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>\ | + | 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>\alpha_{1}</math> , <math>\alpha_{2}</math>, and <math>\alpha_{3}</math> be <math>\ | + | Let <math>\alpha_{1}</math> , <math>\alpha_{2}</math>, and <math>\alpha_{3}</math> be <math>\angle PAB</math>, <math>\angle PBC</math>, <math>\angle PCA</math>, respcetively. |
Using law of sines on <math>\Delta PAB</math> we get: <math>\frac{\left| PA \right|}{sin(A_{2}-\alpha_{2})}=\frac{\left| PB \right|}{sin(\alpha_{1})}</math>, therefore, <math>\frac{\left| PA \right|}{\left| PB \right|}=\frac{sin(A_{2}-\alpha_{2})}{sin(\alpha_{1})}</math> | Using law of sines on <math>\Delta PAB</math> we get: <math>\frac{\left| PA \right|}{sin(A_{2}-\alpha_{2})}=\frac{\left| PB \right|}{sin(\alpha_{1})}</math>, therefore, <math>\frac{\left| PA \right|}{\left| PB \right|}=\frac{sin(A_{2}-\alpha_{2})}{sin(\alpha_{1})}</math> | ||
+ | |||
+ | Using law of sines on <math>\Delta PBC</math> we get: <math>\frac{\left| PB \right|}{sin(A_{3}-\alpha_{3})}=\frac{\left| PC \right|}{sin(\alpha_{2})}</math>, therefore, <math>\frac{\left| PB \right|}{\left| PC \right|}=\frac{sin(A_{3}-\alpha_{3})}{sin(\alpha_{2})}</math> | ||
+ | |||
+ | Using law of sines on <math>\Delta PCA</math> we get: <math>\frac{\left| PC \right|}{sin(A_{1}-\alpha_{1})}=\frac{\left| PA \right|}{sin(\alpha_{3})}</math>, therefore, <math>\frac{\left| PC \right|}{\left| PA \right|}=\frac{sin(A_{1}-\alpha_{1})}{sin(\alpha_{3})}</math> | ||
+ | |||
+ | Multiply all three equations we get: | ||
+ | <math>\frac{\left| PA \right|}{\left| PB \right|}\frac{\left| PB \right|}{\left| PC \right|}\frac{\left| PC \right|}{\left| PA \right|}=\frac{sin(A_{2}-\alpha_{2})}{sin(\alpha_{1})}\frac{sin(A_{3}-\alpha_{3})}{sin(\alpha_{2})}\frac{sin(A_{1}-\alpha_{1})}{sin(\alpha_{3})}</math> | ||
+ | |||
+ | <math>1=\frac{sin(A_{2}-\alpha_{2})}{sin(\alpha_{1})}\frac{sin(A_{3}-\alpha_{3})}{sin(\alpha_{2})}\frac{sin(A_{1}-\alpha_{1})}{sin(\alpha_{3})}</math> | ||
+ | |||
+ | <math>\prod_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}=1</math> | ||
+ | |||
+ | Using AM-GM we get: | ||
+ | |||
+ | <math>\frac{1}{3}\sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}\ge \sqrt[3]{\prod_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}}</math> | ||
+ | |||
+ | <math>\frac{1}{3}\sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}\ge 1</math> | ||
+ | |||
+ | <math>\sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}\ge 3</math>. [Inequality 1] | ||
+ | |||
+ | <math>\sum_{i=1}^{3}\frac{sin(A_{i})cos(\alpha_{i})-cos(A_{i})sin(\alpha_{i})}{sin(\alpha_{i})}\ge 3</math> | ||
+ | |||
+ | <math>\sum_{i=1}^{3}\left[ sin(A_{i})cot(\alpha_{i})-cos(A_{i})\right]\ge 3</math> | ||
+ | |||
+ | Note that for <math>0<\alpha_{i}<180^{\circ}</math>, <math>cot(\alpha_{i})</math> decreases with increasing <math>\alpha_{i}</math> and fixed <math>A_{i}</math> | ||
+ | |||
+ | Therefore, <math>\left[ sin(A_{i})cot(\alpha_{i})-cos(A_{i})\right]</math> decreases with increasing <math>\alpha_{i}</math> and fixed <math>A_{i}</math> | ||
+ | |||
+ | From trigonometric identity: | ||
+ | |||
+ | <math>sin(x)+sin(y)=2sin\left( \frac{x+y}{2} \right)cos\left( \frac{x-y}{2} \right)</math>, | ||
+ | |||
+ | since <math>-1\le cos\left( \frac{x-y}{2} \right) \le 1</math>, then: | ||
+ | |||
+ | <math>sin(x)+sin(y) \le 2sin\left( \frac{x+y}{2} \right)</math> | ||
+ | |||
+ | Therefore, | ||
+ | |||
+ | <math>sin(A_{1}-30^{\circ})+sin(A_{2}-30^{\circ}) \le 2sin\left( \frac{A_{1}+A_{2}-60^{\circ}}{2} \right)</math> | ||
+ | |||
+ | and also, | ||
+ | |||
+ | <math>sin(A_{3}-30^{\circ})+sin(30^{\circ}) \le 2sin\left( \frac{A_{3}}{2} \right)</math> | ||
+ | |||
+ | Adding these two inequalities we get: | ||
+ | |||
+ | <math>sin(30^{\circ})+\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 2\left[ sin\left( \frac{A_{1}+A_{2}-60^{\circ}}{2} \right)+sin\left( \frac{A_{3}}{2} \right) \right]</math> | ||
+ | |||
+ | <math>\frac{1}{2}+\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 2\left[ 2sin\left( \frac{A_{1}+A_{2}+A_{3}-60^{\circ}}{4} \right) \right]</math> | ||
+ | |||
+ | <math>\frac{1}{2}+\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 2\left[ 2sin\left( \frac{180^{\circ}-60^{\circ}}{4} \right) \right]</math> | ||
+ | |||
+ | <math>\frac{1}{2}+\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 4sin\left( 30^{\circ} \right)</math> | ||
+ | |||
+ | <math>\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le \frac{3}{2}</math> | ||
+ | |||
+ | <math>2\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 3</math>. | ||
+ | |||
+ | <math>\sum_{i=1}^{3}\frac{sin(A_{i}-30^{\circ})}{sin(30^{\circ})}\le 3</math>. [Inequality 2] | ||
+ | |||
+ | Combining [Inequality 1] and [Inequality 2] we see the following: | ||
+ | |||
+ | <math>\sum_{i=1}^{3}\frac{sin(A_{i}-30^{\circ})}{sin(30^{\circ})}\le \sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}</math> | ||
+ | |||
+ | This implies that for at least one of the values of <math>i=1</math>,<math>2</math>,or <math>3</math>, the following is true: | ||
+ | |||
+ | <math>\frac{sin(A_{i}-30^{\circ})}{sin(30^{\circ})}\le \frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}</math> | ||
+ | |||
+ | or | ||
+ | |||
+ | <math>\frac{sin(\alpha_{i})}{sin(A_{i}-\alpha_{i})}\le \frac{sin(30^{\circ})}{sin(A_{i}-30^{\circ})}</math> | ||
+ | |||
+ | Which means that for at least one of the values of <math>i=1</math>,<math>2</math>,or <math>3</math>, the following is true: | ||
+ | |||
+ | <math>\alpha_{i} \le 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>. | ||
+ | |||
+ | ~Tomas Diaz, orders@tomasdiaz.com | ||
{{alternate solutions}} | {{alternate solutions}} | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | At least one of <math>\angle ABC, \angle BCA, \angle CAB \ge 60^\circ</math>. Without loss of generality, assume that <math>\angle BCA \ge 60^\circ</math> | ||
+ | |||
+ | If <math>\angle PAB > 30^\circ</math> and <math>\angle PBC > 30^\circ</math> | ||
+ | |||
+ | Draw a circle <math>R</math> centered at <math>O</math> and passing through <math>A, P, B</math>. Since <math>P</math> is an interior point of <math>\triangle ABC</math>, thus <math>C</math> is outside the circle <math>R</math> | ||
+ | |||
+ | Draw two lines <math>CD, CE</math> passing through <math>C</math> and tangent to <math>R</math>. Line <math>CD</math> intersect <math>R</math> at <math>D</math>, and line <math>CE</math> intersect <math>R</math> at <math>E</math>. Choose <math>D</math> near <math>A</math>, and choose <math>E</math> near <math>B</math> | ||
+ | |||
+ | Extends line <math>BC</math>, and intersect <math>R</math> at <math>F</math> other than <math>B</math> when <math>BC</math> is not tangent to <math>R</math>. If <math>BC</math> is tangent to <math>R</math>, we have <math>B = E</math> be the tangent point, and simply let <math>F = B = E</math> | ||
+ | |||
+ | 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, 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 POB = 2 \angle PAB > 60^\circ, \angle POF = 2 \angle PBC > 60^\circ</math>, since we have either <math>\angle POE \ge \angle POB</math> or <math>\angle POE \ge \angle POF</math>, thus <math>\angle POE > 60^\circ = \angle GOE</math> | ||
+ | |||
+ | Thus we have <math>\angle COE \le \angle GOE < \angle POE</math>, then <math>\angle OCE \le \angle GCE < \angle PCE</math> | ||
+ | |||
+ | Because <math>\angle GCE \ge \angle OCE \ge 30^\circ = \angle GEC</math>, thus <math>GC \le GE = OG</math>, and <math>\angle GCO \ge \angle GOC</math> | ||
+ | |||
+ | Finally, <math>\angle PCA = \angle ACE - \angle PCE < \angle ACE - \angle GCE = \angle ACO - \angle GCO</math> | ||
+ | |||
+ | Since <math>\angle ACO \le \angle DCO</math>, and <math>\angle GCO \ge \angle GOC</math>, thus we have <math>\angle PCA < \angle ACO - \angle GCO \le \angle DCO - \angle GOC = 90^\circ - \angle COE - \angle GOC = 90^\circ - \angle GOE = 30^\circ</math> | ||
+ | |||
+ | We have proved that when <math>\angle PAB > 30^\circ</math> and <math>\angle PBC > 30^\circ</math>, the angle <math>\angle PCA</math> must be less than <math>30^\circ</math>. Thus at least one of <math>\angle PAB, \angle PBC, \angle PCA</math> should less than or equal to <math>30^\circ</math> | ||
+ | |||
+ | ~Joseph Tsai, mgtsai@gmail.com | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=1991|num-b=4|num-a=6}} | ||
+ | [[Category:Olympiad Geometry Problems]] | ||
+ | [[Category:Geometry Problems]] |
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 |