Question:

From $f(10+x) = f(10-x)$, we know $f$ is symmetric about $x=10$. From $f(20-x) = -f(20+x)$, substituting $y = 20+x$ gives $f(40-y) = -f(y)$, meaning $f$ is antisymmetric about $x=20$. Combining these: $f(x) = f(20-x)$ (from first condition with $x\to 10-x$) and $f(20-x) = -f(x)$ (from second condition), we get $f(x) = -f(x)$, so $f(x) = 0$ at certain points. More carefully, $f(40-x) = -f(x)$ combined with $f(x+40) = f(40-(x+40)) = -f(-(x)) = -f(x)$ and applying the symmetries again yields $f(x+40) = f(x)$, proving periodicity with period 40. For odd function: $f(-x) = f(20-(-x-20)) = -f(20+(−x−20)) = -f(-x)$ using the antisymmetry property, which means $f$ is odd.