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

## 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

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 $2*1=\boxed{2}$