XLVI OM - III - Problem 2

The diagonals of a convex pentagon divide this pentagon into a pentagon and ten triangles. What is the maximum possible number of triangles with equal areas?
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Let's denote the considered pentagon by $ABCDE$, and the pentagon formed by the intersection points of the diagonals by $KLMNP$ so that the following triangles are those mentioned in the problem:

$\Delta_0$: triangle $LEM$; $\quad \Delta_1$: triangle $EMA$;

$\Delta_2$: triangle $MAN$; $\quad \Delta_3$: triangle $ANB$;

$\Delta_4$: triangle $NBP$; $\quad \Delta_5$: triangle $BPC$;

$\Delta_6$: triangle $PCK$; $\quad \Delta_7$: triangle $CKD$;

$\Delta_8$: triangle $KDL$; $\quad \Delta_9$: triangle $DLE$

(figure 12). We will start by proving the implication:

We assume that the numbering of the triangles is cyclic (that is, we assume: $\Delta_{-1} = \Delta_9$; $\Delta_{10} = \Delta_0$).
Suppose, for example, that triangles $\Delta_4$, $\Delta_5$, and $\Delta_6$ have equal areas. From the equality of the areas of triangles $NBP$ and $BPC$, it follows that segment $BP$ is a median in triangle $BCN$; from the equality of the areas of triangles $BPC$ and $PCK$, it follows that segment $CP$ is a median in triangle $BCK$. Point $P$ would then be the common midpoint of diagonals $BK$ and $CN$ of quadrilateral $BCKN$, which should therefore be a parallelogram - but the lines $BN$ and $CK$ intersect at point $E$. This is a contradiction; implication (1) is proven.
We proceed to the main solution. The example of a regular pentagon shows that it is possible to obtain five triangles $\Delta_i$ with equal areas. We will prove that it is not possible to obtain seven such triangles.
Suppose, therefore, that there exists a seven-element set $Z$, contained in the set $\{0,1,\ldots,9\}$, such that triangles $\Delta_i$ with indices $i \in Z$ have equal areas. Let $k$, $l$, $m$ be three numbers from the set $\{0,1,\ldots,9\}$, not belonging to $Z$. Consider all triples of consecutive indices:

where, as before, the addition $i \pm 1$ should be understood modulo $10$. Each number from the set $\{0,1,\ldots,9\}$ belongs to exactly three triples $T_i$. This applies, in particular, to each of the numbers $k$, $l$, $m$. Therefore, the three-element set $\{k,l,m\}$ has a non-empty intersection with at most nine triples $T_i$. At least one triple $T_j$ remains that does not contain any of the numbers $k$, $l$, $m$, and is therefore contained in the set $Z$. The number $j$ must be even, according to observation (1).
Without loss of generality, assume that $j = 2$. This means that the numbers $1$, $2$, $3$ belong to the set $Z$. Therefore, the numbers $0$ and $4$ cannot belong to it (since, again by observation (1), the triples $\{0,1,2\}$ and $\{2,3,4\}$ are excluded); thus, one of the numbers $k$, $l$, $m$ equals $0$, and another equals $4$. The third of these numbers must be part of the triple $\{6,7,8\}$ (which otherwise would be contained in the set $Z$, contrary to implication (1)). The roles of the numbers $6$ and $8$ are symmetric in this context. We therefore have to consider two essentially different cases:

The set $Z$, respectively, has one of the following forms:

Let $S_i$ be the area of triangle $\Delta_i$ (for $i = 0,1,\ldots,9$). We will show that the implications hold:

This will be a proof that none of the seven-element sets $Z$ listed above can be the set of indices of triangles with equal areas. According to earlier statements, this will also justify the conclusion that among the ten triangles $\Delta_i$, there are never seven triangles with the same area.
Proof of implication (2). From the given equalities of areas:

and

it follows (respectively) that points $N$ and $P$ are the midpoints of segments $BM$ and $BK$, and points $M$ and $L$ are the midpoints of segments $EN$ and $EK$ (figure 13).
Thus, segment $NP$ connects the midpoints of two sides of triangle $BMK$, and segment $ML$ connects the midpoints of two sides of triangle $ENK$, and therefore

- and

The obtained parallelism relations show that quadrilateral $AMKN$ is a parallelogram; hence $|MK| = |AN|$. Line $MK$ is parallel to $NC$, so triangles $ENC$ and $EMK$ are similar in the ratio $|EN| : |EM| = 2$. Therefore, $|NC| = 2 \cdot |MK|$, and consequently

The segments $PC$ and $AN$, lying on the same line, are the bases of triangles $ANB$ and $BPC$ with a common vertex $B$. The ratio of the lengths of these bases is therefore also the ratio of the areas of the triangles: area($BPC$) : area($ANB$) = $3 : 2$, i.e., $S_5 : S_3 = 3 : 2$. The conclusion of implication (2) is thus proven.
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Proof of implication (3). The given equalities of the areas of triangles:

and

show that points $M$ and $N$ divide segment $EB$ into three equal parts, and similarly, points $L$ and $K$ divide segment $EC$ into three equal parts. Therefore, line $BC$ is parallel to $ML$ (i.e., to $AM$). Since $N$ is the midpoint of segment $BM$, triangles $AMN$ and $CBN$ are similar in the ratio $-1$ (figure 14). Hence, area($AMN$) = area($CBN$) $> $ area($CBP$), i.e., $S_2 > S_5$. This completes the proof of implication (3), and thus also the proof of the general theorem: it is not possible to have seven triangles $\Delta_i$ with equal areas.
The reasoning conducted in the last case (proof of (3)) also provides a hint on how to obtain six triangles $\Delta_i$ with equal areas. We take any isosceles triangle $BCE$ where $|EB| = |EC|$. On the sides $EB$ and $EC$, we find points $M$ and $N$ and $L$ and $K$, dividing these sides into three equal parts:

The intersection points of line $ML$ with lines $CN$ and $BK$ are denoted by $A$ and $D$, respectively (figure 14 can still serve as an illustration; one only needs to imagine moving point $E$ to the perpendicular bisector of side $BC$ and appropriately repositioning the other points). The equalities (4) and (5) then hold; and thanks to the assumption that triangle $BCE$ is isosceles (and the resulting symmetry of the entire configuration), all areas appearing in relations (4) and (5) are equal.
Conclusion: six of the triangles $\Delta_i$ can have equal areas, and this number cannot be increased.