Difference between revisions of "2012 AMC 10B Problems/Problem 21"
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label("$b=\sqrt{3}a$", (1/2, sqrt(3)/2)--(2, 0), NE); | label("$b=\sqrt{3}a$", (1/2, sqrt(3)/2)--(2, 0), NE); | ||
</asy> | </asy> | ||
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+ | == Solution 2 == | ||
== See Also == | == See Also == |
Revision as of 16:29, 6 August 2019
Contents
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
Solution 2
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.