Difference between revisions of "1974 IMO Problems/Problem 2"
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The above solution was posted and copyrighted by pontios. The original thread for this problem can be found here: [https://aops.com/community/p365059] | The above solution was posted and copyrighted by pontios. The original thread for this problem can be found here: [https://aops.com/community/p365059] | ||
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+ | == See Also == {{IMO box|year=1974|num-b=1|num-a=3}} |
Latest revision as of 14:56, 29 January 2021
Problem
In the triangle , prove that there is a point on side such that is the geometric mean of and if and only if .
Solution
Let a point on the side . Let the altitude of the triangle , and the symmetric point of through . We bring a parallel line from to . This line intersects the ray at the point , and we know that .
The distance between the parallel lines and is .
Let the circumscribed circle of , and the perpendicular diameter to , such that are on difererent sides of the line .
In fact, the problem asks when the line intersects the circumcircle. Indeed:
Suppose that is the geometric mean of .
Then, from the power of we can see that is also a point of the circle . Or else, the line intersects
where is the altitude of the isosceles .
We use the formulas:
and
So we have
For
Suppose that
Then we can go inversely and we find that the line intersects the circle (without loss of generality; if then is tangent to at )
So, if then for the point we have and
The above solution was posted and copyrighted by pontios. The original thread for this problem can be found here: [1]
See Also
1974 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |