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Centroid and Triangles

  1. Aug 28, 2007 #1
    I found this problem off of mathematics magazine and I want to give it a try solving it, but i'm lost for ideas.

    The problem states the following: Let G be the centroid of triangle ABC. Prove that if angle BAC = 60 degrees and angle BGC = 120 degrees then the triange is equilateral.

    My idea was to use the medians to find G and find the angles that way. I then quickly realized I assumed to many things, so now i'm asking for new outlooks on this problem. Perhaps i've been staring at it for far to long, but i'm stump. Any ideas?
    Last edited: Aug 28, 2007
  2. jcsd
  3. Aug 28, 2007 #2


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    Hint: (because solving it yourself is more fun than being told the answer!)

    Forget about G being the centroid for a while.

    Just think about the points B and C, and the two angles BAC amd BGC. One angle is twice the other one.

    Does that remind you of any theorems? It if does, you can draw something else on the figure that involves all the points A B C G, and gives you a lot more information to work with...
  4. Aug 28, 2007 #3
    aye, I wasn't asking for the answer, just some hints, sorry for the misunderstanding. Yes it does remind me of a certain theorem. I'll go ahead and see what I find. Thanks for the hint.
  5. Aug 28, 2007 #4
    What? Any triangle with a 60 degrees angle respects that condition!
  6. Aug 28, 2007 #5
    So it seems, but i'm trying to prove that it does.

    So far I drew an equilateral triangle and showed that bgc would have to be 120 degrees. However, i'm not sure if what I did is proof.
  7. Aug 28, 2007 #6
    The centroid of a triangle is the intersection of its angles bisectors. This said, the sum of the angles BCG and GBC is half the sum of the angles BCA and CBA (if you represent this with a drawing, it will become obvious). The sum of BCG and GBC is obviously 60 degrees, from that it should be clear that the sum of BCA and CBA is 120 degrees, which when added to BAC is 180 degrees. Had BGC not been 120 degrees, the angles of the triangle ABC wouldn't add up to 180.
  8. Aug 29, 2007 #7


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    No, the centroid is the intersection of the medians.

    But passing over that mistake, if you are talking about the intersection of the angle bisectors, then

    No. Suppose ABC = 60, ACB = 90, BAC = 30

    BCG = 90/2 = 45
    GBC = 60/2 = 30
    BCG + GBC isn't equal to 60.

    I think you assumed what you are trying to prove (i.e. that the triangle is equilateral). For an equliateral triangle the medians DO bisect the angles, of course.

    To give the OP another hint: draw ANY triangle ABC, and draw the circle through A,B,C. If angle BAC is twice angle BGC, where is point G?

    But you also know that G is the centroid....
    Last edited: Aug 29, 2007
  9. Aug 29, 2007 #8
    so I drew this (refer to picture)
    I see that BAC is half of BGC not twice of it. Perhaps I drew a wrong picture. I'm really bad at proving things with geometry, but that's why I picked this problem. Work on your weak points.

    Attached Files:

  10. Aug 29, 2007 #9
    Just thinking about it for a second and I reminder something. If a triangle is circumscribed around by a circle, you can find R, radius by R= abc/4k where a b and c are sides and k is the area. So the circumcenter has to bisect the sides.

    Is that helpful to me?
  11. Aug 29, 2007 #10
    Yes but isn't 30/2 + 90/2 = 60? :wink: Anyway, sorry for inducing PowerIso into error.
  12. Aug 29, 2007 #11


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    Sorry, my bad. You are right BAC is half BGC.

    I'm not too sure what you mean by "circumcenter has to bisect the sides".

    The one thing you haven't actually posted here is "G is the center of the circle" - but I think you know why it is the centre! (The angle in subtended in an arc of a circle is half the angle at the centre).

    Call the 3 medians AD, BE, CF.

    AF = FB, BD = DC, CE = EA because the medians bisect the sides.

    Also GA = GB = GC because G is the centre of the circle.

    So there are lots of similar triangles in the figure.

    To prove ABC is equilateral, you can either prove all the sides are equal, or all the angles are equal. There are lots of ways to prove either of those, using similar triangles.

    Or, you could prove AD bisects angle BAC (using similar triangles again), then use the fact that BAC = 60 to find the other angles in the figure and show angles ABC = 60 and ACB = 60.
  13. Aug 29, 2007 #12
    Sorry I made an error in my math. I'll repost once I figure it out.
    Last edited: Aug 29, 2007
  14. Aug 29, 2007 #13
    oh wow, I'm not sure how I could've missed this. Perhaps it's all the stress I have. I drew the diagram and thought about it for a while, but couldn't figure it out. So, I went on a walk and on the walk, I figured it out. Thank you so much. Since it's kind of hard to show the proof on here i'll delay until I have it typed up on a document.
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