Tuesday, August 18, 2009

Problem 339. Triangle, Angle Bisectors, Perpendiculars, Interior Point, Distances

Proposed Problem
Click the figure below to see the complete problem 339 about Triangle, Angle Bisectors, Perpendiculars, Interior Point, Distances.

 Problem 339. Triangle, Angle Bisectors, Perpendiculars, Interior Point, Distances.
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Complete Problem 339
Level: High School, SAT Prep, College geometry


  1. B:(p,q), A:(0,0) & C:(b,0)
    AB = c, AC = b & BC = a
    Since E divides AB in the ratio b:a
    D:[bp/(a+b), bq/(a+b)]
    Similarly since D divides CB in the ratio b:c
    we've E:[(bp+bc)/(b+c), bq/(b+c)]
    Slope DE =((bq/(a+b))- bq/(b+c)) /(bp/(a+b)- (bp+bc)/(b+c)) = (c-a)q/(c(b-p)+a(c+p))
    or DE is y = (c-a)qx/(cb+ac+pa-pc) + k and k is determined as = bqc/(cb+ac+pa-pc)
    or DE is y=((c-a)qx+bqc)/(cb+ac+pa-pc)---(1)
    Let F:(x,y) be any point on DE satisfying the eqn. above.
    Now AB: y=qx/p or qx-py=0
    AC: y=0
    Slope BC = -q/(b-p)
    BC: y= -qx/(b-p) +k1
    Passes thru’ (b,0) so k1 = qb/(b-p)
    Hence BC: qx+ y(b-p)-qb=0
    Hence, FG = qx-py/c where c^2=p^2+q^2
    FH = (qx +y(b-p)-qb)/a where a^2=(b-p)^2+q^2
    FG and FH are of opposite signs.
    Hence, FG + FH = (qx-py)/c - (qx +y(b-p) -qb)/a
    If this is equated to FM = y then we obtain
    y = ((c-a)qx+bqc)/(cb+ac+pa-pc) which is, in fact, equation (1) above. Therefore for any point F on DE we’ve FG + FH = FM .QED.
    Ajit: ajitathle@gmail.com

  2. let A be the area and a,b,c are lenghts of the sides BC,CA,AB. join B,F and A,F and C,F hence area of triangle ABC is equal to the sum of areas of triangles BFA,BFC,AFC. and given AD,CE are angular bisectors of angles A,C hence BE= ca/(a+b) and BD= ac/(b+c)and A =1/2ac Sin B implies Sin B = 2A/ac, and Ar of Tr BFE= 1/2 BE.FG and Ar of TrBFD= 1/2BD.FH since FG,FH are perpendiculars drawn from Fto the sides BE,BD and area of Tr BED = 1/2 (BD)(BE)(Sin B) =Aac/(a+b)(b+c). but area of Tr BED = Ar of TrBFE + Ar of Tr BFD. so by equating both L.H.S and R.H.S areas we will get the required relation that FM = FG + FH

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  4. http://s2.postimg.org/hpwg8xnpl/pro_339.png1.
    Call Points M1, H1 as the projections of point E over segments AC and BC
    Call Points M2, G1 as the projections of point D over segments AC and AB
    Let x=EF and y=ED
    2. We have EM1=EH1 ( EC is angle bisector of angle ACB)
    DM2=DG1 (AD is angle bisector of angle BAC)
    3. calculate FM as linear interpolation of EM1 and EM2
    FH=x/y*(-EH1) +EH1 ( linear interpolation)
    FG=x/y*(DG1) ( linear interpolation)

    4. FH+FG= x/y*(DG1-EH1)+EH1
    5. Replace DG1=DM2 and EH1=EM1 we get
    FH+FG=x/y*(DM2-EM1)+EH1 = FM per step 3