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Thread: Please help me find two more Equation (to prove that α=β=γ=30)

  1. #11
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    No, JeffM, the problem is perfectly clear:

    Quote Originally Posted by navneet9431 View Post
    The question asks to prove that α=β=γ=30 .
    The goal is to prove that the ONLY triangle that satisfies the conditions in the picture (namely, that the three angles marked as 30 are so) is the equilateral triangle.

    It appears to be a very interesting problem, not a standard geometry problem. In playing with the figure in GeoGebra, I find that it appears to be an extremum problem in disguise, because if two of the angles marked as 30 are so, the other is always less than 30 unless the triangle is equilateral.

    When I follow Jonathan's suggestion, I obtain the following equation: [tex]sin\alpha \cdot sin\beta \cdot sin\gamma = \frac{1}{8}[/tex]. I have not dug into this, but I think 1/8 is the maximum value, and is attained only when the three angles are equal (to 30).

    So, Navneet, can you tell us the source of the problem, and whether a multivariable calculus solution seems appropriate? There may be a geometric inequality that can be applied, but I haven't seen it yet.

  2. #12
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    Quote Originally Posted by tkhunny View Post
    Sorry. Marking all six angles with the same marking MEANS all six angles are congruent. If you don't want it to mean that, you will have to use some other markings.
    No, these little arcs are labeled with measurements or variables. In that case, they just mean "this is the angle I'm referring to".

    You're thinking of marks that stand alone without labels, to indicate congruent pairs by using the same marking. Don't confuse the two usages. (And even if what he is using is not standard on your country, why couldn't it be standard in his? Accept what someone says on his own terms, rather than insist that he is saying nonsense when it is perfectly possible to interpret it to make sense.)

    Incidentally, GeoGebra, which I used to check the problem, marks angles in exactly this way, with an arc together with a label.

  3. #13
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    Quote Originally Posted by Dr.Peterson View Post
    No, JeffM, the problem is perfectly clear:



    The goal is to prove that the ONLY triangle that satisfies the conditions in the picture (namely, that the three angles marked as 30 are so) is the equilateral triangle.

    It appears to be a very interesting problem, not a standard geometry problem. In playing with the figure in GeoGebra, I find that it appears to be an extremum problem in disguise, because if two of the angles marked as 30 are so, the other is always less than 30 unless the triangle is equilateral.

    When I follow Jonathan's suggestion, I obtain the following equation: [tex]sin\alpha \cdot sin\beta \cdot sin\gamma = \frac{1}{8}[/tex]. I have not dug into this, but I think 1/8 is the maximum value, and is attained only when the three angles are equal (to 30).

    So, Navneet, can you tell us the source of the problem, and whether a multivariable calculus solution seems appropriate? There may be a geometric inequality that can be applied, but I haven't seen it yet.
    Ahh, I see.

    The geometric fact that I was looking for is that AK, BK, and CK meet at a common point. Now of course the angle bisectors also meet at a common point, call it L. It may be possible to find a purely geometric proof by contradiction of the assumption that K and L are distinct points. I last studied plane geometry in 1962 and cannot remember enough to attempt a proof by contradiction.

  4. #14
    Quote Originally Posted by Dr.Peterson View Post
    No, JeffM, the problem is perfectly clear:



    The goal is to prove that the ONLY triangle that satisfies the conditions in the picture (namely, that the three angles marked as 30 are so) is the equilateral triangle.

    It appears to be a very interesting problem, not a standard geometry problem. In playing with the figure in GeoGebra, I find that it appears to be an extremum problem in disguise, because if two of the angles marked as 30 are so, the other is always less than 30 unless the triangle is equilateral.

    When I follow Jonathan's suggestion, I obtain the following equation: [tex]sin\alpha \cdot sin\beta \cdot sin\gamma = \frac{1}{8}[/tex]. I have not dug into this, but I think 1/8 is the maximum value, and is attained only when the three angles are equal (to 30).

    So, Navneet, can you tell us the source of the problem, and whether a multivariable calculus solution seems appropriate? There may be a geometric inequality that can be applied, but I haven't seen it yet.
    Thank you Dr.Peterson for making other people here understand this question!
    Actually, i have found this question on this website. The website also provides many solutions of this question. But,i was trying to solve it my own way by discovering at least three different equations in order to get the value of α,β and γ.
    I now realise that it is impossible to solve this question in the way i was trying to solve.

    Thanks!

  5. #15
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    Quote Originally Posted by Dr.Peterson View Post
    No, these little arcs are labeled with measurements or variables. In that case, they just mean "this is the angle I'm referring to".

    You're thinking of marks that stand alone without labels, to indicate congruent pairs by using the same marking. Don't confuse the two usages. (And even if what he is using is not standard on your country, why couldn't it be standard in his? Accept what someone says on his own terms, rather than insist that he is saying nonsense when it is perfectly possible to interpret it to make sense.)

    Incidentally, GeoGebra, which I used to check the problem, marks angles in exactly this way, with an arc together with a label.
    Yes, it is incidental nonstandard notation. It should be explained explicitly if that is the intent. Does GeoGebra make them different colors or use some other method to avoid the standard interpretation as a congruence?
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  6. #16
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    Quote Originally Posted by tkhunny View Post
    Yes, it is incidental nonstandard notation. It should be explained explicitly if that is the intent. Does GeoGebra make them different colors or use some other method to avoid the standard interpretation as a congruence?
    I don't think there is anything nonstandard about it. For another example, see this picture from a Wikipedia page.

    The arcs are all the same style, but the labels distinguish them. GeoGebra likewise labels them; you can change the colors, but they are not so by default, because it is not needed. I see no way to imagine that the three angles here are being called congruent.

  7. #17
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    I like the picture, but it is clear in the picture that they are three entirely different angles. It is not clear at all in the usage seen in this thread.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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