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

(Solution)
(Solution)
 
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label("$1$",(1,0.86602),N);
 
label("$1$",(1,0.86602),N);
 
</asy>
 
</asy>
Where <math>P</math> is the point of tangency. This clearly works, so the smallest distance would be <math>2*1=\boxed{2}</math>
+
Where <math>P</math> is the point of tangency. This clearly works, so the smallest distance would be <math>2\cdot1=\boxed{2}</math>
  
  

Latest revision as of 22:44, 17 November 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

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 $2\cdot1=\boxed{2}$


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 (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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