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Thread: triangle inequality 2

  1. #1
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    triangle inequality 2

    ABCD is a trapezoid. AD//BC.
    |AD|=3 units
    |DC|=5 units
    m(D)>2m(B)
    Find the smallest integer value of |BC|.

    I drew a parallel AK to DC.
    |BK|=a and |AB|=y and |AK|=5.

    There are 2 choices y<5 or y>5

    If I take y as 1 then BK>4 THEN | BC| can be 7. But the answer is 9.
    Last edited by kastamonu; Oct 17th 2017 at 12:30 PM.
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  2. #2
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    Re: triangle inequality 2

    You have a rhombus with sides of different lengths? I think I am misunderstanding something.
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  3. #3
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    Re: triangle inequality 2

    It is a trapezoid.
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  4. #4
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    Re: triangle inequality 2

    If $m(D)>2m(B)$ then consider if $m(D) = 2m(B) = 2\theta$. This gives the angles $m(A) = m(D) = 2\theta$ and $m(B)=m(C) = \theta$, so $\theta+\theta+2\theta+2\theta = 6\theta = 2\pi=360^\circ$. Thus, $\theta = \dfrac{\pi}{3} = 60^\circ$. Now, you know that $2m(B) < 2\theta$ so $m(B) < \theta$. You should know the sides of a 30-60-90 triangle are $a, a\sqrt{3}, 2a$. So, since the hypotenuse is 5, we know $|CK|>\dfrac{5}{2}$. So, $|BC| > \dfrac{5}{2} + 3 + \dfrac{5}{2} = 8$. Therefore, the smallest integer value for $|BC|$ would be 9.
    Last edited by SlipEternal; Oct 17th 2017 at 01:48 PM.
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    Re: triangle inequality 2

    Many Thanks.
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    Re: triangle inequality 2

    1. m(D) < 2*m(B) (given)
    2. m(D) = m(AKC) (opposite angles in a parallelogram)
    3. m(AKC) = m(KAB) + m(B)
    (exterior angle = sum of other 2 interior angles)
    4. m(KAB) + m(B) < 2*m(B) ([1], [2], [3])
    5)m(B)<m(KAB)
    6)BK>5 (because of 5) )
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  7. #7
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    Re: triangle inequality 2

    Quote Originally Posted by socrateszeno View Post
    1. m(D) < 2*m(B) (given)
    2. m(D) = m(AKC) (opposite angles in a parallelogram)
    The given is a trapezoid not a parallelogram.
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  8. #8
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    Re: triangle inequality 2

    Quote Originally Posted by SlipEternal View Post
    If $m(D)>2m(B)$ then consider if $m(D) = 2m(B) = 2\theta$. This gives the angles $m(A) = m(D) = 2\theta$ and $m(B)=m(C) = \theta$, so $\theta+\theta+2\theta+2\theta = 6\theta = 2\pi=360^\circ$. Thus, $\theta = \dfrac{\pi}{3} = 60^\circ$. Now, you know that $2m(B) < 2\theta$ so $m(B) < \theta$. You should know the sides of a 30-60-90 triangle are $a, a\sqrt{3}, 2a$. So, since the hypotenuse is 5, we know $|CK|>\dfrac{5}{2}$. So, $|BC| > \dfrac{5}{2} + 3 + \dfrac{5}{2} = 8$. Therefore, the smallest integer value for $|BC|$ would be 9.
    My suggestion is not actually correct. I do not know that $m(A) = m(D)$ or $m(B)=m(C)$. Instead, I only know that $m(A)+m(B) = 180^\circ$ and $m(C)+m(D) = 180^\circ$. It is easy enough to show that $m(A)>90^\circ$ and $m(C)<90^\circ$. Then, since $180^\circ = m(C) +m(D) > m(C)+2m(B)$ implies $180^\circ -2m(B) > m(C)$. The length of $|BC|$ is minimized when $m(B)$ and $m(C)$ are maximized. Obviously, the bigger we let $m(B)$ be, the smaller $m(C)$ becomes. Similarly, the large we allow $m(C)$ to be, the smaller $m(B)$ becomes. What remains is to show that $|BC|$ is maximized when $m(B) = m(C)$, which yields my suggestion above. I would need to think about that.
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