A surface such that any two triangles on it are not congruent. Does a surface such that any two different triangles on it are not congruent exist?
Added:Suppose that surfaces are defined by some continuous function:$R^2\rightarrow R$ and a triangle is a set of three non collinear points connected by three geodesics.
 A: Assuming by "surface" you mean a graph $\Gamma_f=\{(x_1,x_2,f(x_1,x_2))|\;x_1,x_2\in\mathbb R\}$ of a function $f:\mathbb R^2\to\mathbb R$, as you mention in the comments, and assuming by "different triangles on it are not congruent" you mean what Hagen von Eitzen says in the comments (i.e. that the function $\Phi$ we will define below is injective; edit: this in particular means that triangle should be interpreted as three point metric space, which might be a bit non-standard), the answer is negative.
The reason is as follows: we define the function $\Phi:\mathbb R^6\to\mathbb R^3$ (as suggested by Hagen von Eitzen in the comments) by $$\Phi(X,Y,Z)=(d(F(X),F(Y)),d(F(X),F(Z)),d(F(Y),F(Z))),$$ where $X,Y,Z\in\mathbb R^2$ and $F:\mathbb R^2\to \Gamma_f$ is the homeomorphism $(x,y)\to(x,y,f(x,y))$ (and we identify $\mathbb R^6$ with $\mathbb R^2\times\mathbb R^2\times\mathbb R^2$).
The problem is that $\Phi$ is a continuous function, but a continuous injection $\mathbb R^6\to\mathbb R^3$ cannot exist. This is because such an injection would in particular have to map $$S^5 =\{(x,y,z,u,v,w)\in\mathbb R^6|\;x^2+y^2+z^2+u^2+v^2+w^2=1\}$$ continuously and injectively into $\mathbb R^3$, and thus into $\mathbb R^5$. This would contradict the Borsuk-Ulam theorem.
Added: If by surface you mean $2$-manifold, the same argument works (and the answer is again negative): let $M$ be the $2$-manifold in question. (We assume that there is a metric given on $M$.) Then an arbitrary point in its interior has a neighborhood $U$ homeomorphic to $\mathbb R^2$. This homeomorphism $h:U\to\mathbb R^2$ allows us to define a metric $d_U$ on $\mathbb R^2$ by $d_U(h(p),h(q))=d(p,q)$, where $p,q\in U$ (and $d$ is the metric on $M$). Now define $\Phi:\mathbb R^6\to\mathbb R^3$ analogously to the definition above, i.e. $$\Phi(X,Y,Z)=(d_U(X,Y),d_U(X,Z),d_U(Y,Z)).$$ This is again continuous, and thus cannot be injective by Borsuk-Ulam.
