx+Kx)/2.0
    Ny = (Dy+Ky)/2.0

    #########

    # Residual - get this as close to zero as possible

    # Point M should be (1,2)
    # Point N should be (1,1)

    resM = math.fabs(Mx-1.0) + math.fabs(My-2.0)
    resN = math.fabs(Nx-1.0) + math.fabs(Ny-1.0)

    res = resM + resN

    # If mutation gets residual closer to zero, keep it
    # Otherwise, discard it

    if res < minres:

        minres = res

        Ax = Axm 
        Ay = Aym 
        S = Sm
        theta = thetam

########################################################################################

print minres
print ""
print Ax
print Ay
print ""
print S
print theta

Dx = Ax + 0.0*S*b1x + 1.0*S*b2x
Dy = Ay + 0.0*S*b1y + 1.0*S*b2y

print ""
print Dx      
print Dy
print (Dx+Dy)

I solved this problem using coordinate geometry. First attach a frame $O'x'y'$ to the figure with the origin $O'$ at point A. The $x'$ axis extends to the right along AB, and the $y'$ axis extends along AD upward. Taking the side length of square $ABCD$ to be $s$ , we can express the coordinates of all the points of interest with respect to this frame, as follows, $A' = (0, 0) , B' = (s, 0), C' = (s, s), D'= (0, s)$ . Now from the angle condition we know that $AE = \tan 60^{\circ} AB = \sqrt{3} s$ , hence, $E' = (0, \sqrt{3} s )$ . Next, we can determine the coordinate of points $M , K$ and $N$ , using the midpoint formula,

$M' = \frac{1}{2} (C' + E') = ( \frac{1}{2} s , \frac{1}{2} (1 + \sqrt{3} ) s )$

$K' = \frac{1}{2} (B' + E') = ( \frac{1}{2} s , \dfrac{\sqrt{3}}{2} s )$

$N' = \frac{1}{2} (K' + D') = ( \frac{1}{4} s , \dfrac{1}{2} (1 + \dfrac{\sqrt{3}}{2} ) s )$

Now, coordinates $p$ in the absolute frame $O x y$ are related to the coordinates $p'$ in the $O'x'y'$ frame by

$p = p_0 + R p'$

for some rotation matrix $R$ . We know that $M = (1, 2)$ and that $N = (1, 1)$ , therefore,

$M - N = (1, 2) - (1,1) = (0, 1) = R (M' - N') = R ( \frac{1}{4} s , \dfrac{\sqrt{3}}{4} s )$

Recall that a rotation matrix is of the form $R = \begin{bmatrix} \cos \theta && -\sin \theta \\ \sin \theta && \cos \theta \end{bmatrix}$ . Hence,

$(0, 1) = ( \frac{1}{4} s ( \cos \theta - \sqrt{3} \sin \theta ) , \frac{1}{4} s ( \sin \theta + \sqrt{3} \cos \theta ) )$

The x-coordinate of the above equation, implies that $\theta = \dfrac{\pi}{6}$ , and using this into the y-coordinate, yields $s = 2$ .

Now, we have

$M = (1, 2) = p_0 + R M' = p_0 + R ( 1 , 1 + \sqrt{3} )$

and

$D = (x_d, y_d) = p_0 + R ( 0, 2 )$

Subtracting the first from second yields,

$( x_d - 1 , y_d - 2) = R ( -1, 1 - \sqrt{3} ) = ( - \dfrac{\sqrt{3}}{2} - \frac{1}{2} +\dfrac{\sqrt{3}}{2} , -\frac{1}{2} + \dfrac{\sqrt{3}}{2} - \dfrac{3}{2} ) = ( -\frac{1}{2}, - 2 + \dfrac{\sqrt{3}}{2} )$

Hence,

$( x_d , y_d ) = ( \dfrac{1}{2}, \dfrac{\sqrt{3}}{2} )$

And therefore, the answer is, $\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} \approx \boxed{1.366}$