If you were in 8th grade, would you rather go to MOP or mc nats? I chose to study the former more and got in so was wondering if that was valid given that I'll never make mc nats.

Hi all
Just wondering if any orz or high scorers on contests at young age (which are a lot of u guys lol) can share what your math path has been like?
- school math: you probably finish calculus in 5th grade or something lol then what do you do for the rest of the school? concurrent enrollment? college class? none (focus on math competitions)?
- what grade did you get honor roll or higher on AMC 8, AMC 10, AIME qual, USAJMO qual, etc?
- besides aops do you use another program to study? (like Mr Math, Alphastar, etc)?

You're all great inspirations and i appreciate the answers.. you all give me a lot of motivation for this math journey. Thanks

Let be the number of complex numbers with the properties that and is a real number. Find the remainder when is divided by .

For every subset of , let be the sum of the elements of , with defined to be . If is chosen at random among all subsets of , the probability that is divisible by is , where and are relatively prime positive integers. Find .

Often times, a problem will ask you to construct something subject to certain conditions. You may need to arrange some objects and make some connections so as to satisfy some relation; you may need to place some points on the Euclidean plane, while their mutual distances/angles are constrained in some convoluted way. The request for constructing the thing is usually implicit; you are asked to merely prove that there exists some configuration. More often than not, you can get away with clever existential arguments, but in most cases, you'll have to get your hands dirty.

There are competition problems that blatantly ask you "Hey, arrange these stuff/fill in these grids so that we're happy." Example: many of the past USAMTS problems, say this:

[quote]Fill in the spaces of the grid below with positive integers so that in each square with top left number , top right number , bottom left number , and bottom right number , either or . You do not need to prove that your answer is the only one possible; you merely need to find an answer that satisfies the constraints above. (Note: In any other USAMTS problem, you need to provide a full proof. Only in this problem is an answer without justification acceptable.)

IMAGE[/quote]

Technically, most logic-grid puzzles (read: newspaper puzzles with Japanese names) fall into this category: think of Sudoku, Kakuro, Hitori etc. The previous example can be regarded as a logic grid. Theoretically, you can always write a programs and brute force all possibilities; but usually a solution completely by "human" logic is possible, and more fun to carry out. (As a side note, the ongoing Fortnightly Topic Challenge of Puzzling.SE is . If you love trying to fill grids, go check it out!)

But more frequent are problems where the construction occurs as a necessary, but not entire part of your solution. A recent (nice) example would be IMO 2016 P2. Fortunately, the construction part doesn't require a tremendous amount of out-of-the-box thinking; a simple, regular fill-up works just fine.

But needlessly sadistic problems (or problem-writers) are a thing (math would be boring otherwise, right?).

A Projective Project

Consider this fairly recent problem from the European Mathematical Cup:

[quote]
(EMC, Jr, P4) We will call a pair of positive integers with a if there exists a table consisting of ones and zeros with following properties:
[list][*] In every row there are exactly ones.
[*] For each two rows there is exactly one column such that on both intersections of that column with the mentioned rows, number one is written.[/list]
Solve the following subproblems:
[list=a][*] Let be a divisor of . Determine all remainders that can give when divided by [*] Prove that there exist infinitely many lovely couples.[/list]

Miroslav Marinov, Daniel Atanasov[/quote]

Via some case bashing, it's not hard (but annoying) to prove that the table is more or less 'regular'; in other words, every column contains 's as well (the proof isn't entirely trivial, but I'll leave it out; part a isn't our main concern anyway). Now a quick double counting of the pairs of 's in the same column shows . RIP part (a).

Now let's move on to the more interesting bit, the construction of infinitely many tables. The 's and 's are strongly reminiscents of incidence matrices: let's try to understand what the rows and columns stand for. Let's say that the in the th row and the th column means that the th 'row thing' 'has' the th 'column thing', since this is usually the kind of relationship incidence matrices record. By now, we have drawn a number of interesting conclusions:
[list=1]
[*]Every 'row thing' has exactly 'column things'.
[*]Every 'column thing' occurs in exactly 'row things'.
[*]Every two 'row things' have exactly one 'column thing' in common.
[*]Every two 'column things' occur simultaneously in exactly one 'row thing'.
[/list]
The last point is not so obvious. However, it's easy to see; if they occur in two 'row things', then we must have a 'rectangle' of 's, which is bad. On the other, the relation forces them to occur once at least.

Anyway, notice something eerie in the last two conditions? The 'row things' and the 'column things' play similar roles as 'lines' and 'points' in projective geometry. But the projective plane has way too many points and lines: infinitely many, to be specific. That's a pretty darn large value for : we need to get finite.

Let's then construct something, that's a finite analogue of our vanilla projective plane. But how is our familiar projective plane constructed, anyway? If you're familiar with homogeneous coordinates, you'll remember that all points in are obtained by taking triples of reals such as (except for the finicky all-zero triple), and then stipulating that is actually the same point as for non-zero . The 'ratio'-nal relation is emphasized by writing the point as instead. Similarly we can define lines as the triples , and define the 'incidence ' relation as :
In our case, we need some finite analogue of the reals with some nice properties. All we'll need is that addition, subtraction, multiplication and division (except by zero) work nicely enough. (For those of you care for esoteric names, these structures are often called 'fields'.) Immediately springs to mind the set of residues of integers modulo (some primes); you can add, subtract, multiply them together with no problem. So let's proceed exactly as we did in the real case. For the sake of laziness, let's just call this . Suppose,
[list][*]A point is a triple with . We say that points are basically the same one as long as .
[*] A line is a similar triple written as .
[*] The point and the line are incident when . In , that is.[/list]
And now, for the moment of truth. Does it actually work? As it turns out, it does! The details are pretty easy and boring to work out; I'll just show a bunch of special cases.

To begin with, let's prove that every line contains exactly points. Take a line Since things can be scaled up/down, let's say we're looking for points of the form . (Actually, we still need to deal with the case , but shhh...) So the equation we need to solve is:
When we vary , we get exactly solutions. The proof that two lines intersect exactly once is left to the reader.

This strange beast that we've created just now, is actually called, rather unsurprisingly, a 'Finite Projective Plane'. It's about as bad as constructions on a contest problem can get.

Then I ran, towards Iran
Up next is a cute problem from Iran.

[quote](Iran TST 2015) We call a permutation of the set "good" if for any three natural numbers , find all natural numbers such that there exist a "good" permutation of a set .[/quote]

The set looks natural at first, but once you realize that decreasing all the numbers by the same amount doesn't make any difference, you can take the more natural set instead; after all, it's the complete residue system modulo , so it's more likely to some nice properties. Let's call all that work 'good'. After some experimentations, trial and error (and depending on you nature, maybe some quick snacks, a short chat with someone in another tab, and shooting a reply to an e-mail you got in ), we can finally guess that all powers of two work, and others don't. Something to do with odd prime factor, perhaps?

In a moment of vision, you realize that in the working permutation for , we can pick out the number in the order as they appear in the permutation, scale 'em down by , and have a 'good' ordering for . Therefore is good is good.

If remains to prove that odd primes don't work; which is pretty easy: take , and note that there's a magic number which can't appear anywhere else. Done.

Now for the construction for the case. It's pretty easy to construct inductively, but let's try something else. Specifically, we take the numbers , write them in binary (making sure that each has bits, adding leading zeros if necessary) and reverse them (!), and finally convert them back to decimal. So for , it would look like this:
Why on earth would this completely arbitrary out-of-the-blue construction work? Well, consider the scenario right before converting back to decimal again (after all, a base-to-base conversion doesn't change the number itself). Take any three numbers in this permutation occurring in that order. From the right, consider the first place where the bits are not all equal. Say, if we had chosen , then it would be the second place from the right; because the last digit is the same in all of them. The differing digits must be or ; because before reversing, these numbers were in increasing order. It is easy to see that , so none of them leads to a , and hence is not divisible by , which means we win.

The Hexagonal Conspiracy

[quote](Iranian Geometry Olympiad 2016) Do there exist six points in the plane such that all of the triangles are similar for ?

Proposed by Morteza Saghafian
[/quote]

OK, so this is a barefaced constructive (probably instructive too, for that matter) problem. There does exist such a monster, and we're somehow expected to magically summon it.

If you've seen the official solution, you'd know it looks super-contrived. Here's another solution, found by the PSE user deep thought ( have been replaced by respectively):

[center]IMAGE[/center]


And that's how you use constructions for destruction of a problem.

[center]"One line done!" - zdyzhj[/center]

Continuing the path last travelled in Elegant, not Elephant: Part II, once again I am going to list down some mathematical gems. Admittedly, this post deviates a bit from its prequels, in the sense that elegance has been given somewhat more importance than brevity, and I couldn't resist posting some extremely beautiful proofs in spite of the fact that they are not really one-line. But I have carefully selected only magical, surprising proofs that show the underlying beauty of mathematical notions. And oh, dark magic and sorcery like the following is a big no-no.

[center]IMAGE[/center]
[center]Next, let's assume the decision of whether to take the Axiom of Choice is made by a deterministic process ...[/center]
[right]Image credit: Randall Munroe, Proofs, url: https://xkcd.com/1724/[/right]

The first problem was asked in Tournament of Towns, 1998.

At many train stations, post offices and courier services around the world, the cost of sending a
rectangular box is determined by the sum of its dimensions; that is, length plus width plus height. (So a box with dimensions costs more than sending a because , even though they have the same volume.) Can I ever put a box into a cheaper box, and thus "cheat"?

Solution

The next problem may sound like a boring algebra exercise (and can be solved that way), but the solution that follows definitely deserves mention.

One day, a person went to a horse racing area. Instead of counting the number of humans and horses, he counted heads and legs. How many humans and horses were there?


Solution

The following problem occurs in Mathematical Puzzles: A Connoisseur’s Collection by Peter Winkler.

Fifty coins of different denominations are placed in a row on a table. Alice and Bob play a game with these coins. First, Alice picks a coin placed at any of the ends of the row, and then Bob picks up a coin at any of the ends of the remaining row, and so on, till Bob pockets the last coin. Prove that regardless of the initial position, Alice can get at least as much money as Bob.


Solution

Next, a classic one. I've seen it in both Mathematical Miniatures (Savchev and Andreescu) and Problem-Solving Strategies (Engel), but most probably it's much older than that and more of a chestnut.

The game of Double-chess has the same rules as normal chess, except that player must make two consecutive moves on every turn. Prove that the white player, who begins first, has a non-losing strategy.

Solution

This one is taken from a post by rand al'thor on Puzzling.SE.

A perfectly symmetrical small 4-legged table is standing in a large room with a continuous but uneven floor. Is it always possible to position the table in such a way so that all four legs are touching the floor?

Solution

"Any good idea can be stated in fifty words or less." - Stanislaw Ulam

The purpose of this post is collect some illuminating proofs where a small insight trivializes the seemingly difficult problem. The proofs need not be exactly one-line long; the general (and flexible) criteria for inclusion is

[list][*]The solution must not be obvious; it must be unexpected and surprising.
[*]It should preferably demonstrate the interdisciplinary interconnections in mathematics.
[*]And of course, as noted above, the central idea should be describable in fifty words or less.[/list]

Now have a look at some of the one-liners I have encountered so far:

Alice and Bob are playing a game with the set . They make alternating moves, with Alice going first. At each move, one has to choose a number among the set which has not been chosen by anyone before. If at some point, a player has three numbers adding up to among those (s)he has chosen, (s)he wins. Prove that Alice can force a draw at least.

Solution

A Limp King is a chess piece that can move one square in any direction except for northeast and southwest.
[center]IMAGE IMAGE[/center]
In a standard chessboard, some squares are to be destroyed. To cross the board, the Limp King has to start on any undestroyed square on the North side of the board, and make valid moves passing through undestroyed squares, and end up on an undestroyed square on the South edge of the board. How many ways are there to destroy exactly squares, such that the Limp King cannot cross the board?

Solution

Let be a positive integer greater than . Prove that there exists an irrational number such that is a rational number.

Solution

An isosceles right-angled triangle shaped billiards table has three pockets at the vertices. A ball starts moving from one of the pockets adjacent to hypotenuse. When it reaches to one side then it will reflect its path. Prove that if it falls into a pocket then it is not the pocket whence it started.

Solution

The game of two-move chess follows the usual rules of chess with one exception: each player has to make two consecutive moves at a time. Prove that White (who goes first) has a non-losing strategy.

Solution

I will try to update this list as often as possible. You are welcome to suggest other one-line proofs you know to add to this post.

Sources:
1. An AoPS topic here, 2. A post on Puzzling.SE here, 3. own, here, 4. Iran TST 2007, 5. Mathematical Miniatures, Svetoslav Savchev and Titu Andreescu.

This post is a sequel to my previous blog post: Elegant, not Elephant. As before, I aim to collect and present some more instances of mathematical problems, which are seemingly unassailable but are deceptively simple. The topics vary widely : combinatorics, algebra, geometry -- but a common theme of innate elegance binds them together.

OK, I guess that's enough of a prologue; let's start this off with a rather famous one:

One hundred ants are dropped on a meter stick. Each ant is traveling either to the left or the right with constant speed meter per minute. When two ants meet, they bounce off each other and reverse direction. When an ant reaches an end of the stick, it falls off. Over all possible initial configurations, what is the longest amount of time that you would need to wait to guarantee that the stick has no more ants?

Solution

A rectangle is called special if at least one of its sides has integer length. Suppose a rectangle is tiled by rectangles each of which is special. Prove that the big rectangle itself is special.

Solution

is the median of the triangle , and and are incenters of and respectively. Given that is cyclic, prove that is isosceles.

Solution

Suppose we can place equal disk-shaped coins on a rectangular table such that no two coins overlap and it's impossible to place another coin on the table without causing overlap. Prove that allowing overlap, we can cover the table entirely using no more than coins. Note that a coin can stay on the table iff its center lies on the table.

Solution

roads on a plane, each a straight line, are in general position so that no two are parallel and no three pass through the same point. Along each road walks a traveler at a constant speed. Their speeds, however, may not be the same. It's known that traveler met with Travelers , and each of the other travellers met both Traveller and . Show that any two travellers met each other.

Solution

Appendix: Here's another of my favourite ones; I didn't include it above since this proof is disregarded by many mathematicians.

Hex Theorem : The game of Hex cannot end in a draw.

Proof

--------------

Sources:
6. Su, Francis E., et al. "Ants on a Stick." Math Fun Facts. http://www.math.hmc.edu/funfacts.
7. Stan Wagon, Fourteen Proofs of a Result About Tiling a Rectangle.
8. PDC, India, 2016
9. An AoPS topic here.
10. Four Traveller's Problem at Cut-The-Knot , http://www.cut-the-knot.org/gproblems.shtml
Appendix: http://www.cut-the-knot.org/Curriculum/Games/HexTheory.shtml