1
$\begingroup$

I came up with this but I have not been able to solve it. I would really appreciate any help.

Let $ABC$ be a triangle and let $\omega$ be its circumcircle. Produce the internal angle bisector of $\angle BAC$ to meet $BC$ and $\omega$ in $D$ and $E$, respectively. Let the circle with diameter $DE$ intersect $\omega$ at a second point $F$. Drop a perpendicular from $F$ to "$AC$ produced" and let it meet "$AC$ produced" at $G$. Is it true that line $GF$ is tangent to $\omega$?

(The reason I ask "is it true" is because I don't know if it is.)

$\endgroup$
2
  • $\begingroup$ It seems to be false; could you please attach a figure? $\endgroup$
    – user133281
    Feb 1, 2015 at 1:38
  • $\begingroup$ As pointed out, this is not true. It can also be verified by a moderately accurate drawing using Geogebra. Then, it would be more interesting to ask “If a line from F is drawn to cut AC produced at G, what condition is needed on G such that GF is tangent to ω?” $\endgroup$
    – Mick
    Feb 1, 2015 at 5:41

1 Answer 1

Reset to default
0
$\begingroup$

It is not true, and a single counterexample is enough to show that it is not:

Counterexample

I was interested in classifying those triangles where your assumption actually holds. Without loss of generality, you can choose a coordinate system in such a way that the circumcircle $\omega$ is the unit circle, and the point $A$ has coordinates $(1,0)$. Then you can describe $B$ and $C$ using a rational representation as $B=(b^2-1,2b)/(b^2+1)$ and $C=(c^2-1,2c)/(c^2+1)$. Then the triangles for which your assumption holds must satisfy

$$b^2c + 2bc^2 + c^3 + 4b = 0$$

This was obtained through some advanced computation wich I won't fit here in this post since it's not exactly asked by the question. I haven't found a nice geometric interpretation for this relation, but here is one example triangle:

Example where the claim holds

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .