Evaluate ∠APR if ∠APQ=3∠APR and ∠PAQ=45°

[anemone]

Let $A$ be the midpoint of the side $QR$ on a triangle $PQR$. Given that $\angle APQ=3\angle APR$ and $\angle PAQ=45^\circ$, evaluate $\angle APR$.
This is an unsolved problem I found @ AOPS.

 

[Opalg]

[sp]
[TIKZ]\coordinate [label=below left:{$Q$}] (Q) at (-5,0) ;
\coordinate [label=left:{$P$}] (P) at (-3.5,3.5) ;
\coordinate [label=below right:{$R$}] (R) at (5,0) ;
\coordinate [label=below:{$A$}] (A) at (0,0) ;
\coordinate [label=below:{$N$}] (N) at (-3.5,0) ;
\draw (P) -- node

{$x$} (N) ;
\node [fill=white] at (-3.4,2.85) {$3\theta$} ;
\draw [thick] (P) -- (Q) -- (R) -- cycle ;
\draw [thick] (A) -- (P) ;
\node at (-0.7,0.25) {$45^\circ$} ;
\node at (-2.5,2.85) {$\theta$} ;
\node at (-1.75,-0.25) {$x$} ;[/TIKZ]
Choose a unit of length so that $\overline{QA} = \overline{AR} = 1$. Let $PN$ be the perpendicular from $P$ to $QR$ and let $x = \overline{PN} = \overline{NA}.$ Write $\theta$ for $^\angle APR$, so that $^\angle APQ = 3\theta.$ Then $\tan(^\angle NPR) = \dfrac{1+x}x$, and $$\tan\theta = \tan(^\angle NPR - 45^\circ) = \frac{\frac{1+x}x - 1}{1 + \frac{1+x}x} = \frac1{2x+1}.$$ Similarly, $\tan(^\angle NPQ) = \dfrac{1-x}x$, and $$\tan(3\theta) = \tan(^\angle NPQ + 45^\circ) = \frac{\frac{1-x}x + 1}{1 - \frac{1-x}x} = \frac1{2x-1}.$$ But $\tan(3\theta) = \dfrac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}.$ So (*deep breath*) $$\frac1{2x-1} = \frac{\frac3{2x+1} - \frac1{(2x+1)^3}}{1 - \frac3{(2x+1)^2}}.$$ After multiplying out all the fractions (and in my case, doing so several times until I weeded out all the mistakes), this reduces to $16x^3 - 8x = 0$, for which the only positive solution is $x = \dfrac1{\sqrt2}.$
Thus \(\displaystyle \tan\theta = \frac 1{2x+1} = \frac1{\sqrt2 + 1} = \sqrt2 - 1.\) The answer to the problem is then $^\angle APR = 22.5^\circ$, as you can check by verifying that $\tan(2\theta) = 1$.

Notice that $4\theta = 90^\circ.$ Hence $^\angle QPR$ is a right angle. I assume that there must be an elegant geometric solution to this problem which somehow exploits that fact, but I have no ideas along those lines.
[/sp]​

 

[anemone]

Hi Opalg!

I can't tell you enough how much I appreciate your participation to the two unsolved challenges of mine...and you definitely are on a roll to manage in solving these two hardest challenges that bothered me for quite some time...

I just want to thank you and I also want to tell you that you are my math hero who I hold in the highest esteem! (Blush)(Nod)(Sun)

I thought there might be an elegant geometric way to solve the problem too, and I will definitely try it when I have the time because it is too interesting a problem that one can hardly resist not to attempt at it!(Tmi)

 

[Opalg]

I thought there might be an elegant geometric way to solve the problem too

[sp]As soon as you know that $x = 1/\sqrt2$, it follows (by looking at the triangle $PNA$) that $\overline{AP} = 1$. So $Q,P,R$ all lie on a circle centred at $A$. It then easily follows that $\theta = 22.5^\circ$, and also explains the right angle at $P$. But I don't see how to show that $\overline{AP} = 1$ without using the clunky trigonometric calculation.[/sp]