Problem of the Week

Problem 27 Solution

Congratulations to Riddhiman Kumar from Cambourne Village College for successfully solving (c), Nate for successfully solving (a), and Markus Kuan from Northstowe Secondary College for successfully solving the problem!

Problem of the Week will be paused and return after the New Years. Merry Christmas and Happy Holidays!

Consider the line perpendicular to $y=mx$ that passes through the point $(1,0)$ :

$y = \frac{-1}{m}x+c$

$c = \frac{1}{m}$

$y= \frac{-1}{m}(x-1)$

The intersection of $y=mx$ and $y=\frac{-1}{m}(x-1)$ is found by solving:

$mx = \frac{-1}{m}(x-1)$

$m^2x = 1 - x$

$x(m^2+1) = 1$

$x = \frac{1}{m^2+1}, \quad y = \frac{m}{m^2+1}.$

Call the point $(1,0)$ as $A$ and the vector from $A$ to $\left(\frac{1}{m^2+1}, \frac{m}{m^2+1}\right)$ as $\vec{p}$ .
The reflection of $A$ in the line $y=mx$ is given by $A + 2\vec{p}$ due to symmetry:

$\vec{p} = \left(\frac{-m^2}{m^2+1}, \frac{m}{m^2+1}\right)$

$A + 2\vec{p} = \left(\frac{1-m^2}{1+m^2}, \frac{2m}{1+m^2}\right)$

This reflection does not change distance to the origin. Thus, for the transformed point, it remains 1.
The angle between the $x$ -axis and the line connecting the transformed point to the origin is twice that of the angle between the line $y=mx$ and the $x$ -axis.

Let $t = \tan\left(\frac{\theta}{2}\right)$ . Then the line $y=tx$ forms an angle of $\frac{\theta}{2}$ with the $x$ -axis. This is due to the likeness between $gradient = \frac{change\ in\ y}{change\ in\ x}$ and $\tan{\theta}=\frac{opposite}{adjacent}$ .
By forming a right-angled triangle whose hypotenuse is the line segment between the origin and the transformed point, the coordinates of the transformed point are:

$(\cos \theta, \sin \theta)$

Comparing this with the result from part A gives:

$\sin \theta = \frac{2t}{1+t^2}, \quad \cos \theta = \frac{1-t^2}{1+t^2}$

with $t = \tan\left(\frac{\theta}{2}\right)$ .

From part B, the reflected point has coordinates

$(\cos \theta, \sin \theta),$

where $\theta$ is the angle in degrees between the line connecting the point to the origin and the positive $x$ -axis.

As $t$ increases from $0$ to $\infty$ , $\theta$ increases from $0^\circ$ to $180^\circ$ .
So, the particle moves along the upper half of a circle with radius 1, following a semicircular path.