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 distance formula between two moving points after $n$ seconds.
For two points with starting coordinates $(x_1,y_1)$ and $(x_2,y_2)$, after $n$ steps their positions are $\left(\frac{x_1}{2^n}, y_1 2^n\right)$ and $\left(\frac{x_2}{2^n}, y_2 2^n\right)$. The squared distance between them is
\[ D_n^2 = \left(\frac{x_1-x_2}{2^n}\right)^2 + \left((y_1-y_2)2^n\right)^2 \]

Step 2: Factor out the dominant term for large $n$.
As $n$ grows, $\left(\frac{x_1-x_2}{2^n}\right)^2 \to 0$, while $\left((y_1-y_2)2^n\right)^2$ grows fast unless $y_1=y_2$. So for a large $n$ like 500,
\[ D_n \approx |y_1-y_2| \cdot 2^n \]
This means the x-coordinates stop mattering almost completely once $n$ is large, and only the original difference in y-values decides how far apart two points end up.

Step 3: Since $2^n$ is the same multiplier for every pair, only $|y_1-y_2|$ decides the ranking.
Between two different pairs of points, the one with the smaller starting $|y_1-y_2|$ will always have the smaller distance for large $n$, since both distances get multiplied by the identical factor $2^n$.

Step 4: List the y-values and compute all relevant gaps.
A = 4, B = 0, C = 3, D = 10, E = 7. Checking the answer choices: A,B gives 4; B,C gives 3; A,E gives 3; A,C gives 1. Checking the rest for completeness: A,D gives 6; B,D gives 10; B,E gives 7; C,D gives 7; C,E gives 4; D,E gives 3.

Step 5: The smallest gap wins.
The smallest of all these gaps is 1, between A and C. No other pair starts anywhere near this close in y-value.

Final Answer:
As time goes on, A and C become the closest pair of points, since their y-coordinates started nearest to each other and the x-coordinates become irrelevant. \[ \boxed{\text{A and C}} \]