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 "1989 USAMO Problems/Problem 4"

(→‎Solution)
m (→‎See Also)
 
(2 intermediate revisions by 2 users not shown)
Line 16: Line 16:
  
 
Similarly, <math>I</math> is in <math>ACF</math> and <math>ABE</math>. Thus, <math>I</math> is in their intersection, <math>AOF</math>. From above, we have <math>IO</math> passes through <math>a</math> and <math>c</math>. <math>\blacksquare</math>
 
Similarly, <math>I</math> is in <math>ACF</math> and <math>ABE</math>. Thus, <math>I</math> is in their intersection, <math>AOF</math>. From above, we have <math>IO</math> passes through <math>a</math> and <math>c</math>. <math>\blacksquare</math>
−
{{solution}}
 
  
 
==See Also==
 
==See Also==
  
 
{{USAMO box|year=1989|num-b=3|num-a=5}}
 
{{USAMO box|year=1989|num-b=3|num-a=5}}
 +
{{MAA Notice}}
 +
[[Category:Olympiad Geometry Problems]]

Latest revision as of 19:11, 18 July 2016

Problem

Let $ABC$ be an acute-angled triangle whose side lengths satisfy the inequalities $AB < AC < BC$. If point $I$ is the center of the inscribed circle of triangle $ABC$ and point $O$ is the center of the circumscribed circle, prove that line $IO$ intersects segments $AB$ and $BC$.

Solution

Consider the lines that pass through the circumcenter $O$. Extend $AO$, $BO$, $CO$ to $D$,$E$,$F$ on $a$,$b$,$c$, respectively.

We notice that $IO$ passes through sides $a$ and $c$ if and only if $I$ belongs to either regions $AOF$ or $COD$.


Since $AO = BO = CO = R$, we let $\alpha = \angle OAC = \angle OCA$, $\beta = \angle BAO = \angle ABO$, $\gamma = \angle BCO = \angle CBO$.

We have $c < b < a\implies C < B < A\implies$ $\gamma+\alpha <\beta+\gamma <\alpha+\beta\implies\gamma <\alpha <\beta$

Since $IA$ divides angle $A$ into two equal parts, it must be in the region marked by the $\beta$ of angle $A$, so $I$ is in $ABD$.

Similarly, $I$ is in $ACF$ and $ABE$. Thus, $I$ is in their intersection, $AOF$. From above, we have $IO$ passes through $a$ and $c$. $\blacksquare$

See Also

1989 USAMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1989_USAMO_Problems/Problem_4&oldid=79450"