# Finding complex number for intersection of a line and a circle?

How to find complex number for intersection of a line and a circle?

In particular, if $$M=$$ midpoint $$BC$$, then I want $$AM ∩ (ABC)$$ in complex numbers.

Let me describe why I needed this,

I was doing a problem which I reduced to proving $$4$$ points $$(B,T,M',J)$$ are cyclic where $$ABC$$ is an acute scalene triangle, $$T$$ is the intersection of tangents to $$(ABC)$$ from $$B$$ and $$C$$, $$M' = AM ∩ (ABC)$$ ;($$M =$$ midpoint $$BC$$) , and $$J = AB ∩ CM'$$.

I tried to prove this synthetically but I could not, so I decided to bash. First I tried coordinate but that was too long, so I decided to go with complex because that felt feasible. But then I remembered that I have not yet studied complex numbers (except when I would read EGMO complex numbers for fun, like a storybook). So I opened EGMO and could get complex numbers for everything like how to prove cyclic, intersection, $$T$$ etc. but Point $$M'$$ was a problem. I couldn't find how to get it anywhere so I come to MSE.

Let $$A,B,C$$ lie on the unit circle, with complex coordinates $$a,b,c$$. Since you know EGMO, you should be able to find a proof of the fact that the line through the points $$a,b$$ on the unit circle has equation $$z + ab\overline{z} = a+b$$

Now denoting $$D = AM \cap (ABC)$$ with coordinate $$d$$, the line $$AD$$ is given by $$z + ad \overline{z} = a+d$$.

However you know that the midpoint of $$BC$$, namely $$\frac{b + c}{2}$$ lies on line $$AD$$, so

$$\frac{b + c}{2} + ad \frac{\overline{b} + \overline{c}}{2} = a+d$$

You can now just solve for $$d$$ to find the required intersection.

• so $d = \frac{b+c-2a}{2-2a(\overline b + \overline c)}$ Mar 6, 2021 at 17:27
• thanks so much man, appreciate your help Mar 6, 2021 at 17:28
• That's right! Also note that since $b,c$ are on the unit circle, $\overline{b} = \frac{1}{b}$ and $\overline{c} = \frac{1}{c}$, which gives an expression without conjugates (which in my experience makes complex bashing much easier...) Mar 6, 2021 at 17:38
• woah, thanks so much Mar 6, 2021 at 17:49
• You try to show that it's equal to it's conjugate. The idea behind this is that if $z = a+bi$ is a complex number, then $a+bi = \overline{a+bi} \iff a+bi = a-bi \iff b = 0 \iff z$ is real. Mar 6, 2021 at 18:07

$$AT$$ is the symmedian of $$\triangle ABC$$ through $$A$$. Hence, $$\angle BAT=\angle MAC=\angle M'AC$$ and also $$\angle CAT=\angle M'AB$$.

Now, notice that, $$\angle JAT=\angle BAT=\angle M'AC=\angle M'CT=\angle JCT$$ and thus quadrilateral $$AJTC$$ is cyclic.

Now, $$\angle M'JT=\angle CJT=\angle CAT=\angle BAM'=\angle M'BT$$ and therefore quadrilateral $$BJTM'$$ is cyclic.

P.S. The symmedian through a vertex is the reflection of the median over the internal angle bisector through that vertex. It is provable by euclidean geometry that the symmedian through a vertex goes through the intersection point of the tangents to the circumcircle at the other two vertices. Check out this link for more: https://brilliant.org/wiki/symmedian/#:~:text=A%20symmedian%20of%20a%20triangle,D%3D%E2%88%A0CAM.

• Thanks a lot for your answer, I will try this again after doing EGMO ch. 4 Mar 6, 2021 at 19:55
• @Aditya_math Welcome! Yes, it's a great tool for olympiad preparation. Mar 7, 2021 at 3:51
• I did do this again after completing EGMO chapter 4. Yes, after so long. But you know what, I will get this, I will win a gold at IMO this year. Oct 24, 2022 at 21:45