Rolling the square root graph

First let's get some measurements of $y = \sqrt{x}$ before its roll. Let $O$ be the endpoint at the the origin, let $A$ be the endpoint at $(1, 1)$ , let $AB$ be the tangent to $y$ at $A$ , and let $B$ be on the $y$ -axis, as shown below:

Since $A$ is at $(1, 1)$ , by Pythagorean's Theorem, $AO = \sqrt{1^2 + 1^2} = \sqrt{2}$ . The slope of $AO$ is $\frac{1}{1} = 1$ , so its angle of inclination is $\tan^{-1} 1 = 45°$ , which means $\angle AOB = 90° - 45° = 45°$ .

Also, $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$ , so the slope of $y$ at $A(1, 1)$ (and also the slope of $AB$ ) is $m = \frac{1}{2\sqrt{1}} = \frac{1}{2}$ , so its angle of inclination is $\tan^{-1} \frac{1}{2}$ , which means $\angle ABO = 90° + \tan^{-1} \frac{1}{2}$ . From the triangle sum of $\triangle ABO$ , $\angle BAO = 180° - 45° - (90° + \tan^{-1} \frac{1}{2}) = 45° - \tan^{-1} \frac{1}{2}$ .

The arc length from $A$ to $O$ is $s = \int_{0}^{1} \sqrt{1 + (\frac{dy}{dx})^2} dx = \int_{0}^{1} \sqrt{1 + (\frac{1}{2\sqrt{x}})^2} dx \approx 1.47894$ .

After the curve is rolled, $\triangle ABO$ is mapped to $\triangle A'B'O'$ , where $A'B'$ is on the $y$ -axis, and $A$ is at $(0, 1.47894)$ . Let $C'$ be on the $y$ -axis such that $\angle A'C'O'$ is a right angle, as shown below:

Then $A'O' = AO = \sqrt{2}$ and $\angle C'A'O' = \angle BAO = 45° - \tan^{-1} \frac{1}{2}$ , and by trigonometry $A'C' = \sqrt{2} \cos (45° - \tan^{-1} \frac{1}{2}) \approx 1.34164$ , and $O'C' = \sqrt{2} \sin (45° - \tan^{-1} \frac{1}{2}) \approx 0.44721$ .

Therefore, the coordinates of $O'$ are $(0.44721, 1.47894 - 1.34164)$ or $(0.44721, 0.13730)$ , so $x_0 = 0.44721$ and $y_0 = 0.13730$ , and $\lfloor 1000(x_0 + y_0) \rfloor = \boxed{584}$ .

Let's consider the initial position of the graph, the tangent at $(1,1)$ will roll into the y-axis at the point $(0, L)$ , where $L$ is the length of the curve. When we roll the curve along the y-axis, we're basically applying a rotation and shifting to the frame $(x, y)$ . Let's call the final frame $Ox'y'$ , then

$(x', y') = d + R (x, y)$

where $R$ is a rotation matrix, the angle of rotation is $\theta = \dfrac{\pi}{2} - \phi$ , where $\phi$ is the angle that the tangent at $(1,1)$ makes with the x-axis, that is, $\phi = \tan^{-1}(\dfrac{1}{2})$ ,

from which $\sin \theta = \cos( \tan^{-1}(\dfrac{1}{2}) ) = \dfrac{1}{\sqrt{1 + \dfrac{1}{4} }} = \dfrac{2}{\sqrt{5}}$ , and $\cos \theta = \dfrac{1}{\sqrt{5}}$ . Thus, matrix $R$ is given by,

$R = \begin{bmatrix} \cos \theta && -\sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} = \begin{bmatrix} \dfrac{1}{\sqrt{5}} && -\dfrac{2}{\sqrt{5}} \\ \dfrac{2}{\sqrt{5}} && \dfrac{1}{\sqrt{5}} \end{bmatrix}$

To find the vector $d$ in the equation relating $(x, y)$ and $(x', y')$ , we substitute the point $(x, y) = (1,1 )$ and the corresponding point $(x', y') = (0, L)$ . The curve length $L$ is given by,

$L = \displaystyle \int_0^1 \sqrt{1 + 4 y^2} dy = \dfrac{1}{4} ( 2 \sqrt{5} + \ln(2 + \sqrt{5} ) ) = 1.4789428$

It follows that

$d = (0, L) - R (1, 1) = (0, 1.4789428) - ( - \dfrac{1}{\sqrt{5}} , \dfrac{3}{\sqrt{5}} ) = ( 0.4472136, 0.1373020135 )$

And therefore, the final destination coordinates of the point that was originally at $(0,0)$ are

$(x_0, y_0)= d + R (0, 0) = d = (0.4472136, 0.1373020135)$

making the answer, $\lfloor 1000(x_0 + y_0) \rfloor = \boxed{584}$

Wow, another endearing and charming problem! How do you come up with this stuff?!

The problem itself is not hard at all, which makes it even more delightful. For psychological reasons, I rather work with the square graph than with the square root, reflection across the diagonal of the first quadrant. With the notations introduced in the attached figure, the point we seek is $(x,y)=(L-a,b)$ . Now $a$ and $b$ are just distances, and the arc length is $L=\int_{0}^{1}\sqrt{1+4x^2}dx\approx 1.47894$ . My calculator tells me that $\lfloor 1000(x+y)\rfloor=\boxed{584}$ .

One way of solving the problem is by decomposing the roll movement into a rotation and a displacement. Such movement would have two nice properties:

• The angle of rotation would be such that the tangent line of the curve in the upper tip is now the y axis.
• The upper tip would end up in the coordinate $(0,l)$ where $l$ is the length of the curve from 0 to 1.

To easily calculate such transformations we can use complex algebra. By doing this, the rotation would be a multiplication and the displacement an addition. Rewriting the two properties above explicitly:

• $cotan(\theta) = \left.\frac{d\sqrt{x}}{dx}\right|_{x=1}$
• $(1+i) \left[cos(\theta) + i sin (\theta)\right] + x_0 + y_0 i = i l =i \int_{0}^1 \sqrt{1+\left(\frac{d\sqrt{x}}{dx}\right)^2} dx$

Solving the first equation we obtain that $cotan(\theta) = \frac{1}{2} \Rightarrow cos(\theta) = \frac{1}{\sqrt{5}}, sin(\theta) = \frac{2}{\sqrt{5}}$ . Plugin in the second equation and by solving the integral:

$(1+i) \left(\frac{1}{\sqrt{5}} + i \frac{2}{\sqrt{5}}\right) + x_0 + y_0 i = \frac{i}{8} \left[4\sqrt{5} + ln(9+4\sqrt{5})\right]$

Finally we can solve for the real and imaginary parts independently obtaining the following displacement coordinates:

• $x_0 = \frac{1}{\sqrt{5}}$
• $y_0 = \frac{1}{8} \left[4\sqrt{5} + ln(9+4\sqrt{5})\right] - \frac{3}{\sqrt{5}}$

The solution is thus: $\lfloor 1000 \left( \frac{1}{8} \left[4\sqrt{5} + ln(9+4\sqrt{5})\right] - \frac{2}{\sqrt{5}} \right) \rfloor = 584$

For the arc length we use the formula $L=\int_0^1\sqrt{1+(g'(x))^2}$ . For ease we do not use $f=\sqrt{x}$ , but its inverse $g=x^2$ instead. $\int_0^1{\sqrt{1+4x^2}dx}$ $= \frac{1}{2}\int_0^2{\sqrt{1+u^2}du}$ $= \frac{1}{2}\int_0^{arcsinh(2)}{\sqrt{1+(\sinh(t))^2}d\sinh(t)}$ $= \frac{1}{2}\int_0^{arcsinh(2)}{\cosh^2(t)}dt$ $= \frac{t^*}{4}+\frac{\sinh(2t^*)}{8}$ evaluated at $t^*= arcsinh(2)$ , which gives L = 1.4789...

The point (1,1) is on the line y=x at distance $\sqrt{2}$ from the origin, and the graph makes an angle $\alpha=arctan (1)-arctan\frac{1}{2}$ with that line. Doing some trig, we see our point will be at $(\sqrt{2}\sin(\alpha), L-\sqrt{2}\cos(\alpha))$ which is appoximately at (0.4472, 0.1373), so that the required answer is 584.

Find the two lines, first line:

(1) y=(1/2)(x+1), tangent at (1,1) to curve y=√x,

and second line

(2) y=-2x-0.30692,

perpendicular to tangent line at distance 1.4789 ( length of curve) away at (-0.322768, 0.338616).

Find the distance of (0,0) to above two lines as (0.44721, 0.137258),

Answer=1000(0.44721+0.137258)=584