Difference between revisions of "2001 Pan African MO Problems/Problem 3"
Rockmanex3 (talk | contribs) m |
Rockmanex3 (talk | contribs) m (Changed difficulty to Intermediate Geo.) |
||
Line 79: | Line 79: | ||
}} | }} | ||
− | [[Category: | + | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 10:09, 27 March 2020
Problem
Let be an equilateral triangle and let be a point outside this triangle, such that is an isosceles triangle with a right angle at . A grasshopper starts from and turns around the triangle as follows. From the grasshopper jumps to , which is the symmetric point of with respect to . From , the grasshopper jumps to , which is the symmetric point of with respect to . Then the grasshopper jumps to which is the symmetric point of with respect to , and so on. Compare the distance and . .
Solution
We can use coordinate geometry to solve the problem. Let , , and , making . To calculate the coordinates of , note that since is a kite. Thus, bissects , so . Additionally, because bissects . Thus, the coordinates of are .
By repeatedly applying the Midpoint Formula, we can determine the coordinates of , , , and so on. We can also use the Distance Formula to calculate the distance of , , and so on. The values are shown in the below table.
Coordinates of | ||
1 | ||
2 | ||
3 | ||
4 | ||
5 | ||
6 | ||
7 | ||
8 |
Note that the coordinates of as well as the distance cycle after . Thus, if , if , if , and if .
See Also
2001 Pan African MO (Problems) | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All Pan African MO Problems and Solutions |