Synthetic Geometry Group - Surya Prakash's Proposal

This is my submission to Xuming's Synthetic Geometry Group. These problems are taken from different Olympiads. Try them on your own. I will soon post the solutions. Feel free to post the solutions.

  1. Let Γ\Gamma be the circumcircle of ΔABC\Delta ABC, and let ll be the tangent of Γ\Gamma passing through AA. Let DD, EE be the points on side ABAB and ACAC such that BD:DA=AE:ECBD :DA = AE : EC. Line DEDE meets Γ\Gamma at points FF, GG. The line parallel to ACAC passing DD meets ll at HH, the line parallel to ABAB passing EE meets ll at II. Prove that FF, GG, HH, II are cyclic and BC is tangent to the circle through these points.

  2. In ΔABC\Delta ABC, let HH be the orthocenter of the triangle and MM be the midpoint of the side BCBC. Let the line perpendicular to HMHM through HH meet ACAC and ABAB at EE and FF. Prove that HE=HFHE = HF. (Proposed by Xuming).

  3. Let ΔABC\Delta ABC be an acute triangle with DD, EE, FF the feet of the altitudes lying on BCBC, CACA, ABAB respectively. One of the intersection points of the line EFEF and the circumcircle is PP. The lines BPBP and DFDF meet at point QQ. Prove that AP= AQ.

  4. Let PP be a point inside triangle ΔABC\Delta ABC. Lines APAP, BPBP, CPCP meet the circumcircle of ΔABC\Delta ABC again at points KK, LL, MM respectively. The tangent to the circumcircle at CC meets line ABAB at SS. Prove that MK=ML if and only if SP=SCSP=SC.

Note by Surya Prakash
4 years, 1 month ago

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@Nihar Mahajan @Mehul Arora @Aditya Kumar @Xuming Liang @Calvin Lin @Shivam Jadhav @Swapnil Das @Adarsh Kumar @Sharky Kesa

Please respond to this proposals. and post your proposals too.

Surya Prakash - 4 years, 1 month ago

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This proposal*

Will check it out :)

Mehul Arora - 4 years, 1 month ago

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Nice problems!

Aditya Kumar - 4 years, 1 month ago

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Nice ones man!

Adarsh Kumar - 4 years, 1 month ago

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Nicce problems! I just want to point out that I did not propose that problem. It can be viewed as a simple application of the Butterfly theorem(do you see it?). A "pesudo" generalization of this was utilized in one of my recent problems though.

Xuming Liang - 4 years, 1 month ago

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no problem. by the way I got the solution for that.

Surya Prakash - 4 years, 1 month ago

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is the condition if and only if in question 4 right?

onkar tiwari - 3 years, 10 months ago

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