Question

Given five points A = (7, 4), B = (-10, 0), C = (-10, 3), D = (0, 10) and E = (7, 7). Every second, all the points move by halving their abscissas (x-coordinates) and doubling their ordinates (y-coordinates). This process continues for 500 years. After 500 years, which two points are closest?

• A. A and B
• B. B and C
• C. A and E
• D. A and C

Hint:

After many steps the x-coordinates shrink to near zero while the y-coordinates blow up, so the closest pair is simply the one whose starting y-values are nearest to each other.

Answer 1

The Correct Option is D

Step 1: Write the position of a point after \(n\) steps.
If a point starts at \((x_0, y_0)\), after \(n\) seconds its abscissa has been halved \(n\) times and its ordinate doubled \(n\) times:
\[ x(n) = \frac{x_0}{2^n}, \qquad y(n) = y_0 \cdot 2^n \]

Step 2: See what happens to the x-coordinates as \(n\) grows very large.
For every point, \(x(n) = \frac{x_0}{2^n} \to 0\) as \(n\) becomes large, since 500 years gives a huge number of steps, regardless of the starting \(x_0\). So after a long time, all x-coordinates become tiny, and differences between them shrink to almost nothing.

Step 3: See what happens to the y-coordinates.
Every nonzero \(y_0\) gets doubled again and again, so \(y(n) = y_0 \cdot 2^n\) grows without bound. The gap between two points' y-values also grows: if two points start with y-values \(y_1\) and \(y_2\), their gap after \(n\) steps is \((y_1 - y_2) \cdot 2^n\), scaled up by the same huge factor \(2^n\) for every pair.

Step 4: Compare pairs using only their starting y-values.
Since the x-part becomes negligible and every pair's y-gap is scaled by the same \(2^n\), the distance between two points after a long time is dominated by \(|y_1 - y_2| \cdot 2^n\). The pair that stays closest is the pair with the smallest \(|y_1-y_2|\) at the start, since scaling by the same huge factor keeps the ranking of the gaps unchanged.
The starting y-values are A: 4, B: 0, C: 3, D: 10, E: 7. Checking the given choices:
A and B: \(|4-0| = 4\); B and C: \(|0-3| = 3\); A and E: \(|4-7| = 3\); A and C: \(|4-3| = 1\).

Step 5: Pick the smallest gap.
A and C have the smallest y-gap, just 1, smaller than every other pair, including ones not listed as choices (A-D: 6, B-D: 10, B-E: 7, C-D: 7, C-E: 4, D-E: 3). So A and C stay the closest pair as time goes on.

Final Answer:
The two points that remain closest after many iterations are A and C. \[ \boxed{\text{A and C}} \]