Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1991 IMO Problems/Problem 5"

(→‎Solution)
(→‎Solution)
Line 64: Line 64:
 
<math>\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le \frac{3}{2}</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>. [Inequality 2]
+
<math>2\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 3</math>.
 +
 
 +
<math>\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>
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 12:18, 12 November 2023

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 $\measuredangle CAB$, $\measuredangle ABC$, $\measuredangle BCA$, respcetively.

Let $\alpha_{1}$ , $\alpha_{2}$, and $\alpha_{3}$ be $\measuredangle PAB$, $\measuredangle PBC$, $\measuredangle PCA$, 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:

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

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

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

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

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

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

$\le 3$. 

[Inequality 2]

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

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

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1991_IMO_Problems/Problem_5&oldid=202612"