# Some Thoughts on Olympiad Material Design

(This is a bit of a follow-up to the solution reading post last month. Spoiler warnings: USAMO 2014/6, USAMO 2012/2, TSTST 2016/4, and hints for ELMO 2013/1, IMO 2016/2.)

I want to say a little about the process which I use to design my olympiad handouts and classes these days (and thus by extension the way I personally think about problems). The short summary is that my teaching style is centered around showing connections and recurring themes between problems.

Now let me explain this in more detail.

## 1. Main ideas

Solutions to olympiad problems can look quite different from one another at a surface level, but typically they center around one or two main ideas, as I describe in my post on reading solutions. Because details are easy to work out once you have the main idea, as far as learning is concerned you can more or less throw away the details and pay most of your attention to main ideas.

Thus whenever I solve an olympiad problem, I make a deliberate effort to summarize the solution in a few sentences, such that I basically know how to do it from there. I also make a deliberate effort, whenever I write up a solution in my notes, to structure it so that my future self can see all the key ideas at a glance and thus be able to understand the general path of the solution immediately.

The example I’ve previously mentioned is USAMO 2014/6.

Example 1 (USAMO 2014, Gabriel Dospinescu)

Prove that there is a constant ${c>0}$ with the following property: If ${a, b, n}$ are positive integers such that ${\gcd(a+i, b+j)>1}$ for all ${i, j \in \{0, 1, \dots, n\}}$, then

$\displaystyle \min\{a, b\}> (cn)^n.$

If you look at any complete solution to the problem, you will see a lot of technical estimates involving ${\zeta(2)}$ and the like. But the main idea is very simple: “consider an ${N \times N}$ table of primes and note the small primes cannot adequately cover the board, since ${\sum p^{-2} < \frac{1}{2}}$”. Once you have this main idea the technical estimates are just the grunt work that you force yourself to do if you’re a contestant (and don’t do if you’re retired like me).

Thus the study of olympiad problems is reduced to the study of main ideas behind these problems.

## 2. Taxonomy

So how do we come up with the main ideas? Of course I won’t be able to answer this question completely, because therein lies most of the difficulty of olympiads.

But I do have some progress in this way. It comes down to seeing how main ideas are similar to each other. I spend a lot of time trying to classify the main ideas into categories or themes, based on how similar they feel to one another. If I see one theme pop up over and over, then I can make it into a class.

I think olympiad taxonomy is severely underrated, and generally not done correctly. The status quo is that people do bucket sorts based on the particular technical details which are present in the problem. This is correlated with the main ideas, but the two do not always coincide.

An example where technical sort works okay is Euclidean geometry. Here is a simple example: harmonic bundles in projective geometry. As I explain in my book, there are a few “basic” configurations involved:

• Midpoints and parallel lines
• The Ceva / Menelaus configuration
• Harmonic quadrilateral / symmedian configuration
• Apollonian circle (right angle and bisectors)

(For a reference, see Lemmas 2, 4, 5 and Exercise 0 here.) Thus from experience, any time I see one of these pictures inside the current diagram, I think to myself that “this problem feels projective”; and if there is a way to do so I try to use harmonic bundles on it.

An example where technical sort fails is the “pigeonhole principle”. A typical problem in such a class looks something like USAMO 2012/2.

Example 2 (USAMO 2012, Gregory Galperin)

A circle is divided into congruent arcs by ${432}$ points. The points are colored in four colors such that some ${108}$ points are colored Red, some ${108}$ points are colored Green, some ${108}$ points are colored Blue, and the remaining ${108}$ points are colored Yellow. Prove that one can choose three points of each color in such a way that the four triangles formed by the chosen points of the same color are congruent.

It’s true that the official solution uses the words “pigeonhole principle” but that is not really the heart of the matter; the key idea is that you consider all possible rotations and count the number of incidences. (In any case, such calculations are better done using expected value anyways.)

Now why is taxonomy a good thing for learning and teaching? The reason is that building connections and seeing similarities is most easily done by simultaneously presenting several related problems. I’ve actually mentioned this already in a different blog post, but let me give the demonstration again.

Suppose I wrote down the following:

$\displaystyle \begin{array}{lll} A1 & B11 & C8 \\ A9 & B44 & C27 \\ A49 & B33 & C343 \\ A16 & B99 & C1 \\ A25 & B22 & C125 \end{array}$

You can tell what each of the ${A}$‘s, ${B}$‘s, ${C}$‘s have in common by looking for a few moments. But what happens if I intertwine them?

$\displaystyle \begin{array}{lllll} B11 & C27 & C343 & A1 & A9 \\ C125 & B33 & A49 & B44 & A25 \\ A16 & B99 & B22 & C8 & C1 \end{array}$

This is the same information, but now you have to work much harder to notice the association between the letters and the numbers they’re next to.

This is why, if you are an olympiad student, I strongly encourage you to keep a journal or blog of the problems you’ve done. Solving olympiad problems takes lots of time and so it’s worth it to spend at least a few minutes jotting down the main ideas. And once you have enough of these, you can start to see new connections between problems you haven’t seen before, rather than being confined to thinking about individual problems in isolation. (Additionally, it means you will never have redo problems to which you forgot the solution — learn from my mistake here.)

## 3. Ten buckets of geometry

I want to elaborate more on geometry in general. These days, if I see a solution to a Euclidean geometry problem, then I mentally store the problem and solution into one (or more) buckets. I can even tell you what my buckets are:

1. Direct angle chasing
2. Power of a point / radical axis
3. Homothety, similar triangles, ratios
4. Recognizing some standard configuration (see Yufei for a list)
5. Doing some length calculations
6. Complex numbers
7. Barycentric coordinates
8. Inversion
9. Harmonic bundles or pole/polar and homography
10. Spiral similarity, Miquel points

which my dedicated fans probably recognize as the ten chapters of my textbook. (Problems may also fall in more than one bucket if for example they are difficult and require multiple key ideas, or if there are multiple solutions.)

Now whenever I see a new geometry problem, the diagram will often “feel” similar to problems in a certain bucket. Exactly what I mean by “feel” is hard to formalize — it’s a certain gut feeling that you pick up by doing enough examples. There are some things you can say, such as “problems which feature a central circle and feet of altitudes tend to fall in bucket 6”, or “problems which only involve incidence always fall in bucket 9”. But it seems hard to come up with an exhaustive list of hard rules that will do better than human intuition.

## 4. How do problems feel?

But as I said in my post on reading solutions, there are deeper lessons to teach than just technical details.

For examples of themes on opposite ends of the spectrum, let’s move on to combinatorics. Geometry is quite structured and so the themes in the main ideas tend to translate to specific theorems used in the solution. Combinatorics is much less structured and many of the themes I use in combinatorics cannot really be formalized. (Consequently, since everyone else seems to mostly teach technical themes, several of the combinatorics themes I teach are idiosyncratic, and to my knowledge are not taught by anyone else.)

For example, one of the unusual themes I teach is called Global. It’s about the idea that to solve a problem, you can just kind of “add up everything at once”, for example using linearity of expectation, or by double-counting, or whatever. In particular these kinds of approach ignore the “local” details of the problem. It’s hard to make this precise, so I’ll just give two recent examples.

Example 3 (ELMO 2013, Ray Li)

Let ${a_1,a_2,\dots,a_9}$ be nine real numbers, not necessarily distinct, with average ${m}$. Let ${A}$ denote the number of triples ${1 \le i < j < k \le 9}$ for which ${a_i + a_j + a_k \ge 3m}$. What is the minimum possible value of ${A}$?

Example 4 (IMO 2016)

Find all integers ${n}$ for which each cell of ${n \times n}$ table can be filled with one of the letters ${I}$, ${M}$ and ${O}$ in such a way that:

• In each row and column, one third of the entries are ${I}$, one third are ${M}$ and one third are ${O}$; and
• in any diagonal, if the number of entries on the diagonal is a multiple of three, then one third of the entries are ${I}$, one third are ${M}$ and one third are ${O}$.

If you look at the solutions to these problems, they have the same “feeling” of adding everything up, even though the specific techniques are somewhat different (double-counting for the former, diagonals modulo ${3}$ for the latter). Nonetheless, my experience with problems similar to the former was immensely helpful for the latter, and it’s why I was able to solve the IMO problem.

## 5. Gaps

This perspective also explains why I’m relatively bad at functional equations. There are some things I can say that may be useful (see my handouts), but much of the time these are just technical tricks. (When sorting functional equations in my head, I have a bucket called “standard fare” meaning that you “just do work”; as far I can tell this bucket is pretty useless.) I always feel stupid teaching functional equations, because I never have many good insights to say.

Part of the reason is that functional equations often don’t have a main idea at all. Consequently it’s hard for me to do useful taxonomy on them.

Then sometimes you run into something like the windmill problem, the solution of which is fairly “novel”, not being similar to problems that come up in training. I have yet to figure out a good way to train students to be able to solve windmill-like problems.

## 6. Surprise

I’ll close by mentioning one common way I come up with a theme.

Sometimes I will run across an olympiad problem ${P}$ which I solve quickly, and think should be very easy, and yet once I start grading ${P}$ I find that the scores are much lower than I expected. Since the way I solve problems is by drawing experience from similar previous problems, this must mean that I’ve subconsciously found a general framework to solve problems like ${P}$, which is not obvious to my students yet. So if I can put my finger on what that framework is, then I have something new to say.

The most recent example I can think of when this happened was TSTST 2016/4 which was given last June (and was also a very elegant problem, at least in my opinion).

Example 5 (TSTST 2016, Linus Hamilton)

Let ${n > 1}$ be a positive integers. Prove that we must apply the Euler ${\varphi}$ function at least ${\log_3 n}$ times before reaching ${1}$.

I solved this problem very quickly when we were drafting the TSTST exam, figuring out the solution while walking to dinner. So I was quite surprised when I looked at the scores for the problem and found out that empirically it was not that easy.

After I thought about this, I have a new tentative idea. You see, when doing this problem I really was thinking about “what does this ${\varphi}$ operation do?”. You can think of ${n}$ as an infinite tuple

$\displaystyle \left(\nu_2(n), \nu_3(n), \nu_5(n), \nu_7(n), \dots \right)$

of prime exponents. Then the ${\varphi}$ can be thought of as an operation which takes each nonzero component, decreases it by one, and then adds some particular vector back. For example, if ${\nu_7(n) > 0}$ then ${\nu_7}$ is decreased by one and each of ${\nu_2(n)}$ and ${\nu_3(n)}$ are increased by one. In any case, if you look at this behavior for long enough you will see that the ${\nu_2}$ coordinate is a natural way to “track time” in successive ${\varphi}$ operations; once you figure this out, getting the bound of ${\log_3 n}$ is quite natural. (Details left as exercise to reader.)

Now when I read through the solutions, I found that many of them had not really tried to think of the problem in such a “structured” way, and had tried to directly solve it by for example trying to prove ${\varphi(n) \ge n/3}$ (which is false) or something similar to this. I realized that had the students just ignored the task “prove ${n \le 3^k}$” and spent some time getting a better understanding of the ${\varphi}$ structure, they would have had a much better chance at solving the problem. Why had I known that structural thinking would be helpful? I couldn’t quite explain it, but it had something to do with the fact that the “main object” of the question was “set in stone”; there was no “degrees of freedom” in it, and it was concrete enough that I felt like I could understand it. Once I understood how multiple ${\varphi}$ operations behaved, the bit about ${\log_3 n}$ almost served as an “answer extraction” mechanism.

These thoughts led to the recent development of a class which I named Rigid, which is all about problems where the point is not to immediately try to prove what the question asks for, but to first step back and understand completely how a particular rigid structure (like the ${\varphi}$ in this problem) behaves, and to then solve the problem using this understanding.

(Ed Note: This was earlier posted under the incorrect title “On Designing Olympiad Training”. How I managed to mess that up is a long story involving some incompetence with Python scripts, but this is fixed now.)

Spoiler warnings: USAMO 2014/1, and hints for Putnam 2014 A4 and B2. You may want to work on these problems yourself before reading this post.

## 1. An Apology

At last year’s USA IMO training camp, I prepared a handout on writing/style for the students at MOP. One of the things I talked about was the “ocean-crossing point”, which for our purposes you can think of as the discrete jump from a problem being “essentially not solved” (${0+}$) to “essentially solved” (${7-}$). The name comes from a Scott Aaronson post:

Suppose your friend in Boston blindfolded you, drove you around for twenty minutes, then took the blindfold off and claimed you were now in Beijing. Yes, you do see Chinese signs and pagoda roofs, and no, you can’t immediately disprove him — but based on your knowledge of both cars and geography, isn’t it more likely you’re just in Chinatown? . . . We start in Boston, we end up in Beijing, and at no point is anything resembling an ocean ever crossed.

I then gave two examples of how to write a solution to the following example problem.

Problem 1 (USAMO 2014)

Let ${a}$, ${b}$, ${c}$, ${d}$ be real numbers such that ${b-d \ge 5}$ and all zeros ${x_1}$, ${x_2}$, ${x_3}$, and ${x_4}$ of the polynomial ${P(x)=x^4+ax^3+bx^2+cx+d}$ are real. Find the smallest value the product

$\displaystyle (x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)$

can take.

Proof: (Not-so-good write-up) Since ${x_j^2+1 = (x+i)(x-i)}$ for every ${j=1,2,3,4}$ (where ${i=\sqrt{-1}}$), we get ${\prod_{j=1}^4 (x_j^2+1) = \prod_{j=1}^4 (x_j+i)(x_j-i) = P(i)P(-i)}$ which equals to ${|P(i)|^2 = (b-d-1)^2 + (a-c)^2}$. If ${x_1 = x_2 = x_3 = x_4 = 1}$ this is ${16}$ and ${b-d = 5}$. Also, ${b-d \ge 5}$, this is ${\ge 16}$. $\Box$

Proof: (Better write-up) The answer is ${16}$. This can be achieved by taking ${x_1 = x_2 = x_3 = x_4 = 1}$, whence the product is ${2^4 = 16}$, and ${b-d = 5}$.

Now, we prove this is a lower bound. Let ${i = \sqrt{-1}}$. The key observation is that

$\displaystyle \prod_{j=1}^4 \left( x_j^2 + 1 \right) = \prod_{j=1}^4 (x_j - i)(x_j + i) = P(i)P(-i).$

Consequently, we have

\displaystyle \begin{aligned} \left( x_1^2 + 1 \right) \left( x_2^2 + 1 \right) \left( x_3^2 + 1 \right) \left( x_1^2 + 1 \right) &= (b-d-1)^2 + (a-c)^2 \\ &\ge (5-1)^2 + 0^2 = 16. \end{aligned}

This proves the lower bound. $\Box$

You’ll notice that it’s much easier to see the key idea in the second solution: namely,

$\displaystyle \prod_j (x_j^2+1) = P(i)P(-i) = (b-d-1)^2 + (a-c)^2$

which allows you use the enigmatic condition ${b-d \ge 5}$.

Unfortunately I have the following confession to make:

In practice, most solutions are written more like the first one than the second one.

The truth is that writing up solutions is sort of a chore that people never really want to do but have to — much like washing dishes. So must solutions won’t be written in a way that helps you learn from them. This means that when you read solutions, you should assume that the thing you really want (i.e., the ocean-crossing point) is buried somewhere amidst a haystack of other unimportant details.

## 2. Diff

But in practice even the “better write-up” I mentioned above still has too much information in it.

Suppose you were explaining how to solve this problem to a friend. You would probably not start your explanation by saying that the minimum is ${16}$, achieved by ${x_1 = x_2 = x_3 = x_4 = 1}$ — even though this is indeed a logically necessary part of the solution. Instead, the first thing you would probably tell them is to notice that

$\displaystyle \prod_{j=1}^4 \left( x_j^2 + 1 \right) = P(i)P(-i) = (b-d-1)^2 + (a-c)^2 \ge 4^2 = 16.$

In fact, if your friend has been working on the problem for more than ten minutes, this is probably the only thing you need to tell them. They probably already figured out by themselves that there was a good chance the answer would be ${2^4 = 16}$, just based on the condition ${b-d \ge 5}$. This “one-liner” is all that they need to finish the problem. You don’t need to spell out to them the rest of the details.

When you explain a problem to a friend in this way, you’re communicating just the difference: the one or two sentences such that your friend could work out the rest of the details themselves with these directions. When reading the solution yourself, you should try to extract the main idea in the same way. Olympiad problems generally have only a few main ideas in them, from which the rest of the details can be derived. So reading the solution should feel much like searching for a needle in a haystack.

## 3. Don’t Read Line by Line

In particular: you should rarely read most of the words in the solution, and you should almost never read every word of the solution.

Whenever I read solutions to problems I didn’t solve, I often read less than 10% of the words in the solution. Instead I search aggressively for the one or two sentences which tell me the key step that I couldn’t find myself. (Functional equations are the glaring exception to this rule, since in these problems there sometimes isn’t any main idea other than “stumble around randomly”, and the steps really are all about equally important. But this is rarer than you might guess.)

I think a common mistake students make is to treat the solution as a sequence of logical steps: that is, reading the solution line by line, and then verifying that each line follows from the previous ones. This seems to entirely miss the point, because not all lines are created equal, and most lines can be easily derived once you figure out the main idea.

If you find that the only way that you can understand the solution is reading it step by step, then the problem may simply be too hard for you. This is because what counts as “details” and “main ideas” are relative to the absolute difficulty of the problem. Here’s an example of what I mean: the solution to a USAMO 3/6 level geometry problem, call it ${P}$, might look as follows.

Proof: First, we prove lemma ${L_1}$. (Proof of ${L_1}$, which is USAMO 1/4 level.)

Then, we prove lemma ${L_2}$. (Proof of ${L_2}$, which is USAMO 1/4 level.)

Finally, we remark that putting together ${L_1}$ and ${L_2}$ solves the problem. $\Box$

Likely the main difficulty of ${P}$ is actually finding ${L_1}$ and ${L_2}$. So a very experienced student might think of the sub-proofs ${L_i}$ as “easy details”. But younger students might find ${L_i}$ challenging in their own right, and be unable to solve the problem even after being told what the lemmas are: which is why it is hard for them to tell that ${\{L_1, L_2\}}$ were the main ideas to begin with. In that case, the problem ${P}$ is probably way over their head.

This is also why it doesn’t make sense to read solutions to problems which you have not worked on at all — there are often details, natural steps and notation, et cetera which are obvious to you if and only if you have actually tried the problem for a little while yourself.

## 4. Reflection

The earlier sections describe how to extract the main idea of an olympiad solution. This is neat because instead of having to remember an entire solution, you only need to remember a few sentences now, and it gives you a good understanding of the solution at hand.

But this still isn’t achieving your ultimate goal in learning: you are trying to maximize your scores on future problems. Unless you are extremely fortunate, you will probably never see the exact same problem on an exam again.

So one question you should often ask is:

“How could I have thought of that?”

(Or in my case, “how could I train a student to think of this?”.)

There are probably some surface-level skills that you can pick out of this. The lowest hanging fruit is things that are technical. A small number of examples, with varying amounts of depth:

• This problem is “purely projective”, so we can take a projective transformation!
• This problem had a segment ${AB}$ with midpoint ${M}$, and a line ${\ell}$ parallel to ${AB}$, so I should consider projecting ${(AB;M\infty)}$ through a point on ${\ell}$.
• Drawing a grid of primes is the only real idea in this problem, and the rest of it is just calculations.
• This main claim is easy to guess since in some small cases, the frogs have “violating points” in a large circle.
• In this problem there are ${n}$ numbers on a circle, ${n}$ odd. The counterexamples for ${n}$ even alternate up and down, which motivates proving that no three consecutive numbers are in sorted order.
• This is a juggling problem!

(Brownie points if any contest enthusiasts can figure out which problems I’m talking about in this list!)

## 5. Learn Philosophy, not Formalism

But now I want to point out that the best answers to the above question are often not formalizable. Lists of triggers and actions are “cheap forms of understanding”, because going through a list of methods will only get so far.

On the other hand, the un-formalizable philosophy that you can extract from reading a question, is part of that legendary “intuition” that people are always talking about: you can’t describe it in words, but it’s certainly there. Maybe I would even be better if I reframed the question as:

“What does this problem feel like?”

So let’s talk about our feelings. Here is David Yang’s take on it:

Whenever you see a problem you really like, store it (and the solution) in your mind like a cherished memory . . . The point of this is that you will see problems which will remind you of that problem despite having no obvious relation. You will not be able to say concretely what the relation is, but think a lot about it and give a name to the common aspect of the two problems. Eventually, you will see new problems for which you feel like could also be described by that name.

Do this enough, and you will have a very powerful intuition that cannot be described easily concretely (and in particular, that nobody else will have).

This itself doesn’t make sense without an example, so here is an example of one philosophy I’ve developed. Here are two problems on Putnam 2014:

Problem 2 (Putnam 2014 A4)

Suppose ${X}$ is a random variable that takes on only nonnegative integer values, with ${\mathbb E[X] = 1}$, ${\mathbb E[X^2] = 2}$, and ${\mathbb E[X^3] = 5}$. Determine the smallest possible value of the probability of the event ${X=0}$.

Problem 3 (Putnam 2014 B2)

Suppose that ${f}$ is a function on the interval ${[1,3]}$ such that ${-1\le f(x)\le 1}$ for all ${x}$ and

$\displaystyle \int_1^3 f(x) \; dx=0.$

How large can ${\int_1^3 \frac{f(x)}{x} \; dx}$ be?

At a glance there seems to be nearly no connection between these problems. One of them is a combinatorics/algebra question, and the other is an integral. Moreover, if you read the official solutions or even my own write-ups, you will find very little in common joining them.

Yet it turns out that these two problems do have something in common to me, which I’ll try to describe below. My thought process in solving either question went as follows:

In both problems, I was able to quickly make a good guess as to what the optimal ${X}$/${f}$ was, and then come up with a heuristic explanation (not a proof) why that guess had to be correct, namely, “by smoothing, you should put all the weight on the left”. Let me call this optimal argument ${A}$.

That conjectured ${A}$ gave a numerical answer to the actual problem: but for both of these problems, it turns out that numerical answer is completely uninteresting, as are the exact details of ${A}$. It should be philosophically be interpreted as “this is the number that happens to pop out when you plug in the optimal choice”. And indeed that’s what both solutions feel like. These solutions don’t actually care what the exact values of ${A}$ are, they only care about the properties that made me think they were optimal in the first place.

I gave this philosophy the name Equality, with poster description “problems where looking at the equality case is important”. This text description feels more or less useless to me; I suppose it’s the thought that counts. But ever since I came up with this name, it has helped me solve new problems that come up, because they would give me the same feeling that these two problems did.

Two more examples of these themes that I’ve come up with are Global and Rigid, which will be described in a future post on how I design training materials.

# Against Perfect Scores

One of the pieces of advice I constantly give to young students preparing for math contests is that they should probably do harder problems. But perhaps I don’t preach this zealously enough for them to listen, so here’s a concrete reason (with actual math!) why I give this advice.

## 1. The AIME and USAMO

In the USA many students who seriously prepare for math contests eventually qualify for an exam called the AIME (American Invitational Math Exam). This is a 3-hour exam with 15 short-answer problems; the median score is maybe about 5 problems.

Correctly solving maybe 10 of the problems qualifies for the much more difficult USAMO. This national olympiad is much more daunting, with six proof-based problems given over nine hours. It is not uncommon for olympiad contestants to not solve a single problem (this certainly happened to me a fair share of times!).

You’ll notice the stark difference in the scale of these contests (Tanya Khovanova has a longer complaint about this here). For students who are qualifying for USAMO for the first time, the olympiad is terrifying: I certainly remember the first time I took the olympiad with a super lofty goal of solving any problem.

Now, my personal opinion is that the difference between AIME and USAMO is generally exaggerated, and less drastic than appearances suggest. But even then, the psychological fear is still there — so what do you think happens to this demographic of students?

Answer: they don’t move on from AIME training. They think, “oh, the USAMO is too hard, I can only solve 10 problems on the AIME so I should stick to solving hard problems on the AIME until I can comfortably solve most of them”. So they keep on working through old AIME papers.

## 2. Perfect Scores

To understand why this is a bad idea, let’s ask the following question: how good to you have to be to consistently get a perfect score on the AIME?

Consider first a student averages a score of ${10}$ on the AIME, which is a fairly comfortable qualifying score. For illustration, let’s crudely simplify and assume that on a 15-question exam, he has a independent ${\frac23}$ probability of getting each question right. Then the chance he sweeps the AIME is

$\displaystyle \left( \frac23 \right)^{15} \approx 0.228\%.$

This is pretty low, which makes sense: ${10}$ and ${15}$ on the AIME feel like quite different scores.

Now suppose we bump that up to averaging ${12}$ problems on the AIME, which is almost certainly enough to qualify for the USAMO. This time, the chance of sweeping is

$\displaystyle \left( \frac{4}{5} \right)^{15} \approx 3.52\%.$

This should feel kind of low to you as well. So if you consistently solve ${80\%}$ of problems in training, your chance at netting a perfect score is still dismal, even though on average you’re only three problems away.

Well, that’s annoying, so let’s push this as far as we can: consider a student who’s averaging ${14}$ problems (thus, ${93\%}$ success), id est a near-perfect score. Then the probability of getting a perfect score

$\displaystyle \left( \frac{14}{15} \right)^{15} \approx 35.5\%.$

Which is\dots just over ${\frac 13}$.

At which point you throw up your hands and say, what more could you ask for? I’m already averaging one less than a perfect score, and I still don’t have a good chance of acing the exam? This should feel very unfair: on average you’re only one problem away from full marks, and yet doing one problem better than normal is still a splotchy hit-or-miss.

## 3. Some Combinatorics

Those of you who either know statistics / combinatorics might be able to see what’s going on now. The problem is that

$\displaystyle (1-\varepsilon)^{15} \approx 1 - 15\varepsilon$

for small ${\varepsilon}$. That is, if your accuracy is even a little ${\varepsilon}$ away from perfect, that difference gets amplified by a factor of ${15}$ against you.

Below is a nice chart that shows you, based on this oversimplified naïve model, how likely you are to do a little better than your average.

$\displaystyle \begin{array}{lrrrrrr} \textbf{Avg} & \ge 10 & \ge 11 & \ge 12 & \ge 13 & \ge 14 & \ge 15 \\ \hline \mathbf{10} & 61.84\% & 40.41\% & 20.92\% & 7.94\% & 1.94\% & 0.23\% \\ \mathbf{11} & & 63.04\% & 40.27\% & 19.40\% & 6.16\% & 0.95\% \\ \mathbf{12} & & & 64.82\% & 39.80\% & 16.71\% & 3.52\% \\ \mathbf{13} & & & & 67.71\% & 38.66\% & 11.69\% \\ \mathbf{14} & & & & & 73.59\% & 35.53\% \\ \mathbf{15} & & & & & & 100.00\% \\ \end{array}$

Even if you’re not aiming for that lofty perfect score, we see the same repulsion effect: it’s quite hard to do even a little better than average. If you get an average score of ${k}$, the probability of getting ${k+1}$ looks to be about ${\frac25}$. As for ${k+2}$ the chances are even more dismal. In fact, merely staying afloat (getting at least your average score) isn’t a comfortable proposition.

And this is in my simplified model of “independent events”. Those of you who actually take the AIME know just how costly small arithmetic errors are, and just how steep the difficulty curve on this exam is.

All of this goes to show: to reliably and consistently ace the AIME, it’s not enough to be able to do 95% of AIME problems (which is already quite a feat). You almost need to be able to solve AIME problems in your sleep. On any given AIME some people will get luckier than others, but coming out with a perfect score every time is a huge undertaking.

## 4. 90% Confidence?

By the way, did I ever mention that it’s really hard to be 90% confident in something? In most contexts, 90% is a really big number.

If you don’t know what I’m talking about:

This is also the first page of this worksheet. The idea of this quiz is to give you a sense of just how high 90% is. To do this, you are asked 10 numerical questions and must provide an interval which you think the answer lies within with probability 90%. (So ideally, you would get exactly 9 intervals correct.)

As a hint: almost everyone is overconfident. Second hint: almost everyone is overconfident even after being told that their intervals should be embarrassingly wide. Third hint: I just tried this again and got a low score.

(For more fun of this form: calibration game.)

## 5. Practice

So what do you do if you really want to get a perfect score on the AIME?

Well, first of all, my advice is that you have better things to do (like USAMO). But even if you are unshakeable on your desire to get a 15, my advice still remains the same: do some USAMO problems.

Why? The reason is that going from average ${14}$ to average ${15}$ means going from 95% accuracy to 99% accuracy, as I’ve discussed above.

So what you don’t want to do is keep doing AIME problems. You are not using your time well if you get 95% accuracy in training. I’m well on record saying that you learn the most from problems that are just a little above your ability level, and massing AIME problems is basically the exact opposite of that. You’d maybe only run into a problem you couldn’t solve once every 10 or 20 or 30 problems. That’s just grossly inefficient.

The way out of this is to do harder problems, and that’s why I explicitly suggest people start working on USAMO problems even before they’re 90% confident they will qualify for it. At the very least, you certainly won’t be bored.

# Stop Paying Me Per Hour

Occasionally I am approached by parents who ask me if I am available to teach their child in olympiad math. This is flattering enough that I’ve even said yes a few times, but I’m always confused why the question is “can you tutor my child?” instead of “do you think tutoring would help, and if so, can you tutor my child?”.

Here are my thoughts on the latter question.

## Charging by Salt

I’m going to start by clearing up the big misconception which inspired the title of this post.

The way tutoring works is very roughly like the following: I meet with the student once every week, with custom-made materials. Then I give them some practice problems to work on (“homework”), which I also grade. I throw in some mock olympiads. I strongly encourage my students to email me with questions as they come up. Rinse and repeat.

The actual logistics vary; for example, for small in-person groups I prefer to do every other week for 3 hours. But the thing that never changes is how the parents pay me. It’s always the same: I get $N \gg 0$ dollars per hour for the actual in-person meeting, and $0$ dollars per hour for preparing materials, grading homework, responding to questions, and writing the mock olympiads.

Now I’m not complaining because $N$ is embarrassingly large. But one day I realized that this pricing system is giving parents the wrong impression. They now think the bulk of the work is from me taking the time to meet with their child, and that the homework is to help reinforce what I talk about in class. After all, this is what high school does, right?

I’m here to tell you that this is completely wrong.

It’s the other way around: the class is meant to supplement the homework. Think salt: for most dishes you can’t get away with having zero salt, but you still don’t price a dish based on how much salt is in it. Similarly, you can’t excise the in-person meeting altogether, but the dirty secret is that the classtime isn’t the core component.

I mean, here’s the thing.

• When you take the IMO, you are alone with a sheet of paper that says “Problem 1”, “Problem 2”, “Problem 3”.
• When you do my homework, you are alone with a sheet of paper that says “Problem 1”, “Problem 2”, “Problem 3”.
• When you’re in my class, you get to see my beautiful smiling face plus a sheet of paper that says “Theorem 1”, “Example 2”, “Example 3”.

Which of these is not like the other?

## Active Ingredients

So we’ve established that the main active ingredient is actually you working on problems alone in your room. If so, why do you need a teacher at all?

The answer depends on what the word “need” means. No USA IMO contestant in my recent memory has had a coach, so you don’t need a coach. But there are some good reasons why one might be helpful.

Some obvious reasons are social:

• Forces you to work regularly; though most top students don’t really have a problem with self-motivation
• You have a person to talk to. This can be nice if you are relatively isolated from the rest of the math community (e.g. due to geography).
• You have someone who will answer your questions. (I can’t tell you how jealous I am right now.)
• Feedback on solutions to problems. This includes student’s written solutions (stylistic remarks, or things like “this lemma you proved in your solution is actually just a special case of X” and so on) as well as explaining solutions to problems the student fails to solve.

In short, it’s much more engaging to study math with a real person.

Those reasons don’t depend so much on the instructor’s actual ability. Here are some reasons which do:

• Guidance. An instructor can tell you what things to learn or work on based on their own experience in the past, and can often point you to things that you didn’t know existed.
• It’s a big plus if the instructor has a good taste in problems. Some problems are bad and don’t teach you anything; some (old) problems don’t resemble the flavor of problems that actually appear on olympiads. On the flip side, some problems are very instructive or very pretty, and it’s great if your teacher knows what these are.
• Ideally, also a good taste in topics. For example, I strongly object to classes titled “collinearity and concurrence” because this may as well be called “geometry”, and I think that such global classes are too broad to do anything useful. Conversely, examples of topics I think should be classes but aren’t: “looking at equality cases”, “explicit constructions”, “Hall’s marriage theorem”, “greedy algorithms”. I make this point a lot more explicitly in Section 2 of this blog post of mine.

In short, you’re also paying for the material and expertise. Past IMO medalists know how the contest scene works. Parents and (beginning) students less so.

Lastly, the reason which I personally think is most important:

• Conveys strong intuition/heuristics, both globally and for specific problems. It’s hard to give concrete examples of this, but a few global ones I know were particularly helpful for me: “look at maximal things” (Po-Shen Loh on greedy algorithms), “DURR WE WANT STUFF TO CANCEL” (David Yang on FE’s), “use obvious inequalities” (Gabriel Dospinescu on analytic NT), which are take-aways that have gotten me a lot of points. This is also my biggest criteria for evaluating my own written exposition.

You guys know this feeling, I’m sure: when your English teacher assigned you an passage to read, the fastest way to understand it is to not read the passage but to ask the person sitting next to you what it’s saying. I think this is in part because most people are awful at writing and don’t even know how to write for other human beings.

The situation in olympiads is the same. I estimate listening to me explain a solution is maybe 4 to 10 times faster than reading the official solution. Turns out that writing up official solutions for contests is a huge chore, so most people just throw a sequence of steps at the reader without even bothering to identify the main ideas. (As a contest organizer, I’m certainly guilty of this laziness too!)

Aside: I think this is a large part of why my olympiad handouts and other writings have been so well-received. Disclaimer: this was supposed to be a list of what makes a good instructor, but due to narcissism it ended up being a list of things I focus on when teaching.

## Caveat Emptor

And now I explain why the top IMO candidates still got by without teachers.

It turns out that the amount of math preparation time that students put in doesn’t seem to be a normal distribution. It’s a log normal distribution. And the reason is this: it’s hard to do a really good job on anything you don’t think about in the shower.

Officially, when I was a contestant I spent maybe 20 hours a week doing math contest preparation. But the actual amount of time is higher. The reason is that I would think about math contests more like 24/7. During English class, I would often be daydreaming about the inequality I worked on last night. On the car ride home, I would idly think about what I was going to teach my middle school students the next week. To say nothing of showers: during my showers I would draw geometry diagrams on the wall with water on my finger.

So spiritually, I maybe spent 10 times as much time on math olympiads compared to an average USA(J)MO qualifier.

And that factor of 10 is enormous. Even if I as a coach can cause you to learn two or three or four times more efficiently, you will still lose to that factor of 10. I’d guess my actual multiplier is somewhere between 2 and 3, so there you go. (Edit: this used to say 3 to 4, I think that’s too high now.)

The best I can do is hope that, in addition to making my student’s training more efficient, I also cause my students to like math more.

You can use a wide range of wild, cultivated or supermarket greens in this recipe. Consider nettles, beet tops, turnip tops, spinach, or watercress in place of chard. The combination is also up to you so choose the ones you like most.

— Y. Ottolenghi. Plenty More

In this post I’ll describe how I come up with geometry proposals for olympiad-style contests. In particular, I’ll go into detail about how I created the following two problems, which were the first olympiad problems which I got onto a contest. Note that I don’t claim this is the only way to write such problems, it just happens to be the approach I use, and has consistently gotten me reasonably good results.

[USA December TST for 56th IMO] Let ${ABC}$ be a triangle with incenter ${I}$ whose incircle is tangent to ${\overline{BC}}$, ${\overline{CA}}$, ${\overline{AB}}$ at ${D}$, ${E}$, ${F}$, respectively. Denote by ${M}$ the midpoint of ${\overline{BC}}$ and let ${P}$ be a point in the interior of ${\triangle ABC}$ so that ${MD = MP}$ and ${\angle PAB = \angle PAC}$. Let ${Q}$ be a point on the incircle such that ${\angle AQD = 90^{\circ}}$. Prove that either ${\angle PQE = 90^{\circ}}$ or ${\angle PQF = 90^{\circ}}$.

[Taiwan TST Quiz for 56th IMO] In scalene triangle ${ABC}$ with incenter ${I}$, the incircle is tangent to sides ${CA}$ and ${AB}$ at points ${E}$ and ${F}$. The tangents to the circumcircle of ${\triangle AEF}$ at ${E}$ and ${F}$ meet at ${S}$. Lines ${EF}$ and ${BC}$ intersect at ${T}$. Prove that the circle with diameter ${ST}$ is orthogonal to the nine-point circle of ${\triangle BIC}$.

## 1. General Procedure

Here are the main ingredients you’ll need.

• The ability to consistently solve medium to hard olympiad geometry problems. The intuition you have from being a contestant proves valuable when you go about looking for things.
• In particular, a good eye: in an accurate diagram, you should be able to notice if three points look collinear or if four points are concyclic, and so on. Fortunately, this is something you’ll hopefully have just from having done enough olympiad problems.
• Geogebra, or some other software that will let you quickly draw and edit diagrams.

With that in mind, here’s the gist of what you do.

1. Start with a configuration of your choice; something that has a bit of nontrivial structure in it, and add something more to it. For example, you might draw a triangle with its incircle and then add in the excircle tangency point, and the circle centered at ${BC}$ passing through both points (taking advantage of the fact that the two tangency points are equidistant from ${B}$ and ${C}$).
2. Start playing around, adding in points and so on to see if anything interesting happens. You might be guided by some actual geometry constructions: for example, if you know that the starting configuration has a harmonic bundle in it, you might project this bundle to obtain the new points to play with.
3. Keep going with this until you find something unexpected: three points are collinear, four points are cyclic, or so on. Perturb the diagram to make sure your conjecture looks like it’s true in all cases.
4. Figure out why this coincidence happened. This will probably add more points to you figure, since you often need to construct more auxiliary points to prove the conjecture that you have found.
5. Repeat the previous two steps to your satisfaction.
6. Once you are happy with what you have, you have a nontrivial statement and probably several things that are equivalent to it. Pick the one that is most elegant (or hardest), and erase auxiliary points you added that are not needed for the problem statement.
7. Look for other ways to reduce the number of points even further, by finding other equivalent formulations that have fewer points.

Or shorter yet: build up, then tear down.

None of this makes sense written this abstractly, so now let me walk you through the two problems I wrote.

## 2. The December TST Problem

In this narrative, the point names might be a little strange at first, because (to make the story follow-able) I used the point names that ended up in the final problem, rather than ones I initially gave. Please bear with me!

I began by drawing a triangle ${ABC}$ (always a good start\dots) and its incircle, tangent to side ${BC}$ at ${D}$. Then, I added in the excircle touch point ${T}$, and drew in the circle with diameter ${DT}$, which was centered at the midpoint ${M}$. This was a coy way of using the fact that ${MD = MT}$; I wanted to see whether it would give me anything interesting.

So, I now had the following picture.

Now I had two circles intersecting at a single point ${D}$, so I added in ${Q}$, the second intersection. But really, this point ${Q}$ can be thought of another way. If we let ${DS}$ be the diameter of the incircle, then as ${DT}$ is the other diameter, ${Q}$ is actually just the foot of the altitude from ${D}$ to line ${ST}$.

But recall that ${A}$, ${S}$, ${T}$ are collinear! (Again, this is why it’s helpful to be familiar with “standard” contest configurations; you see these kind of things immediately.) So ${Q}$ in fact lies on line ${AT}$.

This was pretty cool, though not yet interesting enough to be a contest problem. So I looked for most things that might be true.

I don’t remember what I tried next; it didn’t do anything interesting. But I do remember the thing I tried after that: I drew in the angle bisector, line ${AI}$. And then, I noticed a big coincidence: the first intersection of ${AI}$ with the circle with diameter ${DT}$ seemed to lie on line ${DE}$! I was initially confused by this; it didn’t seem like it could possibly be true due to symmetry reasons. But in my diagram, it was indeed correct. A moment later, I realized the reason why this was plausible: in fact, the second intersection of line ${AI}$ with the circle was on line ${DF}$.

Now, I could not see quickly at all why this was true. So I started trying to prove it, but initially failed: however, I managed to show (via angle chasing) that

$\displaystyle D, P, E \text{ collinear} \iff \angle PQE = 90^\circ.$

So, at least I had an interesting equivalent statement.

After another half hour of trying to prove my conjecture, I finally realized what was happening. The point ${P}$ was the one attached to a particular lemma: the ${A}$-bisector, ${B}$-midline, and ${C}$ touch-chord are concurrent, and from this ${MD = MP}$ just follows by some similar triangles. So, drawing in the point ${N}$ (the midpoint of ${AB}$), I had the full configuration which gave the answer to my conjecture.

Finally, I had to clean up the mess that I had made. How could I do this? Well, the points ${N}$, ${S}$ could be eliminated easily enough. And we could re-define ${Q}$ to be a point on the incircle such that ${\angle AQD = 90^\circ}$. This actually eliminated the green circle and point ${T}$ altogether, provided we defined ${P}$ by just saying that it was on the angle bisector, and that ${MD = MP}$. (So while the circle was still implicit in the condition ${MD = MP}$, it was no longer explicitly part of the problem.)

Finally, we could even remove the line through ${D}$, ${P}$ and ${E}$; we ask the contestant to prove ${\angle PQE = 90^\circ}$.

And that was it!

## 3. The Taiwan TST Problem

In fact, the starting point of this problem was the same lemma which provided the key to the previous solution: the circle with diameter ${BC}$ intersects the ${B}$ and ${C}$ bisectors on the ${A}$ touch chord. Thus, we had the following diagram.

The main idea I had was to look at the points ${D}$, ${X}$, ${Y}$ in conjunction with each other. Specifically, this was the orthic triangle of ${\triangle BIC}$, a situation which I had remembered from working on Iran TST 2009, Problem 9. So, I decided to see what would happen if I drew in the nine-point circle of ${\triangle BIC}$. Naturally, this induces the midpoint ${M}$ of ${BC}$.

At this point, notice (or recall!) that line ${AM}$ is concurrent with lines ${DI}$ and ${EF}$.

So the nine-point circle of the problem is very tied down to the triangle ${BIC}$. Now, since I was in the mood for something projective, I constructed the point ${T}$, the intersection of lines ${EF}$ and ${BC}$. In fact, what I was trying to do was take perspectivity through ${I}$. From this we actually deduce that ${(T,K;X,Y)}$ is a harmonic bundle.

Now, what could I do with this picture? I played around looking for some coincidences, but none immediately presented themselves. But I was enticed by the point ${T}$, which was somehow related to the cyclic complete quadrilateral ${XYMD}$. So, I went ahead and constructed the pole of ${T}$ to the nine-point circle, letting it hit line ${BC}$ at ${L}$. This was aimed at “completing” the picture of a cyclic quadrilateral and the pole of an intersection of two sides. In particular, ${(T,L;D,M)}$ was harmonic too.

I spent a long time thinking about how I could make this into a problem. I unfortunately don’t remember exactly what things I tried, other than the fact that I was taking a lot of perspectivity. In particular, the “busiest” point in the picture is ${K}$, so it makes sense to try and take perspectives through it. Especially enticing was the harmonic bundle

$\displaystyle \left( \overline{KT}, \overline{KL}; \overline{KD}, \overline{KM} \right) = -1.$

How could I use this to get a nice result?

Finally about half an hour I got the right idea. We could take this bundle and intersect it with the ray ${AI}$! Now, letting ${N}$ be the midpoint ${EF}$, we find that three of the points in the harmonic bundle we obtain are ${A}$, ${I}$, and ${N}$; let ${S}$ be the fourth point, which is the intersection of line ${KL}$ with ${AI}$. Then by hypothesis, we ought to have ${(A,I;N,S) = -1}$. But from this we know exactly what the point ${S}$. Just look at the circumcircle of triangle ${AEF}$: as this has diameter ${AI}$, we see that ${S}$ is the intersection of the tangents at ${E}$ and ${F}$.

Consequently, we know that the point ${S}$, defined very naturally in terms of the original picture, lies on the polar of ${T}$ to the nine-point circle. By simply asking the contestant to prove this, we thus eliminate all the points ${K}$, ${M}$, ${D}$, ${N}$, ${I}$, ${X}$, and ${Y}$ completely from the picture, leaving only the nine-point circle. Finally, instead of directly asking the contestant to show that ${T}$ lies on the polar of ${S}$, one can rephrase the problem as saying “the circle with diameter ${ST}$ is orthogonal to the nine-point circle of ${\triangle BIC}$”, concealing all the work that went into the creation of the problem.

Fantastic.

# What leads to success at math contests?

I think this is an important question to answer, not the least of reasons being that understanding how to learn is extremely useful both for teaching and learning. [1]

About a year ago [2], I posted my thoughts on what the most important things were in math contest training. Now that I’m done with the IMO I felt I should probably revisit what I had written.

It looks like the main point of my post a year ago was mainly to debunk the idea that specific resources are important. Someone else phrased this pretty well in the replies to the thread

The issue is many people simply ask about how they should prepare for AIME or USAMO without any real question. They simply figure that AOPS has a lot of successful people that excel at both contests, so why not see what they did? Unfortunately, that’s not how it works – that’s what this post is saying. There is no “right” training.

This is so obvious to me now that I’m going to focus more on what I think actually matters. So I now have the following:

1. Do lots of problems.
2. Learn some standard tricks.
3. Do problems which are just above your reach.
4. Understand the motivation behind solutions to problems you do.
5. Know when to give up.
6. Do lots of problems.

Elaboration on the above:

1. Self-explanatory. I can attest that the Contests section on AoPS suffices.
2. One should, for example, know what a radical axis is. It may also help to know what harmonic quadrilaterals, Karamata, or Kobayashi is, for example, but increasingly obscure things are increasingly less necessary. This step can be achieved by using books/handouts or doing lots of problems.
3. Basically, you improve when you do problems that are hard enough to challenge you but reasonable for you to solve. My rule of thumb is that you shouldn’t be confident that you can solve the practice problem, nor confident that you won’t solve it. There should be suspense.

In my experience, people tend to underestimate themselves — probably my biggest regret was being scared of IMO/USAMO #3’s and #6’s until late in my IMO training, when I finally realized I needed to actually start solving some. I encourage prospective contestants to start earlier.

4. I think the best phrasing of this is, “how would I train a student to be able to solve this problem?”, something I ask myself a lot. By answering this question you also understand

a. Which parts of the solution are main ideas and which steps are routine details,
b. Which parts of the problem are the “hard steps” of the problem,
c. How one would think of the hard steps of the solution,
and so on. I usually like to summarize the hard parts of the solution in a few sentences. As an example, “USAMO 2014 #6 is solved by considering the $N \times N$ grid of primes and noting that small primes cannot cover the board adequately”. Or “ELMO 2013 #5 is solved by considering the 1D case, realizing the answer is $cn^k$, and then generalizing directly to the 3D case”.

In general, after reading a solution, you should be able to state in a couple sentences all the main ideas of the solution, and basically know how to solve the problem from there.

5. In 2011, JMO #5 took me two hours. In 2012, the same problem took me 30 seconds and SL 2011 G4 took me two hours. Today, SL 2011 G4 takes me about five minutes and IMO 2011 #6 took me seven hours. It would not have been a good use of my time in 2011 to spend several hundred hours on IMO #6.

This is in part doing (3) correctly by not doing things way, way over your head and not doing things way below your ability. Regardless you should know when to move on to the next problem. It’s fine to try out really hard problems, just know when more time will not help.

In the other direction, some students give up too early. You should only give up on a problem after you’ve made no progress for a while, and realize you are unlikely to get any further than you already are. My rule of thumb for olympiads is one or two hours without making progress.

6. Self-explanatory.

I think the things I mentioned above are at least extremely useful (“necessary” is harder to argue, but I think you could make a case for it). Now is it sufficient? I have no idea.

##### Footnotes
1. The least of reasons is that people ask me this all the time and I should properly prepare a single generic response.
2. It’s only been a year? I could have sworn it was two or three.