Difference between revisions of "2007 USAMO Problems/Problem 6"
m (→Problem) |
(→Problem) |
||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | (''Kiran Kedlaya, Sungyoon Kim'') Let <math>ABC</math> be an acute triangle with <math>\omega</math>, <math>\Omega</math>, and <math>R</math> being its incircle, | + | (''Kiran Kedlaya, Sungyoon Kim'') Let <math>ABC</math> be an acute triangle with <math>\omega</math>, <math>\Omega</math>, and <math>R</math> being its incircle, circumcircle, and circumradius, respectively. The circle <math>\omega_A</math> is tangent internally to <math>\Omega</math> at <math>A</math> and externally tangent to <math>\omega</math>. Circle <math>\Omega_A</math> is internally tangent to <math>\Omega</math> at <math>A</math> and tangent internally to <math>\omega</math>. Let <math>P_A</math> and <math>Q_A</math> denote the centers of <math>\omega_A</math> and <math>\Omega_A</math>, respectively. Define points <math>P_B</math>, <math>Q_B</math>, <math>P_C</math>, <math>Q_C</math> analogously. Prove that |
<cmath>8P_AQ_A \cdot P_BQ_B \cdot P_CQ_C \le R^3,</cmath> | <cmath>8P_AQ_A \cdot P_BQ_B \cdot P_CQ_C \le R^3,</cmath> | ||
with equality [[iff|if and only if]] triangle <math>ABC</math> is equilateral. | with equality [[iff|if and only if]] triangle <math>ABC</math> is equilateral. |
Latest revision as of 07:35, 27 August 2024
Contents
Problem
(Kiran Kedlaya, Sungyoon Kim) Let be an acute triangle with , , and being its incircle, circumcircle, and circumradius, respectively. The circle is tangent internally to at and externally tangent to . Circle is internally tangent to at and tangent internally to . Let and denote the centers of and , respectively. Define points , , , analogously. Prove that with equality if and only if triangle is equilateral.
Solutions
Solution 1
Lemma.
Proof. Note and lie on since for a pair of tangent circles, the point of tangency and the two centers are collinear.
Let touch , , and at , , and , respectively. Note . Consider an inversion, , centered at , passing through , . Since , is orthogonal to the inversion circle, so . Consider . Note that passes through and is tangent to , hence is a line that is tangent to . Furthermore, because is symmetric about , so the inversion preserves that reflective symmetry. Since it is a line that is symmetric about , it must be perpendicular to . Likewise, is the other line tangent to and perpendicular to .
Let and (second intersection).
Let and (second intersection).
Evidently, and . We want: by inversion. Note that , and they are tangent to , so the distance between those lines is . Drop a perpendicular from to , touching at . Then . Then . So . Note that . Applying the double angle formulas and , we get
End Lemma
The problem becomes: which is true because , equality is when the circumcenter and incenter coincide. As before, , so, by symmetry, . Hence the inequality is true iff is equilateral.
Comment: It is much easier to determine by considering . We have , , , and . However, the inversion is always nice to use. This also gives an easy construction for because the tangency point is collinear with the intersection of and .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
- <url>viewtopic.php?t=145852 Discussion on AoPS/MathLinks</url>
2007 USAMO (Problems • Resources) | ||
Preceded by Problem 5 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.