Q: Complex numbers $z_1$, $z_2$, $z_3$ are represented by the points of contact $D$, $E$, $F$ of the in-circle of a triangle $ABC$, with centre $O$ of the incircle taken as the origin. If $BO$ meets $DE$ at $G$, find the complex number represented by $G$?


  • I $\qquad\displaystyle\frac{(z_1-z_2)z_3}{z_2-z_3}$

  • II $\qquad\displaystyle\frac{(z_1-z_2)z_3}{z_2+z_3}$

  • III $\qquad\displaystyle\frac{(\overline z_1- \overline z_2)z_3}{z_2+z_3}$

  • IV None Of These

My Try:

I know that incircle touches the sides, and hence the sides are tangents to the same, with touching points $z_1, z_2, z_3$

So, as $B$ is point of intersection of the $2$ tangents with point of intersection with circle at $z_1$ and $z_2$, $\displaystyle B = \frac{2z_1z_2}{z_1+z_2}$, and likewise for other vertices.

I am stuck after this.

  • $\begingroup$ The point $C$ is the intersection of the tangents in $D=z_1$ and $E=z_2$, please edit if this is so... I suppose the multiple choice problem was stated so that in the special case of some triangle(s) one can distinguish between the answers, for such questions often the proof is more complicated than the guess... So we want the proof?! $\endgroup$
    – dan_fulea
    Apr 17 at 9:51
  • $\begingroup$ The question doesn't specify which of D, E, F lies on which side, but in general, D is on AB, E on BC, and F on CA. $\endgroup$ Apr 17 at 10:04
  • $\begingroup$ The formula $D=2z_1z_2/(z_1+z_2)$ cannot hold, we have $D=z_1$, if i correctly understood the statement. It should be $C=2z_1z_2/(z_1+z_2)$, if $D$ is opposed to $A$ and $E$ to $B$ in the triangle. $\endgroup$
    – dan_fulea
    Apr 17 at 10:22
  • $\begingroup$ The condition that three points $z, v,w\in\Bbb C$ are on a line is equivalent to the vanishing of a determinant:$$0=\begin{vmatrix}1 & z & \bar z\\1 & v &\bar v\\ 1 & w &\bar w\end{vmatrix}\ .$$One can compute or simply check if the given points are on both $BO$ and $DE$. In our case, $\bar z_k=1/z_k$, so if we want we can immediately obtain algebraic expressions involving only $z_1,z_2,z_3$ (and avoiding their conjugates). Here, using a computer, the problem is then a simple check when using a computer algebra system... $\endgroup$
    – dan_fulea
    Apr 17 at 10:27
  • $\begingroup$ Oh yes, my mistake, i meant point B, not D. Thanks for pointing out, edited $\endgroup$ Apr 17 at 10:32

1 Answer 1


Let us do all the computations. We denote by $r$ the common value of $|z_1|=|z_2|=|z_3|$.

mse 4900557 complex numbers in geometry

  • The equation of the tangent in $D=z_1$ to the incircle $z\bar z=r^2=z_1\bar z_1$, also written as $(z-z_1)\overline{(z-z_1)}+(z-z_1)\bar z_1+z_1(\bar z-\bar z_1)=0$, is obtained by "neglecting" the part of degree higher than linear in the local variables $(z-z_1)$ and $\overline{(z-z_1)}$, so it is $(z-z_1)\bar z_1+z_1(\bar z-\bar z_1)=0$, i.e. $z\bar z_1+\bar z z_1= 2r^2=2z_1\bar z_1$.
  • The intersection of the tangents in $D=z_1$ and $F=z_3$ is the point $B$, which is the solution of the linear system: $$ \left\{ \begin{aligned} z\bar z_1+\bar z z_1 &= 2z_1\bar z_1\ ,\\ z\bar z_3+\bar z z_3 &= 2z_3\bar z_3\ ,\\ \end{aligned} \right. $$ and Kramer's rule gives the solution $z$ (of the above linear system in $z$, $\bar z$) as $$ z= \frac {\begin{vmatrix} 2z_1\bar z_1 & z_1 \\ 2z_3\bar z_3 & z_3\end{vmatrix}} {\begin{vmatrix} \bar z_1 & z_1 \\ \bar z_3 & z_3\end{vmatrix}} = 2z_1z_3\frac{\bar z_1-\bar z_3}{\bar z_1 z_3-\bar z_3 z_1} = 2z_1z_3\frac{\frac{r^2}{ z_1 }-\frac{r^2}{ z_3 }}{r^2\left(\frac{ z_3}{ z_1 }-\frac{ z_1}{ z_3 }\right)} =\frac{2z_1z_3}{z_1+z_3}\ . $$
  • We use the same linear algebra to obtain the point $G$. It is the intersection of the lines $BO$ and $DE$. So we need the equations of these lines in terms of $z,\bar z$.
  • The equation of the line through two points $v,w$ is $$ \tag{*} 0= \begin{vmatrix} 1 & z & \bar z\\ 1 & v & \bar v\\ 1 & w & \bar w \end{vmatrix}\ . $$ We apply this formula below.
  • The equation of $BO$ is $$0= \begin{vmatrix} 1 & z & \bar z\\ 1 & 0 &\bar 0\\ 1 & \frac{2z_1z_3}{z_1+z_3} & \frac{2\bar z_1\bar z_3}{\bar z_1+\bar z_3} \end{vmatrix} \ . $$We then compute $$\frac{2\bar z_1\bar z_3}{\bar z_1+\bar z_3} = \frac{\frac{2r^4}{z_1 z_3}}{\frac{r^2}{ z_1}+\frac{r^2}{ z_3}} =\frac {2r^2}{z_1+z_3} \ . $$So the equation is simpler written as $$0= \begin{vmatrix} z & \bar z\\ z_1z_3 & r^2 \end{vmatrix} \ . $$Equivalently, $\bar z =\frac{r^2 z}{z_1z_3}$.
  • The equation of $DE$ is as in $(*)$, with $v,w$ replaced by $z_1,z_2$. We intersect with $BO$, so replace in it the value for $\bar z$, and obtain: $$ 0= \begin{vmatrix} 1 & z & \bar z\\ 1 & z_1 & \bar z_1\\ 1 & z_2 & \bar z_2 \end{vmatrix} = \begin{vmatrix} 1 & z & \frac{r^2 z}{z_1z_3}\\ 1 & z_1 & \frac{r^2}{z_1}\\ 1 & z_2 & \frac{r^2}{z_2} \end{vmatrix} = \begin{vmatrix} 1 & z & \frac{z}{z_1z_3}\\ 1 & z_1 & \frac 1{z_1}\\ 1 & z_2 & \frac 1{z_2} \end{vmatrix} \\ = 1\cdot \begin{vmatrix} z_1 & \frac 1{z_1}\\ z_2 & \frac 1{z_2} \end{vmatrix} -z\cdot \begin{vmatrix} 1 & \frac 1{z_1}\\ 1 & \frac 1{z_2} \end{vmatrix} +z\cdot \begin{vmatrix} 1 & z_1 \\ 1 & z_2 \end{vmatrix} \\ =\frac{z_1^2-z_2^2}{z_1z_2} -z\cdot\frac{z_1-z_2}{z_1z_2} -z\cdot\frac{z_1-z_2}{z_1z_3} \ . $$ We factor in the last expression $(z_1-z_2)$, and force the factor $1/(z_1z_2z_3)$. The resulted equation is: $$(z_1+z_2)z_3=z(z_3+z_2)\ .$$ So the value $z$ corresponds to $$ \bbox[lightyellow]{ \qquad z = \frac{(z_1+z_2)z_3}{z_2+z_3}\ .\qquad} $$

Let us check the formula in particular cases. For $z_1=-i$, $z_2= 1$, $z_3=\frac 15(-4+3i)$ we obtain

mse 4900557 geometry problem complex numbers special case

$$B=\frac{2z_1z_3}{z_1+z_3} =\frac{\frac 25(3+4i)}{-\frac 15(4+2i)} =-\frac{3+4i}{2+i} =-\frac 15(3+4i)(2-i)=-\frac 15(10+5i)=-(2+i)\ . $$ In full agreement with the picture.

From the picture, $BO$ and $DE$ intersect in the point $2+i$. So let us check our answer, and the answers from the multiple choice exercise. The point in I is $\frac 13(2+i)$. The point in II is $1-2i$. The point in III is $-2-i$. Our point is $$ G=\frac{(z_1+z_2)z_3}{z_2+z_3} =\frac{(1-i)\frac 15(-4+3i)}{\frac 15(1+3i)} =\frac 1{10}(-4+3i)(1-i)(1-3i) =-\frac 1{10}(-4+3i)(2+4i) =\frac 1{10}(20 +10i)= 2+i\ . $$

An other case that may quickly let us decide what happens is obtained by forcing $BA=BC$. We consider $D,F$ to be $-4\pm 3i$, and $E=5$.

mse question 4900557 an other special case

Then $B=-25/4$ is immediate, and the computed point $G$ is: $$ G=\frac{(z_1+z_2)z_3}{z_2+z_3} =\frac{(-4-3i+5)(-4+3i)}{-4+3i+5} =\frac{(1-3i)(-4+3i)}{1+3i} =\frac 1{10}(1-3i)^2(-4+3i) =\frac 1{10}(-8 -6i)(-4+3i) =\frac 15(4+3i)(4-3i) =5=E\ . $$ "Unfortunately" some of the offered answers is also leading to this result... They have something special against the students that still can compute something, and have a good idea to work.

It may be that the multiple choice exercise wanted to use some own special case to see which formula may work. Usually, one searches for reasons to dismiss the one or the other answer. What can we do to reject I, II, and/or III?

  • Let $u$ be a unit. Consider the rotation implemented by multiplication with $u$. Then $z_k$ is mapped to $uz_k$, and we expect that the result will also be multiplied by $u$. This rejects III.
  • What happens when $z_1$ is fixed and $z_2$ goes to $-z_1$? Then something happens with the points $A,C$, but we need only $B$, and have a free choice of $z_3$. The line $DE$ passes near $O$, and we expect that the result $DE\cap BO$ to also lie near $O$. So in the limit when $z_1+z_2\to 0$ we expect also a zero result for $G$. For reasons in algebraic geometry, we would even expect it to be a factor... This rejects I, II, III (if there is no algebra that somehow uses that the points are constrained to be on the circle... well, no such reason is transparent).
  • Case II has a denominator $z_2+z_3$. It leads to an infinity point if $z_3$ goes to the opposite of $z_2$. But drawing some figures shows that the point $G$ lives here, in the city, you can walk there quickly, no need to have expensive phone calls. The formula is rejected.

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