2018 Poland Math Olympiad Geometry Problem 5 (Two Solutions)

[@FranziskavonKarma]

Thanks a lot, this video set me on my current path and now I'm in my country's IMO team for this year

[@aa-dz4pl]

great job!

[@aasthasharma3820]

wow I just spent half an hour watching this even though I barely even know what radical axis/projective geometry is......................... haha what a great use of time :) thanks Dr. Nal

[@michaelgreenberg782]

For the "HM" Theorem, here is another way to prove it: Let X' be the perpendicular from H to AM. Then using right angles, AFHX'E is cyclic and HX'MD is cyclic.Also, FEMC is cyclic as all four points lie on the nine-point circle. Thus, by the radical axis theorem, FE, HX', and BC are concurrent at a point K. By right angles, BFEC is cyclic, so KB*KC=KF*KE=KH*KX', so BHX'C is cyclic. Thus X=X', which completes the proof.

[@OsmanNal]

This is a very nice proof. Thank you 🙏

[@michaelgreenberg782]

@@OsmanNal Thanks! I actually did a video with my proof on my KZbin channel here: kzbin.info/www/bejne/iZC1gGmYbLt_bZo

[@matthewchuang6707]

Love the first solution! And thanks to this, I finally know how point circle works!!

[@sparkyroyale6023]

The 2 nd solution it was so nice, but where are the harmonic quadrilatirals?

[@azzamlabib4785]

i wondering why in the second solution, the "last" circle is apollonian. I have read about it, and still don't understand. Does point A be the locus of BC ? i'm very curious

[@vincentwu2241]

Stunning solutions! Just wondering why the triangle's Apollonius circle respective to AB and AC goes through point X and has the center at P. Do you have any link to the explanation? Thanks for your best Math Olympiad videos.

[@tonyha8888]

Thankyou for the lecture. I am very much enjoy both solutions. Very much appreciated!

[@ScoopSav]

Hi osman, I was wondering if you would be able to go through past British maths olympiad round 2 question as I am participating in 2 weeks any help would be greatly appreciated. Thanks man!

[@tonyha8888]

Yes, I am second this. There are NO videos on BMO2 questions on the internet so far.

[@drozfarnyline4940]

First one was sooo... nice point circle is really awesome

[@ducchu7176]

Good job Keep doing this Ur having nice mathcn

[@akashsudhanshu5420]

I found something amazing The diameter subtends 90°at any point on a circle. Now, measure the diameter. Measure the two "legs" which have 90° btw. them . Now it's a right angled triangle . Then A²=B²+C²{A:diameter,B and Care legs } Angle btw. Band C =90° You can take many many points on a circle and most of the time you will get different length of the legs STATEMENT:a²=b²+c² a=constant Then {b,c} can have many values.

[@twinbrain7743]

MATHEMATICS, PHYSICS lover lol is this a joke?

[@tahmidhameemchowdhuryzarif8195]

THE FIRST SOLUTION WAS AWESOME

[@igml1145]

So in the end we didnt really need PA^2=PB*PC because we used the apollonian circle thing...

[@OsmanNal]

ig ml that’s right!

[@mrmathcambodia2451]

ok ,so good solution also.

[@invincible3011]

The proof is much simple of the case when AH IS A DIAMETER. LETS SAY Produced AX to cut the circumcircle of BHC AT N. THEN since /_AXH = 90 Implies/_ HXN= 90. So HN is a diameter of circumcircle of BHC. so ANGLE BCN = angle HCN - angle HCN = 90-(90-B)= B so CN is parallel to AB, similarly angle CBN= ANGLE HBN- angle HBC= 90-(90-C)= C so BN is parallel to AC thus ABNC is a parallelogram so it's diagonals bisects each other, therefore AXN IS A median of triangle ABC🙏🙏😊

[@koksutv6976]

My country!!!

[@dalibormaksimovic6399]

Where do you find these problems?

[@OsmanNal]

Dalibor Maksimovic I pick the problems from aops contest page (artofproblemsolving.com/community/c13_contests) somewhat randomly.

[@ongnguyencong855]

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