Difference between revisions of "2007 USAMO Problems/Problem 6"
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8\prod \frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}} &\leq R^{3} \ | 8\prod \frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}} &\leq R^{3} \ | ||
\frac{2^{9}R^{6}\left[\prod(s-a)\right]^{4}}{r^{3}a^{4}b^{4}c^{4}} &\leq R^{3} \ | \frac{2^{9}R^{6}\left[\prod(s-a)\right]^{4}}{r^{3}a^{4}b^{4}c^{4}} &\leq R^{3} \ | ||
− | \left(\frac{4AR}{abc}\right)^{ | + | \left(\frac{4AR}{abc}\right)^{4}\cdot\left(\frac{A}{rs}\right)^{4}2rR^{2} &\leq R^{3} \ |
2r &\leq R, | 2r &\leq R, | ||
\end{align*}</cmath> | \end{align*}</cmath> |
Revision as of 14:11, 31 August 2017
Contents
[hide]Problem
(Kiran Kedlaya, Sungyoon Kim) Let be an acute triangle with , , and being its incircle, circumcircle, and circumradius, respectively. Circle is tangent internally to at and tangent externally to . Circle is tangent internally 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 |
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