Show if $f,g:S^n \to S^n$ and $|\text{deg}(f)| \neq |\text{deg}(g)|$ there is some $x$ with $f(x), g(x)$ orthogonal. More specifically, I want to show if f and g are maps from $S^{n} \to S^{n}$ with $|\text{deg}(f)| \neq |\text{deg}(g)|$ show that there is some $x \in S^{n}$ with $f(x), g(x)$ orthogonal. 
One way I think the hypothesis could be used is the standard "contrapositive" trick for these sorts of proofs where I assume $h(x) = \langle f(x), g(x)\rangle$ is never zero. Then it either is always positive or negative. But otherwise I'm not really sure where to go from there.
 A: Hint: 
Instead of treating the information as a function $h$, figure out what it means geometrically. Let say $h >0$ (For the case $h<0$, consider $-f$ instead). Then $g(x), f(x)$ both lies in the same hemisphere (treating $g(x)$ as the "north pole"). Can you find a canonical line on the sphere joining $f(x)$ to $g(x)$? If you can do so, then $f$ and $g$ are homotopic and so $\text{deg}( f) = \text{deg}(g)$, hence the contradiction. 
A: Suppose for every $x, \langle f(x), g(x)\rangle\neq 0$. Consider $h:S^n\rightarrow R$ defined by $h(x)=\langle f(x), g(x)\rangle$. Since $S^n$ is connected, we have: 


*

*For every $x\in S^n$, $h(x)>0$

*For every $x\in S^n, h(x)<0$.
Suppose 1.
Define $H(t,x)={{tf(x)+(1-t)g(x)}\over{\|t(x)+(1-t)g(x)\|}}$.
$H$ is well-defined: $tf(x)+(1-t)g(x)=0$ implies that $f(x), g(x)$ are colinear. If $f(x)=g(x)$, then $tf(x)+(1-t)g(x)=tf(x)+(1-t)f(x)=f(x)\neq 0$.  If $f(x)=-g(x)$, then $\langle f(x),g(x)\rangle=-1$. But we have supposed $1$.
$H(0,x)=g(x)$ and $H(1,x)=f(x)$, which implies that $H$ is an homotopy between $f$ and $g$, and therefore $f$ and $g$ have the same degree. Contradiction.
If $2$ is verified define
$H(t,x)={{tf(x)-(1-t)g(x)}\over{\|t(x)-(1-t)g(x)\|}}$.
$H$ is well defined, if $tf(x)+(1-t)g(x)=0$, then $f(x)$ and $g(x)$ are colinear. If $f(x)=-g(x), tf(x)-(1-t)g(x)=tf(x)-(1-t)(-f(x))=f(x)\neq 0$, if $f(x)=g(x), \langle f(x),g(x)\rangle=1$. Contradiction since we have supposed 2.
$H(0,x)=-g(x), H(1,x)=f(x)$ we deduce that $|degree(g)|=|degree(f)|$.  Contradiction.
