# 1973 AHSME Problems/Problem 30

## Problem

Let $[t]$ denote the greatest integer $\leq t$ where $t \geq 0$ and $S = \{(x,y): (x-T)^2 + y^2 \leq T^2 \text{ where } T = t - [t]\}$. Then we have

$\textbf{(A)}\ \text{the point } (0,0) \text{ does not belong to } S \text{ for any } t \qquad$

$\textbf{(B)}\ 0 \leq \text{Area } S \leq \pi \text{ for all } t \qquad$

$\textbf{(C)}\ S \text{ is contained in the first quadrant for all } t \geq 5 \qquad$

$\textbf{(D)}\ \text{the center of } S \text{ for any } t \text{ is on the line } y=x \qquad$

$\textbf{(E)}\ \text{none of the other statements is true}$

## Solution

The region $S$ is a circle radius $T$ and center $(T,0)$. Since $T = t-[t]$, $0 \le T < 1$. That means the area of the circle is less than $\pi$, and since the region can also be just a dot (achieved when $t$ is integer), the answer is $\boxed{\textbf{(B)}}$.

## See Also

 1973 AHSME (Problems • Answer Key • Resources) Preceded byProblem 29 Followed byProblem 31 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 All AHSME Problems and Solutions