# Finding the locus of all the points $C$ so $\angle ACB=\frac{2\pi}{3}$

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: 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?

• 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. – YNK Jan 27 at 14:57
• @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? – vesii Jan 27 at 15:30

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. 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 $$m\angle AXB=\frac{2\pi}3$$.

• +1 for beating me to it. Not +1 for me because I am spending too much time watching American Ninja Warrior! :-S – Oscar Lanzi Jan 27 at 2:34
• @OscarLanzi compared to 10 rep, I think you might be making the better decision. ^_^ – Matthew Daly Jan 27 at 2:35
• why can you assume that $ABC$ is equilateral? It does not have to be like this. – vesii Jan 27 at 11:28
• @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 $AB$. – Matthew Daly Jan 27 at 11:59

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

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

• 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. – vesii Jan 27 at 11:30

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).

• why does this solve the problem? you will have an equation with $x_a,x_b,y_a,y_b$ and not with $x_c,y_c$. – vesii Jan 27 at 11:27
• @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 $C(x_c,y_c)$. – Quanto Jan 27 at 20:36
• So my proof is actually valid? Also, what if $A$ and $B$ are not know points, Isn't it possible to solve the problem? – vesii Jan 27 at 20:38
• @vesii - I do not see a final locus equation for all C in your proof. – Quanto Jan 27 at 20:40
• I got to $AB^2=3CO^2$ so I get $(x_a-x_b)^2+(y_a+y_b)^2=3(x_c^2+y_c^2)=k$ – vesii Jan 27 at 20:45