Difference between revisions of "2012 AMC 10B Problems/Problem 21"
Flamedragon (talk | contribs) (→Problem 21) |
(→Problem) |
||
| Line 3: | Line 3: | ||
<math>\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \pi</math> | <math>\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \pi</math> | ||
| + | [[Category: Introductory Geometry Problems]] | ||
==Solution== | ==Solution== | ||
Revision as of 10:46, 13 August 2014
Problem
Four distinct points are arranged on a plane so that the segments connecting them have lengths
,
,
,
,
, and
. What is the ratio of
to
?
Solution
When you see that there are lengths a and 2a, one could think of 30-60-90 triangles. Since all of the other's lengths are a, you could think that
.
Drawing the points out, it is possible to have a diagram where
. It turns out that
and
could be the lengths of a 30-60-90 triangle, and the other 3
can be the lengths of an equilateral triangle formed from connecting the dots.
So,
, so
See Also
| 2012 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.