Canadian Mathematical Olympiad

Solution:

Let $O$ be the origin. Let $A_{i}$ have position vector $\vec{a}_{i}$ for $i=1,2,3,4$.

Since $OB_{i}$ is parallel and equal in length to $A_{i}A_{i+1}$, we have:

$$\vec{OB}_{i} = \vec{b}_{i} = \vec{a}_{i+1} - \vec{a}_{i}$$

So $B_{i}$ has position vector $\vec{b}_{i} = \vec{a}_{i+1} - \vec{a}_{i}$.

The area of quadrilateral $A_{1}A_{2}A_{3}A_{4}$ is:

$$S_{A} = \frac{1}{2} | (\vec{a}_{2} - \vec{a}_{1}) \times (\vec{a}_{3} - \vec{a}_{1}) + (\vec{a}_{4} - \vec{a}_{1}) \times (\vec{a}_{3} - \vec{a}_{1}) |$$

But for a quadrilateral with consecutive vertices $P_{1}, P_{2}, P_{3}, P_{4}$, the area is:

$$S = \frac{1}{2} | (\vec{P}_{1} \times \vec{P}_{2}) + (\vec{P}_{2} \times \vec{P}_{3}) + (\vec{P}_{3} \times \vec{P}_{4}) + (\vec{P}_{4} \times \vec{P}_{1}) |$$

Apply this to $A_{1}A_{2}A_{3}A_{4}$:

$$S_{A} = \frac{1}{2} | \vec{a}_{1} \times \vec{a}_{2} + \vec{a}_{2} \times \vec{a}_{3} + \vec{a}_{3} \times \vec{a}_{4} + \vec{a}_{4} \times \vec{a}_{1} |$$

Similarly, for $B_{1}B_{2}B_{3}B_{4}$:

$$S_{B} = \frac{1}{2} | \vec{b}_{1} \times \vec{b}_{2} + \vec{b}_{2} \times \vec{b}_{3} + \vec{b}_{3} \times \vec{b}_{4} + \vec{b}_{4} \times \vec{b}_{1} |$$

Recall $\vec{b}_{i} = \vec{a}_{i+1} - \vec{a}_{i}$ for $i=1,2,3,4$ (with $A_{5} = A_{1}$).

Compute $\vec{b}_{1} \times \vec{b}_{2}$:

$$\vec{b}_{1} \times \vec{b}_{2} = (\vec{a}_{2} - \vec{a}_{1}) \times (\vec{a}_{3} - \vec{a}_{2})$$

Expand:

$$= \vec{a}_{2} \times \vec{a}_{3} - \vec{a}_{2} \times \vec{a}_{2} - \vec{a}_{1} \times \vec{a}_{3} + \vec{a}_{1} \times \vec{a}_{2}$$

But $\vec{a}_{2} \times \vec{a}_{2} = 0$, so:

$$= \vec{a}_{2} \times \vec{a}_{3} - \vec{a}_{1} \times \vec{a}_{3} + \vec{a}_{1} \times \vec{a}_{2}$$

Similarly, compute all four terms:

1. $\vec{b}_{1} \times \vec{b}_{2} = \vec{a}_{2} \times \vec{a}_{3} - \vec{a}_{1} \times \vec{a}_{3} + \vec{a}_{1} \times \vec{a}_{2}$
2. $\vec{b}_{2} \times \vec{b}_{3} = (\vec{a}_{3} - \vec{a}_{2}) \times (\vec{a}_{4} - \vec{a}_{3}) = \vec{a}_{3} \times \vec{a}_{4} - \vec{a}_{2} \times \vec{a}_{4} - \vec{a}_{3} \times \vec{a}_{3} + \vec{a}_{2} \times \vec{a}_{3} = \vec{a}_{3} \times \vec{a}_{4} - \vec{a}_{2} \times \vec{a}_{4} + \vec{a}_{2} \times \vec{a}_{3}$
3. $\vec{b}_{3} \times \vec{b}_{4} = (\vec{a}_{4} - \vec{a}_{3}) \times (\vec{a}_{1} - \vec{a}_{4}) = \vec{a}_{4} \times \vec{a}_{1} - \vec{a}_{3} \times \vec{a}_{1} - \vec{a}_{4} \times \vec{a}_{4} + \vec{a}_{3} \times \vec{a}_{4} = \vec{a}_{4} \times \vec{a}_{1} - \vec{a}_{3} \times \vec{a}_{1} + \vec{a}_{3} \times \vec{a}_{4}$
4. $\vec{b}_{4} \times \vec{b}_{1} = (\vec{a}_{1} - \vec{a}_{4}) \times (\vec{a}_{2} - \vec{a}_{1}) = \vec{a}_{1} \times \vec{a}_{2} - \vec{a}_{4} \times \vec{a}_{2} - \vec{a}_{1} \times \vec{a}_{1} + \vec{a}_{4} \times \vec{a}_{1} = \vec{a}_{1} \times \vec{a}_{2} - \vec{a}_{4} \times \vec{a}_{2} + \vec{a}_{4} \times \vec{a}_{1}$

Add all four terms:

Sum:

$$\begin{align*} & [\vec{a}_{2} \times \vec{a}_{3} - \vec{a}_{1} \times \vec{a}_{3} + \vec{a}_{1} \times \vec{a}_{2}] \\ & + [\vec{a}_{3} \times \vec{a}_{4} - \vec{a}_{2} \times \vec{a}_{4} + \vec{a}_{2} \times \vec{a}_{3}] \\ & + [\vec{a}_{4} \times \vec{a}_{1} - \vec{a}_{3} \times \vec{a}_{1} + \vec{a}_{3} \times \vec{a}_{4}] \\ & + [\vec{a}_{1} \times \vec{a}_{2} - \vec{a}_{4} \times \vec{a}_{2} + \vec{a}_{4} \times \vec{a}_{1}] \end{align*}$$

Now, group like terms:

- $\vec{a}_{2} \times \vec{a}_{3}$ appears twice
- $\vec{a}_{3} \times \vec{a}_{4}$ appears twice
- $\vec{a}_{4} \times \vec{a}_{1}$ appears twice
- $\vec{a}_{1} \times \vec{a}_{2}$ appears twice

Negative terms:
- $-\vec{a}_{1} \times \vec{a}_{3}$
- $-\vec{a}_{2} \times \vec{a}_{4}$
- $-\vec{a}_{3} \times \vec{a}_{1}$
- $-\vec{a}_{4} \times \vec{a}_{2}$

But $\vec{a}_{1} \times \vec{a}_{3} + \vec{a}_{3} \times \vec{a}_{1} = 0$ (since $\vec{u} \times \vec{v} = -\vec{v} \times \vec{u}$), so $-\vec{a}_{1} \times \vec{a}_{3} - \vec{a}_{3} \times \vec{a}_{1} = 0$.
Similarly, $-\vec{a}_{2} \times \vec{a}_{4} - \vec{a}_{4} \times \vec{a}_{2} = 0$.

So the sum is:

$$2[\vec{a}_{1} \times \vec{a}_{2} + \vec{a}_{2} \times \vec{a}_{3} + \vec{a}_{3} \times \vec{a}_{4} + \vec{a}_{4} \times \vec{a}_{1}]$$

Therefore,

$$S_{B} = \frac{1}{2} |2[\vec{a}_{1} \times \vec{a}_{2} + \vec{a}_{2} \times \vec{a}_{3} + \vec{a}_{3} \times \vec{a}_{4} + \vec{a}_{4} \times \vec{a}_{1}]| = |\vec{a}_{1} \times \vec{a}_{2} + \vec{a}_{2} \times \vec{a}_{3} + \vec{a}_{3} \times \vec{a}_{4} + \vec{a}_{4} \times \vec{a}_{1}| = 2S_{A}$$

Thus, the area of $B_{1}B_{2}B_{3}B_{4}$ is twice that of $A_{1}A_{2}A_{3}A_{4}$.