Tuesday, December 8, 2020

Geometry Problem 1491: Cyclic Quadrilateral, Diagonal, Incircle, Angle, Measurement

Geometry Problem 1491. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

Geometry Problem 1491: Cyclic Quadrilateral, Diagonal, Incircle, Angle, Measurement

5 comments:

  1. I think there is a typo and O1 is the Incentre of Tr. ABD not ABC

    Let <ABD = <ACD = 2p.

    Its not difficult to infer (considering Tr. ABD) that <AO1D = 90+p = <AO2D (considering Tr. ACD)

    Hence AO1O2D is concylic

    Again considering Tr. AO1D, <ADO1 = 180 - (90+p) - 40 = 50 - p = <AO2O1

    So <O1O2D = (50-p) + (90+p) = 140

    Sumith Peiris
    Moratuwa
    Sri Lanka

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    Replies
    1. If we are given the lengths of the 4 sides of ABCD and the length of either diagonal we can easily calculate the length of O1O2 (irrespective of the angle A)

      Delete
  2. Join A to O1. Quadrilateral AO1O2D cyclic
    => Ang O1O2D=140
    See sketch for details

    ReplyDelete
  3. https://photos.app.goo.gl/BezctJeJzUHUpM9y8

    ReplyDelete
  4. https://photos.app.goo.gl/ZYD4fjg9k2jNDHAF9
    Let BO1 and CO2 meet at I
    I will be the midpoint of arc(AD)
    We have isosceles triangles AID, AIO1, and DIO2
    In circle ABCD, ∡ (BID)= ∡ (BAD)= ∡ (DIO1)= 80 deg.
    In circle I, arc(O1AD)= 360-∡ (DIO1)=280 deg.
    So ∡ (O1O2D)= ½ x 280=140 deg.

    ReplyDelete