Kvant Math Problem 306
Let the removed corner be the unit square with vertices $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$.
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Verdicts: SKIP + SKIP
Solve time: 6m49s
Source on kvant.digital
Problem
From a chessboard ($8\times8$), one corner square ($1\times1$) has been removed (Fig. 1). Into what smallest number of equal-area triangles can the remaining part of the board be dissected?
Figure 1
V. P. Fedotov
Exploration
Let the removed corner be the unit square with vertices $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$. The remaining region has area
$$64-1=63.$$
If it is dissected into $n$ equal-area triangles, each triangle has area
$$\frac{63}{n}.$$
The region is the union of the rectangle
$$R=[1,8]\times[0,8],$$
of area $56$, and the rectangle
$$S=[0,1]\times[1,8],$$
of area $7$.
The narrow rectangle $S$ is important. Any triangle contained in a $1\times7$ rectangle has area at most
$$\frac{1\cdot7}{2}=3.5.$$
Hence a triangle of area exceeding $7$ cannot lie completely inside $S$.
A natural construction appears when the common area is $7$. Since $63=9\cdot7$, one may try to obtain nine triangles of area $7$. The large rectangle $R$ has area $56=8\cdot7$, so if $R$ can be divided into eight triangles of area $7$ and the strip $S$ contributes one more triangle of area $7$, then nine triangles are obtained.
The delicate point is the lower bound. One must exclude the possibility of eight equal triangles. Since eight triangles would have area $63/8=7.875$, larger than the whole area of $S$, every triangle would have to take some area from $R$. The geometry of the strip then forces at lea