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] | ||
+ | |||
+ | == See Also == {{IMO box|year=1974|num-b=1|num-a=3}} |
Latest revision as of 15:56, 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 |