2014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 4

Revision as of 21:58, 27 September 2019 by Someonenumber011 (talk | contribs) (Solution)

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

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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: [asy] 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)); [/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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