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Actually you should have $AO + XB > 2AO  XB$ where $AO$ is the length of the line segment $AO$ and so on. What this means is that you have an algebraic inequality and you can solve it. Subtract the number $AO$ from each side and add $XB$ to each side. But $AO = OB$ so you end up with \[2XB > OB.\] Which points $X$ in the interval $OB$ satisfy this inequality? Does the point $X$ in my diagram above? What about an $X$ close to $B?$ What fraction of the points $X$ between $O$ and $B$ satisfy the inequality? This is the fraction of the points between $O$ and $B$ for which $AX, XB$ and $AO$ can form a triangle. Penny  


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