Difference between revisions of "2012 AMC 10B Problems/Problem 21"
m (→Solution (using the answer choices)) |
|||
| (43 intermediate revisions by 14 users not shown) | |||
| Line 1: | Line 1: | ||
| − | When you see a and 2a, | + | [[Category: Introductory Geometry Problems]] |
| − | Drawing the points out, it is possible to have a diagram where b= | + | ==Problem== |
| − | So, b= | + | Four distinct points are arranged on a plane so that the segments connecting them have lengths <math>a</math>, <math>a</math>, <math>a</math>, <math>a</math>, <math>2a</math>, and <math>b</math>. What is the ratio of <math>b</math> to <math>a</math>? |
| + | |||
| + | <math>\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \pi</math> | ||
| + | |||
| + | |||
| + | ==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{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. | ||
| + | So, <math>b=\sqrt{3}a</math>, so <math>b:a= \boxed{\textbf{(A)} \: \sqrt{3}}</math> | ||
| + | <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> | ||
| + | |||
| + | == Solution 2 == | ||
| + | |||
| + | 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 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)}\sqrt{3}}</math> | ||
| + | |||
| + | ~ Nafer | ||
| + | |||
| + | == 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>. | ||
| + | |||
| + | ==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 == | ||
| + | |||
| + | |||
| + | {{AMC10 box|year=2012|ab=B|num-b=20|num-a=22}} | ||
| + | {{MAA Notice}} | ||
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. 
