Difference between revisions of "1988 USAMO Problems/Problem 4"
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==Solution== | ==Solution== | ||
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Let the circumcenters of <math>\Delta IAB</math>, <math>\Delta IBC</math>, and <math>\Delta ICA</math> be <math>O_c</math>, <math>O_a</math>, and <math>O_b</math>, respectively. It then suffices to show that <math>A</math>, <math>B</math>, <math>C</math>, <math>O_a</math>, <math>O_b</math>, and <math>O_c</math> are concyclic. | Let the circumcenters of <math>\Delta IAB</math>, <math>\Delta IBC</math>, and <math>\Delta ICA</math> be <math>O_c</math>, <math>O_a</math>, and <math>O_b</math>, respectively. It then suffices to show that <math>A</math>, <math>B</math>, <math>C</math>, <math>O_a</math>, <math>O_b</math>, and <math>O_c</math> are concyclic. | ||
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This shows that point <math>O_a</math> is on the circumcircle of <math>\Delta ABC</math>. Analagous proofs show that <math>O_b</math> and <math>O_c</math> are also on the circumcircle of <math>ABC</math>, which completes the proof. <math>\blacksquare</math> | This shows that point <math>O_a</math> is on the circumcircle of <math>\Delta ABC</math>. Analagous proofs show that <math>O_b</math> and <math>O_c</math> are also on the circumcircle of <math>ABC</math>, which completes the proof. <math>\blacksquare</math> | ||
− | ==Solution 2== | + | ===Solution 2=== |
Let <math>M</math> denote the midpoint of arc <math>AC</math>. It is well known that <math>M</math> is equidistant from <math>A</math>, <math>C</math>, and <math>I</math> (to check, prove <math><IAM = <AIM = \frac{<BAC + <ABC}{2}</math>), so that <math>M</math> is the circumcenter of <math>AIC</math>. Similar results hold for <math>BIC</math> and <math>CIA</math>, and hence <math>O_c</math>, <math>O_a</math>, and <math>O_b</math> all lie on the circumcircle of <math>ABC</math>. | Let <math>M</math> denote the midpoint of arc <math>AC</math>. It is well known that <math>M</math> is equidistant from <math>A</math>, <math>C</math>, and <math>I</math> (to check, prove <math><IAM = <AIM = \frac{<BAC + <ABC}{2}</math>), so that <math>M</math> is the circumcenter of <math>AIC</math>. Similar results hold for <math>BIC</math> and <math>CIA</math>, and hence <math>O_c</math>, <math>O_a</math>, and <math>O_b</math> all lie on the circumcircle of <math>ABC</math>. | ||
Revision as of 19:10, 18 July 2016
Problem
is a triangle with incenter . Show that the circumcenters of , , and lie on a circle whose center is the circumcenter of .
Solution
Solution 1
Let the circumcenters of , , and be , , and , respectively. It then suffices to show that , , , , , and are concyclic.
We shall prove that quadrilateral is cyclic first. Let , , and . Then and . Therefore minor arc in the circumcircle of has a degree measure of . This shows that , implying that . Therefore quadrilateral is cyclic.
This shows that point is on the circumcircle of . Analagous proofs show that and are also on the circumcircle of , which completes the proof.
Solution 2
Let denote the midpoint of arc . It is well known that is equidistant from , , and (to check, prove ), so that is the circumcenter of . Similar results hold for and , and hence , , and all lie on the circumcircle of .
See Also
1988 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.