here is ugly solution if sb wants : 1) take CD=a BD= b and AO-CD=p 2)by counting angles we easily find that => triangles DOC and DAC are similar so then the next equality holds $\frac{CD}{AD}=\frac{CO}{AC}$, hence $a(a+b)^2=$ $\cancel{(2a+b)(p+b)^2}$ $(2a+b)(p+a)^2$ we have two cases (a,p)=1 or (a,p)=p 1)then since a(a+b)^2 is divisible by (a+p)^2 so (a+b) is divisible by (p+a) but by triangle inequality in triangle AOC we have $2(p+a)\geq (a+b)$ => a+b=a+p => b=p but by putting this solution to the initial equation we find contradiction .So this case is impossible 2) a=pc for some c positive integer , our initial equation transforms to $c(pc+b)^2=$ $\cancel{p(c+1)^2}$ $p(c+1)^2(2pc+b)$ ohh sorry I found a mistake in the rest of solution (exactly:when c is divisible by p ) but when p doesn't divide c I solved sorry (( can sb post solution?

Actually, you've done quite well up to now to eliminate the impossible cases (except some typos that I edited in your post above). Indeed your case $p\nmid c$ does not lead to a solution either. It remains your case $p\mid c$, which will force $c=pd$ and further $b=de$, and the equation writes $(1+pd)^2(2p^2+e) = d^2(p^2+e)^2$. Notice that $\gcd(1+pd,d^2) = 1$. Now, presumably the case $\gcd(2p^2+e, p^2+e)>1$ could be fully analyzed (it forces $p\mid e$ to start with), but the case $\gcd(2p^2+e, p^2+e)=1$ immediately leads to $p^2+e = 1+pd$ and $2p^2+e = d^2$, which quickly leads to the unique solution $p=3$, $d=5$, $e=7$, thus $c=15$, $b=35$ and $a=45$, which is a solution (it follows $r=OA = 48$). Since the problem just asked to determine such three numbers (not all), an answer is $\boxed{35,45,3}$.

mavropnevma wrote: but the case $\gcd(2p^2+e, p^2+e)=1$ immediately leads to $p^2+e = 1+pd$ and $2p^2+e = d^2$, which quickly leads to the unique solution $p=3$, $d=5$, $e=7$ Can you explain this in detail ?