I did this with a random triangle with nice dimensions and I got 2/3. I don't know if that is even right, but I have no idea how to do this for any given triangle. Does anyone have a good solution for this. This is a really weird problem.

There are a lot of similar triangles in this configuration. I got an idea of reflect $ C$ in the midpoint $ M$ of segment $ AP$ to $ C'$ and tried to prove that triangles $ C'PC$ and $ RAQ$ are similar. BTW, I still failed to solve this problem

Ugly problem $ \frac {AB}{PB} = \frac {BQ}{BC} = \frac {\sin 60^\circ}{\sin 45^\circ} = \sqrt {\frac {3}{2}}$ and $ \angle PBC = \angle PBA + \angle ABC = \angle ABC + \angle CBQ = \angle ABQ$ $ \Longrightarrow$ $ \triangle ABQ \sim \triangle PBC$ by s.a.s. with similarity coefficient $ \sqrt {\frac {3}{2}}.$ April wrote: ... I got an idea of reflect $ C$ in the midpoint $ M$ of segment $ AP$ ... Correct. If $ D$ is reflection of $ C$ in $ M,$ then $ ADPC$ is parallelogram, because its diagonals cut in half at $ M$ $ \Longrightarrow$ $ AD = CP$ and $ \frac {AD}{AQ} = \frac {CP}{AQ} = \sqrt {\frac {3}{2}}.$ $ \angle QAD = \angle QAB + \angle BAP + \angle PAD = \angle CPB + \angle BAP + \angle APC =$ $ \angle BAP + \angle APB = 105^\circ.$ $ \angle CAD = \angle CAQ + \angle QAD = \angle CAQ + \angle RAC = \angle RAQ.$ $ \frac {QR}{DC} = \frac {QR}{2MC} = \frac {\sqrt {6}}{2} = \sqrt {\frac {3}{2}} = \frac {AQ}{AD}$ $ \Longrightarrow$ $ \triangle AQR \sim \triangle ADC$ by S.s.a. $ \Longrightarrow$ $ \frac {AR}{AC} = \sqrt {\frac {3}{2}}.$

This is pretty ridiculous to do synthetically without an actual diagram or looking at specific trivial cases. Once the answer is known, however, it is quite clear what's happening. MellowMelon wrote: Let $ ABP, BCQ, CAR$ be three non-overlapping triangles erected outside of acute triangle $ ABC$. Let $ M$ be the midpoint of segment $ AP$. Given that $ \angle PAB = \angle CQB = 45^\circ$, $ \angle ABP = \angle QBC = 75^\circ$, $ \angle RAC = 105^\circ$, and $ RQ^2 = 6CM^2$, compute $ AC^2/AR^2$. Zuming Feng. Answer. $\tfrac{2}{3}$. Construct parallelogram $CADP$. Claim. $\triangle AQR \sim \triangle ADC$. (Proof) Observe that $\triangle BPA \sim \triangle BCQ$ hence $\triangle BAQ \sim \triangle BPC$. Consequently, $$\tfrac{AQ}{AD}=\tfrac{AQ}{CP}=\tfrac{BP}{BA}=\sqrt{\tfrac{3}{2}}=\tfrac{QR}{DC}.$$Since $\angle RAC=105^{\circ}$ and $\angle QAD=\angle CPA+\angle QAP=180^{\circ}-\angle (CP, AQ)=180^{\circ}-\angle ABP=105^{\circ}$. Now we can use SSA similarity (since $105^{\circ}>90^{\circ}$) so $\triangle AQR \sim \triangle ADC$ as desired. $\blacksquare$ Clearly, $\tfrac{AC^2}{AR^2}=\tfrac{2}{3}$ as claimed.

anantmudgal09 wrote: This is pretty ridiculous to do synthetically without an actual diagram or looking at specific trivial cases. Once the answer is known, however, it is quite clear what's happening. Well, the angles are nice enough that you can (with some effort) construct basically everything aside from $R$ fairly accurately (for $R$ you would need to multiply something by $\sqrt6$, which seems like a bit more guesswork). While I agree this problem is fairly ridiculous, I think for me the parallelogram construction was somewhat motivated: 1. I think that given the conditions on $ABP$ and $CBQ$, it's fairly clear that spiral similarities are going to be a big player here, and thus that's what we should be focusing on. 2. I think given a sufficiently nice diagram it's not too difficult to note that the point $N$, which is where the spiral similarity at $B$ taking $P$ to $A$ sends $M$ off to, is going to be rather important - $MANB$ is cyclic, $\angle{NAP}$ is also equal to $105^\circ$, and with a decent enough diagram it looks like $N$ is on $RQ$. 3. Another calculation now reveals that $RQ = 2NQ$. Given this, and the previous collinearity description, it seems natural to guess that $R$ is the reflection of $Q$ across $N$, which turns out to be true! Note that all the above was not done through any particular special case, although I may have gotten very lucky. Still quite difficult for a TST 1 though, and very amusing.

Let $C'$ denote $C$ reflected over $M$ so $CPC'A$ is a parallelogram. The claim is that $CPBC'\sim QBAR$. To see $\triangle CPB\sim\triangle QBA$, apply the Law of Sines. Moreover, the ratio $QR/CC'=AB/BP$ by Law of Sines and the given, and \[\angle BPC'=60^\circ+\angle APC'=60^\circ+\angle PAC=105^\circ+\angle BAC=\angle BAR,\]so the claim follows. Then $AR=\frac{QR}{CC'}\cdot C'P=\frac{\sqrt{3}}{\sqrt{2}}AC$, which finishes: $AC^2/AR^2=2/3$.

felt very hard for a tst 4 The answer is $2/3$, or $$\frac{AC}{AR}=\frac{2}{\sqrt{6}}.$$Let $C'$ be the reflection of $C$ across $M$. It suffices to show that $$\triangle ACC'\sim\triangle ARQ.$$Note that there is a spiral similarity at $B$ sening $PA$ to $CQ$, so $$\angle BPC=\angle BAQ.$$Thus, $$\angle C'AQ=45+\angle C'AP+\angle BAQ=45+\angle APC+\angle BPC=45+60=105.$$ It then suffices to show that $$\frac{AQ}{RQ}=\frac{AC'}{CC'}$$$$\frac{RQ}{CC'}=\frac{AQ}{AC'}$$$$\frac{AQ}{CP}=\frac{\sqrt{6}}{2}$$ However, this just follows from the spiral similarity at $B$ sending $PC$ to $AQ$, since the scale factor of the spiral similarity is $$\frac{BA}{BP}=\frac{\sin 60}{\sin 45}=\frac{\sqrt{6}}{2}.$$ remark: the motivation I had for reflecting C was quite complicated. First, I felt that it was very surprising that the quantity is invariant in $\triangle ABC$. Of course, one thing you can do in this scenario is to try special cases. The really nice special case is when $\angle A=180-105-45=30$ and $\angle B=180-75-75=30$, since it makes a lot of things collinear, and even has an added bonus of $\triangle ABC$ itself being nice. This case is actually not hard to solve explicitly, and once you notice that $$\frac{AR}{AC}=\frac{\sqrt{6}}{2}$$you realize that the problem has already given a $\sqrt{6}$ ratio, so in a sense $CM$ "needs to be twice as long" for a nice ratio to exist. The reflection is then easily motivated.

Construct $D$ as the reflection of $C$ over $M$. First notice the spiral similarity sending $\triangle ABP \mapsto \triangle QBC$ implies \[\triangle ABQ \sim \triangle PBC \implies \frac{AQ}{CP} = \frac{AB}{PB} = \frac{\sin 60}{\sin 45} = \frac{\sqrt 6}{2}.\] Using the obtuse angle equality \begin{align*} \angle RAQ &= \angle CAQ + 45 + 60 \\ &= \angle CAQ + \angle PAB + \angle APB \\ &= \angle CAQ + \angle PAB + \angle DAP + \angle BAQ \\ &= \angle DAC \end{align*} along with the length ratios \[\frac{AQ}{AD} = \frac{AQ}{CP} = \frac{\sqrt 6}{2} = \frac{RQ}{2 CM} = \frac{RQ}{CD}\] gives the similarity $\triangle ADC \sim \triangle AQR$, so our desired ratio is $\left(\frac{2}{\sqrt 6}\right)^2 = \boxed{\frac 23}$. $\blacksquare$

Let $C'$ be the reflection of $C$ over $M$ and redefine $R$ such that $C'PBC\sim RABQ.$ Then \[\frac{RQ^2}{CM^2}=4\frac{RQ^2}{CC'^2}=4\frac{BQ^2}{BC^2}=6,\]and \[\angle RAC=180^\circ-\angle(RA,C'P)=180^\circ-\angle PBA=105^\circ,\]so $R$ is the same point as given. Now \[\frac{AC}{AR}=\frac{PC'}{AR}=\frac{BC}{BQ}=\sqrt{\frac23}.\]

Let C' be the reflection of C over M. Since $\triangle ABP \sim \triangle QBC$ and we have the spiral similarity sending $\triangle ABP \mapsto \triangle QBC$ we get that $\triangle ABQ \sim \triangle PBC$ $\Rightarrow$ $\frac{AQ}{CP} = \frac{AB}{PB} = \frac{\sin 60}{\sin 45} = \frac{\sqrt 6}{2}$. Now we have that $\angle RAQ = \angle CAQ + 45 + 60 = \angle CAQ + \angle PAB + \angle APB = \angle CAQ + \angle PAB + \angle C'AP + \angle BAQ = \angle C'AC$. Also $\frac{AQ}{AC'} = \frac{AQ}{CP} = \frac{\sqrt 6}{2} = \frac{RQ}{2 CM} = \frac{RQ}{CC'}$ $\Rightarrow$ $\triangle ADC \sim \triangle AQR$, from which we get that $\frac{AC}{AR} = \frac{AC'}{AQ} = \frac{2}{\sqrt 6} = \frac{\sqrt 6}{3}$ $\Rightarrow$ $\frac{AC}{AR} = \frac{\sqrt 6}{3}$.