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

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In the triangle PQR, given that angle APQ equals three times angle APR and angle PAQ is 45 degrees, angle APR is evaluated to be 22.5 degrees. The calculations involve using trigonometric identities and relationships between angles, ultimately leading to the conclusion that x equals 1/sqrt(2). This value indicates that the points Q, P, and R lie on a circle centered at point A, confirming that angle QPR is a right angle. The discussion suggests a potential for a more elegant geometric solution, although the trigonometric approach is currently the focus.
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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.
 
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[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 $$\tan\theta = \frac 1{2x+1} = \frac1{\sqrt2 + 1} = \sqrt2 - 1.$$ 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.
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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)
 
anemone said:
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]
 
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