Difference between revisions of "2014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 4"

(Solution)
(Solution)
Line 32: Line 32:
  
  
The largest distance would be found by first finding the closest place to the edge of a circle to place a line segment with side length <math>1</math>(a side of the triangle), then adding the other two sides outwards like shown:
+
The largest distance would be found by first finding the closest place to the edge of a circle to place a line segment with side length <math>1</math> (a side of the triangle), then adding the other two sides outwards like shown:
 
<asy>
 
<asy>
 
draw(circle((0,0),1));
 
draw(circle((0,0),1));
 
draw(circle((2.73205,0),1));
 
draw(circle((2.73205,0),1));
 
draw((0.86602,0.5)--(0.86602,-0.5),linewidth(.5));
 
draw((0.86602,0.5)--(0.86602,-0.5),linewidth(.5));
draw((1.86602,0.5)--(1.86602,-0.5),linewidth(.5));
+
draw((0.86602,0.5)--(1.73205,0),linewidth(.5))
 +
draw((1.73205,0)--(0.86602,-0.5),linewidth(.5));
 +
draw((0,0)--(0.86602,-0.5),linewidth(.5));
 +
draw((0,0)--(0.86602,0.5),linewidth(.5));
 +
dot((0.86602,0.5));
 +
dot((0.86602,-0.5));
 +
dot((1.73205,0));
 
</asy>
 
</asy>
  

Revision as of 21:38, 27 September 2019

Problem

Find the smallest and largest possible distances between the centers of two circles of radius $1$ such that there is an equilateral triangle of side length $1$ with two vertices on one of the circles and the third vertex on the second circle.

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

The smallest distance would be found if the two circles were externally tangent, so testing that and messing around with it yields: [asy] draw(circle((0,0),1)); draw(circle((2,0),1)); draw((1,0)--(0.5,0.86602),linewidth(.5)); draw((0,0)--(0.5,0.86602),linewidth(.5)); draw((1,0)--(1.5,0.86602),linewidth(.5)); draw((2,0)--(1.5,0.86602),linewidth(.5)); draw((0.5,0.86602)--(1.5,0.86602),linewidth(.5)); dot((0,0)); dot((2,0)); dot((1,0)); dot((0.5,0.86602)); dot((1.5,0.86602)); label("$P$",(1,0),SW); label("$1$",(0.25,0.43301),NW); label("$1$",(1.75,0.43301),NE); label("$1$",(0.75,0.43301),SW); label("$1$",(1.25,0.43301),SE); label("$1$",(1,0.86602),N); [/asy] Where $P$ is the point of tangency. This clearly works, so the smallest distance would be $2*1=\boxed{2}$


The largest distance would be found by first finding the closest place to the edge of a circle to place a line segment with side length $1$ (a side of the triangle), then adding the other two sides outwards like shown:

draw(circle((0,0),1));
draw(circle((2.73205,0),1));
draw((0.86602,0.5)--(0.86602,-0.5),linewidth(.5));
draw((0.86602,0.5)--(1.73205,0),linewidth(.5))
draw((1.73205,0)--(0.86602,-0.5),linewidth(.5));
draw((0,0)--(0.86602,-0.5),linewidth(.5));
draw((0,0)--(0.86602,0.5),linewidth(.5));
dot((0.86602,0.5));
dot((0.86602,-0.5));
dot((1.73205,0));
 (Error compiling LaTeX. d856f942c24e46cd45e1bf8b0904d19a39cae561.asy: 9.1: syntax error
error: could not load module 'd856f942c24e46cd45e1bf8b0904d19a39cae561.asy')

See also

2014 UNM-PNM Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10
All UNM-PNM Problems and Solutions
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