# CMOQR 2011

Here I’m going to go over some of my solutions to the CMOQR (Canadian Math Olympiad Qualifying Repechage) — the qualifying contest for the CMO. Of the 8 questions, I managed to solve the first six but I didn’t manage to get the last two. My solutions here are going to be rather concise.

### Problem 1

We know angle BOC so we know angle BAC. Use the cosine law on BOC to get $BC=\sqrt{21}$. Next use the cosine law again on BAC, but set x = AB and x+1 = AC. Solving for x, we get x=4.

### Problem 2

No two of the sums are equal. Therefore no two of a,b,c,d,e are equal, so a<b<c<d<e. (letting abc mean a+b+c to avoid too many plus symbols) in set {a,b,c,d}, we have abc<abd<acd<bcd and similarly bcd<bce<bde<cde, therefore abc<abd<acd<bcd<bce<bde<cde. But we don’t have abe, ace, ade yet so abc<abd<abe<ace<ade<bde<cde.

In either case, abc is the smallest, abd is the second smallest, etc. Then abc=0, abd+3, bde=14, and cde=19. Thus we know d=c+3, b=c-5, e=a+11, so we have {a,b,c,d,e} = {a,c-5,c,c+3,a+11}.

As abc=0, we write everything in terms of c: a=5-2c and e=16-2c. Now we have {5-2c,c-5,c,c+3,16-c}. By substitution, acd=8 and bce=11. Since acd=8, abe=4 (not by deduction, rather it’s the only choice).

From a+b+e=4, we can solve for c and get c=4. It follows that the five numbers are {-3.-1,4,7,8}.

### Problem 3

Adding the two equations and factoring gives (x+y+3)(x+y)=18. Solving the quadratic for x+y, we have x+y = -6 or x+y=3. Suppose x+y=-6: substituting back into the original two equations gives x=y=-3. Suppose that x+y=3. Substituting back gives x=0, y=3 or x=3,y=0.

### Problem 4

Alphonse can always win. It turns out that he has a ‘strategy stealing’ strategy: first split into $2^{2011}$ and $5^{2011}$. Beryl has to make some move now affecting one of the two piles (never both), and Alphonse can simply make the symmetrical move for the other pile. Obviously Beryl runs out of moves first.

### Problem 5

It suffices to prove that for any 6 vertices of a 11-gon, there exists at least one pair of congruent triangles (either there are 6 black vertices, or there are 6 gold vertices). Enumerating all possible congruent triangles in a 11-gon is the same as enumerating all sets of integers adding to 11 — that is, {1,1,9}, {2,2,7} and so on. There are 10 such sets, so any 11 triangles must contain a pair of congruent ones.

Given 6 vertices we can form $\binom{6}{3} = 20$ triangles, so one pair must be congruent.

### Problem 6

Let us write b for angle FBE and a for angle EBD. Obviously triangles ABF and EBF are congruent, so $b = \frac{90-a}{2}$. We want to prove that 30<a<45 or 22.5<b<30. To prove this we show that for a=30, b=30 then x>y, and then for a=45, b=22.5 then x<y, because then the value for which x=y must exist in that range.

The first case is simple, as if a=b=30 then we have 3 congruent triangles, and x=2y. The second case, EB=ED and (assuming the radius is 1), the area of BED is $\frac{1}{2}$. The area in the circle is $\frac{ \pi}{8}$ so $y = \frac{1}{2} - \frac{\pi}{8}$. On the other hand x is twice of FBE minus $\frac{\pi}{8}$ and using some trigonometry we get $x = \tan 22.5 - \frac{\pi}{8}$. To prove x<y for this case, it is sufficient to prove $\tan 22.5 < \frac{1}{2}$, which can be done using half angle formulas.