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Find the point P on the x axis such that the angle APB in below figure is maximal. enter image description here

I was trying to use $$ \tan (\phi-\theta)=\frac{\tan\phi-\tan\theta}{1+\tan\phi\tan\theta}$$ but the end result was a crazy equation that did not seem right. i really appreciate an explanation as well.

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  • $\begingroup$ sorry it was a typo which I fixed up there. $\endgroup$
    – jax
    Dec 4 '13 at 23:10
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Your approach is sound. If $P$ is the point $(-x,0)$, then $\tan\phi = \frac2x$ and $\tan\theta=\frac1{x+1}$. So $\tan(\phi-\theta) = \frac{x+2}{x^2+x+2}$.

Differentiating this with respect to $x$ gives $\frac{-x(x+4)}{(x^2+x+2)^2}$. Thus there are two local extrema, at $P=(0,0)$ and $P=(4,0)$. (Yes, this surprised me too!)

The first of these has $\tan(\phi-\theta) = 1$, and the second has $\tan(\phi-\theta) = -\frac17$. So angle $APB$ is greatest (in absolute value) at $P=(0,0)$.

By the way, these two solutions correspond to the two circles through $A$ and $B$ that are tangent to the $x$-axis at $P$. This has a nice geometrical explanation, which you might like to think about.

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