Difference between revisions of "2012 AMC 10B Problems/Problem 21"
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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>. | 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{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\text{'s}</math> can be the lengths of an equilateral triangle formed from connecting the dots. | 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\text{'s}</math> can be the lengths of an equilateral triangle formed from connecting the dots. | ||
| − | So, <math>b=\sqrt{3}a</math>, so <math>b:a= \sqrt{3} | + | So, <math>b=\sqrt{3}a</math>, so <math>b:a= \boxed{\sqrt{3} \: \textbf{(A)}}</math> |
<asy>draw((0, 0)--(1/2, sqrt(3)/2)--(1, 0)--cycle); | <asy>draw((0, 0)--(1/2, sqrt(3)/2)--(1, 0)--cycle); | ||
draw((1/2, sqrt(3)/2)--(2, 0)--(1,0)); | draw((1/2, sqrt(3)/2)--(2, 0)--(1,0)); | ||
Revision as of 01:15, 12 February 2017
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 
![[asy]draw((0, 0)--(1/2, sqrt(3)/2)--(1, 0)--cycle); draw((1/2, sqrt(3)/2)--(2, 0)--(1,0)); label("$a$", (0, 0)--(1, 0), S); label("$a$", (1, 0)--(2, 0), S); label("$a$", (0, 0)--(1/2, sqrt(3)/2), NW); label("$a$", (1, 0)--(1/2, sqrt(3)/2), NE); label("$b=\sqrt{3}a$", (1/2, sqrt(3)/2)--(2, 0), NE); [/asy]](http://latex.artofproblemsolving.com/c/1/6/c169ac2474df3176c192ec03c2214eff85ecce50.png)
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. 
