The following solution of APMO q.4 is a claimed official solution posted by a New Zealanded in MathLinks. (Please forgive for any mistakes resulted during the process of copying.)

Label the houses according to their “distance” from (1,c): say that house (i, j) is at level t if . Let d(t) be the number of houses at level t defended by time t, p(t) the number of houses at level t + 1 or greater defended by time t. Clearly.(*)

Let s(t) be the number of houses at level t which are not burning at time t. We prove by induction that for . The base case t = 1 is clearly true, so suppose it is true when t = k. At time k + 1, consider the set M(k+1) of the s(k+1) – d(k+1) houses at level k+1 that do not burn but are not themselves defended. These houses survive because all their neighbours at level k do not burn. There are at least s(k+1) – d(k+1) houses at level k that are neighbours of at least one house in M(k+1), so there must be at least s(k+1) – d(k+1) houses at level k that do not burn; that is, . But by the inductive hypothesis , so . Hence, using (*), also.

for all t, so from that it follow that for all . Summing, we find that at most houses at levels n-1 or below can be defended. It is easy to see that the v-shaped algorithm described by Severius and others saves exactly this many houses at levels n-1 and below, and every house at level n and above, so this must be the optimal algorithm, and at most houses can be saved.