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.

2 comments:

  1. 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

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  2. 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

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