Difference between revisions of "2014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 4"
(→Solution) |
(→Solution) |
||
Line 37: | Line 37: | ||
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((0.86602,0.5)--(1.73205,0),linewidth(.5)) | + | draw((0.86602,0.5)--(1.73205,0),linewidth(.5)); |
draw((1.73205,0)--(0.86602,-0.5),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)); |
Revision as of 21:58, 27 September 2019
Problem
Find the smallest and largest possible distances between the centers of two circles of radius such that there is an equilateral triangle of side length 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: Where is the point of tangency. This clearly works, so the smallest distance would be
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 (a side of the triangle), then adding the other two sides outwards like shown:
See also
2014 UNM-PNM Contest II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNM-PNM Problems and Solutions |