Problem 1117

The products $s_{i} d_{i}$ in the task are exactly twice the areas of those triangles, one of whose vertices is $P$, and the opposite side to that vertex is one of the medians of the original triangle. Essentially, we need to show that these three areas can be signed in such a way that their signed sum is 0.

Let the vectors from $P$ to the vertices of the original triangle be denoted by $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$. Then the vectors to the midpoints of the sides are $\frac{\mathbf{a}+\mathbf{b}}{2}$, $\frac{\mathbf{b}+\mathbf{c}}{2}$, and $\frac{\mathbf{c}+\mathbf{a}}{2}$. Thus, $\frac{\mathbf{a}+\mathbf{b}}{2} \times \mathbf{c}$, $\frac{\mathbf{b}+\mathbf{c}}{2} \times \mathbf{a}$, and $\frac{\mathbf{c}+\mathbf{a}}{2} \times \mathbf{b}$ are vectors that are parallel to each other, and their magnitudes are $s_{1} d_{1}$, $s_{2} d_{2}$, and $s_{3} d_{3}$ in some order. Since for any vectors $\mathbf{x}$ and $\mathbf{y}$, it is true that $\mathbf{x} \times \mathbf{y} + \mathbf{y} \times \mathbf{x} = \mathbf{0}$, we have

$$\begin{gathered} \left(\frac{\mathbf{a}+\mathbf{b}}{2} \times \mathbf{c}\right) + \left(\frac{\mathbf{b}+\mathbf{c}}{2} \times \mathbf{a}\right) + \left(\frac{\mathbf{c}+\mathbf{a}}{2} \times \mathbf{b}\right) = \\ = \frac{1}{2}(\mathbf{a} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{b} + \mathbf{b} \times \mathbf{a} + \mathbf{a} \times \mathbf{b}) = \mathbf{0} \end{gathered}$$

This, however, means that among the products $s_{1} d_{1}$, $s_{2} d_{2}$, and $s_{3} d_{3}$, the largest is equal to the sum of the other two.

Based on the work of Gergely Naszódi (Fazekas M. Főv. Gyak. Gimn., 11th grade)

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