Difference between revisions of "1988 USAMO Problems/Problem 4"
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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. | ||
− | We shall prove that quadrilateral <math>ABO_aC</math> is cyclic first. Let <math>\angle BAC=\alpha</math>, <math>\angle CBA=\beta</math>, and <math>\angle ACB=\gamma</math>. Then <math>\angle ICB=\gamma/2</math> and <math>\angle IBC=\beta/2</math>. Therefore minor arc <math>\ | + | We shall prove that quadrilateral <math>ABO_aC</math> is cyclic first. Let <math>\angle BAC=\alpha</math>, <math>\angle CBA=\beta</math>, and <math>\angle ACB=\gamma</math>. Then <math>\angle ICB=\gamma/2</math> and <math>\angle IBC=\beta/2</math>. Therefore minor arc <math>\overarc{BIC}</math> in the circumcircle of <math>IBC</math> has a degree measure of <math>\beta+\gamma</math>. This shows that <math>\angle CO_aB=\beta+\gamma</math>, implying that <math>\angle BAC+\angle BO_aC=\alpha+\beta+\gamma=180^{\circ}</math>. Therefore quadrilateral <math>ABO_aC</math> is cyclic. |
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> |
Revision as of 12:43, 9 August 2015
Contents
Problem
is a triangle with incenter . Show that the circumcenters of , , and lie on a circle whose center is the circumcenter of .
Solution
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.