ka March Highlights and 2025 AoPS Online Class Information
jlacosta0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.
Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!
Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.
Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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I think I know why this problem was rejected by IMO PSC several times...
mshtand10
4 minutes ago
Source: Ukrainian Mathematical Olympiad 2025. Day 2, Problem 11.8
Exactly country leaders arrived at the IMO. At the final session, the IMO chairperson wants to introduce some changes to the regulations, which the leaders must approve. To pass the changes, the chairperson must gather at least of the votes "FOR" out of the total number of leaders. Some leaders do not attend such meetings, and it is known that there will be exactly leaders present. The chairperson must seat them in a square-shaped conference hall of size , where each leader will be seated in a designated cell. It is known that exactly of these leaders will surely support the chairperson, i.e., they will always vote "FOR." All others will vote as follows: At the last second of voting, they will look at how their neighbors voted up to that moment — neighbors are defined as leaders seated in adjacent cells (sharing a side). If the majority of neighbors voted "FOR," they will also vote "FOR." If there is no such majority, they will vote "AGAINST." For example, a leader seated in a corner of the hall has exactly neighbors and will vote "FOR" only if both of their neighbors voted "FOR."
(a) Can the IMO chairperson arrange their supporters so that they vote "FOR" in the first second of voting and thereby secure a "FOR" vote from at least of all leaders?
(b) What is the maximum number of "FOR" votes the chairperson can obtain by seating their 28 supporters appropriately?
This problem has unintended solution, found by almost all who solved it :(
mshtand10
17 minutes ago
Source: Ukrainian Mathematical Olympiad 2025. Day 2, Problem 11.7
Given a triangle , an arbitrary point is chosen on the side . In triangles and , the angle bisectors and are drawn, respectively. The point is the circumcenter of . Prove that the second intersection point of the circumcircles of triangles and lies on the line .
Best geo on this year UKR MO, even contestants call it marvellous
mshtand11
N20 minutes ago
by khina
Source: Ukrainian Mathematical Olympiad 2025. Day 2, Problem 10.8
In an acute-angled triangle with altitude , the following points are marked: - the orthocenter, - the circumcenter, - the midpoint of side . Inside the triangle , there is a point such that the following equality holds: Prove that Proposed by Vadym Solomka
isn't it strange if we apply the engel form to
we don't get the desired result...we get a weak bound.
u have to multiply by
Multiplying by isn't necessary. It's enough to multiply by . (But makes things a bit more easier)
Then it remains to show that
Which follows from AM-GM
PS. Congrats to Erken.
OFFICIAL SOLUTION:
LEMMA: Suppose that , and is a permutation of .Then .
Proof: Since , we get and ;moreover, . , Q.E.D.
Denote by the left-hand part of the desired inequality.Since is invariant under the cyclical permutation of variables, we can assume or . In both cases, by LEMMA we get: . Hence, , Q.E.D.
Actually if anyone saw my link above it shows that the following one holds for all satisfying
And it's easily proven using only Cauchy-Schwarz inequality.
Let's write the as , equivalently (since ). Then using Cauchy-Schwarz inequality we have So, we have to show that Expanding and using AM-GM inequality, we get So, the inequality is proven.
i think the last inequality x^4 + y^4 + z^4 >= x^3y +y^3z +z^3x.... (it's true according to rearrangement) but in this question a.b.c = 1
so i think we can't work
Actually if anyone saw my link above it shows that the following one holds for all satisfying And it's easily proven using only Cauchy-Schwarz inequality.
Let , then by Titu
So we have to show that
By Rearrangement inequality,
By Vasc inequality,
So we're done.
This post has been edited 1 time. Last edited by Takeya.O, Apr 7, 2018, 2:15 AM Reason: adding explanation
Since we can use the Ravi substitution ,, where . Now the LHS rewrites as , furthermore: Note that AM-GM implies that and also thatSumming and we get the inequality in , so we're done.
Since we can use the Ravi substitution ,, where . Now the LHS rewrites as , furthermore: Note that AM-GM implies that and also thatSumming and we get the inequality in , so we're done.