Difference between revisions of "2012 AMC 10B Problems/Problem 21"
m (→Solution 2) |
m (→Solution 2) |
||
Line 28: | Line 28: | ||
If exactly <math>3</math> points are collinear, the only possibility that works is when a <math>30^{\circ}-90^{\circ}-60^{\circ}</math> triangle is formed. | If exactly <math>3</math> points are collinear, the only possibility that works is when a <math>30^{\circ}-90^{\circ}-60^{\circ}</math> triangle is formed. | ||
− | Thus <math>b=\sqrt{3}a</math>, or <math>\frac{b}{a}=\boxed{\mathrm{(A) | + | Thus <math>b=\sqrt{3}a</math>, or <math>\frac{b}{a}=\boxed{\mathrm{(A)}\sqrt{3}}</math> |
~ Nafer | ~ Nafer |
Latest revision as of 22:57, 15 May 2021
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
Solution (using the answer choices)
We know that form a triangle. From triangle inequality, we see that . Then, we also see that there is an isosceles triangle with lengths . From triangle inequality: . The only answer choice that holds these two inequalities is: .
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012amc10b/271
~dolphin7
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.