1989 USAMO Problems/Problem 4

Revision as of 23:53, 22 April 2010 by Moplam (talk | contribs) (Solution)


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 $BC$.


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 $c$. $\blacksquare$ This problem needs a solution. If you have a solution for it, please help us out by adding it.

See Also

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