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.

7 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

    ReplyDelete
  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




    ReplyDelete
  4. This solution is based on identifying AOHC as concyclic points and the rest is proved by angle tracing

    Let the foot of the altitudes from A,B & C be denoted as P,Q&R respectively. ABP & CBR are 30-60-90 triangles
    => m(PHC)=60 ----------(1)
    and hence m(AHC)=m(AOC)=120 => AOHC are concyclic -----(2)
    Connect AO & OC and consider the isosceles triangle AOC (Since O is the circumcenter), we have m(OAC)=m(OCA)=30 -------(3)
    From (2) & (3), we have m(OHA)=30
    From (1), we have m(PHC)=m(RHA)=60 implying DH is the angle bisector of m(RHA) and triangle ADH is a 30-120-30 isosceles triangle (AD=DH), similaly HE=EC-----------(4)

    Hence m(HDR)=60 => BDE is equilateral

    Now let m(OAB)=m(OBA)=x => m(OBC)=60-x --------(5)
    and m(OAH)=30-x and from (2) we have m(OCH)=30-x
    Hence m(C) = m(OCA)+m(OCH)+m(HCP)=30+30-x+30=90-x---------(6)
    From right triangle BQC, we have m(CBQ)=x----------(7)
    => m(OBH)=60-2x ---------(8)
    In the triangle OEB, m(OEB)=60=> m(BOE)=60+x and from (8), we have m(BHO)=60+x

    Hence BOH is isosceles => BO=BH

    Since BD=BE, m(DBO)=m(HBE)=x (from (7)) and BO=BH, the triangles DBO and EBH are congruent and hence DO=HE, similarly the triangles BDH and BEO are congruent and hence DH=OE ----------(9)

    As BDE is equilateral, BD=DE=DH+HE=AD+CE (From (4))

    From (9), we have OE=DH
    => OH+HE=AD (from (4))
    => OH=AD-CE (From (4))

    Also AB=BD+DA
    => AB=BE+DH
    => AB=BE+DO+OH
    => AB=(BE+EC)+OH
    => OH=AB-BC

    Hence OH=AB-BC=AD-CE

    ReplyDelete
  5. Solution 2

    Let AH meet BC at X and CH meet AB at Y

    Since < BAX = 30, < BHC = 120 = < AOC
    Hence AOHC is concyclic.
    Since <OAC = 30, <YHD = 30 and so <BDE = 60 and Tr.BDE is equilateral.

    Further since <DBO = 90-C = <EBH, Tr.s BDO and BEH are congruent ASA
    So DO = HE

    Now ADH and CEH are both 30-30-120 Tr.s with AD = DH and CE = HE
    So BD = DE = DH + HE = AD + CE

    Also OH =DH = DO = AD - HE = AD -CE
    Also OH = AD - CE = (AD + BD) -(CE + BE) = AB -BC

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete