Intersection lies on AM

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Intersection lies on AM

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Source: Junior Olympiad of Malaysia Shortlist 2015 G2

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#1 h Jul 17, 2015, 1:54 PM • 1 Y

Y by Adventure10

Let $ABC$ be a triangle, and let $M$ be midpoint of $BC$ . Let $I_b$ and $I_c$ be incenters of $AMB$ and $AMC$ . Prove that the second intersection of circumcircles of $ABI_b$ and $ACI_c$ distinct from $A$ lies on line $AM$ .

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#2 h Jul 17, 2015, 2:44 PM • 4 Y

Y by huynguyen, enhanced, Adventure10, Mango247

My solution :

Let $T$ be the point on the segment $AM$ such that $MB=MT=MC$ .

From $MT=MB \Longrightarrow \measuredangle ATB=90^{\circ}+\tfrac{1}{2} \measuredangle AMB=\measuredangle AI_bB \Longrightarrow T \in \odot (ABI_b)$ .
Similarly, we can prove $T$ lie on $\odot (ACI_c) \Longrightarrow T \equiv \odot (ABI_b) \cap \odot (ACI_c) \in AM$ .

Q.E.D

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#3 h Aug 16, 2018, 4:39 AM • 2 Y

Y by Adventure10, Mango247

Let $X$ be the second intersection of the circumcircle of $ABI_b$ and $ACI_c$ . Note that $\angle XBI=\angle IAX=\angle IAB=\angle BXI$ $\implies$ $BI=IX.$ Also, $\angle BIM=90^{\circ}+\frac{BAM}{2}$ and $\angle ABX=\angle ABI-\angle XBI=\frac{\angle B}{2}-\frac{\angle BAM}{2}$ .
$\angle XIM$
$=\angle AIM-\angle XIA$

$=(90^{\circ}+\frac{\angle B}{2})-\angle XBA$

$=(90^{\circ}+\frac{\angle B}{2})-(\frac{\angle B}{2}-\frac{\angle BAM}{2})$

$=90^{\circ}+\frac{\angle BAM}{2}$

$=\angle BIM$
Hence, by SAS, $\triangle BIM \cong \triangle XIM$ so $\angle IXM=\angle IBM=\angle IBA$ . Because $\angle AXI=180^{\circ}-\angle ABI$ , $\angle AXI+\angle IXM =180^{\circ},$ as desired. $\blacksquare$

This post has been edited 1 time. Last edited by Jzhang21, Aug 16, 2018, 4:40 AM
Reason: Typo

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#4 h Aug 16, 2018, 5:08 AM • 2 Y

Y by Adventure10, Mango247

zschess wrote:

Let $ABC$ be a triangle, and let $M$ be midpoint of $BC$ . Let $I_b$ and $I_c$ be incenters of $AMB$ and $AMC$ . Prove that the second intersection of circumcircles of $ABI_b$ and $ACI_c$ distinct from $A$ lies on line $AM$ .

This is the so--called "sparrow's lemma": if the circle passes through th points $A$ and $I$ (icenter) in intersects lines $AB$ and $AC$ in points $X$ and $Y$ , then $BX+CY=BC$ .
Using this lemma, consider the intersiction points $T_1$ and $T_2$ of two circumcircles with $AM$ . Hence $0 + MT_1=MB=MB=MT_2+0.$ So $T_1=T_2$ and circumcircles pass through one point $T=T_1=T_2$ on $AM$ .

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#5 h Feb 9, 2022, 7:28 AM

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Let $ABI_b$ meet AM at S. $\angle MSB = \angle 180 - \angle ASB = \angle 180 - \angle AI_bB = \angle 90 - \angle AMB/2$ so $\angle SBM = \angle SMB = \angle 90 - \angle AMB/2$ so $MB = MS = MC$ .
$\angle AI_cC = \angle 180 - (\angle 90 - \angle AMC/2) = \angle 180 - \angle CSM = \angle ASC$ so $ACI_cC$ is cyclic.

This post has been edited 1 time. Last edited by Mahdi_Mashayekhi, Feb 23, 2022, 11:36 AM

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Mutse

#6 h Feb 9, 2022, 11:02 AM

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Inversion works here. Invert respect circle centered at $A$ .

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