In triangle $ABC$, the (internal) angle bisectors from vertices $A$, $B$, and $C$ intersect the opposite sides at points $D$, $E$, and $F$, respectively.

What is the necessary and sufficient condition for the areas of triangles $BDF$ and $EDF$ to be equal?
I. solution. With the usual notations in triangles, based on the angle bisector division ratio,

$$
B D=a \cdot \frac{c}{b+c}, \quad B F=c \cdot \frac{a}{a+b}
$$

thus the area of the $B D F$ triangle is:

$$
\frac{1}{2} B D \cdot B F \cdot \sin \beta=\frac{a^{2} c^{2} \sin \beta}{2(a+b)(b+c)}
$$

and the ratio of the areas of the $B D F$ and $A B C$ triangles is:

$$
\frac{a^{2} c^{2} \sin \beta}{2(a+b)(b+c)}: \frac{a c \sin \beta}{2}=\frac{a c}{(a+b)(b+c)}
$$

$1985-11-376-1 . e p s$

Similarly, the ratios of the areas of the $A E F$ and $C D E$ triangles to the area of the original triangle are

$$
\frac{c b}{(c+a)(a+b)}, \quad \frac{b a}{(b+c)(c+a)}
$$

therefore, the ratio of the areas of the $D E F$ and $A B C$ triangles, simplified further, is

$$
1-\left(\frac{a c}{(a+b)(b+c)}+\frac{c b}{(c+a)(a+b)}+\frac{b a}{(b+c)(c+a)}\right)=\frac{2 a b c}{(a+b)(b+c)(c+a)}
$$

(no denominator can become zero).

The desired equality holds if and only if expressions (2) and (1) are equal. Simplifying this,

$$
\frac{2 b}{c+a}=1, \quad \text { i.e., if } \quad b=\frac{a+c}{2}
$$

In words: if the side $b$ opposite the distinguished vertex $B$ is the arithmetic mean of the other two, i.e., if the sides $a, b, c$ form an arithmetic sequence in this order.

Agnes Csermely (Veszprem, Lovassy L. Gymnasium, 4th grade)

The following solution includes some geometric insight along with the calculations.

II. solution. Considering their common base $D F$, the areas of the triangles in question are equal if and only if the distances of the points $B, E$ from the line $D F$ are equal. This condition is equivalent to the following: drawing a line parallel to $D F$ through $E$, this line intersects the rays $B C, B A$ at points $D_{1}, F_{1}$ such that $B D_{1}=2 \cdot B D$ and $B F_{1}=2 \cdot B F$.

Consider then these two points, as well as the intersection point $U$ of the line $D_{1} F_{1}$ with $A C$, and seek the condition for $U$ to coincide with $E$. For the time being, assume that $D_{1}$ is not identical to $C$ - which is obviously true if $A B \neq A C$ - and $F_{1}$ is not identical to $A$, i.e., $C B \neq B A$. The point $U$ will lie between $C$ and $A$ if and only if $D_{1}$ and $F_{1}$ lie on different sides of the line $A C$.

1. Choose the labeling such that $B D_{1}=2 B D>A C$, and $B F_{1}=2 B F>C D$; by the angle bisector division ratio - $B A>C A$, and on the other hand, $B F<A F$ implies $B C<C A$, so $C A$ is the middle side in terms of length.

From the triangles $U C D_{1}$ and $U A F_{1}$,

$$
\sin \angle C U D_{1}=\frac{C D_{1}}{C U} \sin \angle C D_{1} U=\frac{C D_{1}}{C U} \sin \angle B D F,
$$

and

thus their ratio is

$$
\sin \angle C U D_{1}=\sin \angle A U F_{1}=\frac{A F_{1}}{A U} \sin \angle A F_{1} U=\frac{A F_{1}}{A U} \sin \angle B F D,
$$

$$
\frac{C D_{1}}{A F_{1}} \cdot \frac{A U}{C U} \cdot \frac{\sin \angle B D F}{\sin \angle B F D}=1
$$

However, again by the division ratio,

$$
\frac{\sin \angle B D F}{\sin \angle B F D}=\frac{B F}{B D}=\frac{a c}{a+b}: \frac{c a}{c+b}=\frac{b+c}{b+a},
$$