Difference between revisions of "2012 AMC 10B Problems/Problem 21"
(→Solution 2) |
(→Solution 2) |
||
Line 21: | Line 21: | ||
== Solution 2 == | == Solution 2 == | ||
− | For any <math>4</math> | + | For any <math>4</math> non-collinear points with the given requirement, notice that there must be a triangle with side lengths <math>a</math>, <math>a</math>, <math>2a</math>, which is not possible as <math>a+a=2a</math>. Thus at least <math>3</math> of the <math>4</math> points must be collinear. |
If all <math>4</math> points are collinear, then there would only be <math>3</math> lines of length <math>a</math>, which wouldn't work. | If all <math>4</math> points are collinear, then there would only be <math>3</math> lines of length <math>a</math>, which wouldn't work. |
Revision as of 16:41, 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
For any non-collinear points with the given requirement, notice that there must be a triangle with side lengths
,
,
, which is not possible as
. Thus at least
of the
points must be collinear.
If all points are collinear, then there would only be
lines of length
, which wouldn't work.
If exactly points are collinear, the only possibility that works is when a
triangle is formed.
Thus , or
~ Nafer
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.