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Let $A$ and $B$ be two different points in the plane. Find the Locus of all the points $C$ so $\angle ACB=\frac{2\pi}{3}$.

What I tried to do:

enter image description here

Also the center of the circle is in $O(0,0)$. We can find the distance of $CO$: $$ CO=\sqrt{x_c^2+y_c^2} $$ Also we know that $CO=AO=BO$. If $\angle ACB=\frac{2\pi}{3}$ then due to being an inscribed angle we conclude that $\angle AOB=\frac{2\pi}{3}$.

Now I'm stuck. I understand that I need to get some equation that contains only $x_c$ and $y_c$ in order to get the locus. I though of using the Law of cosines: $$ AB^2=AO^2+BO^2-2AO\cdot BO\cdot\cos120=2CO^2+CO^2=3CO^2 $$ But how can I represent $AB$ with only $x_c$ and $y_c$ so I could use the distance theorem and solve it?

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  • $\begingroup$ You have already express $AB$ as a function of $x_c$ and $y_c$. I mean, $AB^2=3\left(x_c^2+y_c^2\right)$. There is no other way do that. Actually, you have already solved your problem. The solution is $x_c^2+y_c^2=\frac{k^2}{3}$, where $k$ is the distance between the two points $A$ and $B$. This is the equation of the red circle you have drawn. $\endgroup$ – YNK Jan 27 at 14:57
  • $\begingroup$ @YNK Hi, thanks for your reply. I have learnt that in order to find all the points that fulfil a condition the answer should be dependent only on the want point (which is $C$ and not $A$ and $B$). Why can I use $A$ and $B$ in the final answer? $\endgroup$ – vesii Jan 27 at 15:30
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You are right that you essentially want to figure out where the origin is relative to $A$ and $B$ (and its reflection across $\overline{AB}$). Here is the construction you are looking for.

enter image description here

Construct $C$ such that $\triangle ABC$ is equilateral. Bisect both $\angle BAC$ and $\angle ABC$, and let $D$ be their point of intersection. Note that $m\angle BAD=m\angle ABD=\frac\pi6$, so $m\angle ADB=\frac{2\pi}3$. Draw the minor arc of a circle with center $D$ connecting $A$ and $B$. Extend $\overline{CD}$ to intersect this arc at $E$, and draw another minor arc with center $E$ connecting $A$ and $B$. Those two arcs (excluding $A$ and $B$ themselves) are the locus of all points $X$ such that $m\angle AXB=\frac{2\pi}3$.

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  • $\begingroup$ +1 for beating me to it. Not +1 for me because I am spending too much time watching American Ninja Warrior! :-S $\endgroup$ – Oscar Lanzi Jan 27 at 2:34
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    $\begingroup$ @OscarLanzi compared to 10 rep, I think you might be making the better decision. ^_^ $\endgroup$ – Matthew Daly Jan 27 at 2:35
  • $\begingroup$ why can you assume that $ABC$ is equilateral? It does not have to be like this. $\endgroup$ – vesii Jan 27 at 11:28
  • $\begingroup$ @vesii Because that's where $C$ was constructed to be. There is certainly a point that is the same distance from $A$ and $B$ as $AB$. $\endgroup$ – Matthew Daly Jan 27 at 11:59
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Hint

I believe we need to fix constant values to $A(m,n);B(r,s)$

so that we can use cosine rule

Otherwise

If the gradient of $CA,BC$ are $m,n$ respectively

$$\tan\dfrac{2\pi}3=\pm\dfrac{m-n}{1+mn}$$

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  • $\begingroup$ If I use the cosine rule, I get an equation with $AB$. My problem is to find an expression to $AB$ with only $C$ coordinates. $\endgroup$ – vesii Jan 27 at 11:30
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Let the coordinates of the points $A(x_a, y_a)$ and $B(x_b, y_b)$. Then, the corresponding midpoint of $AB$ is $M(x_m,y_m)=M(\frac{x_a+x_b}2, \frac{y_a+y_b}2)$ and the directional vector perpendicular to the chord $AB$ is $(y_b-y_a,-x_b+x_a)$. The center of the locus circle lies on the line passing through the midpoint $M$ and perpendicular to $AB$, which can be parametrized as,

$$(x,y) = (x_m,y_m) +t(y_b-y_a,-x_b+x_a)$$

Knowing that the triangle $AOM$ is an 30-60-90 right triangle, we have

$$OM = \frac12 AB \>\cot 60=\frac1{2\sqrt3}AB$$

where

$$OM^2= t^2[(x_b-x_a)^2+(y_b-y_a)^2]$$ $$AB^2= (x_b-x_a)^2+(y_b-y_a)^2$$ Substitute OM and AB into above equation to get the center parameter $t=\pm\frac1{2\sqrt3}$. Thus, the center of the circle is

$$(x_0,y_0) = (\frac{x_a+x_b}2\pm\frac{y_b-y_a}{2\sqrt3},\> \frac{y_a+y_b}2\mp\frac{x_b-x_a}{2\sqrt3})\tag 1$$

and its radius is $R = \frac12AB\csc60$, or

$$R^2=\frac13[(x_b-x_a)^2+(y_b-y_a)^2]\tag 2$$

As a result, the equation of the locus, expressed in terms of known coordinates $(x_a, y_a)$ and $(x_b, y_b)$, is

$$(x-x_0)^2+(y-y_0)^2 = R^2$$

where the center $(x_0,y_0) $ and the radius $R$ are given by (1) and (2), respectively. Note that there are two circles as shown by the two centers in (1).

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  • $\begingroup$ why does this solve the problem? you will have an equation with $x_a,x_b,y_a,y_b$ and not with $x_c,y_c$. $\endgroup$ – vesii Jan 27 at 11:27
  • $\begingroup$ @vesii - The problem asks for the locus equation of the point C that satisfies the requirement. Since A and B are the known points in terms of their coordinates in the plane, you express the result in terms of the known $(x_a,y_a)$ and $(x_b,y_b)$. $(x,y)$ in the locus circle equation in the answer represents all points of $C(x_c,y_c)$. $\endgroup$ – Quanto Jan 27 at 20:36
  • $\begingroup$ So my proof is actually valid? Also, what if $A$ and $B$ are not know points, Isn't it possible to solve the problem? $\endgroup$ – vesii Jan 27 at 20:38
  • $\begingroup$ @vesii - I do not see a final locus equation for all C in your proof. $\endgroup$ – Quanto Jan 27 at 20:40
  • $\begingroup$ I got to $AB^2=3CO^2$ so I get $(x_a-x_b)^2+(y_a+y_b)^2=3(x_c^2+y_c^2)=k$ $\endgroup$ – vesii Jan 27 at 20:45

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