Question

Area of $\triangle PBC$ : Area of $\triangle PCA$ : Area of $\triangle PAB$ = $w_1:w_2:w_3$

Prove that: $\begin{cases} x = \frac{w_1x_1 + w_2x_2 + w_3x_3}{w_1 + w_2 + w_3} \\ y = \frac{w_1y_1 + w_2y_2 + w_3y_3}{w_1 + w_2 + w_3} \end{cases}$

Answer 1

$\begin{cases} x = \frac{w_1x_1 + w_2x_2 + w_3x_3}{w_1 + w_2 + w_3} \\ y = \frac{w_1y_1 + w_2y_2 + w_3y_3}{w_1 + w_2 + w_3} \end{cases}$

Answer 2

The problem requires proving the formula for the coordinates of a point P inside a triangle ABC, given the ratio of the areas of the three triangles formed by P and the vertices of ABC.

Let the vertices of $\triangle ABC$ be $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$. Let $P(x, y)$ be a point inside $\triangle ABC$.

We are given the ratio of areas: Area($\triangle PBC$) : Area($\triangle PCA$) : Area($\triangle PAB$) = $w_1 : w_2 : w_3$.

Let Area($\triangle PBC$) = $k w_1$, Area($\triangle PCA$) = $k w_2$, and Area($\triangle PAB$) = $k w_3$ for some positive constant $k$.

Step 1: Find the ratio in which the side BC is divided by the line segment AP.

Draw a line segment from vertex A through P, intersecting side BC at point D.

Consider triangles $\triangle PAB$ and $\triangle PCA$. They share the same vertex P. The ratio of their areas is given by: $\frac{\text{Area}(\triangle PAB)}{\text{Area}(\triangle PCA)} = \frac{w_3}{w_2}$.

Now, consider triangles $\triangle ABD$ and $\triangle ACD$. They share the same height from vertex A to the base BC. Thus, the ratio of their areas is equal to the ratio of their bases: $\frac{\text{Area}(\triangle ABD)}{\text{Area}(\triangle ACD)} = \frac{BD}{DC}$.

Similarly, consider triangles $\triangle PBD$ and $\triangle PCD$. They share the same height from vertex P to the base BC. Thus: $\frac{\text{Area}(\triangle PBD)}{\text{Area}(\triangle PCD)} = \frac{BD}{DC}$.

From the above two relations, we have: $\frac{\text{Area}(\triangle ABD)}{\text{Area}(\triangle ACD)} = \frac{\text{Area}(\triangle PBD)}{\text{Area}(\triangle PCD)}$. This implies: $\frac{\text{Area}(\triangle ABD) - \text{Area}(\triangle PBD)}{\text{Area}(\triangle ACD) - \text{Area}(\triangle PCD)} = \frac{\text{Area}(\triangle ABD)}{\text{Area}(\triangle ACD)}$. The numerator $\text{Area}(\triangle ABD) - \text{Area}(\triangle PBD)$ is $\text{Area}(\triangle PAB)$. The denominator $\text{Area}(\triangle ACD) - \text{Area}(\triangle PCD)$ is $\text{Area}(\triangle PCA)$. So, $\frac{\text{Area}(\triangle PAB)}{\text{Area}(\triangle PCA)} = \frac{\text{Area}(\triangle ABD)}{\text{Area}(\triangle ACD)} = \frac{BD}{DC}$.

Therefore, $\frac{BD}{DC} = \frac{w_3}{w_2}$. This means that point D divides the side BC in the ratio $w_3 : w_2$. Using the section formula, the coordinates of D are: $D = \left( \frac{w_2 x_2 + w_3 x_3}{w_2 + w_3}, \frac{w_2 y_2 + w_3 y_3}{w_2 + w_3} \right)$.

Step 2: Find the ratio in which the line segment AD is divided by P.

Now, consider the line segment AD. Point P lies on AD.

Consider $\triangle PAB$ and $\triangle PBD$. They share the same height from B to AD. $\frac{\text{Area}(\triangle ABP)}{\text{Area}(\triangle DBP)} = \frac{AP}{PD}$. Similarly, $\triangle ACP$ and $\triangle DCP$ share height from C to AD. $\frac{\text{Area}(\triangle ACP)}{\text{Area}(\triangle DCP)} = \frac{AP}{PD}$.

Therefore, $\frac{AP}{PD} = \frac{\text{Area}(\triangle ABP) + \text{Area}(\triangle ACP)}{\text{Area}(\triangle DBP) + \text{Area}(\triangle DCP)} = \frac{\text{Area}(\triangle PAB) + \text{Area}(\triangle PCA)}{\text{Area}(\triangle PBC)}$. We know Area($\triangle PAB$) = $k w_3$, Area($\triangle PCA$) = $k w_2$, and Area($\triangle PBC$) = $k w_1$. So, $\frac{AP}{PD} = \frac{k w_3 + k w_2}{k w_1} = \frac{w_2 + w_3}{w_1}$. This means that point P divides the line segment AD in the ratio $(w_2 + w_3) : w_1$.

Step 3: Use the section formula to find the coordinates of P.

Point P divides AD in the ratio $(w_2 + w_3) : w_1$. Using the section formula for point P: $x = \frac{w_1 \cdot x_A + (w_2 + w_3) \cdot x_D}{w_1 + (w_2 + w_3)}$ $y = \frac{w_1 \cdot y_A + (w_2 + w_3) \cdot y_D}{w_1 + (w_2 + w_3)}$

Substitute $x_A = x_1$, $y_A = y_1$ and the coordinates of D: $x = \frac{w_1 x_1 + (w_2 + w_3) \left( \frac{w_2 x_2 + w_3 x_3}{w_2 + w_3} \right)}{w_1 + w_2 + w_3}$ $x = \frac{w_1 x_1 + w_2 x_2 + w_3 x_3}{w_1 + w_2 + w_3}$

Similarly for the $y$-coordinate: $y = \frac{w_1 y_1 + (w_2 + w_3) \left( \frac{w_2 y_2 + w_3 y_3}{w_2 + w_3} \right)}{w_1 + w_2 + w_3}$ $y = \frac{w_1 y_1 + w_2 y_2 + w_3 y_3}{w_1 + w_2 + w_3}$

Thus, the given relations are proven.

The point P is the barycenter (or center of mass) of the system of masses $w_1, w_2, w_3$ placed at $A, B, C$ respectively.