1. ## 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

2. ## Re: Triangles

Hints:

triangle is equilateral, sides = 1

a = 4 is out of bounds

3. ## Re: Triangles

Originally Posted by socrateszeno
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?

4. ## Re: Triangles

Yes I can solve it . Many Thanks.

5. ## 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.