Issac's door

PHYSICAL SOLUTION

Ok, Let's see.... I used some trigonometry in this(first time it has actually helped me out)

Firstly,it's easy to observe that the locus is ${ x }^{ 2 }+y^{ 2 }=1$ upto $x=\frac { 1 }{ \sqrt { 2 } }$ (since the tip is always at $1$ unit from the origin)

Now, observe that the curve after $x=\frac { 1 }{ \sqrt { 2 } }$ has the right stick always tangent to it. This means that we have to figure out the curve whose tangent is the second stick.

$\quad$

If we take the apex of the sticks as $(\cos { t } ,\sin { t })$ , then,

the x-intercept of stick 2= $(2\cos { t } ,0)$

slope of stick 2 $= -\tan { t }$

$\quad$

Assume $(h,k)$ to be a point on the required locus (when slope of stick 2 is $=-\tan { t }$ ),

$\Rightarrow \frac { k-0 }{ h-2\cos { t } } =-\tan { t } .........................................(1)$

$\Rightarrow k = -h\tan { t } +2\sin { t }$

$\quad$

Now, since $(h,k)$ lies on the locus, differentiating $k$ w.r.t. $h$ will give the slope at $(h,k)$

$\Rightarrow \frac { dk }{ dh } =-\tan { t } =-\tan { t } -h\frac { dt }{ dh } { (\sec { t } ) }^{ 2 }+2\frac { dt }{ dh } \cos { t }$

on solving, this gives: $\frac { dt }{ dh } (2{ (\cos { t } ) }^{ 3 }-h)=0$

$\Rightarrow h=2{ (\cos { t } ) }^{ 3 }$

from $(1)\Rightarrow k=2{ (\sin { t } ) }^{ 3 }$

$\quad$

Look! We just found the locus! In parametric form:

$(2{ (\cos { t } ) }^{ 3 },2{ (\sin { t } ) }^{ 3 })$

and in cartesian form:

${ x }^{ 2/3 }+y^{ 2/3 }={ 2 }^{ 2/3 } ........................................(2)$

$\quad$

We have to integrate this locus $(2)$ from $x=\frac { 1 }{ \sqrt { 2 } }$ to $x=2$ . I took a bit of help here, looked up basics on such curves.You can do that too: Astroid

The required area is: $\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ \sqrt { 1-{ x }^{ 2 } } } dx + Area(astroid)\\ \\ = \frac { 2+\pi }{ 8 } + 0.339049\\ = 0.981749$

Therefore, the required answer is $\boxed{981}$

Shortly after I posted this problem, I was notified that this problem is already existent on the internet.

So for the sake of variety, I will post a solution without the use of trigonometry:

Im going to use enveloping

First lets define $F(x, y, k)$ for the second purple line (the one that reaches 2), we shall call this $F_{0}(x, y, k)$ . In order to define this, it would be easier to define $F(x, y, k)$ for a sliding ladder (Which would form an astroid), we shall call this $F_{1}(x, y, k)$ .

So to derive $F_{1}(x, y, k)$ , note that the lie would pass through points $(k,0)$ and $(0, \sqrt { 1-{ k }^{ 2 } })$ , $0\le k\le 1$ . Since the equation of the line is linear, we can derive the equation of the line to be $ax+b$ where $a=-\frac{\sqrt{1-k^2}}{k}$ and $b=\sqrt{1-k^2}$ . Therefore, $F_{1}(x, y, k) = ax + b -y = 0$

Now, let the topmost point of the second line be Point $A$ . Point $A$ can be derived to be moving in the path described by the equation ${x}^{2}+{y}^{2}=1$ where $y$ and $x$ are positive. From this, we can derive from $F_{1}(x, y, k)$ that $F_{0}(x, y, k) = ax+b-y+\sqrt{1-k^2}$ .

Using Enveloping, we can find the equation of the path needed in this question from $\sqrt{0.5}\le x\le2$ :

$F_{ 0 }(x,y,k)=ax+b-y+\sqrt{1-k^2}=-\frac { x\sqrt { 1-k^{ 2 } } }{ k } +2\sqrt { 1-k^{ 2 } } -y=0----(1)$ $\frac { \partial }{ \partial k } F_{ 0 }(x,y,k)=\frac { x-2{ k }^{ 3 } }{ { k }^{ 2 }\sqrt { 1-k^{ 2 } } } =0----(2)$

From equations $(1)$ and $(2)$ , we can derive this:

$x=2k^{ 3 }$ $y=2\left( 1-k^{ 2 } \right) ^{ 1.5 }$

Therefore, we have the parametric equation for the equation where $\sqrt{0.5}\le x \le 2$ :

$\left(2k^3,2\left(1-k^2\right)^{1.5}\right)$

$\sqrt { 0.5 } \le k \le 1$

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For the equation (that this question require) when $0\le x\le \sqrt { 0.5 }$ is just a circular arc: ${x}^{2}+{y}^{2}=1$ where $y$ is positive and $\sqrt { 0.5 } \le k \le 1$

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Now what is left is to integrate:

$\int _{ 0 }^{ \sqrt { 0.5 } }{ \sqrt { 1-{ x }^{ 2 } } dx } +\int _{ \sqrt { 0.5 } }^{ 1 }{ \left( 6{ k }^{ 2 } \right) } \left( 2\left( 1-k^{ 2 } \right) ^{ 1.5 } \right) dk=0.981748...=A$

Therefore:

$\left\lfloor 1000A \right\rfloor = \boxed{\boxed{981}}$

Solution to problem: ask Julian for answer IRL

The apex of the triangle will be on the curve when the angle made by the left purple line with y-axis is between 0 and pi/4. For 2 > x > 1/sqrt(2) the curve would be part of (x/2)^(2/3) + (y/2)^(2/3)=1, which comes from a maximization problem. Solving the integration would give the answer