Jutaro wrote: Let $\Gamma$ and $\Gamma_1$ be two circles internally tangent at $A$, with centers $O$ and $O_1$ and radii $r$ and $r_1$, respectively ($r>r_1$). $B$ is a point diametrically opposed to $A$ in $\Gamma$, and $C$ is a point on $\Gamma$ such that $BC$ is tangent to $\Gamma_1$ at $P$. Given that $O_1A'$ is parallel to $AP$, find the ratio $r/r_1$. what is $A'$?

$ A' $ is midpoint of BC.

I'm really sorry. Thanks for the correction

Where is the mistake? It is clear that $B-O-O_1-A$.Let $BA'=a $ and $A'P=x$. Also, let the intersection of $BA$ with $ \Gamma_{1} $ be $R$. It is clear that $OR=r-2r_1$. From $BP^{2}= BR*BA$ we get $BPR \sim BAP$. Also, since $ O_{1}A' $ is parallel to $AP$ we get $B O_{1}A' \sim BAP \sim BPR$. Since $BPR \sim BO_{1}A' $ we have $\frac{a+x}{r}=\frac{2r-2r_1}{a}$ $=> (a+x)^{2}= (\frac{(2r-2r_1)r}{a})^{2}$ (1) Since $BP^{2}= BR*BA => (a+x)^{2}= (2r-2r_1)2r$ $=> (\frac{(2r-2r_1)r}{a})^{2}= (2r-2r_1)2r => {a}^{2} = (r-r_1)*r$ From $B O_{1}A' \sim BAP$ we get $\frac{2r-r_1}{2r}= \frac{a}{a+x}$ which reduces to $\frac{r_1}{2r}=\frac{x}{a+x}$ $=> x= \frac{ar_1}{2r-r_1}$ Finally in 1 we had $ a+x= \frac{(2r-2r_1)r}{a}$ $=> {a}^{2}+ax= 2r^{2}-2rr_1$ $=> r^{2}-rr_1+ \frac{(r-r_1)*r_1*r}{2r-r_1}= 2r^{2}-2rr_1$ $=> \frac{(r-r_1)*r_1*r}{2r-r_1}= r^{2}-rr_1$ $=> \frac{(r-r_1)*(r_1)}{2r-r_1}= r-r_1$ => $r*r_1-r_1^{2}= 2r^{2}-3rr_1+r_1^{2}$ $=> 2r^{2}-4rr_1+2r_1^{2} => (r-r_1)^{2}= 0 $ So $r=r_1$? That cant happen..

Let $AP$ cuts $\Gamma$ again at $D$. Then $D$ halves arc $BC$, so $D,A',O$ are collinear. Since $A'OO_1 \sim DOA$ we have $x=A'O= OA = r-r_1$. Now $BA'O \sim BPO_1$ so \[ {x\over r_1} = {r\over 2r-r_1}\] Write $k= {r\over r_1}$ and we get $2k^2-4k+1=0$. Since $k>1$ we get $k=1+{\sqrt{2}\over 2}$ is solution.

A simple solution. $OA'\perp BC\perp O_1P$ so $OA'\parallel O_1P$ and we have $\frac{r}{r-r_1}=\frac{BA'}{A'P}=\frac{BO_1}{O_1A}=\frac{2r-r_1}{r_1}$ and we can slowly solve it. ($\frac{r}{r_1}=1+\frac{1}{\sqrt2}$)

let AO1=x and OO1=y.. its very easy that AC||PO1||OA1 so AO1/O1b=y/x+2y = PA1/A1B so let PA1=k*x and A1B=k*(x+2y) and AO1/OO1 = CP/PA1 so PC=k*x*x/y.. but CA1=A1B . from here (k*x*x/y)+k*x=k*(x+2y) |=> x*x=2*y*y and we know that x=r1 and x+y=r... I don't know how write radical (:D:D) but answer is very simply