Saturday, June 1, 2019

Geometry Problem 1438: Triangle, Incenter, Centroid, Circumcenter, Parallel, 90 Degree, Art, Poster

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 1438: Triangle, Incenter, Centroid, Circumcenter, Parallel, 90 Degree, Art, Poster, iPad apps, Tutoring.

12 comments:

  1. Extend AI to meet the circle in M, the midpoint of smaller arc BC. Further, Join M to O and produce it to meet the circumcircle in N. Let the in-circle touch AB in L. From M draw a perpendicular to BC meeting the latter in P. Now we can see that Tr. AIL///Tr. NBM; AI*MB=IL*NM=2r*R. We further observe that MP=r since GI//BC and thus r/MB=sin(A/2) as also r/AI=sin(A/2) which implies that AI=MB & therefore AI²=2r*R. We now recall by Euler's Formula : OI²=R² -2r*R or OI²=R²-AI² or AI²+OI²=R² or /_OIA =90° by the converse of Pythagoras Theorem.

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  2. Borrowing from Ajit’s proof

    When we show that MP = r (not clear how this is so) AI = MI = BM

    So I is the mid point of chord AM and so OI is perpendicular to AM

    Ajit - I would like u to clarify how MP = r

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    Replies
    1. Sumith,
      I'm afraid my proof is incomplete. Must figure out a way to prove my assumption. I hope Antonio will help.

      Delete
  3. Adding to Ajit's proof,
    Let touch point of incircle at at AC is T.
    Then AT=s-a=(b+c-a)/2, since b+c=2a, AT=a/2
    AT=BP=a/2,Angle IAT=Angle MBP=A/2, and Angle ATI =Angle BPM=90
    Hence Triangle ATI is congruent to Triangle BPM. And AI=BM=CM=MI,
    Hence I is midpoint of chord AM and Angle AIO=90.

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    Replies
    1. Pradyumna indeed as I have mentioned above b+c = 2a but it must first be proved

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    2. Yes, thanks Pradyumna. AT =a/2; that is the crux of the matter. But where and how have we utilised the fact that IG//BC

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  4. Let AI meet BC at U and the circle at M.Let MO cut BC at P, the midpoint of BC

    Let BM = BI = p (since < MBI = < MIB = (A+B)/2)
    Let AI = 2q so that IU = q since GP = AG/2 and IG//UP

    Now AB/BU = AI/IU = 2q/q = 2 so that BU = c/2

    Triangles MBU & ABM are similar so AM/BM = AB/BU hence (p+2q)/p = c/(c/2) = 2

    Therefore p + 2q = 2p and so p = 2q

    Hence AI = MI and I is the midpoint of AM and OI is perpendicular to AI

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  5. dear Pradyumna Agashe.

    referring to line 3 of your comment, I wonder how do you get b+c= 2a.
    please explain
    Peter Tran

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    Replies

    1. HiPeter,
      Here is proof for b+c=2a
      Let altitude AH to BC is equal to h, If we draw perpendicular to BC from G, then it is equal to (1/3)h, since IG is parallel to BC, perpendicular to BC from I is r=h/3,
      r=S/s
      h/3=(1/2)ah/s, we get 2s=3a or a+b+c=3a
      Or b+c=2a.

      Delete