Difference between revisions of "1988 USAMO Problems/Problem 4"
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===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>. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | Extend <math>CI</math> to point <math>L</math> on <math>(ABC)</math>. By The Incenter-Excenter Lemma, B, I, A are all concyclic. Thus, L is the circumcenter of triangle <math>IAB</math>. In other words, <math>L=O_c</math>, so <math>O_c</math> is on <math>(ABC)</math>. Similarly, we can show that <math>O_a</math> and <math>O_b</math> are on <math>(ABC)</math>, and thus, <math>A,B,C,O_a,O_b,O_c</math> are all concyclic. It follows that the circumcenters are equal. | ||
==See Also== | ==See Also== |
Revision as of 19:06, 19 February 2018
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
.
Solution 3
Extend to point
on
. By The Incenter-Excenter Lemma, B, I, A are all concyclic. Thus, L is the circumcenter of triangle
. In other words,
, so
is on
. Similarly, we can show that
and
are on
, and thus,
are all concyclic. It follows that the circumcenters are equal.
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.