# 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.

# Random Math Problems (1)

Probably any of these problems would be too small to include in a blog post on their own, so it might be more efficient (?) to put several of them together into one post.

The problems are not mine, and the solutions are not mine. They are just interesting problems I’ve come across when studying for math olympiads.

I hope to publish more collections of these random math problems in the future. That’s why I’m appending (1) to this post.

### Problem 1

(From the 2010 Fermat contest)

Spheres can be stacked together into a tetrahedron (triangular pyramid). Here’s a visual (not created by me, from Wikipedia):

Each of the spheres in the tetrahedron has a number written on it. The top sphere is 1, and the number of any sphere is the sum of the numbers on the spheres in the layer above.

An internal sphere is defined as a sphere that is not on the outside, or in other words, touching exactly three spheres in the layer above.

In a tetrahedron with 13 layers, what is the sum of the numbers on all of the internal spheres?

#### Solution

Let’s draw out a few layers of the tetrahedron.

Look for patterns.

There’s a few things that can be noticed-

1. The numbers at the three corners of a layer are 1
2. Any layer is highly symmetrical: in the last diagram the sequence (1 4 6 4 1) appears in the bottom row, as well as the left and right diagonals. We can rotate the tetrahedron and the numbers would still be the same.
3. The numbers on the outside form Pascal’s triangle. The sequence (1 3 3 1) is the 4th row of Pascal’s triangle; (1 4 6 4 1) is the 5th, and so on.

The most important thing to look for is the fact that Pascal’s triangle appears in the tetrahedron.

Each row of Pascal’s triangle is one longer than the previous row. To calculate a row, overlap the previous row with the new row, like this:

Here the number in a purple box is the number to the top-right of it plus the number to the top-left of it:

Simple.

Notice that each number in the top gets added into exactly two boxes at the bottom.

Therefore the sum of the numbers in the bottom is twice the sum of the numbers in the top. $1 + 3 + 3 + 1 = 8$, and $1 + 4 + 6 + 4 + 1 = 16$.

This applies to any two rows. The sum of any row is twice the sum of the previous row.

Since the sum of the first row is 1, the sum of the second is 2, the sum of the third 4, and so on. We can even express this as a formula. The sum of the nth row is:

$2^{n-1}$

A similar idea applies to the layers of the tetrahedron.

In Pascal’s triangle, each number gets added to two places in the next row.

The tetrahedron is like Pascal’s triangle, in three dimensions. Each sphere in the tetrahedron sits on top of three spheres in the layer under it. So each number gets added to three different numbers in the next row.

Thus the sum of the numbers in any layer of the tetrahedron is 3 times the sum of numbers in the previous layer.

A formula for the sum of the nth layer of the tetrahedron:

$3^{n-1}$

What we really want is a formula for the middle numbers, or internal spheres. The internal spheres are the spheres that are not on the outer layer.

We know the formula for the sum of the whole layer, and we know the formula for any of the three ‘sides’ of a layer.

The internal numbers would be the sum of the entire layer minus three times the sum of a side. To cover up for the three 1’s we’ve counted more than once, we add 3 to the result.

The formula for the sum of the internal numbers on the nth layer:

$3^{n-1} - 3 \cdot 2^{n-1} + 3$

The problem asks for the sum of the first to the thirteenth layer. Well, the first three layers don’t have any internal spheres so we start from the forth:

$\begin{array}{l} \displaystyle\sum_{n=4}^{13} 3^{n-1} - 2^{n-1} + 3 \\ = (3^3 + 3^4 + \cdots + 3^{12}) \\ \quad - 3(2^3 + 2^4 + \cdots + 2^{12}) + 3 \cdot 10 \end{array}$

We can use the sum of geometric series formula, or more easily calculate it by brute force. The answer would be 772626.

### Problem 2

In quadrilateral $ABCD$, $CM$ splits the quadrilateral into two parts of equal area. $AN$ also splits the quadrilateral into two parts of equal area. $AB$ is the longest side of the quadrilateral.

Prove that $MN$ bisects $DB$.

#### Solution

I’m going to use the equal sign to mean two figures with equal areas, not two figures that are congruent.

The triangles $\triangle ANB$ and $\triangle MCB$ are both half of $ABCD$, so their areas are equal. Then $\triangle ANB = \triangle MCB$.

$\triangle ANB$ can be split into $\triangle ANM + \triangle MNB$, and similarly $\triangle MCB$ can be split into $\triangle MCN + \triangle MNB$.

We know that $\triangle ANB = \triangle MCB$. By subtraction, $\triangle AMN = \triangle MCN$.

The triangles $\triangle AMN$ and $\triangle MCN$ share a common base, $MN$. Because their areas are equal, the perpendicular from $A$ to $MN$ is equal to the perpendicular from $C$ to $MN$. Therefore $AC || MN$.

Now we draw a line through $D$, parallel to $AC$ and $MN$. We extend $AB$ and $BC$ to meet this line (and we draw a couple of additional lines):

Because $XY || AC$, $\triangle ADC = \triangle AXC = \triangle AYC$.

Here $\triangle MCB = ADCM$.

We can write $ADCM$ as $\triangle ADC + \triangle ACM$. Since $\triangle ADC = \triangle AXC$, $ADCN = \triangle XCM$.

It follows that $\triangle MCB = \triangle XCM$, and $XM = MB$.

Similarly, $ADCN = \triangle AYN$. Then $\triangle ANB = ADCN = \triangle AYN$, and $YN = NB$.

Because $XM = MB$ and $YN = NB$, the result follows that $DK = KB$.

Problem 3

(From the 2010 USAMO)

This problem actually appears in both the junior and senior USAMO taken about a week ago from time of writing. It’s problem 4 in the senior and problem 6 in the junior.

$\triangle ABC$ is a right triangle. $BD$ bisects $\angle ABC$, and $EC$ bisects $\angle ACB$.

Prove that the lines $[AB, AC, BI, ID, CI, IE]$ cannot all have integer lengths.

#### Solution

I was really surprised at how short and simple the solution can be for this problem.

Because $\angle BAC = 90$, $\angle ABC + \angle ACB = 90$ and $\angle IBC + \angle ICB = 45$ because of the angle bisectors.

Then $\angle BIC = 180-45 = 135$.

First we apply the pythagorean theorem:

$BC^2 = AB^2 + AC^2$

We then apply the Law of Cosines:

$BC^2 = IB^2 + IC^2 - 2 \cdot IB \cdot IC \cdot \cos 135$

Combining the two, and substituting $-\frac{1}{\sqrt{2}}$ for $\cos 135$:

$\begin{array}{c} AB^2 + AC^2 = IB^2 + IC^2 + 2 \cdot IB \cdot IC \cdot \frac{1}{\sqrt{2}} \\ \frac{2 \cdot IB \cdot IC}{AB^2 + AC^2 - IB^2 - IC^2} = \sqrt{2} \end{array}$

This is absurd, if AB, AC, IB, and IC are all integers and nonzero.