Thursday, April 14, 2016

Geometry Problem 1207: Triangle, Circle, Incenter, Circumcenter, Excenter, Circumradius, Perpendicular

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


Click the figure below to view more details of problem 1207.

Geometry Problem 1207: Triangle, Circle, Incenter, Circumcenter, Excenter, Circumradius, Perpendicular

4 comments:

  1. http://s22.postimg.org/nia3yt2fl/pro_1207.png

    Let L, M, N , P are points as shown on the sketch
    Observe that AL= NC= half perimeter of triangle ABC- BC
    P is the midpoint of arc AC => M is the midpoint of AC and LN
    In trapezoid ILND , MO is the mid-base => O is the midpoint of ID
    Triangle IOP similar to triangle IDE ..( case AA)
    Since O is the midpoint of ID so DE= 2 x OP= 2.R

    ReplyDelete
  2. Lets assume touch point of incircle and excircle are F and G, and Midpoint of AC is M. It is easy to see that FM=GM=(a-c)/2 Also IF,OM and DG are parallel to each other ( all are perpendicular to AC), hence O is midpoint of ID.
    B, I and E are collinear, join BE and let it intersects circumcircle at point H, and IH=HE, thus H is midpoint of IE.
    Consider triangle IED, O is midpoint of ID and H is midpoint of IE,
    Hence DE=2*OH, since OH=R, DE=2R.

    ReplyDelete
  3. Join BIE. Let it cut circle(O)at M. Angle ECI is a right angle.
    M is the midpoint of arc AMC. So OM bisects AC at right angles.
    We are done if we can show O is the midpoint of ID.
    Let X, Y, Z be the projections of I. O. D on AC respectively.
    It is easy to see that XY = b/2 - (s -a) = (c -a)/2 = YZ.
    IX, OY, DZ being //, OYM is a midline // to DE in Triangle IDE.
    Hence DE = 2 OM = 2R.
    N Vijaya Prasad
    Rajahmundry - INDIA.

    ReplyDelete
  4. Bring IK⊥AC,MN⊥AC, EL⊥AC(M is midpoint arc AC).Is <MIC=<ABC/2+<ACB/2 , <MCI=<MCA+<ACI=<ABC/2+<ACB/2.Therefore <MIC=<MCI.So MI=MC.But IC⊥CE.
    Then <MIC+<MEC=90 and <MCI+<MCE =90 therefore MI=MC=ME.So MO//ED
    Is IO=OD. Therefore ED=//2OM=2R.

    ReplyDelete