TITLE: Computing van Kampen diagrams
QUESTION [15 upvotes]: If G is a finitely presented group  (with generating set X) and w is a word over X such that
w=1 in G, then the latter can be witnessed by a so called van Kampen diagram for w, which is
a planar diagram where for each region the boundary cycle is labelled with a group relator,
and the boundary of the whole diagram is labelled with the word w.
Is there a finitely presented group G (with generating set X) such that the following hold?

G has polynomial Dehn function
G has a polynomial time word problem
There is no polynomial time algorithm for the following problem:
INPUT: A word w over X such that w=1 in G 
OUTPUT: A van Kampen diagram for w

In other words: We can efficiently check whether w=1 in G but we cannot compute efficiently a witness for this fact (but we know that a small witness exists).
If the answer to the above question is positive, one certainly needs some complexity theoretic assumption. A reasonable starting point might be to assume that a nondeterministic
polynomial time Turing machine M exists such that:

One can check in deterministic polynomial time whether an input w is accepted by M.
There is no polynomial time Turing machine that computes for a given word w (that is
accepted by M) an accepting computation of M for input w.

REPLY [6 votes]: This is equivalent to $P\ne NP$ by the result of the paper : Birget, J.-C.; Olʹshanskii, A. Yu.; Rips, E.; Sapir, M. V. Isoperimetric functions of groups and computational complexity of the word problem. Ann. of Math. (2) 156 (2002), no. 2, 467–518. 
Update See the discussion here.