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Thread: Triangles

  1. #1
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    Triangles

    I couldn't manage to solve the question below:
    ABC is a triangle
    |AB|=1 units
    |BC|=a units
    |CA|=SQUAREroot(a)
    How many integer values can 6a assume?

    I tried this:
    square root(a)=t
    there are 2 options t>1 or 0<t<1
    and I found t as (sqrroot(5)-1)/2<t<(sqroot(5)+1)/2
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  2. #2
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    Re: Triangles

    Hints:

    triangle is equilateral, sides = 1

    a = 4 is out of bounds
    Thanks from topsquark
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  3. #3
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    Re: Triangles

    Quote Originally Posted by socrateszeno View Post
    I couldn't manage to solve the question below:
    ABC is a triangle
    |AB|=1 units
    |BC|=a units
    |CA|=SQUAREroot(a)
    How many integer values can 6a assume?
    Using the rule of cosines:
    $ \begin{align*}|BC|^2&=|CA|^2+|BA|^2-2|CA||BA|\cos(A) \\a^2&=(\sqrt a)^2+1^2-2\sqrt a\cdot 1\cos(A)\\\cos(A)&=\dfrac{a^2-a-1}{2\sqrt a} \end{align*}$ NOW $ \begin{align*}-1&\le\cos(A)\le 1 \\-1&\le\dfrac{a^2-a-1}{2\sqrt a}\le 1 \end{align*}$

    Can you solve now?
    Thanks from topsquark
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  4. #4
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    Re: Triangles

    Yes I can solve it . Many Thanks.
    Last edited by socrateszeno; Oct 24th 2017 at 12:09 PM.
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  5. #5
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    Re: Triangles

    I found the answer as 13 but I used this equation:a+sqrroota>1
    I took a as x^2 and found the answer but I couldn't solve the equation above.
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