Badly formed question? $\|x\|=1=\|y\|$ and $\|x+y\|=\|x\|+\|y\|$, there is a line segment in the unit sphere Show that if a normed linear space $X$ contains linearly independent vectors $x$ and $y$ such that $\|x\|=1$ and $\|y\|=1$ with $\|x+y\|=\|x\|+\|y\|$, then there is a line segment contained in the unit sphere of $X$.

Now since the unit sphere in $X$ is defined as $X=\{a\in X: \|a\|=1\}$, and this means that $x,y$ both lie on the unit sphere of $X$. But $\|x+y\|=2$ and hence they must lie on opposite ends of the sphere, and hence form a straight line through the origin. But then these two vectors aren't linearly independent?
Is this question badly formed or what am I missing?

HINT:
You have $\|\frac{1}{2}x + \frac{1}{2}y\| = 1$. Let's show for instance that
$\|\frac{1}{3}x + \frac{2}{3}y\| = 1$. Note that 
$$\left\|\frac{1}{3}x + \frac{2}{3}y\right\| \le \frac{1}{3}\|x\| + \frac{2}{3}\|y\|=1$$
If we had $\|\frac{1}{3}x + \frac{2}{3}y\| <1$ then 
$$\left\|\frac{1}{2}x + \frac{1}{2}y\,\right\| = \left\|\frac{1}{4} x + \frac{3}{4}\left( \frac{1}{3}x + \frac{2}{3}y\right)\, \right\| \le \frac{1}{4}\left\|x\right\| + \frac{3}{4}\cdot \left\|\frac{1}{3}x + \frac{2}{3}y\,\right\| < 1$$ 
contradiction.
