2024 BxMO P3

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beansenthusiast505

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beansenthusiast505

#1 h Apr 28, 2024, 12:01 AM • 1 Y

Y by Rounak_iitr

Let $ABC$ be a triangle with incentre $I$ and circumcircle $\Omega$ such that $\left|AC\right|\neq\left|BC\right|$ . The internal angle bisector of $\angle CAB$ intersects side $BC$ at $D$ and the external angle bisectors of $\angle ABC$ and $\angle BCA$ intersect $\Omega$ at $E$ and $F$ respectively. Let $G$ be the intersection of lines $AE$ and $FI$ and let $\Gamma$ be the circumcircle of triangle $BDI$ . Show that $E$ lies on $\Gamma$ if and only if $G$ lies on $\Gamma$ .

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896 posts

#2 h Apr 28, 2024, 12:47 AM

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Whenever I draw it neither point lies on the circle

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1519 posts

#3 h Apr 28, 2024, 2:04 AM

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This took longer than I'd like. Nice-ish problem.

First suppose that $BEDI$ is cyclic. Then by radical axis on $(BEDI)$ , $\Omega$ it follows that $IDFC$ is cyclic. Let $I_A$ be the $A$ -excenter, note that $(BII_AC)$ is cyclic. Then note that $E - D - F$ are collinear since
$\measuredangle EDI = \measuredangle EBI = \measuredangle I_ABI = \measuredangle I_ACI = \measuredangle FCI = FDI.$ As such, $\measuredangle GED = \measuredangle AEF = \measuredangle ACF = \measuredangle I_ACB = \measuredangle FCD = \measuredangle FID$ which finishes.

Now consider the case where $BEGI$ is cyclic. Define $D' = (BGEI) \cap (ICF)$ . Then by radaxis on $(BGEI)$ , $\Omega$ , and $(ICF)$ , it follows that $D'$ is on $AI$ . Then we can angle chase $E-D'-F$ again. Then by a similar angle chase we get $\measuredangle GID = \measuredangle FID = \measuredangle GED = \measuredangle FCB$ .

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#4 h Dec 7, 2024, 8:05 AM

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Just note that both conditions are equivalent to $I$ being orthocentre of $\triangle AEF$ thru simple angle chase.

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GeorgeMetrical123

7 posts

GeorgeMetrical123

#5 h Mar 30, 2025, 9:39 AM

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Suppose $(BEDI)$ is a cyclic quad. Then, by radical axis on $(BEDI)$ and $\Gamma$ we get that $I_A$ lies on the radical axis.From there, it follows that $(IDCF)$ is a cyclic quad. From there, we angle chase that $\angle DEF = 0$ :
$\begin{align*} \angle DEI_A &= 180^\circ - \angle BED \\ & = \angle BID \\ &= 180^\circ - \angle AIB \\ &= 180^\circ - (90^\circ + \gamma) \\ & = 90^\circ - \gamma \\ & = \angle BAF \\ & = 180^\circ - \angle BEF \\ &= \angle I_AEF \end{align*}$ Where $\gamma = \frac{1}{2}\angle ACB$ .
From there, we angle chase again:
$\begin{align*} \angle GEB &= \angle AEB \\ &= \angle ACB \\ &= \angle BCF - \angle BCT \\ &= \angle DCF - \angle BIT \\ &= 180^\circ - \angle DIF - \angle BID \\ &= 180^\circ - \angle BIF \\ &= \angle GIB \end{align*}$ voila!

For the other part, we angle chase again. We will first prove that $(AGDF)$ is a cyclic quad.
$\begin{align*} \angle FGD &= \angle IGD \\ &= \angle IBD \\ &= \beta \\ &= 90^\circ - \alpha - \gamma \\ &= \angle BAF - \angle BAD \\ &= \angle DAF \end{align*}$ Now, we prove that $(IDCF)$ is a cyclic quad and from there we finish like in the last proof.
$\begin{align*} \angle FID &= \angle GFD \\ &= \angle GAD \\ &= \angle EAD \\ &= \angle EAC - \angle DAC \\ &= 90^\circ - \beta - \alpha \\ &= \gamma \\ &= \angle ICB \\ &= \angle ICD \end{align*}$ Which proves that $(IDCF)$ is a cyclic quad and by using the same method as above, we are finished! Q.E.D.

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