Christmas gift

Let $p$ be the length of the rope between $(5, 2)$ and the hook. Since the rope has a length of $15$ , the length of the rope between $(-3, 8)$ and the hook will be $15 - p$ .

The locus of points for the hook are then $(x - 5)^2 + (y - 2)^2 = p^2$ and $(x + 3)^2 + (y - 8)^2 = (15 - p)^2$ . Eliminating $p$ gives the ellipse $644x^2 - 3208x + 756y^2 -7944y + 384xy - 6661 = 0$ .

Due to gravity, the hook will fall to the lowest point of this ellipse, which is a $y$ -value with only one $x$ -value. Solving the ellipse equation for $x$ gives $x = \frac{802 -96y \pm 75\sqrt{5(-4y^2 + 40y + 61)}}{322}$ , which has one $x$ -value when $-4y^2 + 40y + 61 = 0$ , or when $y = 5 \pm \frac{1}{2}\sqrt{161}$ .

The lower point solves to $(x_0, y_0) = (1 + \frac{24}{161}\sqrt{161}, 5 - \frac{1}{2}\sqrt{161})$ , which makes $\lfloor 10^3(x_0 + y_0) \rfloor = \boxed{1547}$ .

Let $\vec{F_1}=(-3,8)$ and $\vec{F_2}=(5,2)$ . If we trace all the points where the hook can be keeping tense the rope, we get an ellipse with foci $\vec{F_1}$ and $\vec{F_2}$ , and center $\vec{C} = \dfrac{1}{2}\left(\vec{F_1}+\vec{F_2}\right)=(1,5)$ , and we just want its lowest point. We will find a parametric form of the ellipse first.

Let $a$ , $b$ and $c$ be the semi-major axis length, semi-minor axis length and semi focal distance, respectively. We see that $2c=\left\lVert \vec{F_2}-\vec{F_1} \right\rVert = 10$ , so $c=5$ . We also have $a=\frac{15}{2}$ , and $b=\sqrt{a^2-c^2}=\frac{5\sqrt{5}}{2}$ .

The next step is to find an orthonormal basis $\{\hat{e_1}, \hat{e_2}\}$ , where $\hat{e_1}$ is a unit vector in the direction of the major axis and $\hat{e_2}$ points in the direction of the minor axis. We see that $\hat{e_1}=\dfrac{\vec{F_2}-\vec{F_1}}{\left\lVert \vec{F_2}-\vec{F_1} \right\rVert}=\dfrac{1}{5}(4,-3)$ and $\hat{e_2}=\dfrac{1}{5}(3,4)$ .

Then, we are ready to write the parametric form of the ellipse: $\vec{r}(t) = \vec{C} + \hat{e_1}a\cos(t) + \hat{e_2}b\sin(t) = (1,5) + \dfrac{3}{2}(4,-3)\cos(t) + \dfrac{\sqrt{5}}{2}(3,4)\sin(t)$ , where $0 \leq t \leq 2\pi$ .

In the lowest and highest points of the ellipse, for some value of $t=t_0$ , we have that the $y$ -component of $\vec{r}'(t_0)$ is zero. In this case $\vec{r}'(t)=\dfrac{3}{2}(-4,3)\sin(t) + \dfrac{\sqrt{5}}{2}(3,4)\cos(t)$ , so we have to solve $\dfrac{9}{2}\sin(t_0)+2\sqrt{5}\cos(t_0)=0$ , which gives $\tan(t_0)=-\dfrac{4\sqrt{5}}{9}$ .

We want to find $\cos(t_0)$ and $\sin(t_0)$ , and we get two values for each of them: $\cos(t_0) = \pm \dfrac{9}{\sqrt{161}}$ and $\sin(t_0) = \mp\dfrac{4\sqrt{5}}{\sqrt{161}}$ .

Finally, we get two points: $\vec{r}(t_0)=(1,5) \pm \dfrac{(48,-161)}{2\sqrt{161}}$ . The lowest one has the least value of the $y$ -coordinate, and that is $(x_0,y_0)=\left(1+\dfrac{24}{\sqrt{161}},5-\dfrac{\sqrt{161}}{2}\right) \approx (2.89146,-1.34428)$ , making the answer $\boxed{1547}$ .

The coordinates of the painting hook are $\vec{H} = (x_0,y_0)$ . Call the nail points $\vec{P}$ and $\vec{Q}$ . We know that the sum of the lengths of the hook to $\vec{P}$ and the hook to $\vec{Q}$ must be $15$ . We also know that the tension in the rope is uniform, and that the sum of the horizontal forces must be zero. Therefore, the unit vector from $\vec{H}$ to $\vec{P}$ must have an equal and opposite $x$ component relative to the the unit vector from $\vec{H}$ to $\vec{Q}$ . Mathematically, the solution criteria are:

$|\vec{P} - \vec{H}| + |\vec{Q} - \vec{H}| = 15 \\ \frac{P_x - H_x}{|\vec{P} - \vec{H}|} + \frac{Q_x - H_x}{|\vec{Q} - \vec{H}|} = 0$

Solving results in $(x_0,y_0) \approx (2.8915,-1.3443)$

Here is the code for a hill-climbing solution (since the question is from a programming contest):

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import math
import random

Px = -3.0
Py = 8.0

Qx = 5.0
Qy = 2.0

x0 = -5.0 + 10.0 * random.random()
y0 = -5.0 + 10.0 * random.random()

delta = 0.001

minres = 99999999.0

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

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

    x0m = x0 + delta * (-1.0 + 2.0 * random.random())
    y0m = y0 + delta * (-1.0 + 2.0 * random.random())

    DP = math.hypot(x0m-Px,y0m-Py)
    DQ = math.hypot(x0m-Qx,y0m-Qy)

    D = DP + DQ

    uPx = (Px - x0m) / DP
    uQx = (Qx - x0m) / DQ

    res1 = math.fabs(D - 15.0)
    res2 = math.fabs(uPx + uQx)

    res = res1 + res2 

    if res < minres:
        minres = res

        x0 = x0m
        y0 = y0m


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

if minres < 0.001:

    print minres
    print ""
    print x0
    print y0
    print ""
    print math.floor(1000.0*(x0+y0))
    print "###########################"
    print ""