Difference between revisions of "2012 AMC 10B Problems/Problem 21"
RainbowsHurt (talk | contribs) (→See Also) |
|||
Line 6: | Line 6: | ||
==Solution== | ==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 <math>b=\sqrt{ | + | 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 <math>b=\sqrt{3}a</math>. |
− | Drawing the points out, it is possible to have a diagram where <math>b=\sqrt{ | + | Drawing the points out, it is possible to have a diagram where <math>b=\sqrt{3}a</math>. It turns out that <math>a,</math> <math>2a,</math> and <math>b</math> could be the lengths of a 30-60-90 triangle, and the other 3 <math>"a's"</math> can be the lengths of an equilateral triangle formed from connecting the dots. |
− | So, <math>b=\sqrt{ | + | So, <math>b=\sqrt{3}a</math>, so <math>b:a= \sqrt{3}=(A)</math> |
== See Also == | == See Also == |
Revision as of 15:24, 20 March 2013
Problem 21
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 |