1961 IMO Problems/Problem 4
Revision as of 01:23, 8 February 2009 by Brut3Forc3 (talk | contribs) ((Previous edit was to slightly reword problem) Added solution.)
Problem
In the interior of triangle a point
is given. Let
be the intersections of
with the opposing edges of triangle
. Prove that among the ratios
there exists one not larger than
and one not smaller than
.
Solution
Since triangles and
share the base
, we have
, where
denotes the area of triangle
. Similarly,
. Adding all of these gives
, or
We see that we must have at least one of the three fractions less than or equal to
, and at least one greater than
. These correspond to ratios
being less than or equal to
, and greater than or equal to
, respectively, so we are done.
1961 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
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