Difference between revisions of "1991 IMO Problems/Problem 5"
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Draw the segment <math>OE</math>, and choose a point <math>G</math> on <math>R</math> such that <math>\angle GOE = 60^\circ</math>. There are two possible points, we choose <math>G</math> near point <math>P</math>. Draw segments <math>OG, GE</math>, thus <math>\triangle GOE</math> is an equilateral triangle | Draw the segment <math>OE</math>, and choose a point <math>G</math> on <math>R</math> such that <math>\angle GOE = 60^\circ</math>. There are two possible points, we choose <math>G</math> near point <math>P</math>. Draw segments <math>OG, GE</math>, thus <math>\triangle GOE</math> is an equilateral triangle | ||
− | Draw segments <math>OP, OC, OB, OF, PB</math> | + | Draw segments <math>OP, OC, OB, OF, PB, GC</math> |
<math>\angle OCE = \dfrac{1}{2} \angle DCE \ge \dfrac{1}{2} \angle BCA \ge 30^\circ</math>. Then we have <math>\angle COE = 90^\circ - \angle OCE \le 60^\circ = \angle GOE</math> | <math>\angle OCE = \dfrac{1}{2} \angle DCE \ge \dfrac{1}{2} \angle BCA \ge 30^\circ</math>. Then we have <math>\angle COE = 90^\circ - \angle OCE \le 60^\circ = \angle GOE</math> |
Revision as of 03:57, 23 January 2024
Contents
[hide]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.
Solution 2
At least one of . Without loss of generality, assume that
If and
Draw a circle centered at and passing through . Since is an interior point of , thus is outside the circle
Draw two lines passing through and tangent to . Line intersect at , and line intersect at . Choose near , and choose near
Extends line , and intersect at other than when is not tangent to . If is tangent to , we have be the tangent point, and simply let
Draw the segment , and choose a point on such that . There are two possible points, we choose near point . Draw segments , thus is an equilateral triangle
Draw segments
. Then we have
, since we have either or , thus
Thus we have , then
Because , thus , and
Finally,
Since , and , thus we have
We have proved that when and , the angle must be less than . Thus at least one of should less than or equal to
~Joseph Tsai, mgtsai@gmail.com
See Also
1991 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |