Consider the internal points of a triangle $ABC$ such that the distances from these points to the sides can form a triangle. What fraction of the triangle's area do these points occupy?

 untranslated portion:
Tekintsük egy $A B C$ háromszög azon belső pontjait, amelyeknek az oldalaktól mért távolságaiból háromszög szerkeszthető. A háromszög területének hányadrészét töltik ki ezek a pontok?
Let the sides of the triangle be $a, b, c$, and a suitable interior point $P$, with distances from the sides being $d_{a}, d_{b}, d_{c}$ (considering the distances from the sides as the distances from the side lines). For the sought points $P$, $d_{a}, d_{b}, d_{c}$ must satisfy the triangle inequalities. Let the internal angle bisectors of the triangle intersect the sides at points $A^{\prime}, B^{\prime}$, and $C^{\prime}$ as shown in the figure.

# 1993-04-164-1.eps 

First, we show that for any point $Q$ on the segment $B^{\prime} C^{\prime}$, the distances $x, y, z$ from the side lines satisfy $y+z=x$. Since $C^{\prime}$ is a point on the angle bisector from $C$, $C^{\prime} K = C^{\prime} K^{\prime} = k$, and from the similar triangles $C^{\prime} B^{\prime} K^{\prime}$ and $Q B^{\prime} Y$,

$$
\frac{Q B^{\prime}}{B^{\prime} C^{\prime}} = \frac{y}{k}, \quad \text{so} \quad y = \frac{Q B^{\prime}}{B^{\prime} C^{\prime}} k
$$

Similarly, we get

$$
z = \frac{Q C^{\prime}}{B^{\prime} C^{\prime}} h
$$

From the similar triangles $C^{\prime} Q L$ and $C^{\prime} B^{\prime} M$,

$$
\frac{Q C^{\prime}}{x-k} = \frac{Q B^{\prime}}{h-x}
$$

Solving for $x$ from (3): $x = \frac{Q B^{\prime}}{B^{\prime} C^{\prime}} k + \frac{Q C^{\prime}}{B^{\prime} C^{\prime}} h$. Therefore, from (1) and (2), $x = y + z$, as we stated.

Now consider the central similarity centered at $A$. If its ratio is greater than 1, and the image of the previous point $Q$ is $P$, then in the enlargement, the segments $y$ and $z$ increase, while $x$ decreases, so the inequality $d_{a} < d_{b} + d_{c}$ will hold for point $P$. Similarly, we can show that for any point on $A^{\prime} B^{\prime}$ or $A^{\prime} C^{\prime}$, when mapped by a central similarity from $C$ or $B$ with a ratio greater than 1, the inequalities $d_{c} < d_{a} + d_{b}$ and $d_{b} < d_{a} + d_{c}$ will hold.

All three inequalities hold if and only if point $P$ is an interior point of triangle $A^{\prime} B^{\prime} C^{\prime}$. Therefore, the set of sought points $P$ is the interior of triangle $A^{\prime} B^{\prime} C^{\prime}$.

It remains to determine the ratio of the area of triangle $A^{\prime} B^{\prime} C^{\prime}$, $T_{(A^{\prime} B^{\prime} C^{\prime})}$, to the area of triangle $A B C$, $T_{(A B C)}$. It is clear that

$$
T_{(A^{\prime} B^{\prime} C^{\prime})} = T_{(A B C)} - T_{(A B^{\prime} C^{\prime})} - T_{(C A^{\prime} B^{\prime})} - T_{(B C^{\prime} A^{\prime})}
$$

Thus,

$$
\frac{T_{(A^{\prime} B^{\prime} C^{\prime})}}{T_{(A B C)}} = 1 - \frac{T_{(A B^{\prime} C^{\prime})}}{T_{(A B C)}} - \frac{T_{(C A^{\prime} B^{\prime})}}{T_{(A B C)}} - \frac{T_{(B C^{\prime} A^{\prime})}}{T_{(A B C)}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_98a9e0084cbd35009001g-1.jpg?height=98&width=1797&top_left_y=1835&top_left_x=153)
$T_{(A B^{\prime} C^{\prime})} = \frac{1}{2} \frac{b^{2} c^{2}}{(a+b)(a+c)} \sin \alpha$

Thus,

$$
\frac{T_{(A B^{\prime} C^{\prime})}}{T_{(A B C)}} = \frac{\frac{1}{2} \frac{b^{2} c^{2}}{(a+b)(a+c)} \sin \alpha}{\frac{1}{2} b c \sin \alpha} = \frac{b c}{(a+b)(a+c)}
$$

Similarly, we can calculate the other ratios in (4). Therefore, (4) becomes:

$$
\begin{aligned}
\frac{T_{(A^{\prime} B^{\prime} C^{\prime})}}{T_{(A B C)}} = 1 - \frac{b c}{(a+b)(a+c)} & - \frac{a b}{(c+a)(c+b)} - \frac{a c}{(b+a)(b+c)} = \\
& = \frac{2 a b c}{(a+b)(b+c)(c+a)}
\end{aligned}
$$

Note: Solutions to the problem can be found in the book "Középiskolai Matematikai Versenyek 1985 -1987" on pages 153-159.