Synthetic Geometry Group - Alan Yan's Proposal

Note that although these questions are marked as "Synthetic", you are allowed to solve them with any method you want.

  1. (Medium Difficulty) Let ABCABC be a triangle and let ω \omega be its incircle. Denote by D1D_1 and E1E_1 the points where ω\omega is tangent to sides BCBC and ACAC, respectively. Denote by D2D_2 and E2E_2 the points on sides BC and AC, respectively, such that CD2=BD1CD_2 = BD_1 and CE2=AE1CE_2 = AE_1, and denote by PP the point of intersection of segments AD2AD_2 and BE2BE_2. Circle ω\omega intersects segment AD2AD_2 at two points, the closer of which to the vertex A is denoted by Q. Prove that AQ=D2PAQ = D_2P.

  2. (Easy Difficulty) Let triangle ABCABC satisfy 2BC=AB+AC2BC = AB + AC and have incenter I and circumcircle ω\omega. Let D be the intersection of AIAI and ω\omega (with A, D distinct). Prove that I is the midpoint of AD.

  3. (Hard Difficulty) Let the incircle ω \omega of triangle ABCABC touch BC,CA,BC, CA, and ABAB at D,ED , E and FF, respectively. Let Y1,Y2,Z1,Z2Y_1 , Y_2, Z_1, Z_2 and MM be the midpoints of BF,BD,CE,CDBF , BD, CE, CD and BCBC, respectively. Let the intersection of Y1Y2Y_1Y_2 and Z1Z2Z_1Z_2 be XX. Prove that MXBCMX \perp BC .

If you want solutions posted, just comment down below.

Note by Alan Yan
4 years, 2 months ago

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Q2. It is clear that ABDCABDC is cyclic. DBC=DAC=DAB=DCB\angle DBC = \angle DAC = \angle DAB = \angle DCB. So, ΔDCB\Delta DCB is isosceles and DC=DBDC = DB. But, DIB=DBI\angle DIB = \angle DBI(by some angle chasing). So, DI=DB=DCDI = DB =DC. But, since ABDCABDC is cyclic, using ptolemy's theorem we get AD=2DB=2DIAD = 2DB = 2 DI, which implies that AI=IDAI = ID. So, II is mid-point of ADAD.

Surya Prakash - 4 years, 2 months ago

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Does anyone want solutions?

Alan Yan - 4 years, 2 months ago

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yes i want soutions please post the solutions.

onkar tiwari - 3 years, 11 months ago

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