Friday, October 5, 2018

Geometry Problem 1391: Two Triangles, Ratio of Areas, Circumcircle

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

Details: Click on the figure below.

Geometry Problem 1391: Two Triangles, Ratio of Areas, Circumcircle, Tutoring.

Posted by Antonio Gutierrez at 2:21 PM
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2 comments:

  1. Alexey VorobyovOctober 5, 2018 at 3:32 PM

    According to the Law of Sines (https://en.wikipedia.org/wiki/Law_of_sines) a/sinA = 2R, so, S1 = 1/2*b*c*sinA = 1/2*a*b*c/2R = (a*b*c)/(4*R). Similarly S2 = (d*e*f)/(4*R) and S1/S2 = (a*b*c)/(d*e*f), Q.E.D.

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  2. Sumith PeirisOctober 6, 2018 at 10:44 AM

    Let h be the height of the perpendicular from A to BC.

    If R is the circumradius, from similar triangles,
    (c/2)/R = h/b and S1 = ha/2
    Eliminating h, S1 = abc/4R
    Similarly S2 = def/4R

    Dividing, S1/S2 = abc/def

    Sumith Peiris
    Moratuwa
    Sri Lanka

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