Problem 1633

Let's break down the problem into two parts and solve each part step-by-step.

### Part (a)
Determine the ratio between the area of the trapezium $ABB'A'$ and the area of the triangle $A'B'X$.

1. Understanding the given conditions:
- The points $X, Y, Z$ are marked on the sides $AB, BC, AC$ of the triangle $ABC$, respectively.
- Points $A', B', C'$ are on the sides $XZ, XY, YZ$ of the triangle $XYZ$, respectively, such that:
$$\frac{AB}{A'B'} = \frac{BC}{B'C'} = 2$$
- $ABB'A'$, $BCC'B'$, and $ACC'A'$ are trapezoids with the sides of the triangle $ABC$ as bases.

2. Using the given ratio:
- Since $\frac{AB}{A'B'} = 2$, it implies that $A'B'$ is half the length of $AB$.

3. **Area of trapezium $ABB'A'$:**
- The area of a trapezium is given by:
$$\text{Area} = \frac{1}{2} \times ( \text{sum of parallel sides} ) \times \text{height}$$
- Here, the parallel sides are $AB$ and $A'B'$, and the height is the same as the height of $\triangle ABC$ from $A$ to $BC$.

4. **Area of triangle $A'B'X$:**
- Since $A'B'$ is half the length of $AB$, and the height from $X$ to $A'B'$ is also half the height from $A$ to $BC$ (due to similar triangles), the area of $\triangle A'B'X$ is:
$$\text{Area of } \triangle A'B'X = \frac{1}{2} \times A'B' \times \text{height from } X \text{ to } A'B'$$
- Since both the base and height are halved, the area of $\triangle A'B'X$ is:
$$\text{Area of } \triangle A'B'X = \frac{1}{4} \times \text{Area of } \triangle ABC$$

5. Ratio of areas:
- The area of trapezium $ABB'A'$ is:
$$\text{Area of } ABB'A' = \frac{1}{2} \times (AB + A'B') \times \text{height}$$
- Since $A'B' = \frac{1}{2} AB$, we have:
$$\text{Area of } ABB'A' = \frac{1}{2} \times \left( AB + \frac{1}{2} AB \right) \times \text{height} = \frac{1}{2} \times \frac{3}{2} AB \times \text{height} = \frac{3}{4} \times \text{Area of } \triangle ABC$$
- Therefore, the ratio between the area of the trapezium $ABB'A'$ and the area of the triangle $A'B'X$ is:
$$\frac{\text{Area of } ABB'A'}{\text{Area of } \triangle A'B'X} = \frac{\frac{3}{4} \times \text{Area of } \triangle ABC}{\frac{1}{4} \times \text{Area of } \triangle ABC} = 3$$

### Part (b)
Determine the ratio between the area of the triangle $XYZ$ and the area of the triangle $ABC$.

1. Using homothety:
- The homothety centered at $O$ (the circumcenter of $\triangle ABC$) with a ratio of $\frac{1}{2}$ maps $\triangle ABC$ to $\triangle A'B'C'$.

2. Area ratio:
- Since the homothety ratio is $\frac{1}{2}$, the area ratio between $\triangle A'B'C'$ and $\triangle ABC$ is:
$$\left( \frac{1}{2} \right)^2 = \frac{1}{4}$$

3. **Area of $\triangle XYZ$:**
- Since $\triangle A'B'C'$ is similar to $\triangle XYZ$ with a ratio of $\frac{1}{2}$, the area of $\triangle XYZ$ is:
$$\text{Area of } \triangle XYZ = \frac{1}{2} \times \text{Area of } \triangle ABC$$

The final answer is: