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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

Problem

Let $\,ABC\,$ be a triangle and $\,P\,$ an interior point of $\,ABC\,$. Show that at least one of the angles $\,\angle PAB,\;\angle PBC,\;\angle PCA\,$ is less than or equal to $30^{\circ }$.

Solution

Let $A_{1}$ , $A_{2}$, and $A_{3}$ be $\angle CAB$, $\angle ABC$, $\angle BCA$, respectively.

Let $\alpha_{1}$ , $\alpha_{2}$, and $\alpha_{3}$ be $\angle PAB$, $\angle PBC$, $\angle PCA$, respcetively.

Using law of sines on $\Delta PAB$ we get: $\frac{\left| PA \right|}{sin(A_{2}-\alpha_{2})}=\frac{\left| PB \right|}{sin(\alpha_{1})}$, therefore, $\frac{\left| PA \right|}{\left| PB \right|}=\frac{sin(A_{2}-\alpha_{2})}{sin(\alpha_{1})}$

Using law of sines on $\Delta PBC$ we get: $\frac{\left| PB \right|}{sin(A_{3}-\alpha_{3})}=\frac{\left| PC \right|}{sin(\alpha_{2})}$, therefore, $\frac{\left| PB \right|}{\left| PC \right|}=\frac{sin(A_{3}-\alpha_{3})}{sin(\alpha_{2})}$

Using law of sines on $\Delta PCA$ we get: $\frac{\left| PC \right|}{sin(A_{1}-\alpha_{1})}=\frac{\left| PA \right|}{sin(\alpha_{3})}$, therefore, $\frac{\left| PC \right|}{\left| PA \right|}=\frac{sin(A_{1}-\alpha_{1})}{sin(\alpha_{3})}$

Multiply all three equations we get: $\frac{\left| PA \right|}{\left| PB \right|}\frac{\left| PB \right|}{\left| PC \right|}\frac{\left| PC \right|}{\left| PA \right|}=\frac{sin(A_{2}-\alpha_{2})}{sin(\alpha_{1})}\frac{sin(A_{3}-\alpha_{3})}{sin(\alpha_{2})}\frac{sin(A_{1}-\alpha_{1})}{sin(\alpha_{3})}$

$1=\frac{sin(A_{2}-\alpha_{2})}{sin(\alpha_{1})}\frac{sin(A_{3}-\alpha_{3})}{sin(\alpha_{2})}\frac{sin(A_{1}-\alpha_{1})}{sin(\alpha_{3})}$

$\prod_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}=1$

Using AM-GM we get:

$\frac{1}{3}\sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}\ge \sqrt[3]{\prod_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}}$

$\frac{1}{3}\sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}\ge 1$

$\sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}\ge 3$. [Inequality 1]

$\sum_{i=1}^{3}\frac{sin(A_{i})cos(\alpha_{i})-cos(A_{i})sin(\alpha_{i})}{sin(\alpha_{i})}\ge 3$

$\sum_{i=1}^{3}\left[ sin(A_{i})cot(\alpha_{i})-cos(A_{i})\right]\ge 3$

Note that for $0<\alpha_{i}<180^{\circ}$, $cot(\alpha_{i})$ decreases with increasing $\alpha_{i}$ and fixed $A_{i}$

Therefore, $\left[ sin(A_{i})cot(\alpha_{i})-cos(A_{i})\right]$ decreases with increasing $\alpha_{i}$ and fixed $A_{i}$

From trigonometric identity:

$sin(x)+sin(y)=2sin\left( \frac{x+y}{2} \right)cos\left( \frac{x-y}{2} \right)$,

since $-1\le cos\left( \frac{x-y}{2} \right) \le 1$, then:

$sin(x)+sin(y) \le 2sin\left( \frac{x+y}{2} \right)$

Therefore,

$sin(A_{1}-30^{\circ})+sin(A_{2}-30^{\circ}) \le 2sin\left( \frac{A_{1}+A_{2}-60^{\circ}}{2} \right)$

and also,

$sin(A_{3}-30^{\circ})+sin(30^{\circ}) \le 2sin\left( \frac{A_{3}}{2} \right)$

Adding these two inequalities we get:

$sin(30^{\circ})+\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 2\left[ sin\left( \frac{A_{1}+A_{2}-60^{\circ}}{2} \right)+sin\left( \frac{A_{3}}{2} \right) \right]$

$\frac{1}{2}+\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 2\left[ 2sin\left( \frac{A_{1}+A_{2}+A_{3}-60^{\circ}}{4} \right) \right]$

$\frac{1}{2}+\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 2\left[ 2sin\left( \frac{180^{\circ}-60^{\circ}}{4} \right) \right]$

$\frac{1}{2}+\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 4sin\left( 30^{\circ} \right)$

$\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le \frac{3}{2}$

$2\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 3$.

$\sum_{i=1}^{3}\frac{sin(A_{i}-30^{\circ})}{sin(30^{\circ})}\le 3$. [Inequality 2]

Combining [Inequality 1] and [Inequality 2] we see the following:

$\sum_{i=1}^{3}\frac{sin(A_{i}-30^{\circ})}{sin(30^{\circ})}\le \sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}$

This implies that for at least one of the values of $i=1$,$2$,or $3$, the following is true:

$\frac{sin(A_{i}-30^{\circ})}{sin(30^{\circ})}\le \frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}$

or

$\frac{sin(\alpha_{i})}{sin(A_{i}-\alpha_{i})}\le \frac{sin(30^{\circ})}{sin(A_{i}-30^{\circ})}$

Which means that for at least one of the values of $i=1$,$2$,or $3$, the following is true:

$\alpha_{i} \le 30^{\circ}$

Therefore, at least one of the angles $\,\angle PAB,\;\angle PBC,\;\angle PCA\,$ is less than or equal to $30^{\circ }$.

~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 $\angle ABC, \angle BCA, \angle CAB \ge 60^\circ$. Without loss of generality, assume that $\angle BCA \ge 60^\circ$

If $\angle PAB > 30^\circ$ and $\angle PBC > 30^\circ$

Draw a circle $R$ centered at $O$ and passing through $A, P, B$. Since $P$ is an interior point of $\triangle ABC$, thus $C$ is outside the circle $R$

Draw two lines $CD, CE$ passing through $C$ and tangent to $R$. Line $CD$ intersect $R$ at $D$, and line $CE$ intersect $R$ at $E$. Choose $D$ near $A$, and choose $E$ near $B$

Extends line $BC$, and intersect $R$ at $F$ other than $B$ when $BC$ is not tangent to $R$. If $BC$ is tangent to $R$, we have $B = E$ be the tangent point, and simply let $F = B = E$

Draw the segment $OE$, and choose a point $G$ on $R$ such that $\angle GOE = 60^\circ$. There are two possible points, we choose $G$ near point $P$. Draw segments $OG, GE$, thus $\triangle GOE$ is an equilateral triangle

Draw segments $OP, OC, OB, OF, PB, GC$

$\angle OCE = \dfrac{1}{2} \angle DCE \ge \dfrac{1}{2} \angle BCA \ge 30^\circ$. Then we have $\angle COE = 90^\circ - \angle OCE \le 60^\circ = \angle GOE$

$\angle POB = 2 \angle PAB > 60^\circ, \angle POF = 2 \angle PBC > 60^\circ$, since we have either $\angle POE \ge \angle POB$ or $\angle POE \ge \angle POF$, thus $\angle POE > 60^\circ = \angle GOE$

Thus we have $\angle COE \le \angle GOE < \angle POE$, then $\angle OCE \le \angle GCE < \angle PCE$

Because $\angle GCE \ge \angle OCE \ge 30^\circ = \angle GEC$, thus $GC \le GE = OG$, and $\angle GCO \ge \angle GOC$

Finally, $\angle PCA = \angle ACE - \angle PCE < \angle ACE - \angle GCE = \angle ACO - \angle GCO$

Since $\angle ACO \le \angle DCO$, and $\angle GCO \ge \angle GOC$, thus we have $\angle PCA < \angle ACO - \angle GCO \le \angle DCO - \angle GOC = 90^\circ - \angle COE - \angle GOC = 90^\circ - \angle GOE = 30^\circ$

We have proved that when $\angle PAB > 30^\circ$ and $\angle PBC > 30^\circ$, the angle $\angle PCA$ must be less than $30^\circ$. Thus at least one of $\angle PAB, \angle PBC, \angle PCA$ should less than or equal to $30^\circ$

~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
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