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Source: Indian RMO 2013 Paper 1 Problem 1

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#1 h Feb 1, 2014, 5:21 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be an acute-angled triangle. The circle $\Gamma$ with $BC$ as diameter intersects $AB$ and $AC$ again at $P$ and $Q$ , respectively. Determine $\angle BAC$ given that the orthocenter of triangle $APQ$ lies on $\Gamma$ .

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#2 h Feb 1, 2014, 11:12 PM • 2 Y

Y by Adventure10, Mango247

See that $\stackrel{\frown}{PQ}=\stackrel{\frown}{PB}+\stackrel{\frown}{CQ}\implies\angle BAC=45^\circ$ .

Best regards,
sunken rock

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#3 h Jul 3, 2014, 1:33 PM • 2 Y

Y by Adventure10, Mango247

Let, $H$ be the orthocentre of $\triangle APQ$ .
$\therefore \angle A=180^{\circ}-\angle PHQ=\angle PCQ=90^{\circ}-\angle A \Rightarrow 2\angle A=90^{\circ}\Rightarrow \angle A=45^{\circ}$

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aayush-srivastava

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aayush-srivastava

#4 h Nov 22, 2015, 6:18 AM • 6 Y

Y by purumehta, Evan-Chen123, bond98, Dytto, Adventure10, Mango247

Let $H$ be the $orthocentre$
As $\angle PHQ=180-$ $\angle A$ . As $quadrilateral BPHQ$ is $cyclic$ so $\angle PBQ=A$ . $\angle BQA=90$ (angle subtended by diameter)
So $\angle A=45$
Q.E.D

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Ari004

#6 h Oct 11, 2018, 2:59 PM • 2 Y

Y by Adventure10, Mango247

$J$ is the orthocenter of $\bigtriangleup APQ \implies \angle PJQ=180^{\circ}-\angle A$ .Since, $J$ lies on $\Gamma,\angle PJQ=180^{\circ}-(90^{\circ}-\angle A) \implies \angle A=45^{\circ}$ .

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KKT

#7 h Sep 27, 2019, 6:21 PM • 2 Y

Y by Adventure10, Mango247

sunken rock wrote:

See that $\stackrel{\frown}{PQ}=\stackrel{\frown}{PB}+\stackrel{\frown}{CQ}\implies\angle BAC=45^\circ$ .

Best regards,
sunken rock

How did you get these.....

This post has been edited 2 times. Last edited by KKT, Sep 27, 2019, 6:22 PM

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