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I returned as an Instructor for the week of June 23. School, which also illustrate techniques relevant to Olympiad problemĢ008 United States Math Olympiad Summer Program Highlighted lectures include topics that I encountered during graduate I returned as an Instructor for the week of June 14, to teach
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Lecture notes are below.Ģ009 United States Math Olympiad Summer Program To the Math Olympiad Summer Program for a week, teaching International Mathematical Olympiad, in Astana, Kazakhstan, and the Team Leader at the Initiative (sponsored by a new grant from the National Science Foundation) toĬonnect Olympiad mathematics with research mathematics. This time, inĪddition to teaching several courses in Combinatorics, I also directed a Returned to the Math Olympiad Summer Program for two weeks. International Mathematical Olympiad, in Amsterdam, Netherlands. Up a substantial lead over all other countries on this problem, but Which was the most challenging problem on Day 1 (and highlightedīy Terry Tao on his blog). Probabilistic and algebraic methods in combinatoricsĪ strong combinatorics background came in handy on problem 3 of the IMO, Our research produced a paper, joint with Jenny Mathematical Olympiad, in Mar del Plata, Argentina. Graham, Linus Hamilton, and Ariel Levavi, which we have Our research produced a paper, joint with Ronald Mathematics, supervising fast-paced undergraduate research projects inĬombinatorics, in addition to teaching several courses to high-school Initiative to bridge the gap between Olympiad training and research
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Olympiad Summer Program, I led an NSF-supported Mathematical Olympiad, in Santa Marta, Colombia. I was the Deputy Team Leader for the United States at To the International Mathematical Olympiad, which will be held in 2014 in Starting this season, I will be the Leader for the United States delegation Please let me know if you send us an application for undergraduateĪdmission. Olympiad Summer Program, or if I met you at an international competition, For example, if you were in my classes at the Math
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In the possibility of joining our program, please feel free to contact meįor further discussion. If you enjoyed the high-school Olympiad competitions, and are interested Individually customized to suit each scholar's background and aspiration,Īnd all students are personally mentored by myself and John.
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Program for extremely motivated and advanced students. In 2010, I and my Associate Department Head John Mackey, with theĭirect support of Carnegie Mellon President Jared Cohon and two significantĪlumni donors, launched a new “ultra-honors” talent incubator This means $PQ \parallel FG$, and we can then conclude that $DE\parallel FG$, as required.Math Olympiad teaching notes Math Olympiad teaching notes Thus, $PF=P'F'=QG$, so $PFGQ$ is a cyclic quadrilateral with two opposite sides $PF$ and $QG$ having equal length. Therefore, $P'QGF'$ is a cyclic quadrilateral with parallel opposite sides $P'Q$ and $F'G$. Let $P'$ and $F'$ be the images of $P$ and $F$, respectively, under this rotation. Now, rotate the line $OP$ about the point $O$ until this line coincides with the line $OQ$ in such a way that the image of the line $MF$ under this rotation is exactly the line $MG$ (this is possible because $d_1=d_2$). Note that $DE$ is perpendicular to the (internal) angular bisector $\ell$ of $\angle BAC$. Suppose that the perpendicular bisectors of the line segments $BD$ and $CF$ intersect at $M$. The perpendicular bisectors of the sides $AB$ and $AC$ meet the minor arcs $AB$ and $AC$ at $P$ and $Q$, respectively.