# 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: $[asy] size(300 pt); draw(circle((0,0),1)); draw(circle((2,0),1)); draw((1,0)--(0.5,0.86602),linewidth(.5)); draw((0,0)--(0.5,0.86602),linewidth(.5)); draw((1,0)--(1.5,0.86602),linewidth(.5)); draw((2,0)--(1.5,0.86602),linewidth(.5)); draw((0.5,0.86602)--(1.5,0.86602),linewidth(.5)); dot((0,0)); dot((2,0)); dot((1,0)); dot((0.5,0.86602)); dot((1.5,0.86602)); label("P",(1,0),SW); label("1",(0.25,0.43301),NW); label("1",(1.75,0.43301),NE); label("1",(0.75,0.43301),SW); label("1",(1.25,0.43301),SE); label("1",(1,0.86602),N); [/asy]$ 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: 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: $[asy] size(300pt); draw(circle((0,0),3)); draw(circle((8.19615,0),3)); draw((2.59807,1.5)--(2.59807,-1.5),linewidth(.5)); draw((2.59807,1.5)--(5.19615,0),linewidth(.5)); draw((5.19615,0)--(2.59807,-1.5),linewidth(.5)); draw((0,0)--(2.59807,-1.5),linewidth(.5)); draw((0,0)--(2.59807,1.5),linewidth(.5)); draw((0,0)--(8.19615,0),linewidth(.5)); dot((2.59807,1.5)); dot((2.59807,-1.5)); dot((5.19615,0)); dot((8.19615,0)); dot((0,0)); label("1",(4.09807,0.5),N); label("\frac {\sqrt3}{2}",(4.09807,0),SW); label("1",(6.69615,0),S); label("A",(0,0),SW); label("B",(2.59807,0),SW); label("C",(5.19615,0),SSW); [/asy]$ The segments $AB=BC=\frac {\sqrt3}{2}$ because of $30-60-90$ triangles. From this diagram, we can see that the distance between the centers is $\boxed{1+\sqrt3}$.

## See also

 2014 UNM-PNM Contest II (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 All UNM-PNM Problems and Solutions
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