Generalization to A Serbian Geometry Problem

by XmL, May 24, 2016, 5:06 AM

Generalization: Let $C'\in AB,B'\in AC$ such that $B,C,B',C'$ are concyclic. Suppose $\omega, \omega'$ are two circles through $B,C$ and $C',B'$ respectively, such that they are corresponding circles between similar triangles $ABC, AB'C'$. If $\omega \cap \omega'=X,Y$, prove that $AX,AY$ are isogonals wrt $\angle BAC$.

Note: Another way to characterize $X,Y$ is the property \[\angle BXC+\angle B'XC'=180^{\circ}, \angle BYC+\angle B'YC'=180^{\circ}\]
Solution 1(Indirect):

Let $\phi $ denote the inversion transformation $\phi (A, \sqrt {AB'\cdot AC})$, then $BC\mapsto C'B'$ under $\phi$.

Suppose $\phi (Y)=Y'$. Let $X'$ denote the isogonal conjugate of $Y'$ wrt $ABC$ and $\phi (X')=X''$. We can show through angle chasing that $\angle BX'C=\angle BYC\implies X'\in \omega$. By more angle chasing we can obtain $X''\in (BY'C)$ or $X''\in \phi (\omega')$. Since $X''\in \phi (\omega)$ (by inversion's preservation of concylic points), therefore $X''=\phi (\omega)\cap \phi (\omega')\ne Y'$. Hence $X'=\omega\cap \omega'\ne Y$, so $X'=X$ and $AX,AY$ are isogonals. $\Box$

Solution 2(Direct, at last):

Lemma: Two circles $(O_1,r_1), (O_2,r_2)$ intersect at two points $A,B$. Given a point $P$ that satisfies $\frac {PO_1}{PO_2}=\frac {r_1}{r_2}$ and doesn't lie on $O_1O_2$. Prove that $PA,PB$ are isogonals wrt $\angle O_1PO_2$.

Proof: Let $D$ denote the foot of the bisector of $\angle O_1PO_2$ on $O_1O_2$. Thus the $P$-Apollonius circle of $\triangle PO_1O_2$ (aka the circle of similitude of $(O_1),(O_2)$ passes through $P,A,D,B$. Since $AD=BD$, $PD$ also bisects $\angle PAB$. Hence the lemma is proven.

Main proof: Since $ABC\cup \omega \sim AB'C'\cup \omega'$, therefore if $\omega =(O, r), \omega' =(O',r')$, then $\frac {AO}{AO'}=\frac {r}{r'}\implies A$ lies on the circle of similitude of $\omega, \omega'$. By the Lemma above, this means $AX,AY$ are isogonals wrt $\angle OAO'$. Becuase $AO,AO'$ are isogonals wrt $\angle BAC$, therefore $AX,AY$ are isogonals wrt $\angle BAC$. $\Box$
This post has been edited 1 time. Last edited by XmL, May 24, 2016, 5:07 AM
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Still an immature proof writer :)

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  • Good blog! I hope to see more posts here!

    by WizardMath, Feb 3, 2018, 6:57 PM

  • Thanks Harry Potter :D

    by XmL, Jul 30, 2014, 10:47 PM

  • Hey dude. Nice blog. I like your work.

    by gamjawon, Jul 30, 2014, 7:43 AM

  • Thanks...although most of the posts here won't be about geometry...rightnow I'm trying to improve my skills in other areas such as number theory and combo.

    by XmL, Jun 5, 2014, 1:11 AM

  • Hi Geometer,,nice blog..:D

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