2001 Pan African MO Problems/Problem 3
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 . .
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.
Note that the coordinates of as well as the distance cycle after . Thus, if , if , if , and if .
|2001 Pan African MO (Problems)|
|1 • 2 • 3 • 4 • 5 • 6||Followed by|
|All Pan African MO Problems and Solutions|