Difference between revisions of "1989 USAMO Problems/Problem 4"
(→Problem) |
m (→See Also) |
||
(3 intermediate revisions by 2 users not shown) | |||
Line 3: | Line 3: | ||
==Solution== | ==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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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 19:11, 18 July 2016
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.