TITLE: Intersection of three period functions
QUESTION [2 upvotes]: Let $f(x)=\frac{1}{2}-|\frac{\sqrt{3}}{2}x-1/2|$ for $x\in [0,\frac{2}{\sqrt{3}}]$ and $f(x+\frac{2}{\sqrt{3}})=f(x)$ for all $x\in\mathbb{R}$.
$g(x)=\frac{1}{2}-|\frac{1}{2}x-1/2|$ for $x\in [0,2]$ and $g(x+2)=g(x)$ for all $x\in\mathbb{R}$.
$h(x)=\frac{1}{2}-|\sqrt{2}x-1/2|$ for $x\in [0,\frac{1}{\sqrt{2}}]$ and $f(x+\frac{1}{\sqrt{2}})=f(x)$ for all $x\in\mathbb{R}$.
Are there real $t>0$ such that $f(t)=g(t)=h(t)$ ?
If it true, find the minimum value of $t$.
Thanks in advance.

REPLY [1 votes]: Let's assume there exists such $t>0$. Then for some $p,q,r\in\mathbb N$,
$$t-\frac2{\sqrt3} p\in\left[0,\frac2{\sqrt3}\right],\quad t-2q\in\left[0,2\right],\quad t-\frac1{\sqrt2}q\in\left[0,\frac1{\sqrt2}\right]$$
then
$$f\left(t-\frac2{\sqrt3} p\right)=g\left(t-2q\right)=h\left(t-\frac1{\sqrt2}r\right)$$
And
\begin{align}
f(x)=\frac{\sqrt3}2x\quad \text{or} \quad 1-\frac{\sqrt3}2x&\rightarrow f\left(t-\frac2{\sqrt3} p\right)=\frac{\sqrt3}2t-p\quad \text{or} \quad1-\frac{\sqrt3}2t+p\\
g(x)=\frac12x\quad \text{or} \quad 1-\frac12x&\rightarrow g\left(t-2q\right)=\frac12t-q\quad \text{or} \quad1-\frac12t+q\\
h(x)=\sqrt2x\quad \text{or} \quad 1-\sqrt2x&\rightarrow h\left(t-\frac1{\sqrt2}r\right)=\sqrt2t-r\quad \text{or} \quad1-\sqrt2t+r
\end{align}
From $f\left(t-\frac2{\sqrt3} p\right)=g\left(t-2q\right)$,
$$t=\frac{2(p-q)}{\sqrt3-1}\quad \text{or}\quad t=\frac{2(p+q+1)}{\sqrt3+1}$$
From $g\left(t-2q\right)=h\left(t-\frac1{\sqrt2}r\right)$,
$$t=\frac{2(r-q)}{2\sqrt2-1}\quad \text{or}\quad t=\frac{2(r+q+1)}{2\sqrt2+1}$$
i) If $t=\frac{2(p-q)}{\sqrt3-1}=\frac{2(r-q)}{2\sqrt2-1}$, and $t\ne0$, then
$\frac{p-q}{r-q}=\frac{\sqrt3-1}{2\sqrt2-1}$.
ii) If $t=\frac{2(p-q)}{\sqrt3-1}=\frac{2(r+q+1)}{2\sqrt2+1}$, then
$\frac{p-q}{r+q+1}=\frac{\sqrt3-1}{2\sqrt2+1}$.
iii) If $t=\frac{2(p+q+1)}{\sqrt3+1}=\frac{2(r-q)}{2\sqrt2-1}$, and $t\ne0$, then
$\frac{p+q+1}{r-q}=\frac{\sqrt3+1}{2\sqrt2-1}$.
iv) If $t=\frac{2(p+q+1)}{\sqrt3+1}=\frac{2(r+q+1)}{2\sqrt2+1}$, then
$\frac{p+q+1}{r+q+1}=\frac{\sqrt3+1}{2\sqrt2+1}$.
Any of these four cases is not possible because LHS's are rational and RHS's are irrational.
Therefore, such $t$ does not exist.
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