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Problem 1841

National Olympiad, first round

Geometry Difficulty 6.3 Prove it Fourth STARS OF MATHEMATICS COMPETITION · Romania · 2010

Let $ABCD$ be a square, and the points $M \in [BC]$ , $N \in [CD]$ , $P \in [DA]$ , such that
$\angle(\overrightarrow{AB}, \overrightarrow{AM}) = x, \quad \angle(\overrightarrow{BC}, \overrightarrow{MN}) = 2x, \quad \angle(\overrightarrow{CD}, \overrightarrow{NP}) = 3x.$

i) Show that, for any $x \in [0, \pi/8]$ , such a configuration uniquely exists, and $P$ ranges over the entire segment $[DA]$ ;
ii) Determine the number of angles $x \in [0, \pi/8]$ for which $\angle(\overrightarrow{DA}, \overrightarrow{PB}) = 4x$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

i) Assume $AB = 1$ , and denote $t = \tan x$ . Since
$1 = \tan \frac{\pi}{4} = \frac{2 \tan \frac{\pi}{8}}{1 - \tan^2 \frac{\pi}{8}}$
it follows $\tan \frac{\pi}{8}$ is the positive root of $t^2 + 2t - 1 = 0$ , and so $\tan \frac{\pi}{8} = \sqrt{2} - 1$ .

Now $BM = t$ , hence $M$ takes once and only once all values of interval $[0, \sqrt{2}-1]$ , while $CN = (1-t) \tan 2x = \frac{2t}{1+t}$ , therefore $N$ takes all values of the interval $[0, 2 - \sqrt{2}]$ . Lastly, we also get $DP = \left(1 - \frac{2t}{1+t}\right) \tan 3x = \frac{1-t}{1+t} \cdot \frac{t(3-t^2)}{1-3t^2} \ge 0,$
whence $PA = 1 - \frac{1-t}{1+t} \cdot \frac{t(3-t^2)}{1-3t^2} = \frac{(t^2+2t-1)(t^2+1)}{(t+1)(3t^2-1)} \ge 0$ , hence $P$ takes all values of the full interval $[0, 1]$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.