A point is taken arbitrarily inside a triangle, and lines parallel to the sides of the triangle are drawn through this point. These lines divide the area of the triangle into six parts. What is the area of the given triangle if the areas of the three resulting triangles are given as: $t_{1}, t_{2}, t_{3}$?
I. Solution: Let the sides of the resulting sub-triangles parallel to the side $a$ of the given triangle be $a_{1}, a_{2}, a_{3}$, and the heights corresponding to these sides be $m_{1}, m_{2}, m_{3}$. It is obvious that $a_{1}+a_{2}+a_{3}=a$, and it is also easily verifiable that $m_{1}+m_{2}+m_{3}=m$ (Figure 1), where $m$ is the height of the given triangle corresponding to the side $a$.

![](https://cdn.mathpix.com/cropped/2024_05_02_496f84e55b98213e1718g-1.jpg?height=393&width=615&top_left_y=340&top_left_x=739)

Figure 1

Thus, denoting the area of the given triangle by $T$,

$$
\begin{gathered}
2 T=a m=\left(a_{1}+a_{2}+a_{3}\right)\left(m_{1}+m_{2}+m_{3}\right)= \\
a_{1} m_{1}+a_{2} m_{2}+a_{3} m_{3}+\left(a_{1} m_{2}+a_{2} m_{1}\right)+\left(a_{1} m_{3}+a_{3} m_{1}\right)+\left(a_{2} m_{3}+a_{3} m_{2}\right)
\end{gathered}
$$

But $2 t_{1}=a_{1} m_{1}$ and $2 t_{2}=a_{2} m_{2}$; the product of these two equalities is

$$
4 t_{1} t_{2}=a_{1} m_{1} a_{2} m_{2}=\left(a_{1} m_{2}\right)\left(a_{2} m_{1}\right)
$$

Since the sub-triangles, due to the equality of angles, are all similar to the given triangle and thus also to each other,

$$
\frac{a_{1}}{a_{2}}=\frac{m_{1}}{m_{2}}, \quad \text { or } \quad a_{1} m_{2}=a_{2} m_{1}
$$

Thus, based on (2),

$$
\left(a_{1} m_{2}\right)^{2}=\left(a_{2} m_{1}\right)^{2}=4 t_{1} t_{2}
$$

from which

$$
a_{1} m_{2}=a_{2} m_{1}=2 \sqrt{t_{1} t_{2}}, \quad \text { or } \quad a_{1} m_{2}+a_{2} m_{1}=4 \sqrt{t_{1} t_{2}}
$$

Similarly, it can be shown that

$$
a_{1} m_{3}+a_{3} m_{1}=4 \sqrt{t_{1} t_{3}} \quad \text { and } \quad a_{2} m_{3}+a_{3} m_{2}=4 \sqrt{t_{2} t_{3}} .
$$

Substituting these values into (1) and dividing by 2,

$$
T=t_{1}+t_{2}+t_{3}+2 \sqrt{t_{1} t_{2}}+2 \sqrt{t_{1} t_{3}}+2 \sqrt{t_{2} t_{3}}=\left(\sqrt{t_{1}}+\sqrt{t_{2}}+\sqrt{t_{3}}\right)^{2}
$$

Remark: If the areas of the 3 resulting parallelograms are denoted by $T_{1}, T_{2}, T_{3}$, as shown in the figure, then it is directly readable from the figure that

$$
a_{1} m_{2}\left(=2 \sqrt{t_{1} t_{2}}\right)=T_{3}, \quad a_{3} m_{2}\left(=2 \sqrt{t_{2} t_{3}}\right)=T_{1}
$$

and it is easily verifiable (after a small area transformation) that

$$
a_{1} m_{3}=\left(=2 \sqrt{t_{1} t_{3}}\right)=T_{2}
$$

Thus, the above result follows from the equality $T=t_{1}+t_{2}+t_{3}+T_{1}+T_{2}+T_{3}$.

The proof of the relationships used here (without using similarity) is the basis of the following solution.

II. Solution: First, consider the part of the figure consisting of $t_{1}, t_{3}$, and $T_{2}$. The labeling is shown in Figure 2.

![](https://cdn.mathpix.com/cropped/2024_05_02_496f84e55b98213e1718g-2.jpg?height=710&width=634&top_left_y=125&top_left_x=732)

Figure 2

Extend the triangle $A B_{1} C_{1}$ to a parallelogram by reflecting it over the midpoint of the side $B_{1} C_{1}$, and extend the segments $B_{2} P$ and $C_{2} P$ to the opposite sides of the parallelogram. Then

$$
B_{1} C_{1} A_{\triangle} \cong C_{1} B_{1} A_{\triangle}^{\prime}, \quad P B_{1} C_{2 \Delta} \cong B_{1} P B_{2 \Delta}^{\prime}, \quad C_{1} P B_{2 \triangle} \cong P C_{1} C_{2 \Delta}^{\prime}
$$

and thus the area $T_{2}$ of the parallelogram $C_{2} A B_{2} P$ is equal to the area of the parallelogram $B_{2}^{\prime} A^{\prime} C_{2}^{\prime} P$. The height of the former to the side $P B_{2}=c_{3}$ is equal to the height $m_{1}^{\prime}$ of the triangle $t_{1}$ to the side $c_{1}$; the height of the latter to the side $P B_{2}^{\prime}=c_{1}$ is equal to the height $m_{3}^{\prime}$ of the triangle $t_{3}$ to the side $c_{3}$. Therefore,

$$
T_{2}=c_{3} m_{1}=c_{1} m_{3}^{\prime}
$$

from which

$$
T_{2}^{2}=c_{1} m_{1}^{\prime} c_{3} m_{3}^{\prime}=4 t_{1} t_{3}, \quad \text { or } \quad T_{2}=2 \sqrt{t_{1} t_{3}}
$$

Similarly, it follows that

$$
T_{1}=2 \sqrt{t_{2} t_{3}}, \quad T_{3}=2 \sqrt{t_{1} t_{2}}
$$

and thus

$$
T=t_{1}+t_{2}+t_{3}+2 \sqrt{t_{2} t_{3}}+2 \sqrt{t_{3} t_{1}}+2 \sqrt{t_{1} t_{2}}=\left(\sqrt{t_{1}}+\sqrt{t_{2}}+\sqrt{t_{3}}\right)^{2}
$$

III. Solution: The simplest way to reach the goal is to use the well-known theorem that the areas of similar triangles are in the same ratio as the squares of the corresponding sides.

Thus,

$$
\begin{aligned}
& \sqrt{t_{1}}: \sqrt{T}=a_{1}:\left(a_{1}+a_{2}+a_{3}\right) \\
& \sqrt{t_{2}}: \sqrt{T}=a_{2}:\left(a_{1}+a_{2}+a_{3}\right) \\
& \sqrt{t_{3}}: \sqrt{T}=a_{3}:\left(a_{1}+a_{2}+a_{3}\right)
\end{aligned}
$$

Adding these,

$$
\frac{\sqrt{t_{1}}}{\sqrt{T}}+\frac{\sqrt{t_{2}}}{\sqrt{T}}+\frac{\sqrt{t_{3}}}{\sqrt{T}}=\frac{a_{1}+a_{2}+a_{3}}{a_{1}+a_{2}+a_{3}}=1
$$

from which,

$$
\sqrt{t_{1}}+\sqrt{t_{2}}+\sqrt{t_{3}}=\sqrt{T}
$$

or

$$
T=\left(\sqrt{t_{1}}+\sqrt{t_{2}}+\sqrt{t_{3}}\right)^{2}
$$