Wednesday, February 22, 2012

Problem 730: Triangle, Altitude, Orthocenter, Perpendicular, Concurrent Lines

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 730 details.

Online Geometry Problem 730: Triangle, Altitude, Orthocenter, Perpendicular, Concurrent Lines.


    For clarity see sketch on the left of above picture
    We will use Ceva’s theorem in this case.
    Quadrilateral AA3A1A2 is cyclic with diameter AA1
    We have ∠ (AA3A2)= ∠ (AA1A2)= ∠ (ABA1)= ∠B ( angles face the same arc AA2)
    Similarly ∠ (AA2A3)= ∠AA1A3)= ∠ (ACA1)= ∠C
    So sin(A4AA3)/sin(A4AA2)=cos(AA3A2)/cos(AA2A3)=cos(B)/cos(C) (angles complement) ----(1)
    Similarly sin(B4BB3)/in(B4BB2)=cos(C )/cos(A) ----- (2)
    And sin(C4CC3)/sin(C4CC2)=cos(A)/cos(B) -----(3)
    Multiply expressions (1) x(2) x(3) we get
    RHS= cos(B)/cos(C) x cos(C )/cos(A) x cos(A)/cos(B) =1
    So Cevians AA4, BB4 and CC4 will concurrent per Ceva’s theorem

  2. Join C₁A₁,A₁B₁,B₁C₁.
    ∆BC₁A₁~∆BCA with ∠BA₁C₁=A=∠B₃B₂B,∠BC₁A₁=C=∠B₂B₃B
    So C₁A₁∥B₃B₂. Given B₂B₃⊥BB4. Thus C₁A₁⊥BB4.
    Let C₁A₁^BB4 = Y, A₁B₁^CC4 = Z, B₁C₁^AA4 = X
    (^ denotes intersection)
    BB4 divides C₁A₁ (at Y) in the ratio C₁X:XA₁
    =BXcotC:BXcotA = cotC:cotA.
    CC4 divides A₁B₁(at Z)in the ratio cotA:cotB,
    AA4 divides B₁C₁(at X)in the ratio cotB:cotC.
    Product of the above three ratios being +1, it follows by Ceva,
    AA4X, BB4Y, CC4Z are concurrent.

  3. To Pravin:
    Refer to your solution .
    "Product of the above three ratios being +1" is not applicable to Ceva's theorem in this case.
    please clarify.

  4. Thanks Peter. I have recast my proof as follows.

    Let BB4 be extended to meet AC at Y’
    (similarly CC4 to Z’ on AB, AA4 to X’ on BC)
    BB4 ⊥ B₃B₂
    So ∠ABY’=∠B₃BB4=90°-∠B₂B₃B=90°-C and similarly ∠CBY’=90°-A
    AY’:Y’C=(ABY’)/(BY’C)where (...)denotes area.
    = [AB.BY’.sin∠ABY’]: [BC.BY’.sin∠CBY’]
    = (AB:BC)(cosC:cosA)
    = (sinC:sinA)(cosC:cosA)
    = sin2C:sin2A
    In the same way, CX’:X’B = sin2B:sin2C and BZ’: Z’A = sin2A:sin2B
    So(AY’:Y’C).(BZ’:Z’A).(CX’:X’B) = 1
    Hence by Ceva’s theorem applied to ∆ABC
    the cevians AA4X’, BB4Y’, CC4Z’ are concurrent.

  5. Hint:It's cut at circumcenter