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Given two points on a circumference of radius $R$, $P_0$ and $P_1$ subtended by an angle $\theta$ at the center of the circumference, what is the angle at which a generic point $P_m$ inside the circle 'sees' the two points $P_0$ and $P_1$?

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It depends. If $P_m$ is on the circle, the angle is $\frac12\theta$ on one arc and (with correct orientation) $\pi+\frac12\theta$ on the other arc. For an interior point, any value inbetween is possible. –  Hagen von Eitzen Sep 12 '12 at 9:32
    
@Hagen von Eitzen: $P_m$ can be in any point inside the radius $R$ that means: $P_m=P_m(x,y)$ with $x^2+y^2 \leq R$ –  Riccardo.Alestra Sep 12 '12 at 9:39
    
@Riccardo.Alestra , I think Hagen did understand that and thus his answer: for a point $\,P_m\,$ on the circle, the angle is either $\,\theta/2\,\,or\,\,\theta/2+\pi\,$ , depending on the relative position of the three points. For a point inside the disk enclosed by the circle, any value in between occurs. –  DonAntonio Sep 12 '12 at 11:58
    
@DonAntonio: the problem is: given a point $P_m(x,y)$ and two points on the circunference $P_0$ and $P_1$, what is the formula linking the angle $\theta$ to the angle 'seen' by $P_m(x,y)$? $\theta \leq \frac{\pi}{2}$ –  Riccardo.Alestra Sep 12 '12 at 12:45
    
@Riccardo.Alestra . I think we all understood that from the beginning, and $\,P_m\,$ is inside the circle, as you wrote. –  DonAntonio Sep 12 '12 at 17:09

1 Answer 1

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With $P_0, P_1$ as given and $O$ as center of the circle, extend the line $P_0P_m$ until it meets the circle in $Q$. Let us assume that $P_m$ is on the same side of $P_0P_1$ as $O$. Then by the inscribed angle theorem we have $\theta = \angle P_0OP_1 = 2\angle P_0QP_1$. In triangle $P_1QP_m$, we have $\angle QP_1P_m + \angle P_mQP_1+\angle P_1P_mQ=2\pi$ and we also have $\angle P_0P_mP_1+\angle P_1P_mQ=\angle P_0P_mQ=2\pi$, hence $\angle P_0P_mP_1=\angle QP_1P_m + \angle P_mQP_1> \angle P_mQP_1= \angle P_0QP_1 = \frac12 \theta$.

There is a slight subtlety involved when $P_m$ is on the other side of $P_0P_1$ and also beware of orientation. All in all you will find that the range $\frac12\theta<\angle P_0P_mP_1<\frac12\theta+\pi$ is possible.


Here's a proof variant with coordinate calculations: Without loss of generality, $R=1$, the center is $(0,0)$, $P_0=(c, -s)$, $P_1=(c,s)$ with $0\le c=\cos\frac\theta2<1$, $0<s=\sin\frac\theta2\le 1$ and $P_m=(x,y)$ with $r^2:=x^2+y^2<1$. We can find the sine of $\alpha:=\angle P_0P_mP_1$ by calculating the $z$ coordinate of the crossproduct $\overrightarrow{P_mP_0}\times \overrightarrow{P_mP_1}$ and dividing by the lengths of the factors. The $z$ coordinate of the cross product is $(c-x)(s-y)-(-s-y)(c-x)=2s(c-x) $. Hence $$\sin\alpha = \frac{2s(c-x)}{|P_mP_0||P_mP_1|}.$$ We can also find the cosine of $\alpha$ via the scalar product: $$\cos\alpha=\frac{\overrightarrow{P_mP_0}\cdot \overrightarrow{P_mP_1}}{|P_mP_0||P_mP_1|}= \frac{(c-x)^2+y^2-s^2}{|P_mP_0||P_mP_1|}$$ The lngths in the denominators are a bit clumsy, but we can ignore them if we only need to test for the sign of $\sin\alpha$ and $\cos\alpha$ and compute the tangent of $\alpha$: $$\tan\alpha = \frac{\sin\alpha}{\cos\alpha}=\frac{2s(c-x)}{(c-x)^2+y^2-s^2}.$$ First note that $\sin\theta=2cs$ and $\cos\theta=c^2-s^2$ by the double angle formulas, hence $\tan\theta=\frac{2cs}{c^2-s^2}$. I claim that one of the follwoing cases holds:

  • $\cos\alpha>0, \sin\alpha>0, \tan\alpha>\tan\theta$
  • $\cos\alpha\le 0, \sin\alpha\ge 0$
  • $\cos\alpha<0, \sin\alpha<0, \tan\alpha<\tan\theta$

For each of these cases you can manipulate the expressions given for the trigonometric functions to verify the inequality in the given situation. It is a bit lengthy, and I'm a bit tired, though

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I suppose you have written $P_1$ and $P_2$ rather than $P_0$ and $P_1$ –  Riccardo.Alestra Sep 12 '12 at 14:38
    
In your proof there is something I don't understand. Suppose $P_m$ very close to the circumference, that means $P_{mx}^2+P_{my}^2=R-\epsilon$ with $\epsilon$ very little. Suppose $\theta$ less than $\frac{\pi}{2}$. Following your reasoning we can have the angle $\angle P_0 Q P_1$ shoud be less than $\frac{\pi}{4}$ when it's close to $\pi$ –  Riccardo.Alestra Sep 12 '12 at 14:48
    
@Riccardo.Alestra: $\angle P_0QP_1$ is always either $\frac12\theta$ or $\pi+\frac12\theta$, depending on the arc $Q$ is on. And if $P_m$ is very close to the circumference than $\angle P_0P_mP_1$ is close to one of these two values, but definitely between. –  Hagen von Eitzen Sep 12 '12 at 16:26
    
@von Heitzen: I don't see any dependence of the angle $\angle P_0 P_m P_1$ from the coordinates $x,y$ of the point $P_m$ –  Riccardo.Alestra Sep 13 '12 at 6:14
    
I argued in classic geometry. I'll add an answer purely using cartesian coordinates. –  Hagen von Eitzen Sep 13 '12 at 10:56

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