Thursday, October 20, 2016

Geometry Problem 1274 Isosceles Triangle, 80-20-80 Degrees, Metric Relations, Measurement

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

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click on the figure below to view more details of problem 1274.

Geometry Problem 1274: Isosceles Triangle, 80-20-80 Degrees, Metric Relations, Measurement

Posted by Antonio Gutierrez at 6:54 AM
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5 comments:

  1. Pradyumna AgasheOctober 20, 2016 at 12:32 PM

    Consider point D on segment BC such that AD=b. Tr. ACD is similar to Tr. ABC, we have CD=b^2/a
    BD=a-(b^2/a)
    In triangle BAD, Angle BAD=60. By cosine law a^2+b^2-ab =BD^2
    We have
    a^2+b^2-ab = [a-(b^2/a)]^2
    On simplifying we get the desired result.
    Trigonometric Solution.
    sin(10)=b/2a,
    sin (30)=1/2=3sin(10)-4sin(10)^3, replacing sin(10) by b/2a, we get the required expression.

    ReplyDelete
    Replies
    1. UnknownOctober 22, 2016 at 3:36 PM

      If you let AD = b, then D = C, and you cannot form triangle ACD in your assumption.

      Delete
    2. Peter TranOctober 22, 2016 at 7:10 PM

      In my opinion, Pradyumna has an excellent solution for this problem.
      See below for the link to the sketch of his solution.

      https://goo.gl/photos/oi63XrYgLS88ZzocA

      Peter Tran

      Delete
    3. Pradyumna AgasheOctober 23, 2016 at 6:19 AM

      Thanks Peter for your kind words and diagram.

      -Pradyumna

      Delete
    4. Sumith PeirisOctober 24, 2016 at 5:25 AM

      Agree with Peter re your solution Pradyumna which was the one I used too.
      Like your trigonometry solution too.

      However there are other ways this could be done

      Delete

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