# Show that if f is continuous on [0,1] with f(0)=f(1), there must exist x,y in [0,1] with |x−y|=1/2 and f(x)=f(y) I've been working on this for a while, and can't seem to figure out where to start. Any hints would be appreciated

Show that if $f$ is continuous on [0,1] with $f\left(0\right)=f\left(1\right)$, there must exist $x,y\in \left[0,1\right]$ with $|x-y|=\frac{1}{2}$ and $f\left(x\right)=f\left(y\right)$
I've been working on this for a while, and can't seem to figure out where to start. Any hints would be appreciated
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farbhas3t
Let $g$ be the function defined at $\left[0,\frac{1}{2}\right]$ by
$g:t↦f\left(t\right)-f\left(t+\frac{1}{2}\right)$
we have
$g$g is continuous at $\left[0,\frac{1}{2}\right]$
and
$g\left(0\right).g\left(\frac{1}{2}\right)=-\left(f\left(0\right)-f\left(\frac{1}{2}\right){\right)}^{2}\le 0$ since $f\left(0\right)=f\left(1\right)$.
then
$\mathrm{\exists }x\in \left[0,\frac{1}{2}\right]\phantom{\rule{thickmathspace}{0ex}}$ such that $g\left(x\right)=0$ or
$f\left(x\right)=f\left(x+\frac{1}{2}\right)=f\left(y\right)$
with $y=x+\frac{1}{2}$ satisfying
$|y-x|=\frac{1}{2}$