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There is something misunderstanding with that question that I think it to have inadequte context or information (obviously for my little knowledge). So I couldn't solve the problem.

SOURCE: BANGLADESH MATH OLYMPIAD

$PE$ is a tangent of the below diagram. The diameter of small circle is equal to the radius of the large circle and $BC$ is the diameter of the small circle. If $PA$ = $12$, $\frac{AB}{CD}$ = $\frac{1}{2}$, then what is the value of $PC$ ?

I denoted the center of small circle $H$ and frew three altitude lines $AG$, $HE$ and $DI$ from the vertices $A$ , $H$ and $DI$ and also denoted $AB$ = $x$ and $BH$ = $y$.

So, $HE$ = $y$. $\triangle AGP$ $\sim$ $\triangle HEP$ $\sim$ $\triangle DIP$. So, I showed the relation of $AG$ and $DI$ with $HI$ with the help of their proportion of length because of their similarity. But I was unable to show relation with the rest property and I couldn't find the value of $x$ and $y$.

I think I went into messed situation for solving the above problem. I don't think so even my used method is going to the right direction.

It will be very helpful for me if someone please tells me in which way should I go or where is my mistake to find the right way. Thanks in advance.

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Let $AB = x$ and $CD = 2x$

You can use the power of the point with respect to both circle and point $P$:

$$ PA\cdot PD (=PE^2)= PB\cdot PC$$ $$\implies 12(12+x+2r+2x)= (12+x)(12+x+2r)$$

so $$x+2r = 12\implies PC = 12+x+2r = 24$$


Note: Information about relation betwen both radius of the circles is irrelevant.

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  • $\begingroup$ Why is it true your condition? I don’t understand $\endgroup$ – Federico Fallucca Feb 14 at 16:52
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    $\begingroup$ Do you know the power of the point? @FedericoFallucca $\endgroup$ – Aqua Feb 14 at 16:54
  • $\begingroup$ no sorry, I don’t know this concept $\endgroup$ – Federico Fallucca Feb 14 at 16:54
  • $\begingroup$ Google it or wiki $\endgroup$ – Aqua Feb 14 at 16:55
  • $\begingroup$ @greedoid You are absolutely right. The relation is totally irrelevant to the problem because I made the relation according to my own view and there was no description about the relation of radius and diameter of the small and the larger circle. I will try to notice the fact and I gave the information to provide some context. Pardon me for my fault. $\endgroup$ – Anirban Niloy Feb 14 at 17:19

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