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Source: Iberoamerican Problem 4

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#1 h Sep 22, 2017, 7:18 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be an acute triangle with $AC > AB$ and $O$ its circumcenter. Let $D$ be a point on segment $BC$ such that $O$ lies inside triangle $ADC$ and $\angle DAO + \angle ADB = \angle ADC$ . Let $P$ and $Q$ be the circumcenters of triangles $ABD$ and $ACD$ respectively, and let $M$ be the intersection of lines $BP$ and $CQ$ . Show that lines $AM, PQ$ and $BC$ are concurrent.

Pablo Jaén, Panama

This post has been edited 1 time. Last edited by juckter, Aug 10, 2019, 4:36 PM

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ilai55

#2 h Sep 23, 2017, 3:01 AM • 2 Y

Y by Adventure10, Mango247

Denote the angles $\angle{BAC}=\alpha$ , $\angle{ABC}=\beta$ , $\angle{BCA}=\gamma$ . Note that the condition on the angles is equivalent to $2\angle{ADB} + \angle{DAB} = 180$ . Let's find angle $\angle DAB$ :
$\begin{align*} 2\angle{ADB} + \angle{DAB} = 180\\ \Rightarrow 2\angle{ADB} + \angle{CAD} - \angle{CAO} = 180\\ \Rightarrow 2(180-\beta -\angle{DAB}) + \angle{CAD} - (90-\beta) = 180\\ \Rightarrow \angle{CAD} - 2\angle{DAB} = 180 - \beta + 90 + 2\beta - 360\\ \Rightarrow \angle{CAD} - 2\angle{DAB} = \beta - 90\\ \Rightarrow \alpha - \angle{DAB} - 2\angle{DAB} = \beta - 90\\ \Rightarrow \angle{DAB} = 30 + \frac{\alpha-\beta}{3}. \end{align*}$
Similarly, $\angle{DAC} = 30 + \frac{\alpha-\gamma}{3}$ .

Now let's find $\angle{ADB}$ :

$\begin{align*} \angle{DAB} + \beta + \angle{BDA}=180\\ \Rightarrow 30 + \frac{\alpha + 2\beta}{3} + \angle{BDA} = 180\\ \Rightarrow\angle{BDA} = 150 - \frac{\alpha + 2\beta}{3} = 90 + \frac{\gamma-\beta}{3}\\ \end{align*}$
Now let's find $MBA$ and $MCA$ . Remember $P$ , $Q$ are the circumcenters of $\triangle{ADB}$ y $\triangle{ADC}$ respectively:

$\begin{align*} \angle{MBA}=90-\angle{ADB}=\frac{\beta - \gamma}{3}\\ \angle{MCA} = \angle{ADC} - 90 = (180 - \angle{ADB}) - 90 = \frac{\beta - \gamma}{3}.\\ \end{align*}$
Note that $\angle{MBA}=\angle{MCA} \Rightarrow ABCM$ is cyclical.

Let $X=AM \cap BC$ . Let's find angle $\angle{XAD}$ :

$\begin{align*} \angle{XAD} = 180-\angle{MAD} = 180 -\angle{MAB} + \angle{DAB}\\ = \angle{MCD} + \angle{DAB} = \angle{QCA} + \gamma + \angle{DAB}\\ = 30 + \frac{2\gamma + \alpha}{3} = 90 + \frac{\gamma - \beta}{3}. \end{align*}$
Note that $\angle{XAD} = \angle{BDA}$ , the triangle $\triangle{AXD}$ is isosceles and its prependicular bisector $PQ$ is the bisector of angle $\angle{AXD}$ . $\blacksquare$

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primesarespecial

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primesarespecial

#3 h Jun 21, 2021, 12:32 PM • 3 Y

Y by Mango247, Mango247, Mango247

An alternate solution.
Firstly as above ,we angle chase like above to get $ABCM$ as cyclic .
Then using the given condition , we get quadrilaterals $AMPQ$ and $BCPQ$ as cyclic .
Then radical axis on $(AMPQ),(BCPQ),(ABC)$ just finishes it.

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Mahdi_Mashayekhi

690 posts

Mahdi_Mashayekhi

#4 h Mar 8, 2022, 5:15 PM • 1 Y

Y by AlexCenteno2007

Nice
Claim1 : $AMCB$ is cyclic.
Proof : we have $\angle PBD = \angle 90 - \angle DAB$ and $\angle QCD = \angle 90 - \angle DAC$ so $\angle BMC = \angle DAB + DAC = \angle BAC$ .

Claim2 : $BPQC$ is cyclic.
Proof : $\angle BPQ = \angle BPD + \angle DPQ = 2\angle DAB + \angle ABD$ and $QCD = \angle 90 - \angle DAC$ so we need to prove $\angle 2DAB + \angle ABD = \angle 90 + \angle DAC$ . we have $\angle ADC = \angle ADB + \angle DAO$ so $\angle DAB + \angle ABC = \angle 180 - \angle DAB - \angle ABC + \angle DAC - \angle 90 + \angle ABC = \angle 90 - \angle DAB + \angle DAC \implies\angle 2DAB + \angle ABD = \angle 90 + \angle DAC$ .

Claim3 : $AMQP$ is cyclic.
By simple angle chasing we have $APB$ and $AQC$ are similar so $\angle DAP = \angle CAQ \implies \angle PAQ = \angle BAC = \angle PMQ$

Now By Radical Axis Theorem on $AMCB$ , $AMQP$ and $BPQC$ we have $AM$ , $PQ$ and $BC$ are concurrent.

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