Let $f:[0,1]\to \mathbb R$ be a continuous function such that $f(0)=f(1)$.
Prove that $f(x)=f\left(x+\frac12\right)$ has a solution for $x\in [0,\frac12]$.
This question has to do with continuity and the intermediate value theorem.
I observed that $f(0)=f\left(\frac12\right)=f(1)$ but I don't see how to show that the function go through zero (i.e has a solution) for all we know it can be a straight line parallel to the $x$ axis in $[0,1]$.
Hint
Let $$h:x\mapsto f(x)-f\left(x+\frac 1 2\right)$$
Prove that $$h(0)h\left(\frac12\right)\le0$$ and use the intermediate value theorem.
If $f(0) = f(\frac{1}{2})$ we have done; otherwise let $g(x) = f(x) - f(x+\frac{1}{2})$, $g$ is continous.
Suppose $f(\frac{1}{2}) > f(0)$, then $g(0)<0$ and $g(\frac{1}{2}) = f(\frac{1}{2}) - f(1) = f(\frac{1}{2}) - f(0) > 0 \Rightarrow $ intermediate value theorem and we have done.
If $f(\frac{1}{2}) < f(0)$ the solution is similar.
Hint: put $g(x)=f(x)-f(x+\dfrac{1}{2})$ which is also continous .
Then $g(0)=f(0)-f(\dfrac{1}{2})$ anf $g(\dfrac{1}{2})=f(\dfrac{1}{2})-f(1)=-(f(0)-f(\dfrac{1}{2}))$ so that $g(0)g(\dfrac{1}{2}) <0$
