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> | + | 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>\angle IAM = \angle AIM = \frac{\angle BAC + \angle 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=== | ===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. | 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. | ||
+ | |||
+ | ==Solution 4== | ||
+ | Let the centers be <math>T, R, S</math>. We want to show that <math>ABC</math> and <math>TRS</math> have the same circumcircle. By Fact 5 we know that <math>CATB</math> lie on a circle and similarly with the others. Thus the two triangles have the same circumcircle. | ||
+ | ~coolmath_2018 | ||
==See Also== | ==See Also== |
Latest revision as of 18:25, 14 October 2021
Contents
[hide]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.
Solution 4
Let the centers be . We want to show that and have the same circumcircle. By Fact 5 we know that lie on a circle and similarly with the others. Thus the two triangles have the same circumcircle. ~coolmath_2018
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.