Monday, April 16, 2012

Problem 741: Scalene Triangle, Incenter, Circumcircle, Midpoint, Arc, Congruence

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 741 details.

Online Geometry Problem 741: Scalene Triangle, Incenter, Circumcircle, Midpoint, Arc, Congruence.

17 comments:

  1. http://img37.imageshack.us/img37/4078/problem741.png
    Since D is the midpoint of arc AC , so DA=DC and ∠ ABD=∠DBC=∠CAD…. ( angles face same arc DC)
    In triangle AID , ∠IAD=∠IAC+∠CAD=∠A/2+∠B/2
    In triangle AIB, external angle ∠AID=∠BAI+∠IBA=∠A/2+∠B/2 => Triangle AID is isosceles
    So DA=DC=DI

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  2. Problem 741 - Corollary:
    ∆AID is equilateral ⇔ ∠C = 60°

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  3. Let /_ACB = C. Now /_ADI = /_ADB = C (angles in the same sector) whereas /_ACI = C/2 simce I is the incentre of Tr. ABC. Thus, AI subtends twice the angle at D than at C which means D is the circumcentre of Tr. AIC or DA = DI = DC

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  4. To Ajit

    In my opinion, the statement "AI subtends twice the angle at D than at C which means D is the circumcentre of Tr. AIC or DA = DI = DC" may not true in general case.
    It will be true in the case DA=DI . if DA=DI , we can draw a circle center D , radius DA=DI . Since angle(ICA)=1/2angle(ADI) => C will be on the circle and DA=DI=DC

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  5. Ref. Problem 741 (A Poser):
    Let D, E, F be the midpoints of the arcs AC, AB, BC respectively
    Show that
    (i)I is the ortho-centre of ∆DEF
    (ii)∠ODI = ∠C - ∠B

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  6. To Peter,
    See Problem #8 - Solution by Nilton Lapa.
    Would this solution, in your opinion, be invalid?
    To Antonio,
    I'd very much like a word by way of an explanation from you.

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    Replies
    1. To Ajit, I don't see the solution by Nilton of problem #8.

      Delete
    2. I meant Problem #9. Sorry abt the typo.

      Delete
  7. Actually, I agree with Peter that just the fact that /_ADI=2*/_ACI is insufficient to prove that D is the circumcentre of triangle ACI but this fact coupled with the fact that D lies on the perpendicular bisector of AC since DA=DC is sufficient to say that D is the circumcentre of triangle ACI and hence etc.

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  8. To Peter and Ajit:
    Looked in another way,
    ∠AIC = ∠A/2 + ∠C/2 + ∠B = 90° + ∠B/2
    reflex∠ADC = 360° - ∠ADC = 180° + (180° - ∠ADC) = 180° + ∠B = 2∠AIC
    Follows D is the centre of the circle through A, I, C

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  9. To Pravin:
    About your suggested problem and questions:
    In my opinion, question (i) "I is the ortho-centre of ∆DEF" is OK but question (ii)
    "(ii)∠ODI = ∠C - ∠B" may not correct. I got ∠ODI = 1/2(∠C - ∠A ).
    Please check it

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  10. You are right Peter! Thank You.
    ∠ODI = ∠ODC - ∠IDC = (90° - ∠B/2) - ∠A
    = ∠A/2 + ∠C/2 - ∠A = (∠C - ∠A) /2
    (or)
    ∠ODI = ∠ODB = ∠OBD = ∠ABD - ∠ABO = ∠B/2 - (90 - ∠C)
    = ∠C + ∠B/2 - ∠A/2 - ∠B/2 - ∠C/2 = (∠C - ∠A) /2

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  11. To Ajit: What is the problem which solution sended by me is (perhaps) invalid? I'd like to review that solution.
    Thanks.

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  12. The solution is uploaded to the following link:

    https://docs.google.com/open?id=0B6XXCq92fLJJenVNQ1NYZWJKSHM

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  13. Video solution http://youtu.be/Asa1SJnv_v0

    Greetings =)

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  14. Hi Alejandro, I was wander why did you put into the problem conditions that D is mid point, indeed this is obviously satisfied by any configuration.

    May be you should restating by puting "prove that: 1- D is mid piont of arc AC, 2- AD=DI=DC.

    MReyes

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  15. Simply < ADC = <C + <A and so < DAC = 90 - 1/2(A+C). But < IAC = A/2 and so IAD = 90 - C/2 = < AID since < ADI = < C.

    So AD = DI = DC

    This solution assumes that DIB is collinear which must be the case since AD = DC and IB bisects < ABC

    As others have pointed out earlier D is hence the circumcentre of Tr. AIC

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete