Difference between revisions of "1989 USAMO Problems/Problem 4"

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 $c$ . $\blacksquare$

@@ Line 20: / Line 20: @@
 {{USAMO box|year=1989|num-b=3|num-a=5}}
+{{MAA Notice}}
+[[Category:Olympiad Geometry Problems]]

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

Page

Toolbox

Search

Difference between revisions of "1989 USAMO Problems/Problem 4"

Latest revision as of 18:11, 18 July 2016

Problem

Solution

See Also