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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.

Please help me, Thanks!

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2 Answers 2

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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.

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  • $\begingroup$ so $d = \frac{b+c-2a}{2-2a(\overline b + \overline c)}$ $\endgroup$ Mar 6, 2021 at 17:27
  • $\begingroup$ thanks so much man, appreciate your help $\endgroup$ Mar 6, 2021 at 17:28
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    $\begingroup$ 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...) $\endgroup$ Mar 6, 2021 at 17:38
  • $\begingroup$ woah, thanks so much $\endgroup$ Mar 6, 2021 at 17:49
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    $\begingroup$ 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. $\endgroup$ Mar 6, 2021 at 18:07
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Triangle ABC

$AT$ 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.

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    $\begingroup$ Thanks a lot for your answer, I will try this again after doing EGMO ch. 4 $\endgroup$ Mar 6, 2021 at 19:55
  • $\begingroup$ @Aditya_math Welcome! Yes, it's a great tool for olympiad preparation. $\endgroup$
    – Limestone
    Mar 7, 2021 at 3:51
  • $\begingroup$ 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. $\endgroup$ Oct 24, 2022 at 21:45

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