Friday, March 6, 2020

Geometry Problem 1457: Altitudes, Circles, Similarity, Product of the Inradii Lengths

Interactive step-by-step animation using GeoGebra. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

Geometry Problem 1457: Altitudes, Circles, Similarity, Product of the Inradii Lengths, iPad.

Posted by Antonio Gutierrez at 6:51 PM
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2 comments:

  1. Sumith PeirisMarch 7, 2020 at 11:35 AM

    Let BC1 = BH = BA1 = p, AC1 = AH = AB1 = q and CA1 = CH = CB1 = u where I have used my proof of Problem 1456

    Then r1/r4 = q/p, r3/r6 =u/q and r5/r2 = p/u

    Multiplying, we get r1.r3.r5 = r2.r4.r6

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  2. Peter TranOctober 22, 2020 at 12:09 PM

    https://photos.app.goo.gl/kU6aZQZw5cy48XQh9
    Note that triangles AHB1 and BA1H are isosceles and similar ( case AA)
    So ratio of inradius of similar triangles = ratio of altitudes => r1/r4= AHb/BHa
    So r1/r4 x r3/r6 x r5/r2= 1 Per Ceva’s theorem

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