Difference between revisions of "2012 AMC 10B Problems/Problem 21"
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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 | ||
| − | == Solution (using the answer choices) == | + | == Solution 3 (using the answer choices) == |
We know that <math>a-b-2a</math> form a triangle. From triangle inequality, we see that <math>b>a</math>. Then, we also see that there is an isosceles triangle with lengths <math>a-a-b</math>. From triangle inequality: <math>b<2a</math>. The only answer choice that holds these two inequalities is: <math>\sqrt{3}</math>. | We know that <math>a-b-2a</math> form a triangle. From triangle inequality, we see that <math>b>a</math>. Then, we also see that there is an isosceles triangle with lengths <math>a-a-b</math>. From triangle inequality: <math>b<2a</math>. The only answer choice that holds these two inequalities is: <math>\sqrt{3}</math>. | ||
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==Video Solution by Richard Rusczyk== | ==Video Solution by Richard Rusczyk== | ||
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~dolphin7 | ~dolphin7 | ||
| + | |||
| + | (Direct Youtube Link) | ||
| + | https://www.youtube.com/watch?v=6sL7bhkVivo | ||
| + | |||
| + | ~lukiebear | ||
== See Also == | == See Also == | ||
Latest revision as of 19:18, 23 October 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 
![[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)
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 3 (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
(Direct Youtube Link) https://www.youtube.com/watch?v=6sL7bhkVivo
~lukiebear
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. 
