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Circle inscribed in triangle

  1. Mar 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider a triangle ABC, where angle A = 60o. Let O be the inscribed circle of triangle ABC, as shown in the figure. Let D, E and F be the points at which circle O is tangent to the sides AB, BC and CA. And let G be the point of intersection of the line segment AE and the circle O. Set x = AD. Find the ratio of area of triangle ADF and (AG.AE)
    untitled-5.jpg

    2. Relevant equations
    Area of triangle = 1/2 ab sin α
    radius is perpendicular to tangent
    sine formula
    cosine formula

    3. The attempt at a solution
    I don't have any ideas to begin with. The answer in the manual is √3 / 4

    Thanks
     
  2. jcsd
  3. Mar 22, 2012 #2
    What do you mean by (AG.AE)? Do you mean their multiplication? i.e. [itex]\overline{AG} \cdot \overline{AE}[/itex]? In any case, you have enough information to solve for the area of [itex]\triangle ADF[/itex], that should be a good start.
     
  4. Mar 22, 2012 #3
    Yes it is their multiplication.

    Let me try. AD = x and AF also = x, so area of ADF = 1/2 x2 sin 60o = 1/4 √3 x2.

    Then, how to find AG and AE? Thanks
     
  5. Mar 27, 2012 #4
    anyone please?
     
  6. Mar 27, 2012 #5

    NascentOxygen

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    Staff: Mentor

    You might get inspiration if you mark in things you know.

    If you join each of D,F and E all to O you have equal radii of a circle. Furthermore, you are told that D, E and F are points where the circle is tangent to the sides of the triangle. Tangent? What can you mark in that shouts "I'm a tangent here".

    I'm sure that at least some of these will be useful. :wink:
     
  7. Mar 27, 2012 #6
    I'm not sure what you mean but maybe you are referring that OD, OF and OE are perpendicular to theirs respective tangents?

    ADF is equilateral triangle. But I still don't know how to find AG and AE...
     
  8. Mar 28, 2012 #7

    rcgldr

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    Homework Helper

    I think he means that the next step is to find the radius of the circle versus the chord length of DF.
     
  9. Mar 28, 2012 #8
    Oh ok. Let me try again:

    DF = x and by using cosine rule:
    x2 = r2 + r2 - 2.r.r.cos 120o
    x2 = 2r2 + r2
    r = (1 / √3) x

    Then, how to relate the radius to find AE or AG? Thanks
     
  10. Mar 29, 2012 #9

    rcgldr

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    I'm not sure what to do next. Hopefully NascentOxygen can offer another hint.
     
  11. Mar 30, 2012 #10

    NascentOxygen

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    I haven't solved it to completion, but I believe I have all the information needed.

    Focus on ΔEOA. Find OE and OA both in terms of x. You don't know any of the angles in ΔEOA, so nominate one as angle θ.

    With some effort, you should be able to express EA and EG in terms of x and θ.

    Hopefully, there will be complete cancellation of terms involving θ when you find the expression for AG·AE

    Good luck! :wink:
     
  12. Apr 3, 2012 #11

    NascentOxygen

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    I had time to finish this today, so it is solved.
    Label ∠OEG as θ. Express EO and OA in terms of x and θ, and hence determine EA in terms of x and θ.

    In ΔOEG determine EG in terms of x and θ.
    In the expression, the terms involving θ are found to be equal but opposite, so are eliminated. https://www.physicsforums.com/images/icons/icon14.gif [Broken]

    Good luck! :wink:
     
    Last edited by a moderator: May 5, 2017
  13. Apr 8, 2012 #12
    Sorry but how to find OA in term of θ and x? In Δ OEA, OE = r and ∠OEA = θ, then there is no other information ....

    I think I can continue the working if I am able to find OA

    Thanks
     
    Last edited by a moderator: May 5, 2017
  14. Apr 8, 2012 #13

    NascentOxygen

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    All angles of ∆DOA are known, so express OA in terms of x.
     
  15. Apr 8, 2012 #14
    Wow, it actually works. Despite that in the middle of working the equation becomes complicated, in the end the terms cancel out. Thanks for the help :smile:
     
  16. Apr 8, 2012 #15

    NascentOxygen

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    My pleasure. :cool:

    And here's your reward for persistence: bb64i.gif
     
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