Сумма векторов в треугольнике с использованием середины

Problem

In right isosceles triangle $ABC$ with $\angle B = 90^\circ$ and $BA = BC = \sqrt{2}$, eight points $P_1, P_2, \ldots, P_8$ divide hypotenuse $AC$ into $9$ equal segments; find the magnitude of $\overrightarrow{BP_1} + \overrightarrow{BP_2} + \cdots + \overrightarrow{BP_8}$.

Step 1: Place the triangle on coordinates

Put $B$ at the origin, $A$ on the $x$-axis, and $C$ on the $y$-axis. Then

$$B=(0,0), \quad A=(\sqrt{2},0), \quad C=(0,\sqrt{2}).$$

Since $B$ is the origin, each vector $\overrightarrow{BP_k}$ is just the position vector of $P_k$.

Step 2: Use the midpoint of $AC$

The points $P_1$ through $P_8$ are evenly spaced on $AC$, so their average position is the midpoint $M$ of $AC$. Therefore,

$$\overrightarrow{BP_1}+\overrightarrow{BP_2}+\cdots+\overrightarrow{BP_8}=8\,\overrightarrow{BM}.$$

The midpoint of $AC$ is

$$M=\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right).$$

Step 3: Find the magnitude

So

$$8\,\overrightarrow{BM}=8\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)=(4\sqrt{2},4\sqrt{2}).$$

Its magnitude is

$$\sqrt{(4\sqrt{2})^2+(4\sqrt{2})^2} = \sqrt{32+32} = \sqrt{64} = 8.$$

Answer

The magnitude of the vector sum is $8$.