Given that ω1 and ω2 are two circles which intersects at points X and Y. Let P be an arbitrary points on ω1. Suppose that the lines PX and PY meet ω2 again at points A and B, respectively.
Prove that the circumcircles of all triangles PAB have the same radius.
#Geometry
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We can see that as we move the point P on the circumference of the circle[excluding X and Y],the ∠XPY=∠XP1Y remains constant.So this shows that AB=A1B1.Now we use extended sin rule to complete the problem.
Let the circum-radius of △PAB be R and △P1A1B1 be R1.
In △PAB,sin∠PAB=2R and in △P1A1B1,sin∠P1A1B1=sin∠PAB=2R1. Therefore 2R=2R1⇒R=R1.Hence Proved.
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Very good solution. Very much clear also. I understood it now.
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Thanks...My fav. is geometry.I see that u like olympiad problems.Can u give some links of ur problems?did u get selected in INMO??
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I have that books in PDF.
Actually I am attending KVS INMO camp for INMO 2017. I have to give INMO directly.
By the way, from which book you do geometry problems?
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I am currently out of Delhi. I will send you after 5 Nov.
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My email id : harshus99@gmail.com
rmo dehli right
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yeah.Are you appearing for this year's rmo?
A problem copied from IF Sharygin Plane geometry. This is bad.
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Why will I copy? This is RMO Delhi's problem
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What I am saying and what are you understanding.
I am blaming RMO people that they have took this problem from a popular book which usually does not happen.
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This is disappointing :(
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This is there in IF Sharygin, plane geometry.
@Harsh Shrivastava, @Ayush Rai,
I have 36 Olympiad books related to IMO. Do you want all of them?
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yup.sure.send it to my email.
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Ok no problem.
Can you tell me the author of co-exter book?
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Yeah I also want em all.
@Ayush Rai and @Harsh Shrivastava
I have sent IMO books and KVS RMO 2016 paper to your e-mails.
Check your mail.
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Thanks a lot bro.