1952 AHSME Problems/Problem 31

Revision as of 14:18, 28 February 2019 by Olivera (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Given $12$ points in a plane no three of which are collinear, the number of lines they determine is:

$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 54 \qquad \textbf{(C)}\ 120 \qquad \textbf{(D)}\ 66 \qquad \textbf{(E)}\ \text{none of these}$

Solution

Since no three points are collinear, every two points must determine a distinct line. Thus, there are $\dbinom{12}{2} = \frac{12\cdot11}{2} = 66$ lines.

Therefore, the answer is $\fbox{(D) 66}$

See also

1952 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 30
Followed by
Problem 32
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 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png