1975 IMO Problems/Problem 3
Problems
On the sides of an arbitrary triangle , triangles are constructed externally with . Prove that and .
Solution 1
Consider and so that and . Furthermore, let and intersect at . Now, this means that and , so . Thus, and , so is equilateral. Similarly is . Yet, it is well-known that the intersection of and , which is , must be the Fermat Point of if and are equilateral. Now, . Similarly, , so , so a spiral similarity maps . This implies that . Similarly, , so . Then, . We also realize that . Now, is a rotation of degrees around and is the same rotation of around , so maps to from this rotation, so . It follows that .
The above solution was posted and copyrighted by The QuattoMaster 6000. The original thread for this problem can be found here: [1]
Solution 2
Let be the equilateral triangle constructed such that and are on the same side. . We have from similarity. Also we have . So . Then and . Similar calculations for . We will have and . Also from the question we have . So and .
The above solution was posted and copyrighted by xeroxia. The original thread for this problem can be found here: [2]
Solution 3
Define to be a point such that is directly similar to .
Then, it is trivial to show that and that ; that is, and is a right isosceles triangle. If we prove that is similar to , then we will be done.
According to the property of spiral similarity, it suffices to prove that
Since , we have , and this gives . It remains to prove that .
As , we must prove that . From Law of Sines on , we have . Since , it remains to prove that , which is easily verified. We are done.
The above solution was posted and copyrighted by fantasylover. The original thread for this problem can be found here: [3]
See Also
1975 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |