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"

(New page: ==Problem== ==Solution== {{solution}} ==See Also== {{USAMO box|year=1989|num-b=3|num-a=5}})
 
m (→‎See Also)
 
(4 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
 +
Let <math>ABC</math> be an acute-angled triangle whose side lengths satisfy the inequalities <math>AB < AC < BC</math>. If point <math>I</math> is the center of the inscribed circle of triangle <math>ABC</math> and point <math>O</math> is the center of the circumscribed circle, prove that line <math>IO</math> intersects segments <math>AB</math> and <math>BC</math>.
 +
 +
==Solution==
 +
Consider the lines that pass through the circumcenter <math>O</math>. Extend <math>AO</math>, <math> BO</math>, <math>CO</math> to <math>D</math>,<math>E</math>,<math>F</math> on <math>a</math>,<math>b</math>,<math>c</math>, respectively.
 +
 +
We notice that <math>IO</math> passes through sides <math>a</math> and <math>c</math> if and only if <math>I</math> belongs to either regions <math>AOF</math> or <math>COD</math>.
 +
  
  
 +
Since <math>AO = BO = CO = R</math>, we let <math>\alpha = \angle OAC = \angle OCA</math>, <math>\beta = \angle BAO = \angle ABO</math>, <math>\gamma = \angle BCO = \angle CBO</math>.
  
−
==Solution==
+
We have <math> c < b < a\implies C < B < A\implies</math> <math>\gamma+\alpha <\beta+\gamma <\alpha+\beta\implies\gamma <\alpha <\beta </math>
 +
 
 +
Since <math>IA</math> divides angle <math>A</math> into two equal parts, it must be in the region marked by the <math>\beta</math> of angle <math>A</math>, so <math>I</math> is in <math>ABD</math>.
  
−
{{solution}}
+
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>
  
 
==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 18:11, 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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