Difference between revisions of "1973 AHSME Problems/Problem 30"

Latest revision as of 13:04, 20 February 2020

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 $\boxed{\textbf{(B)}}$ .

1973 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 29	Followed by Problem 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

@@ Line 18: / Line 18: @@
 ==See Also==
-{{AHSME 35p box|year=1973|num-b=29|num-a=31}}
+{{AHSME 30p box|year=1973|num-b=29|num-a=31}}
 [[Category:Introductory Geometry Problems]]

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Difference between revisions of "1973 AHSME Problems/Problem 30"

Latest revision as of 13:04, 20 February 2020

Problem

Solution

See Also