Problem of the Week

Problem 31 Solution

Congratulations to Matthew Whitten from The Cavendish School, Markus Kuan from Northstowe Secondary College, and Ishaan Agarwal from Cambourne Village College for successfully solving the problem!

a) $\pi(N)$ can be considered either undefined or $\infty$ , thus demonstrating the correspondence between the number line with infinity adjoined and a circle.

The line connecting $N$ to a point on $\Sigma$ will only intersect the x-axis once.
Moreover, the line connecting $N$ to a point on the x-axis will only intersect $\Sigma$ at one point other than $N$ .

b) $\angle OAN = 90^\circ$
$\angle ONA + \angle OAN + \angle AON = 180^\circ$
$\therefore \angle ONA + \angle AON = 90^\circ$

$\angle AON = \angle ONB$ – as in reflecting across the diameter perpendicular to $ON$
$\angle ONA + \angle ONB = 90^\circ$ as required

c) Let $\angle ONA = \theta$
$\angle ONA' = \theta$
From part b, $\angle ONB = 90^\circ - \theta$
$\angle ONB' = 90^\circ - \theta$

$\angle NOA' = \angle NOB' = 90^\circ$
$\angle OA'N = 90^\circ - \theta$ and $\angle OB'N = \theta$
$\therefore \triangle ONA'$ and $\triangle ONB'$ are similar

$\frac {OA'}{ON} = \frac {ON}{OB'}$
$\pi(A) = \frac {1}{\pi(B)}$

$\pi(B) = \frac {1}{x}$

d) With $\angle ONA$ as $\theta$
$OA' = \pi(A) = \tan(\theta)$

$OB' = \pi(B) = \tan(\angle ONB')$
$\pi(B) = \tan(90^\circ - \theta)$

From part c, $\pi(B) = \frac {1}{\pi(A)}$
$\therefore \tan(90 - \theta) = \frac{1}{\tan \theta}$

Problem of the Week

Problem 31 Solution

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