Friday, March 2, 2012

Problem 735: Triangle, Altitude, Perpendicular, Semiperimeter of the Orthic Triangle, Congruence

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

Click the figure below to see the problem 735 details.

Online Geometry Problem 735: Triangle, Altitude, Perpendicular,  Semiperimeter of the Orthic Triangle, Congruence.

6 comments:

  1. B,D,B1,E lie on the circle on BB1 as diameter.
    By sine rule DE = BB1.sin(B)in ∆BDE.
    2∆ = b. BB1
    DE:2∆ = sin(B):b = 1:2R
    DE = ∆:R = 2R.sin(A).sin(B).sin(C)
    In ∆AB1C1, by sine rule
    B1C1:sin(A)=AB1:sin(C)
    =c.cos(A):sin(C)
    =2R.cos(A)
    ∴a1=B1C1=2R.cos(A).sin(A)=R.sin(2A)
    Similar expressions for b1, c1.
    ∴ a1+b1+c1=R[(sin(2A)+sin(2B)+sin(2C)]
    =R.4.sin(A).sin(B).sin(C)= 2.DE

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  2. Let be P and Q on ray B1E and B1D such that A1P=A1B1 and C1Q=C1B1. Note that D and E are midpoints of QB1 and B1P because triangles QC1B1 and PC1B1 are isosceles and C1D with A1E are its respectively altitude.

    Then 2DE=QP, but QP=QC1 + C1A1 + A1P=a1+b1+c1, so DE=(a1+b1c1)/2.

    Greetings.

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  3. Greetings to Eder Contreras Ordenes.Your proof is nice.

    Let X,Y be the midpoints of B₁C₁,B₁A₁ resp
    Extend YX to meet BA at D'
    YXD'∥A₁C₁implies ∠C₁D'X = ∠BC₁A₁ = C
    Also YXD'∥A₁C₁ and B₁D'∥CC₁ imply ∠B₁D'X =∠CC₁A =90°-C
    Follows (by adding the angles)
    ∠B₁D'C₁ = 90°, B₁D'⊥ AB and so D’ coincides with D
    So ∆B₁C₁D is right angled at D and DX=XB₁=XC₁=B₁C₁/2=a₁/2
    Similarly EY=A₁B₁/2=c₁/2
    Already XY=A₁C₁/2=b₁/2
    Hence DE=DX+XY+YE=(a₁+b₁+c₁)/2

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  4. Here the video of my solution (in spanish):

    http://www.youtube.com/watch?v=rbZq7dsmHnM

    Greetings Go-solvers!

    ReplyDelete
  5. Let DE intersect CB and AB at X and Y respectively.
    Let H denote the orthocentre of ∆ABC
    A,B₁,H,C₁ are concyclic.
    So ∠AC₁B₁ = ∠AHB₁
    AA₁∥B₁E
    So ∠AHB₁ = ∠HB₁E = ∠BB₁E
    B,D,B₁,E are concyclic
    So ∠BB₁E = ∠BDE
    It follows ∠AC₁B₁ = ∠BDE
    Same as ∠DC₁X = ∠C₁DX
    Therefore DX = C₁X.
    Similarly DX = B₁X
    So X is the midpoint of (the hypotenuse) B₁C₁ (of the right ∆B₁DC₁)
    DX = (1/2)B₁C₁
    Similarly Y is the midpoint of A₁B₁; EY = A₁Y = B₁Y =(1/2)A₁B₁
    Also XY = (1/2)A₁C₁ (X, Y being the midpoints of B₁C₁,B₁A₁ resp.)
    Hence DE = DX + XY + YE = (1/2)(B₁C₁ + A₁B₁ + A₁C₁)= (a₁+b₁+c₁)/2.

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