Circumscribed hexagon problem

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#1 h Sep 27, 2018, 3:46 PM • 5 Y

Y by Pluto1708, buratinogigle, MNJ2357, Adventure10, Mango247

Consider any triangle $ABC$ . Let $A$ - excircle $\omega_A$ of $ABC$ is tangent to $BC$ at $A'$ . Similarly define $\omega_B$ , $\omega_C$ , $B'$ , $C'$ . Consider two tangent lines $\ell_a$ , $\ell_a'$ from $A'$ to $\omega_B$ , $\omega_C$ which are different from $BC$ . Similarly define $\ell_b$ , $\ell_b'$ , $\ell_c$ , $\ell_c'$ . Then $\ell_a$ , $\ell_a'$ , $\ell_b$ , $\ell_b'$ , $\ell_c$ , $\ell_c'$ form a hexagon which has an incircle.

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#3 h Sep 28, 2018, 7:14 PM • 3 Y

Y by buratinogigle, Adventure10, Mango247

I have a complete proof of this fact and will post it soon. Now I'll prove that there exists an inconic of a hexagon formed with $\ell_a$ ,..., $\ell_c'$ . To prove this first consider projective transformation sending $\omega_a$ to a circle and a pole of $AA'$ wrt $\omega_a$ to infinity. So from the fact that new $AA'$ is symmetry axes of $ABC$ and from the concurrency of $AA'$ , $BB'$ , $CC'$ we get that the tangents from $B'$ , $C'$ to $\omega_a$ are concurrent on $AA'$ . So the same fact is true for original $ABC$ before the projective transform. Now from the similar facts for $BB'$ , $CC'$ we get that from the reverse Brianchon's theorem there exists an inconic of a hexagon formed with $\ell_a$ ,..., $\ell_c'$ .

This post has been edited 1 time. Last edited by Mosquitall, Sep 28, 2018, 7:20 PM

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#4 h Oct 1, 2018, 8:33 AM • 3 Y

Y by buratinogigle, Adventure10, Mango247

$\textbf{Solution :}$

Lemma 1

Lemma 2

Let the second tangent from $A'$ to $\omega_b$ meet $AB$ at $A_b$ and also the second tangent from $A'$ to $\omega_c$ meet $AC$ at $A_c$ . Similarly define $B_a$ , $B_c$ , $C_a$ , $C_b$ . Then $AA_b$ , $A'A_b$ , $A'C$ , $AC$ are tangent to $\omega_b$ , so $|AA_b| + |A'A_b| = |AC| + |A'C| = \frac{|AB|+|BC|+|CA|}{2}$ . And similarly $|AA_c|+|A'A_c| = \frac{|AB|+|BC|+|CA|}{2}$ . So $|AA_b| + |A'A_b| =|AA_c|+|A'A_c|$ and so the exists a circle $\Omega_a$ which is tangent to $AB$ , $AC$ , $A'A_b$ , $A'A_c$ . Similarly define $\Omega_b$ , $\Omega_c$ . From Lemma 2 (b) we get that $BC$ , $B_aC_a$ , $B'C'$ are concurrent. But $B_a$ is the internal homothety center of $\omega_a$ and $\Omega_b$ , and similar for $C_a$ . So from Monge's theorem we conclude that $B_aC_a\cap BC = B'C'\cap BC$ - external homothety center of $\Omega_b$ , $\Omega_c$ . From the same facts for $AB\cap A'B'$ , $AC\cap A'C'$ and from Lemma 1 we get that $\ell_a$ , $\ell_a'$ , $\ell_b$ , $\ell_b'$ , $\ell_c$ , $\ell_c'$ form a hexagon which has an incircle. Done

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PROF65

#5 h Oct 1, 2018, 1:32 PM • 2 Y

Y by Adventure10, Mango247

for the lemma1 :
when $A_1$ is sent to $A_2$ by the homothety that sends $\Omega_1$ to $\Omega_2$ three of the six tangents are parallel and the others also hence in this case there is no incircle !

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#6 h Oct 2, 2018, 8:36 AM • 3 Y

Y by Mosquitall, Adventure10, Mango247

The center of the incircle of the hexagon is http://faculty.evansville.edu/ck6/encyclopedia/ETCPart3.html#X4882 (barycentrics):

X(4882) = (a (r + 8 R) - 8 R s : b (r + 8 R) - 8 R s : c (r + 8 R) - 8 R s),

where r and R are the radii of the incircle and circumcircle, and s is the semiperimeter of ABC.

Equation of the incircle of the hexagon formed by the lines $\ell_a$ , $\ell_a'$ , $\ell_b$ , $\ell_b'$ , $\ell_c$ , $\ell_c'$ , see https://amontes.webs.ull.es/otrashtm/HGT2018.htm#HG021018

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#7 h Oct 2, 2018, 9:23 AM • 2 Y

Y by Adventure10, Mango247

PROF65 wrote:

for the lemma1 :
when $A_1$ is sent to $A_2$ by the homothety that sends $\Omega_1$ to $\Omega_2$ three of the six tangents are parallel and the others also hence in this case there is no incircle !

I think that in your case we get infinite incircle. But this is only a degenerate case, and we can always make small deformations in order to omit it.

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