Art of Problem Solving
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

1991 IMO Problems/Problem 5: Difference between revisions

← Older editNewer edit →
Tomasdiaz (talk | contribs)
editor
1,840 edits
Tomasdiaz (talk | contribs)
editor
1,840 edits
Line 19: Line 19:


<math>\prod_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}=1</math>
<math>\prod_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}=1</math>
Using <math>AM-GM</math> we get:
<math>\frac{1}{3}\sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}\le \sqrt[3]{\prod_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}}</math>




{{alternate solutions}}
{{alternate solutions}}

Revision as of 11:30, 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:

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


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

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=1991_IMO_Problems/Problem_5&oldid=202583"