1. ## triangle inequality 2

|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.

2. ## Re: triangle inequality 2

You have a rhombus with sides of different lengths? I think I am misunderstanding something.

3. ## Re: triangle inequality 2

It is a trapezoid.

4. ## 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.

Many Thanks.

6. ## 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) )

7. ## Re: triangle inequality 2

Originally Posted by socrateszeno
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.

8. ## Re: triangle inequality 2

Originally Posted by SlipEternal
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.