Difference between revisions of "1989 USAMO Problems/Problem 4"
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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> | ||
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==See Also== | ==See Also== | ||
{{USAMO box|year=1989|num-b=3|num-a=5}} | {{USAMO box|year=1989|num-b=3|num-a=5}} |
Revision as of 23:53, 22 April 2010
Problem
Let be an acute-angled triangle whose side lengths satisfy the inequalities . If point is the center of the inscribed circle of triangle and point is the center of the circumscribed circle, prove that line intersects segments and .
Solution
Consider the lines that pass through the circumcenter . Extend , , to ,, on ,,, respectively.
We notice that passes through sides and if and only if belongs to either regions or .
Since , we let , , .
We have
Since divides angle into two equal parts, it must be in the region marked by the of angle , so is in .
Similarly, is in and . Thus, is in their intersection, . From above, we have passes through and .
See Also
1989 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |