Given a triangle ABC let D, E, F be orthogonal projections from A, B, C to the opposite sides respectively. Let X, Y, Z denote midpoints of AD, BE, CF respectively. Prove that perpendiculars from D to YZ, from E to XZ and from F to XY are concurrent.
Problem
Source: 2nd European Mathematical Cup
Tags: geometry, circumcircle, geometry unsolved
16.12.2013 22:28
vyfukas wrote: Given a triangle ABC let D, E, F be orthogonal projections from A, B, C to the opposite sides respectively. Let X, Y, Z denote midpoints of AD, BE, CF respectively. Prove that perpendiculars from D to YZ, from E to XZ and from F to XY are concurrent. By Karno theorem in triangle XYZ it is enough to prove: FX2−EX2+EZ2−DZ2+DY−FY2=0 (1). Using: FX2−EX2=14(AF2+FD2−AE2−ED2). (1) is easy.
16.12.2013 22:51
IMI-Mathboy wrote: By Karno theorem in triangle XYZ it is enough to prove: FX2−EX2+EZ2−DZ2+DY−FY2=0 (1). Using: FX2−EX2=14(AF2+FD2−AE2−ED2). (1) is easy. What is Karno theorem?
16.12.2013 23:48
(KARNO).Let A1,B1 and C1 be points on the sides BC,AC and AB of △ABC. The lines through A1,B1 and C1, and perpendicular to BC,AC and AB respectively are concurrent if only if AB21−B1C2+CA21−A1B2+BC21−AC21=0
17.12.2013 07:12
To avoid calculations we can use the following lemma: Let ABC,XYZ be two triangles. We say that the perpendiculars from A to YZ, from B to XZ and from C to XY are concurrent iff the perpendiculars from X to BC, from Y to AC and from Z to AB are concurrent. İn some instances, like this one is far easier to prove the reverse. The proof of lemma uses the before mentıoned Carnot lemma. Now to problem: we want to show that perpendiculars: from X to EF, from Y to FD and from Z to DE are concurrent. Let H be the orthocenter of ΔDEF and O the circumcenter of ΔABC. Because the radii from A,B,C of circle ⊙(ABC) are perpendicular to EF,FD,DE respectively, we infer that our perpendiculars pass through the midpoint of OH. Best regards, sunken rock
29.01.2025 14:33
Using the Carnot's theorem but nice trigono calculation.Let's change the problem into following perspective, Let Ia, Ib and Ic be the excenters of the △ABC. And let D,E,F be the midpoint of AIa, BIb, CIc, respectively. Then prove that perpendiculars from D,E,F to the sides BC,AC,AB are concurrent. Proof: Let's 2α,2β,2γ be the angles of △ABC, and let R be the circumradius. Then we have that: IbB=4Rcos(γ)cos(β) ⟹ BE=2Rcos(γ)cos(β). By the cosine's theorem on the ABE we have that: AE2=AB2+BE2−2AB∗BE∗cos(α)). By the cosine's theorem on the ACE we have that: CE2=CB2+BE2−2CB∗BE∗cos(α)). Thus: AE2−CE2=(AB2−AC2)+(2BEcos(α)(BC−AC)=(AB2−AC2)+4Rcos(γ)cos(β)cos(α)(BC−AC). ∑AE2−CE2=∑(AB2−AC2)+4Rcos(γ)cos(β)cos(α)∑(BC−AC)=0, which means we are done by Carnot's theorem.