Let \(ABC\) be an acute-angeled triangle and let \(H\) be its orthocenter. For any point \(P\) on the circumcircle of triangle \(ABC\), let \(Q\) be the point of the intersection of the line \(BH\) with the line \(AP\) . Show that there is a unique point \(X\) on the circumcircle of \(ABC\) such that for every point \(P\not=A,B\) the circumcircle of \(HQP\) pass through \(X\).

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TopNewestQuite similar to one which came in Rajasthan paper, though more difficult. – Satvik Golechha · 1 year, 10 months ago

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can someone give me the solution to this problem? plzz – Nihar Mahajan · 1 year, 10 months ago

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How many are you getting right? (Outta 6, right?) – Satvik Golechha · 1 year, 10 months ago

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– Aneesh Kundu · 1 year, 10 months ago

Dunno whether they are right or wrongLog in to reply

– Aneesh Kundu · 1 year, 10 months ago

I solved 4Log in to reply

I got this! I hope it is right. – Ranjana Kasangeri · 1 year, 10 months ago

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– Aneesh Kundu · 1 year, 10 months ago

How?Log in to reply

– Ranjana Kasangeri · 1 year, 10 months ago

Take any other point say, \(Z\) and let \(AZ\) intersect \(BH\) at \(Y\). And consider circle \(HYZ\) meeting circle \(ABC\) at \(X.\) Now,prove \(HQPX\) is a cyclic quad!Log in to reply