Regional Math Olympiad (India) Geometry Problems

(This is a work in progress. More problems will be added soon).

  1. Let ABC be an acute-angled triangle in which ∠ABC is the largest angle. Let O be its circumcentre. The perpendicular bisectors of BC and AB meet AC at X and Y respectively. The internal bisectors of ∠AXB and ∠BY C meet AB and BC at D and E respectively. Prove that BO is perpendicular to AC if DE is parallel to AC. (RMO 2014, Mumbai Region)
  2. Let D, E, F be the points of contact of the incircle of an acute-angled triangle ABC with BC, CA, AB respectively. Let I1, I2, I3 be the incentres of the triangles AF E, BDF, CED, respectively. Prove that the lines I1D, I2E, I3F are concurrent. (RMO 2014, Mumbai Region)
  3. Let ABC be an isosceles triangle with AB = AC and let Γ denote its circumcircle. A point D is on arc AB of Γ not containing C. A point E is on arc AC of Γ not containing B. If AD = CE prove that BE is parallel to AD. (RMO 2013, Mumbai Region)
  4. In a triangle ABC, points D and E are on segments BC and AC such that BD = 3DC and AE = 4EC. Point P is on line ED such that D is the midpoint of segment EP. Lines AP and BC intersect at point S. Find the ratio BS/SD (RMO 2013, Mumbai Region)
  5. Let ABC be a triangle and D be a point on the segment BC such that DC = 2BD. Let E be the mid-point of AC. Let AD and BE intersect in P. Determine the ratios BP/P E and AP/P D. (RMO 2012)
  6. Let ABC be a triangle. Let BE and CF be internal angle bisectors of ∠B and ∠C respectively with E on AC and F on AB. Suppose X is a point on the segment CF such that AX ⊥ CF; and Y is a point on the segment BE such that AY ⊥ BE. Prove that XY = (b + c − a)/2 where BC = a, CA = b and AB = c. (RMO 2012)
  7. Let ABC be a triangle. Let D, E, F be points respectively on the segments BC, CA, AB such that AD, BE, CF concur at the point K. Suppose BD/DC = BF/F A and ∠ADB = ∠AF C. Prove that ∠ABE = ∠CAD. (RMO 2011)
  8. Let ABC be a triangle and let BB1, CC1 be respectively the bisectors of ∠B, ∠C with B1 on AC and C1 on AB. Let E, F be the feet of perpendiculars drawn from A onto BB1, CC1 respectively. Suppose D is the point at which the incircle of ABC touches AB. Prove that AD = EF. (RMO 2011)

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