Sunday, November 1, 2015

Geometry Problem 1161: Scalene Triangle, 60 Degrees, Angle, Orthocenter, Circumcenter, Equilateral Triangle, Congruence

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below for more details.

Online Math: Geometry Problem 1161: Scalene Triangle, 60 Degrees, Angle, Orthocenter, Circumcenter, Equilateral Triangle, Congruence.

5 comments:

  1. Let AO meet the circumcircle at N
    1)
    Then < HCB = 30 = < CBN since AN is a diameter. So HC//BN

    Also < BAN = 90-C = < BCN = < CBH so BH //NC

    So BHCN is a parellelogram

    Now Tr. ONC is equilateral so NC = BH = BO = R where R is the circumradius

    2) Easily < CAN = 30 = <ACO hence < AOC = 120 = AHC. So AOHC is cyclic and so < AHD = 30.

    So < BDE = 60 and hence Tr. BDE is equilateral

    3) This implies that Tr.s BDH and BOE are congruent ASA hence DH = OE and so DO = HE

    4) Let BD = DE = BE = p and if AH meets BC at M let ME = q and let CH meet AB at S

    BS = a/2 since Tr. BCS is 30-60-90. So DP = p - a/2 = DH/2 hence DH = 2p -a and so HE = p - (2p - a) = a - p = DO

    Therefore OH = 3p - 2a

    Now q = HE / 2 = (a-p)/2 and BM = c/2 = p - q. So c/2 = p - (a-p)/2 and hence 3p = c + a. (1)

    So AD + CE = c-p + a-p = p from (1 ) above and is therefore = to BD

    5) AB -BC = c-a = (c-p) - (a-p) = AD - CE = a + c -2a = 3p - 2a = OH

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  2. Replies
    1. Ajit sir I need help for the following problem
      http://www.artofproblemsolving.com/community/u271471h1157160p5494288
      plz reply me at hrishiashudesh@gmail.com

      Delete
    2. I've solved the problem using physics.

      Delete
  3. GG Problem 1161.
    Let X be the midpoint of AC.
    BH = 2OX = 2R cos B = R = BO.
    So <BOH = <BHO ......(i).
    Extend BH to cut AC at K and circle at L.
    Extend BO to cut the circle at M.
    Connect CM.
    <ABM = 90 - <AMB = 90 - <ACB = 90 – C = <KBC.
    (i.e)<DBO = < EBH ....(ii).
    Subtract (ii) from (i) to get
    <BDE = <BED = 60 and ΔBDE is equilateral.
    Note that ΔBDO and ΔBEH are congruent and
    ADH is a 30-120-30 triangle.
    So AD = DH.
    Similarly EC = EH.
    Hence AD + CE = DH + EH = DE = DB




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