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1.2: Lámina Triangular Plano

Última actualización

30 oct 2022
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- 1.1: Introducción y algunas definiciones
- 1.3: Áreas de plano

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Definición: Una mediana de un triángulo es una línea desde un vértice hasta el punto medio del lado opuesto.

Teorema I. Las tres medianas de un triángulo son concurrentes (se encuentran en un punto único y único) en un punto que es dos tercios de la distancia desde un vértice hasta el punto medio del lado opuesto.

Teorema II. El centro de masa de una lámina triangular uniforme (o el centroide de un triángulo) se encuentra en el encuentro de las medianas.

La prueba de I se puede hacer con un buen argumento vectorial (Figura I.1):

Que $\bf{\text{A}}$ , $\bf{\text{B}}$ sean los vectores $\text{OA}$ , $\text{OB}$ . Entonces $\bf{\text{A+B}}$ es la diagonal del paralelogramo de los cuales $OA$ y $\text{OB}$ son dos lados, y el vector de posición del punto $\text{C}_{1}$ is $\frac{1}{3}(\bf{\text{A+B}})$ .

Para obtener $\text{C}_{2}$ , we see that

${\bf C}_2 = {\bf A} + \dfrac{2}{3}(\text{AM}_2) = {\bf A} + \dfrac{2}{3}({\bf M_2 - A}) = {\bf A} + \dfrac{2}{3}(\dfrac{1}{2}{\bf B- A}) = \dfrac{1}{3}( {\bf A+B})$

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Así los puntos $\text{C}_{1}$ and $\text{C}_{2}$ are identical, and the same would be true for the third median, so Theorem I is proved.

Now consider an elemental slice as in Figure I.2. The centre of mass of the slice is at its mid-point. The same is true of any similar slices parallel to it. Therefore the centre of mass is on the locus of the mid-points - i.e. on a median. Similarly, it is on each of the other medians, and Theorem II is proved.

alt

Eso solo necesitaba algo de geometría vectorial. Pasamos ahora a algún cálculo.

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