Difference between revisions of "1991 IMO Problems/Problem 5"
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Therefore, at least one of the angles <math> \,\angle PAB,\;\angle PBC,\;\angle PCA\,</math> is less than or equal to <math> 30^{\circ }</math>. | Therefore, at least one of the angles <math> \,\angle PAB,\;\angle PBC,\;\angle PCA\,</math> is less than or equal to <math> 30^{\circ }</math>. | ||
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+ | ~Tomas Diaz, orders@tomasdiaz.com | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 12:29, 12 November 2023
Problem
Let be a triangle and
an interior point of
. Show that at least one of the angles
is less than or equal to
.
Solution
Let ,
, and
be
,
,
, respectively.
Let ,
, and
be
,
,
, respcetively.
Using law of sines on we get:
, therefore,
Using law of sines on we get:
, therefore,
Using law of sines on we get:
, therefore,
Multiply all three equations we get:
Using AM-GM we get:
. [Inequality 1]
Note that for ,
decreases with increasing
and fixed
Therefore, decreases with increasing
and fixed
From trigonometric identity:
,
since , then:
Therefore,
and also,
Adding these two inequalities we get:
.
. [Inequality 2]
Combining [Inequality 1] and [Inequality 2] we see the following:
This implies that for at least one of the values of ,
,or
, the following is true:
or
Which means that for at least one of the values of ,
,or
, the following is true:
Therefore, at least one of the angles is less than or equal to
.
~Tomas Diaz, orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.