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anonymous

  • one year ago

Let Γ be the circumcircle of ΔABC and let D be the midpoint of the arc BC. Prove that the incenter I of ΔABC lies on the circle centered at D that goes through B.

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  1. ganeshie8
    • one year ago
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    Do you really mean "lies on the curcumference of circle centered at D" ?

  2. anonymous
    • one year ago
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    Well, lies on the boundary, edge, I guess |dw:1437879668554:dw|

  3. ganeshie8
    • one year ago
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    Because after playing a bit, im getting that " the incenter lies interior to the circle centered at D" Not completely sure though.. still working..

  4. anonymous
    • one year ago
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    I checked it on a program and it seems to work. Ill try and get the screenshot.

  5. ganeshie8
    • one year ago
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    Basically we want to show \(DB=DI=DC\) |dw:1437879935160:dw|

  6. anonymous
    • one year ago
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    Wish I knew how to upload it faster, lol http://i201.photobucket.com/albums/aa120/ApocalypticDeity/6_zpsq2dhiksj.png

  7. ganeshie8
    • one year ago
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    Interesting, do we know if AID lie on a straight line from some theorem or somehting ? |dw:1437880349134:dw|

  8. anonymous
    • one year ago
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    I dont know actually. That picture I posted was the original picture plus me trying out a few things. It looked like they may be collinear, so I tried it and they were. No explanation that I could give, though.

  9. ganeshie8
    • one year ago
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    btw i need to start over, i made a stupid mistake in my earlier attempt

  10. ganeshie8
    • one year ago
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    |dw:1437880661407:dw|

  11. anonymous
    • one year ago
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    Well, we want to show that DI would be the same as DB and DC. I assumed that was the way to try and tackle it, see if I can show that DI is a radii. I didn't find any similarity with the triangles where the two circles overlap. The fact A, I, and D were collinear was really interesting, but even if it was useful Id still have to state why theyre collinear, which I dont know.

  12. ganeshie8
    • one year ago
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    @dan815

  13. dan815
    • one year ago
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    |dw:1437881020931:dw|

  14. dan815
    • one year ago
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    |dw:1437881129726:dw|

  15. dan815
    • one year ago
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    |dw:1437881183330:dw|

  16. dan815
    • one year ago
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    equaliteral triangle so we are gone i think

  17. dan815
    • one year ago
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    done

  18. dan815
    • one year ago
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    oh wait that assumes the incenter of the triangle is the center of the circumcircle too which is not true is it?

  19. ganeshie8
    • one year ago
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    I is incenter dear, not circumcenter

  20. anonymous
    • one year ago
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    Yeah, I is incenter and I would mark O as circumcenter. The picture I posted has the circumcenter on it for reference if you think it may be helpful.

  21. dan815
    • one year ago
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    |dw:1437881398303:dw|

  22. anonymous
    • one year ago
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    No idea if the circumcenter would help. I don't think there's a way to set up an inscribed angle theorem. I don't know if there is a more proper name for the theorem, but I was also trying to see if I could use power of the point. I know we can relate area to the inradius or use extended law of sines to have a relation to the radius of the circumcircle, but I don't think any of that is useful, lol. Just random connections and theorems and such.

  23. ganeshie8
    • one year ago
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    Gotcha!

  24. ganeshie8
    • one year ago
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    cyclic quadrilateral does the job !

  25. anonymous
    • one year ago
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    I was considering that also, but is it cyclic?

  26. ganeshie8
    • one year ago
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    |dw:1437881873780:dw|

  27. anonymous
    • one year ago
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    Oh, different quadrilateral than what I was looking at, lol.

  28. ganeshie8
    • one year ago
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    |dw:1437881954866:dw|

  29. anonymous
    • one year ago
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    That angle is arbitrarily inscribed on that circle or is it in a specific spot?

  30. ganeshie8
    • one year ago
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    bottom is an arbitrarily inscribed angle top is fixed, incenter

  31. dan815
    • one year ago
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    ok got it

  32. anonymous
    • one year ago
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    Okay, that definitely is cyclic then :P And yeah, that is just the circumcircle, I is not the center of it. This was the picture from my program, dan. The actual center of the top circle I have marked as O. It's kind of hidden, though http://i201.photobucket.com/albums/aa120/ApocalypticDeity/6_zpsq2dhiksj.png

  33. ganeshie8
    • one year ago
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    That proves BIC lie on the same circle since 3 points uniquely determine a circle, it must be the case that DB=DI=DC

  34. anonymous
    • one year ago
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    Are we just observing that the one angle subtends an arc have the measure that angle A does?

  35. ganeshie8
    • one year ago
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    Hmm didn't get you

  36. anonymous
    • one year ago
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    Well, in the labeled picture we have angles marked as A/2, B/2, C/2. I was trying to make sure how we knew those were half the measures of angles A, B, and C.

  37. ganeshie8
    • one year ago
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    Ahh okay, let me put the complete proof, it shouldn't be long

  38. anonymous
    • one year ago
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    B/2 and C/2 are from the angle bisectors, but A/2 wasnt as direct. I assumed it was the same logic, but wanted to confirm.

  39. ganeshie8
    • one year ago
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    \(\angle DBC\) is the inscribed angle of arc \(DC\), by inscribed angle formula we have \(\angle DBC = \dfrac{\text{arc}DC}{2} = \dfrac{A}{2}\) |dw:1437882784370:dw|

  40. ganeshie8
    • one year ago
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    Also, it is important that we don't assume \(I\) is on the circle in the start

  41. anonymous
    • one year ago
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    Right :)

  42. ganeshie8
    • one year ago
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    so that in the end we can use cyclic thingy to conclude that \(I\) lies on the circle that passes through remaining 3 vertices of the quadrilateral

  43. ganeshie8
    • one year ago
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    let me know if it is not clear enough, il try writing the complete proof in one reply..

  44. anonymous
    • one year ago
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    Will do.

  45. ganeshie8
    • one year ago
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    wonder if there is an easy way to prove AID are colliner |dw:1437884264926:dw|

  46. anonymous
    • one year ago
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    Whoa, how'd you make that?

  47. ganeshie8
    • one year ago
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    familiar with geogebra ?

  48. anonymous
    • one year ago
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    That's what I used to make my picture, but today is the first day I've ever messed with it, so no idea of its capabilities yet

  49. ganeshie8
    • one year ago
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    its pretty good and very easy to learn

  50. ganeshie8
    • one year ago
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    let me upload the file to geogebratube so that you can play with it

  51. anonymous
    • one year ago
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    Alright, cool, thanks :)

  52. ganeshie8
    • one year ago
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    Here it is http://tube.geogebra.org/m/1437801

  53. ganeshie8
    • one year ago
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    that is very much like a geogebra session, you can do everything that you can do with geogebra on ur computer.. all commands work just fine

  54. anonymous
    • one year ago
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    Gotcha. I have to get familiar with Geometer's Sketchpad really soon, too. It's the required software for my course when we get to that part of the class. But yeah, I think I found a way to show the quadrilateal is cyclic, but not sure it's the best way.

  55. ganeshie8
    • one year ago
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    if psble pls share, id like to see

  56. anonymous
    • one year ago
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    Since \(BI\) and \(CI\) are the angle bisectors of B and C, \(\angle IBC = B/2\) and \(\angle ICB = C/2\). The sum of the angles of a triangle sum to 180 degrees, thus \(\angle BIC = 180° - \frac{B + C}{2}\). \(OB\) and \(OC\) are both radii and thus have the same measure. This implies \(\angle OCB = \angle OBC\). By the inscribed angle theorem, \(\angle OCB = \angle OCB = A/2\). Furthermore, \(\angle BOC = 180° - A\). Draw the lines BE and CE where E is an arbitrary point on the circle of radius D and on the arc counterclockwise from B to C (Or however you would phrase that). By inscribed angle theorem, \(\angle BEC = \angle BOC/2 = 90° - A/2\). From here, \(\angle BEC + \angle BIC = 180° - \frac{B + C}{2} + 90° - A/2 = 270° - \frac{A + B + C}{2} = 180°\). Since opposite angles of a quadrilateral sum to 180°, quadrilateral IBEC is cyclic. Bleh, seems chunky, but idk, lol.

  57. ganeshie8
    • one year ago
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    looks okay to me!

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