Difference between revisions of "1988 USAMO Problems/Problem 4"
(Created page with "==Problem== <math>\Delta ABC</math> is a triangle with incenter <math>I</math>. Show that the circumcenters of <math>\Delta IAB</math>, <math>\Delta IBC</math>, and <math>\Delta ...") |
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==Solution== | ==Solution== | ||
− | {{ | + | 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>\arc{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> | ||
==See Also== | ==See Also== |
Revision as of 11:45, 16 May 2012
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 $\arc{BIC}$ (Error compiling LaTeX. Unknown error_msg) 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.
See Also
1988 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |