The incircle of $\triangle ABC$ is tangent to sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Point $X$ is chosen inside $\triangle ABC$ so that the incircle of $\triangle XBC$ is tangent to $BC$ at $D$, to $CX$ at $Y$, and to $XB$ at $Z$.

Prove that $EFZY$ is a cyclic quadrilateral by inversion.

I drew the mappings of all points and lines in the original diagram under inversion, but now I am stuck. Could someone provide a hint (not too many spoilers please..)?

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  • 4
    Where is Jack? He can solve every geometry problem :) – rae306 Apr 29 '16 at 13:15
up vote 6 down vote accepted

Circular inversion is not really needed, Ceva's theorem is enough. Let $O_X$ be the centre of the incircle of $XBC$. Clearly $O_X$ lies on the perpendicular to $BC$ through $D$, and $CD=CY=CE$ as well as $BD=BZ=BF$. $EFZY$ is a cyclic quadrilateral iff the perpendicular bisectors of $FZ,YE$ and $EF$ concur (the last line is just the angle bisector of $\widehat{BAC}$, since $AF=AE$).

These three lines cut the sides of $ABC$ at three points that are easy to describe in terms of $a,b,c$ (the side lengths of $ABC$) and $x=O_X D$. A little algebra easily proves the previous concurrency. The trigonometric version of Ceva's theorem gives an even faster proof.

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Here it is a trigonometric proof. Let $q=O_X D$. Since $BD=\frac{a-b+c}{2}$ and $DC=\frac{a+b-c}{2}$ we have that the sine of the angle between the $AC$ side and the perpendicular bisector of $YE$ is $\sin\left(\frac{\widehat{C}}{2}-\arctan\frac{2q}{a+b-c}\right)$, while the angle between the $BC$-side and the perpendicular bisector of $YE$ is $\sin\left(\frac{\widehat{C}}{2}+\arctan\frac{2q}{a+b-c}\right)$. Since $\frac{\sin(\theta+x)}{\sin(\theta-x)}=\frac{1+\tan(x)\cot(\theta)}{1-\tan(x)\cot(\theta)}$,

$$ \frac{\sin\left(\frac{\widehat{C}}{2}+\arctan\frac{2q}{a+b-c}\right)}{\sin\left(\frac{\widehat{C}}{2}-\arctan\frac{2q}{a+b-c}\right)}=\frac{(a+b-c)+2q\frac{a+b-c}{2r}}{(a+b-c)-2q\frac{a+b-c}{2r}}=\frac{r+q}{r-q}$$ and the concurrency easily follows.

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Now the alternative way: let we consider a circle $\Gamma$ through the incenter $I$ of $ABC$, having its centre at $D$. Let $U'$ be the inverse of $U$ with respect to $\Gamma$. This circular inversion maps the incircles of $ABC$ and $BCX$ into two parallel lines, so $E'F'Z'Y'$ is a trapezoid. On the other hand, the inverse of the line through $Z'$ and $F'$ is a circle through $Z,F,D$, whose centre lies at $B$, since $BD=BZ=BF$. The last circle and the incircle of $ABC$ are orthogonal at $F$, hence $Z'F'\perp F'E'$. In a similar way, $Y'E'\perp E'F'$, hence $E'F'Z'Y'$ is a rectangle.

  • Jack! Great answer! I am currently working through a book on inversion problems, so I was specifically looking for the solution of the inverted problem. Could you please add that one to your answer too? – rae306 Apr 29 '16 at 15:50
  • I can't figure out how to describe the sides in terms of $a$, $b$, $c$ and $x$, so I tried Ceva's trig form, but how can I prove that $\angle ACK=\angle LBA$ (i.e. that $KLCB$ is cyclic)? Could you collaborate a bit since it's my first time to use Ceva – rae306 Apr 29 '16 at 17:04
  • @rae306: I'll rework this solution through Ceva tomorrow, and add the solution through circular inversion. I am sorry but I am quite in a hurry now, please notify me if I forget to do it. – Jack D'Aurizio Apr 29 '16 at 18:49
  • @rae306: re-work complete ;) – Jack D'Aurizio Apr 30 '16 at 14:41
  • @rae306: many thanks to you! – Jack D'Aurizio Apr 30 '16 at 15:46

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