IMO 2010: Problem 5

Problem 5 of the IMO was definitely the most interesting problem in the contest, although one of the harder ones. A mere 37 people out of 511 managed to solve it completely.

What made it so different was that it did not require any particular knowledge on olympiad math theorems, indeed such knowledge would be useless for this problem. It required only creative insight, and quite a bit of it. Technically it’s a combinatorics problem, but no specific combinatorics experience is needed to solve the problem.

Also this problem was the target for a mini-polymath project organized by Terry Tao, which took about two and a half hours to get from start to solving the problem. The solution I’ll give is mostly based on the polymath solution, although I came up with my own finishing steps which I think is less awkward.


In each of six boxes B_1,B_2,B_3,B_4,B_5,B_6 there is initially one coin. There are two types of operation allowed:

Type 1: Choose a nonempty box B_j with 1 \leq j \leq 5. Remove one coin from B_j and add two coins to B_{j+1}.

Type 2: Choose a nonempty box B_k with 1 \leq k \leq 4. Remove one coin from B_k and exchange the contents of (possibly empty) boxes B_{k+1} and B_{k+2}.

Determine whether there is a finite sequence of such operations that results in boxes B_1,B_2,B_3,B_4,B_5 being empty and box B_6 containing exactly 2010^{2010^{2010}} coins. (Note that a^{b^c}= a^{(b^c)}.)

Initial observations

It is not entirely obvious how this problem should be approached; thus we will begin by playing with it and jotting down any interesting observations.

We begin with one coin in each box. For convenience, I will express this as

(1,1,1,1,1,1) .

We need to have


Again for convenience, let us substitute

D = 2010^{2010^{2010}}.

Now we can express the two types of moves in our new notation. First, Type 1 moves (we shall call it Move 1):

(a,b) \to (a-1,b+2).

Next, Type 2 moves:

(a,b,c) \to (a-1,c,b).

Notice how applying Move 1 adds one coin to the system as a whole, while Move 2 takes away one coin.

By applying Move 2 over and over, it is easy to remove arbitrarily large amounts of coins from the system (as long as the coins are in the first four boxes).

On the other hand, it isn’t so clear how do add large amounts of coins to the system.

Applying Move 1 adds a coin to the system, so logically adding coins should be done by repeated usage of Move 1.

Given (a,b), applying Move 1 results in (a-1,b+2); applying it again results in (a-2,b+4). We can repeat this as many times as we like, or until the first box is depleted.

This creates a compound move for a pair, executed by applying Move 1 repeatedly:

(a,b) \to (0,b+2a).

What if we apply this on all six boxes? We first get


Next we have


A few more iterations and we will have:


We are far from our target: we have 63 coins and we’re stuck. There doesn’t seem to be any other way to introduce any more coins into the system.

The problem seems hopeless.

Building up a repository of compound moves

We first need a way of getting a large amount of coins into the system. From there is is easy to reduce the coins.

The compound move explained in the previous section is a first step. Let’s call it Compound Move 1:

(a,b) \to (0,b+2a).

The next compound move is less obvious. Suppose that we start with


Applying Move 1 gives us


which, using Compound Move 1, can be used to get


Now we apply Move 2. Using up a coin to switch the two boxes, we have


Or equivalently,


We can repeat this: switching, doubling, switching again, and so on, until the first box is depleted. This gives us


So here we have Compound Move 2:

(a,0,0) \to (0,2^a,0).

This is a lot better than what we had before. However, it’s still a while towards our goal.

With Compound Move 1, each additional coin in the left most box adds 2 coins to the next box.

But with Compound Move 2, each additional coin doubles the result in the next box.

Now this problem has a Towers of Hanoi feel to it. It turns out that we can take this one step further, which, when examined closely, looks remarkably similar to the previous case.

Suppose we have


in four consecutive boxes. Applying Move 1:


Applying Compound Move 2 using the last three boxes, we get


Now we apply Move 2 and switch the boxes:


If we do this again, we get


So, repeating this many times until the left box is empty gives us


No doubt the number we have is sufficiently big now. We will call this move Compound Move 3, which can only be expressed reasonably with Knuth’s up-arrow notation:

(a,0,0,0) \to (0, 2 \uparrow \uparrow a,0,0).

Putting it together

The configuration we want is


which is equivalent to


It’s relatively easy to get most of the boxes empty except for the third. From the starting position, Move 1 gives us


Applying Move 2 three times gives us


Move 1 then gives


We invoke Compound Move 3, getting

(0,0,0,2 \uparrow \uparrow 7,0,0).

Finally we can waste coins by consecutively executing Move 2 until we have


And we’re done.

An additional note

It is easy to show, albeit informally, that 2 \uparrow \uparrow 7 > \frac{D}{4}. If we take the binary log (I’ll write it as \log x) of both sides, we have

2 \uparrow \uparrow 6 > 2010^{2010} \cdot \log 2010 - 2.

It’s okay to ‘throw away’ the 2, considering the magnitude of the numbers. Taking the log again:

2 \uparrow \uparrow 5 > 2010 \cdot \log (2010 \cdot \log 2010).

Considering that 2 \uparrow \uparrow 5 = 2^{65536}, and the right hand side evaluates to just a few thousand, it’s fairly safe to say which is bigger.

IMO 2010: Problem 2

In this year’s IMO, there were two geometry problems. Problem 4 is generally considered the easier one of the two. Problem 2 is the other problem, being a bit harder. Of the 517 contestants, 366 of them received a full mark on problem 4; only 162 students got a full mark for problem 2.

The Problem

Let I be the incentre of triangle ABC and let \Gamma be its circumcircle. Let the line AI intersect \Gamma again at D. Let E be a point on the arc \widehat{BDC} and F a point on the side BC such that

\angle BAF = \angle CAE < \frac{1}{2} \angle BAC .

Finally, let G be the midpoint of the segment IF. Prove that the lines DG and EI intersect on \Gamma.


It turns out that several solutions exist. Several of the solutions use Menelaus’s theorem, which I’ve never even heard of; others rely on properties of excircles. I’ll give the solution that I think is the most elegant (although it doesn’t seem to be the standard solution).

We start by working backwards.

Let us denote K to be the point of intersection between DG and EI; we wish to prove that K lies on \Gamma. If K lies on \Gamma, then quadrilateral AKDE is cyclic, and \triangle IKD \sim \triangle IAE.

Obviously \angle KID = \angle AIE. If we can show that \angle GDI = \angle AEI, then triangles IKD and IAE are similar and our result follows.

Let M be the midpoint of AI.

We will prove that \angle GDI = \angle AEI by proving that \triangle MGD and \triangle AIE are similar (these are the triangles highlighted in red).

It is given that \angle CAE = \angle BAF, so angles \angle DAE and \angle DAF are also equal (since AD passes through the incenter I and thus bisects \angle A).

It is also given that FG = GI; now IM = MA by definition so triangles IFA and IGM are similar. Then \angle GMD = \angle IAE.

All that remains to prove the similarity of \triangle MGD and \triangle AIE (and thus the result) is to prove that

\frac{MG}{MD} = \frac{AI}{AE}

or equivalently

MG \cdot AE = AI \cdot MD .

As \triangle IFA \sim \triangle IGM, and IG = \frac{1}{2}IF, it follows that MG = \frac{1}{2}AF. By substitution,

\frac{1}{2}AF \cdot AE = AI \cdot MD .

It can be shown that the product AF \cdot AE is constant no matter where E is on the circle:

Draw the line EC; the triangle formed, \triangle ACE, is similar to \triangle AFB since \angle CAE = \angle CAF and \angle ABF and \angle AEC subtend the same arc.


\frac{AF}{AB} = \frac{AC}{AE}


AE \cdot AF = AB \cdot AC .

Since AB \cdot AC is constant with respect to E (and F), so is AE \cdot AF.

Consequentially we can prove that \frac{1}{2}AE \cdot AF = AI \cdot MD for all values of E and F by proving it for one value of E and F, since \frac{1}{2} AE \cdot AF is constant and obviously AI \cdot MD is constant.

We prove the case for \angle CAE = 0, or in other words when E coincides with C and F coincides with B:

So here we need to prove that

\frac{1}{2} AB \cdot AC = AI \cdot MD .

This is equivalent to proving that K lies on \Gamma in this instance, as that would prove the above equation too for this instance.

Obviously \angle ACK = \angle ADK, also \angle BCK = \angle BDK. As \angle ACK = \angle BCK since CK passes through the incircle and is an angle bisector, it follows that \angle ADK = \angle BDK.

Given BG=GI, this is sufficient to prove \triangle BDI to be isosceles.

So \angle BDG = \angle GDI, and DG meets \Gamma at the midpoint of minor ark \widehat{AB}. Obviously CI meets \Gamma at the same point, and we are done. QED.

Checking the solution

Just to make sure, we can trace out steps back to get from the equation \frac{1}{2} AE \cdot AF = AI \cdot MD to our result that K lies on \Gamma. I’m going to go through it very briefly:

From the equation we get \frac{MG}{MD} = \frac{AI}{AE}, proving triangles MGD and AIE to be similar.

Then \angle GDI = \angle AEI and triangles KDI and AEI are similar. Finally AKDE is cyclic, leading to the result.

It is also valid to, starting with the result, arrive at the equation, which is what we implicitly did near the end of the solution.

IMO 2010: Problem 4

The International Olympiad of Mathematics took place on July 8 and 9, and the six questions were made available the second day. This was a two day contest, three questions per day, and four and a half hours per day, giving a total of nine hours for the six questions. The international contest included questions on geometry, algebra, combinatorics, and number theory.

Contest questions may be downloaded here.

This year, it seems that problems 1 and 4 were the easiest (by comparison). I’ve only looked at problem 4 in detail (probably because it’s geometry).


Let P be a point inside the triangle ABC. The lines AP, BP and CP intersect the circumcircle \Gamma of triangle ABC again at the points K, L and M respectively. The tangent to \Gamma at C intersects the line AB at S. Suppose that SC = SP. Prove that MK = ML.


As often with geometry problems, a variety of different solutions are possible. Some are simple, while others rely on homothetic transformations and inversions. I’m going to go with the method I came up with (albeit with a few peeks at the Art of Problem Solving forums).

From the Power of a Point theorem, we have

SC^2 = SA \cdot SB

But it’s given that SP = SC, so

SP^2 = SA \cdot SB


\frac{SP}{SA} = \frac{SB}{SP}

This is enough to show that triangles SAP and SPB are similar, since they also share an angle.


\frac{SA}{AP} = \frac{SP}{BP}


\frac{AP}{BP} = \frac{SA}{SP}

Since SP = SC,

\frac{AP}{BP} = \frac{SA}{SC} \; \; \; (1)

Triangles SCA and SBC are similar, since they share an angle and \angle SCA = \angle ABC. Thus,

\frac{SA}{AC} = \frac{SC}{BC}


\frac{SA}{SC} = \frac{AC}{BC}

From (1),

\frac{AP}{BP} = \frac{AC}{BC}

Or equivalently,

\frac{AC}{AP} = \frac{BC}{BP} \; \; \; (2)

Notice that triangles APC and MPK are similar (\angle CMK and \angle CAK subtending the same arc); so,

\frac{AC}{AP} = \frac{MK}{MP}

Similarly, triangles MPL and BPC are also similar, giving

\frac{BC}{BP} = \frac{ML}{MP}

From (2),

\frac{MK}{MP} = \frac{ML}{MP}

This shows that MK = ML, as desired. QED.