Saturday, April 4, 2015

Problem 1107: Triangle, Circumcircle, Circle, Radius, Tangent, Chord, Congruence

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

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1107: Triangle, Circumcircle, Circle, Radius, Tangent, Chord, Congruence.

Posted by Antonio Gutierrez at 6:03 AM
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8 comments:

  1. Peter TranApril 4, 2015 at 7:44 PM

    http://s25.postimg.org/ufbka0myn/pro_1107.png

    Let circle D cut AB at E and F
    Since DC^2=DA.DB => Circle D is Apollonius circle of AB
    And PE and CE are bisectors of angles APB and ACB
    So PA/PB=EA/EB=CA/CB (1)
    Triangles PBC similar to PC1B1 => B1C1= (PC1/PB). BC…..(2)
    Triangles PAC similar to PC1A1=> C1A1= (PC1/PA). CA …..(3)
    Divide (2) to (3) and using (1) we get B1C1= C1A1

    ReplyDelete
  2. Sumith PeirisSeptember 28, 2016 at 4:18 AM

    Let AB cut circle D at Q.
    Let < BCD = p, <BPD = q, < QPC = r and < QCP = s

    Then <CAD = p and < PAD = q since DP (=DC) is a tangent to circle APB at P.
    Further < PDA = 2s and < CDA = 2r so that
    < PQD = < QPD = 90-s …..(1) and
    < QCD = CQD = 90-r…….(2)

    From (2) < ACQ = 90-p-r …..(3)
    From (1) < APQ = 90-q-s ……(4)

    Now < ABB1 = 2s+q & …..(5)
    And < CBA = 2r+p ……(6)

    Further < PCB = 90-p-r-s from which
    < A1B1C1 = 90-p-q-r-s …..(7) since < PCB = < PB1C1 and < BB1A = BAA1 = q

    Also < B1A1C1 = < PA1C1 - <PA1B1 = < ACC1 - < ABB1 = (90-p-r+s) – (2s+q) =
    90-p-q-r-s…(8)

    From (7) and (8), < A1B1C1 = < B1A1C1 and therefore C1A1 = C1B1

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. Ivan BazarovDecember 17, 2016 at 5:30 PM

    Dear Antonio Gutierrez,
    I noticed that you deleted my solution in Russian to this problem. Will it be acceptable in the future
    if I post side by side English-Russian translations of my solutions so that Russian readers can understand them?

    ReplyDelete
    Replies
    1. Antonio GutierrezDecember 17, 2016 at 6:04 PM

      Thanks Ivan,
      Problem 1107: Could you post your solution in Russian again? I will publish it

      Thanks again

      Delete
    2. Ivan BazarovDecember 17, 2016 at 6:07 PM

      This comment has been removed by the author.

      Delete
  4. Barth StewarsonFebruary 18, 2017 at 12:32 PM

    Here is my solution, though rushed you should be able to deduct from my reasoning, and if necessary I may repost the solution in case the wording is difficult to understand. http://prnt.sc/eag3ym

    ReplyDelete
  5. Peter TranFebruary 18, 2017 at 7:20 PM

    Dear Barth
    Referring to line 3 of your solution " from triangle A1B1C angle A1 is 90 degrees... "

    Per the problem statement , P is any point on circle center D, radius DC. It may exist a particular point P so that C, O and B1 are collinear, but in general these 3 points are not collinear and angle A1 in triangle A1B1C is not equal 90 degrees. please provide explanation.

    Peter Tran

    ReplyDelete
  6. Ivan BazarovJanuary 12, 2018 at 6:14 AM

    1. PD^2=DC^2=DB*DA, so triangles DPB and DAP are similar so <DPB=<DAP=e.
    2. Circle D cuts AB at E. Call angles in triangle ABC a,b,c respectively.
    <ACD=c+a, so <ADC=180-2a-c so <ECD=a+c/2 so <ECB=<ECD-<BCD=(a+c/2)-(a)=c/2.
    Call <ECP=d. <ACB1=<B1BA=<PDB+<DPB=2d (since <ECP subtends arc EP of circle D)+e.
    <B1CC1=<ACC1-<ACB1=(c/2+d)-(2d+e)=c/2-d-e.
    <A1CC1=<BCC1-<BCA1=(<BCE-d)-(<BAA1)=(c/2-d)-(e)=c/2-d-e.
    From last 2 lines, C1A1=C1B1.

    ReplyDelete

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