Difference between revisions of "1991 IMO Problems/Problem 5"
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Let <math> \,ABC\,</math> be a triangle and <math> \,P\,</math> an interior point of <math> \,ABC\,</math>. Show that at least one of the angles <math> \,\angle PAB,\;\angle PBC,\;\angle PCA\,</math> is less than or equal to <math> 30^{\circ }</math>. | Let <math> \,ABC\,</math> be a triangle and <math> \,P\,</math> an interior point of <math> \,ABC\,</math>. Show that at least one of the angles <math> \,\angle PAB,\;\angle PBC,\;\angle PCA\,</math> is less than or equal to <math> 30^{\circ }</math>. | ||
− | == Solution == | + | == Solution 1 == |
Let <math>A_{1}</math> , <math>A_{2}</math>, and <math>A_{3}</math> be <math>\angle CAB</math>, <math>\angle ABC</math>, <math>\angle BCA</math>, respectively. | Let <math>A_{1}</math> , <math>A_{2}</math>, and <math>A_{3}</math> be <math>\angle CAB</math>, <math>\angle ABC</math>, <math>\angle BCA</math>, respectively. | ||
Latest revision as of 04:01, 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 1
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 |