Let $\frac{AB}{OB}=\frac{AC}{PO}=2$ and also $AB=AC$ then prove that $\frac{OP}{MO}=\frac{1+\sqrt{5}}{2}$

Question

Let $\dfrac{AB}{OB}=\dfrac{AC}{PO}=2$ and also $AB=AC$ then prove that $\dfrac{OP}{MO}=\dfrac{1+\sqrt{5}}{2}$

My Try :

I know that by Thales's theorem we have $\dfrac{AB}{OB}=\dfrac{AC}{PO}=\dfrac{OP}{BC}=2$ and by $AB=AC$ we have $AB=AC=BC=2$ and $BO=PC=AP=AO =OP=1$ Now what ?

Answer 1

Let $AB = s$, $PN = a$, $OP = b$.

Use the power of the point $P$ with respect to the circle:

$$ (a+b)\cdot a = ({s\over 2})^2$$

Note that since $APO$ is equlateral we have $s=2b$. Now write $q= b/a$ and we get$$ q^2-q-1=0\implies q={1\pm \sqrt{5}\over2}$$

Answer 2

$\frac{AB}{OB}\geq1$ but you say it is $\frac{1}{2}$. You applied Thales' theorem incorrectly. $\dfrac{AB}{OB}=\dfrac{AC}{PO}$ does not guarantee that $OP\parallel BC$. I think it will be $\dfrac{AB}{OB}=\dfrac{AC}{PC}$.

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