Saturday, October 9, 2010

Problem 526: Equilateral Triangle, Chord, Measurement

Proposed Problem
Click the figure below to see the complete problem 526 about Equilateral Triangle, Chord, Measurement.

Problem 526: Equilateral Triangle, Chord, Measurement.
See also:
Complete Problem 526

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 11:21 AM
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8 comments:

  1. Peter TranOctober 9, 2010 at 11:44 AM

    Area(BDE)+Area(EDC)=Area(BDC)
    ½*e.d.sin(60)+1/2*d.f.sin(60)=1/2*e.f*sin(120)
    So e.d+d.f=e.f
    Divide both side by e.d.f we get the result

    Peter Tran

    ReplyDelete
  2. AjitOctober 10, 2010 at 4:13 AM

    If we must eschew trigonometry then, by Ptolemy's Theorem, we've: AB*f+ AC*e=AD*BC or AD = e + f----(1). Further,/_BED=/_ABC+/_BAD=/_ACB +/_BCD=/_ACD and /_EBD=/_CAD or Tr.CAD /// Tr.BED ----> f/d=AD/e or 1/d=AD/(e*f) ---(2). Eqns. (1) & (2) give us the necessary result viz. 1/d = 1/e + 1/f
    Ajit

    ReplyDelete
  3. AnonymousOctober 11, 2010 at 2:02 AM

    Let BE = m and EC = n. Let a side of ABC
    Tr. ACE similar to Tr. BDE -> a/e = n/d
    Tr. AEB similar to Tr. CDE -> a/f = m/d
    Then, a/e + a/f = (m+n)/d -> 1/e + 1/f = 1/d.

    Migue.

    ReplyDelete
  4. sourav mishraNovember 1, 2010 at 5:16 AM

    since ABCD is a cyclic quadrilateral, by ptolemy's theorem, we have AD*BC = AB*f + AC*e ... (i)
    since AB = BC = CA as triangle ABC is equilateral,
    we have AD = e + f ... (ii)
    triangle BDE ~ triangle ADC.
    which implies that BD/AD = DE/DC or DE = DC*BD/AD
    using symbols, f for DC, d for DE and e for BD, and (ii), we get d = f*e/(f+e) or 1/d = (f+e)/f*e
    or 1/d = 1/f + 1/e
    Q. E. D.

    ReplyDelete
  5. AnonymousNovember 23, 2010 at 12:59 PM

    ABC = ADC = 60
    BDA = ACB = 60
    S(BDC) = S(BDE) + S(DCE)
    e.f.1/2.sen120 = e.d.1/2.sen60 + d.f.sen60.1/2
    e.f.1/2.sen120 = d.sen60.1/2(e+f)
    ef = d(e+f)
    d = ef/e+f
    d = 1/e +1/f

    ReplyDelete
  6. Sumith PeirisOctober 1, 2015 at 1:49 AM

    Draw BF // to AD where F is on CD extended. Tr. BFD is easily seen to be equilateral and from similar Tr.s e/d = (e+f)/f. Divide both sides of the equation by def and the result follows.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  7. Sumith PeirisOctober 1, 2015 at 1:50 AM

    Applying Prolemy to cyclic quad ABDC we can also that AD = e + f

    ReplyDelete
  8. Benjamin LeisMay 14, 2019 at 5:02 PM

    Yet another way to do it:
    1. ABDC is a cyclic quad so via Ptolemy e + f = AD = d + AE also d.AE = BE.EC
    2. Angle chasing the inscribed arcs shows BDE = CDE = 60. So AD is an angle bisector. Therefore: d^2 = e.f - BE.EC or e.f = d^2 + BE.EC = d^2 + d.AE
    3. Combine 1/e + 1/f = (e + f) / ef = (d + AE)/(d^2 + d.AE) = 1 / d

    ReplyDelete

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