## CMOQR 2011

January 13, 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.

## AHSMC 2010 Part I

November 16, 2010

The first math contest of this year! The AHSMC, has 16 multiple choice problems. 20 free marks, then 5 marks for every correct answer, 2 marks for every blank answer, and 0 for every wrong answer, so possible scores range from 20 to 100. The contest is written in 80 minutes, without a calculator.

The answers I got were: bacb bedd ccbe acbe. I’ll post my solutions here.

#### Problem 1

The number of positive integers $n$ such that $4n$ has 2 digits is

(a) 21; (b) 22; (c) 23; (d) 24; (e) 25

As $4n$ is between 10 and 99, $3 \leq n \leq 24$, giving 22 values. The answer is (b).

#### Problem 2

A 4×6 plot of land is divided into 1×1 lots by fences parallel to the edges (with fences along the edges too). The total length of the fences is

(a) 58; (b) 62; (c) 68; (d) 72; (e) 96

Count the vertical fences separately from the horizontal fences. So let’s suppose the grid is 4 columns and 6 rows, then we have 5 vertical fences and 7 horizontal fences; each vertical fence is 6 units long and each horizontal fence 4 units long.

Therefore the total length is $5 \times 6 + 7 \times 4$ or $58$. The answer is (a).

#### Problem 3

The GCD of two positive numbers is 1, and the LCM is 10. If neither of them are 10, their sum is

(a) 3; (b) 6; (c) 7; (d) 11; (e) none of these

Obviously the numbers are 2 and 5, as no other two coprime numbers both divide into 10. So their sum is 7. The answer is (c).

#### Problem 4

How many non-negative solutions $(x,y)$ are there to the equation $3x+2y=27$?

(a) 4; (b) 5; (c) 8; (d) 9; (e) 10

Solving for x, we get $x = \frac{27-2y}{3}$ or $x = 9 - \frac{2y}{3}$. So in order for x to be non-negative, both $3 | 2y$ and $\frac{2y}{3} \leq 9$, so then $y \leq 13$. Of the numbers y between 0 and 13 inclusive, 5 of them are divisible by 3. The answer is (b).

#### Problem 5

The sequence 1,2,3,4,6,7,8,9,… is obtained by deleting multiples of 5 from the positive integers. What is the 2010th term?

(a) 2511; (b) 2512; (c) 2513; (d) 2514; (e) none of these

Notice that the 4th term is 4, the 5th term is 9, the 12th term is 14, and so on. So the pattern is that the $4n \mathrm{th}$ term is $5n-1$.

Therefore term 2008 would be $5(502) - 1 = 2509$; then term 2009 is 2511 (skipping 2510) and term 2010 is 2512. The answer is (b).

#### Problem 6

5 people in a building are on floors 1, 2, 3, 21, and 40. In order to minimize their total travel distance, what floor should they get together on?

(a) 18; (b) 19; (c) 20; (d) 21; (e) none of these

Suppose we choose floor 19. Then the total distance is 18+17+16+2+21 or 74. If we choose 17, the total distance is 17+16+15+3+22 or 73, which is smaller.

In fact we can repeat this several times: at floor 3, the total is 2+1+0+18+37 or only 58 floors in total. The answer is (e).

#### Problem 7

9 holes are arranged in a 3×3 configuration. Two pigeons each choose a hole at random (possible the same one). The probability that they choose two holes on the opposite side of an interior wall is

(a) $\frac{1}{18}$; (b) $\frac{1}{9}$; (c) $\frac{4}{27}$; (d) $\frac{8}{27}$; (e) $\frac{1}{3}$

Let the first pigeon choose a random hole. Then we split the problem into 3 cases:

• If it’s in one of the 4 corners, then the next pigeon has a $\frac{2}{9}$ chance of landing in the correct spot, so the probability here is $\frac{4}{9} \times \frac{2}{9}$.
• If it’s in one of the 4 edges, then the next pigeon has a $\frac{3}{9}$ chance of landing in the correct spot. This is a probability of $\frac{4}{9} \times \frac{3}{9}$.
• If it’s in the center hole, then the next pigeon may land in 4 possible places, so the probability here is $\frac{1}{9} \times \frac{4}{9}$.

The total is $\frac{4}{9} \times \frac{2}{9} + \frac{4}{9} \times \frac{3}{9} + \frac{1}{9} \times \frac{4}{9}$ which is equal to $\frac{8}{27}$. The answer is (d).

#### Problem 8

The set of real x where $\frac{1}{x} \leq -3 \leq x$ is:

(a) $\{x \leq - \frac{1}{3}\}$; (b) $\{-3 \leq x \leq - \frac{1}{3}\}$; (c) $\{ -3 \leq x < 0 \}$; (d) $\{ - \frac{1}{3} \leq x < 0 \}$; (e) none of these

I solved this graphically:

#### Problem 9

In quadrilateral $ABCD$, $AB || DC$, $DC = 2AB$, $\angle ADC = 30^\circ$, $BCD = 50^\circ$. M is the midpoint of CD. The measure of $\angle AMB$ is

(a) 80; (b) 90; (c) 100; (d) 110; (e) 120

Because $DC || AB$ and $DM = AB = MC$, both ABMD and ABCM are parallelograms. Opposite angles in parallelograms are equal, so $\angle MAB = 50^\circ$, $\angle MBA = 30^\circ$. Thus $\angle AMB = 100^\circ$. The answer is (c).

#### Problem 10

We construct isosceles but non-equilateral triangles with integer side lengths between 1 and 9 inclusive. The number of such non-congruent triangles is

(a) 16; (b) 36; (c) 52; (d) 61; (e) none of these

Let the sides be a, b, c with $a \geq b \geq c$. There are 2 cases, one where $a=b$ and the other when $b=c$.

First, the case $a=b$. If $c=1$, a can be from 2 to 9; if $c=2$ then a can be from 3 to 9, and so on. So the possibilities are 8+7+6+…+1 = 36.

Next, the case $b=c$. We have $a > 2b$ so for $a=9$ we have b = 1, 2, 3, 4, and if $a=8$ then b = 1, 2, 3, and so on. Then the possibilities are 4+3+3+2+2+1+1 or 16.

The combined possibilities are 36+16 = 52. The answer is (c).

#### Problem 11

Which of the following is the largest?

(a) $2^{2^{2^{2^3}}}$; (b) $2^{2^{2^{3^2}}}$; (c) $2^{2^{3^{2^2}}}$; (d) $2^{3^{2^{2^2}}}$; (e) $3^{2^{2^{2^2}}}$

Immediately we know that B>A because $3^2 > 2^3$. Next, B>C because 512 > 81. Comparing B and D, we compare $2^{512}$ with $3^{16}$. Obviously B is bigger.

Finally we compare B with E. $B=2^{2^{512}}$ and $E = 3^{2^{16}}$. But B can be written as $4^{2^{511}}$ which is obviously bigger. The answer is (b).

#### Problem 12

A gold number is one expressible in the form $ab + a + b$ for positive integers a and b. The number of gold numbers between 1 and 20 inclusive is

(a) 8; (b) 9; (c) 10; (d) 11; (e) 12

Write $ab + a + b$ as $a(b+1) + b$. Take this modulo $b+1$, so $n \equiv b \mod b+1$. Then $n+1 \equiv 0 \mod b+1$ or $b+1 | n+1$, with $b+1 > 1$. Now if $n+1$ is composite this is possible, but if n+1 is prime then this is impossible (if $b=n$ then $a=0$, a contradiction). Therefore a gold number is any number that’s not one less than a prime.

Below 20, the primes are 2, 3, 5, 7, 11, 13, 17, 19, so 8 numbers between 1 and 20 are not gold numbers. Then 12 are gold numbers. The answer is (e).

#### Problem 13

In tetrahedron ABCD, edges DA, DB, DC are perpendicular. If $DA=1$, $DB=DC=2$, then the radius of a sphere passing through A, B, C, D is:

(a) $\frac{3}{2}$; (b) $\frac{\sqrt{5}+1}{2}$; (c) $\sqrt{3}$; (d) $\sqrt{2} +\frac{1}{2}$; (e) none of these

Put the tetrahedron on a 3D cartesian grid with D being at $(0,0,0)$, A at $(1,0,0)$, B at $(0,2,0)$, and C at $(0,0,2)$. The equation of a sphere is $(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2$, and since we know four points on the sphere, we can get four equations:

$a^2 + b^2 + c^2 = r^2$

$(1-a)^2 + b^2 + c^2 = r^2$

$a^2 + (2-b)^2 + c^2 = r^2$

$a^2 + b^2 + (2-c)^2 = r^2$

Solving for a, b, c, we get $a = \frac{1}{2}$, $b = 1$, $c = 1$. So the radius, or distance from origin is $\sqrt{(\frac{1}{2})^2 + 1^2 + 1^2}$ or $\frac{3}{2}$. The answer is (a).

#### Problem 14

Let $f(x) = x^2$, $g(x) = x^4$. We apply f and g alternatively: $f(x)$, $g(f(x))$, $f(g(f(x)))$, etc. When we apply f 50 times and g 49 times, the answer is $x^n$ where n is

(a) 148; (b) 296; (c) $2^{148}$; (d) $2^{296}$; (e) none of these

Rather than looking at the numbers themselves, we look at the exponents. Then $f(x)$ multiplies the exponent by 2 and $g(x)$ multiplies the exponent by 4. Applying f 50 times and g 49 times gives an exponent of $2^{50} \cdot 4^{49}$ or $2^{148}$. The answer is (c).

#### Problem 15

Triangle ABC has area 1. X and Y are on AB such that $XY = 2AX$, and Z is a point on AC such that $XZ || YC$ and $YZ || BC$. The area of $XYZ$ is

(a) $\frac{1}{27}$; (b) $\frac{2}{27}$; (c) $\frac{1}{9}$; (d) $\frac{2}{9}$; (e) $\frac{1}{3}$

As $XZ || YC$, it follows that triangles $AXZ$ and $AYC$ are similar, and $AY : AX = 3:1$ so $AC:AZ = 3:1$ and $AB:AY = 3:1$.

The area of $ABC$ is 1, so $AYC$ is $\frac{1}{3}$ and $AYZ$ is $\frac{1}{9}$, and $XYC$ is $\frac{2}{27}$. The answer is (b).

#### Problem 16

The number of integers n for which $2n+1 | n^3 -3n + 2$ is

(a) 3; (b) 4; (c) 5; (d) 6; (e) 8

Notice the left side is always odd, and the right side is always even. Therefore, it is equivalent to count solutions to $2n+1 | 8(n^3 - 3n + 2)$.

Now $8(n^2 - 3n + 2) = 8 n^3 - 24n + 16$ and by long division we have $2n+1 | (2n+1) (4n^2 - 2n + 11) + 27$, or $2n + 1 | 27$.

27 has 4 positive factors (1, 3, 9, 27) and 4 negative factors, all odd. Thus there are 8 solutions. The answer is (e).

## 2010 Euclid Contest

April 13, 2010

Today quite a few people from my school took the Euclid contest, again from the University of Waterloo.

This contest is really meant for grade 12 students, and it’s probably the most important of the contests, as it can be used to apply for universities and scholarships.

While most people took the contest last wednesday (April 7), my school took it on April 12.

Surprisingly I haven’t found anybody uploading the full solutions, or even all the questions of the euclid. Having took the contest, I’ll do so now. On to the first question!

### Question 1

1. a) $3^x = 27$, find $3^{x+2}$.

Solving for x, we get $x=3$. So $x+2 = 5$, and $3^5$ = 243.

b) $2^5 \cdot 3^{13} \cdot 5^9 x = 2^7 \cdot 3^{14} \cdot 5^9$, find x

Solve for x:

$\begin{array}{rcl} x &=& \frac{2^7 \cdot 3^{14} \cdot 5^9}{2^5 \cdot 3^{13} \cdot 5^9} \\ &=& 2^2 \cdot 3 \\ &=& 12 \end{array}$

The value for x is 12.

c)

The equation of line AB is $y=x+2$, and the equation of BC is $-\frac{1}{2}x+2$. Determine the area of $\triangle ABC$.

Because A and C are on the x axis and B is on the y axis, it’s fairly simple to figure out the coordinates of all three points. They are:

$\begin{array}{l} A = (-2,0) \\ B=(0,2) \\ C=(4,0) \end{array}$

We have a base and a height, so the area can be calculated to be 6.

### Question 2

2. a) Maria has a red, blue, and green package.

The sum of the masses of all three packages, is 60kg, of the red and green packages is 25kg, and of the green and blue packages is 50kg.

What is the mass of the green package?

Using R, G, and B for the masses of the red, green, and blue packages respectively, we can put together this system of equations:

$\begin{array}{rrrrrcl} R&+&G&+&B&=&60 \\ R&+&G&&&=&25 \\ &&G&+&B&=&50 \end{array}$

Solve this by elimination and you should get (R,G,B) = (10,15,35). So the weight of the green package is 15 kg.

b) A palindrome is a positive integer reading the same forwards and backwards (151, for example). What is the largest palindrome less than 200 that can be written as the sum of three consecutive integers?

Any number that can be written as a sum of three consecutive integers must be divisible by 3, because if n is an integer then the sum is $n + (n+1) + (n+2)$ or $3n + 3$.

Palindromes less than 200 are 191, 181, 171, etc. The first one in the sequence that’s divisible by 3 is 171.

c) If $(x+1) (x-1) = 8$, determine the value of $(x^2+x) (x^2-x)$.

Solving the first quadratic for x:

$\begin{array}{l} (x+1)(x-1)=8 \\ x^2-1=8 \\ x^2-9=0 \\ (x+3)(x-3) = 0 \\ x = \pm3 \end{array}$

Substituting either of the two roots into the second expression gives the value as 72.

### Question 3

3. a) Bea (a bee) starts from H (the hive), flies south for 1 hour to F (field), spends 30 minutes at F, flies 45 minutes to G (garden), spends 1 hour at G, then flies back to H. She flies at a constant speed in a straight line. What is the total time (in minutes) that she’s away from her hive?

All the times are given, except for the hypotenuse HG, which can easily be calculated using the pythagorean theorem:

The sum of the amount of times spent in each segment is 270 minutes.

b) $\triangle OPB$ is right angled. Determine all possible values of p.

Since $\triangle OPB$ is right angled, P is on a circle with OB as diameter (Thales’ theorem). We can let M be the center of the circle at (5,0):

The y coordinate of P is 4, so the equation is $y=4$, intersecting with the circle at two points given by this system of equations:

$\begin{array}{l} y=4 \\ (x-5)^2 + y^2 = 25 \end{array}$

$\begin{array}{rcl} (x-5)^2 + 16 = 25 \\ (x-5)^2 = 9 \\ x-5 = \pm 3 \\ x = 5 \pm 3 \end{array}$

So the possible values of x, or p, are 2 and 8.

4. a) Toy goats cost $19 each and toy helicopters$17 each. Thurka spent $201 on integral amounts of goats and helicopters. How many of each did she buy? Because $201 \equiv 14 \; (\textrm{mod} \; 17)$ and $19 \equiv 2 \; (\textrm{mod} \; 17)$, 201 – 7*19 is divisible by 17. Thus the only solution is 7 goats and 4 helicopters. b) Determine all values of x where $(x+8)^4 = (2x+16)^2$ Rewrite the equation as $(x+8)^4 = 4(x+8)^2$ and let $t=(x+8)^2$: $\begin{array}{l} t^2 = 4t \\ t^2 - 4t = 0 \\ t (t-4) = 0 \\ t=0, t=4 \end{array}$ Substituting the values of t and solving for x: $\begin{array}{l} (x+8)^2 = 0 \rightarrow x = -8 \\ (x+8)^2 = 4 \rightarrow x+8 = \pm 2 \rightarrow x = -8 \pm 2 \end{array}$ So the possible values for x are -10, -8, and -6. ### Question 5 5. a) If $f(x) = 2x+1$ and $g(f(x)) = 4x^2+1$, determine an expression for $g(x)$. $g(x)$ must be a quadratic function in the form $ax^2 + bx + c$. We know $f(x)$, so we can express $g(f(x))$ as such: $\begin{array}{rcl} g(f(x)) &=& g(2x+1) \\ &=& a(2x+1)^2 + b(2x+1) + c \\ &=& (4a)x^2 + (4a+2b)x + (a+b+c) \end{array}$ In $g(f(x))$, a=4, b=0, and c=1. Thus we can solve this system of equations: $\begin{array}{l} 4a=4 \\ 4a+2b=0 \\ a+b+c=1 \end{array}$ The solution is (a,b,c) = (1,-2,2), so g(x) = x^2-2x+2. b) A geometric sequence has 20 terms. The sum of the first two terms is 40, of the first three is 76, and of the first four is 130. Determine how many terms are integers. Because the difference between the sum of the first three and the sum of the first two is known, the forth term is 54 by subtraction. Similarly, the third term is 36. We can calculate the ratio between terms to be $\frac{54}{36} = \frac{3}{2}$. Then the second term can be interpolated by multiplying the third term by $\frac{2}{3}$, and it is 24. The first term is 16. So the sequence goes like this: $16, 24, 36, 54, 81, \cdots$ Notice that in each term the number of 2-factors decreases by 1. For example the first term, 16, is divisible by $2^4$, and the second term is divisible by $2^3$, and so on. So 81 is divisible by $2^0$, and the next term and all following terms are not integers. The sequence contains 5 terms. ### Question 6 6. a) AB is 1cm, find AH. The ratio of AB to AC is $\cos 30$ or $\frac{\sqrt{3}}{2}$. Or $AC = \frac{2}{\sqrt{3}} \cdot AB$. The ratio of AD to AC is the same, and so is the ratio of AE to AD, all because of similar triangles. AH is just the ratio applied six times. It’s $(\frac{2}{\sqrt{3}})^6$ or 64/27. b) AF=4 and DF=2. Determine the area of BCEF. All the triangles: $\triangle AFB$, $\triangle AFD$, and $\triangle DFE$ are similar because they are all right triangles, all having a right angle plus another angle shared. The ratio of the longer side to the shorter side is 4:2 or 2:1, so BF=8 and FE=1. The area of AFB is 16, the area of AFD is 4, and the area of DFE is 1: The area of ABD is 20, which is equal to BCD. Since DFE=1, BCEF must be 19. ### Question 7 7. a) Determine all solutions for this equation: $3^{x-1} \cdot 9^{\frac{3}{2x^2}} = 27$ Using some logarithms, it’s not too difficult to solve: $\begin{array}{rcl} 3^{x-1} \cdot 3^{2 \cdot \frac{3}{2x^2}} &=& 3^3 \\ (x-1) + \frac{3}{x^2} &=& 3 \\ x + 4 + \frac{3}{x^2}&=& 0 \\ x^3 - 4x^2 + 3 &=& 0 \\(x-1) (x^2-3x-3) &=& 0 \end{array}$ Using the quadratic formula we can get the solutions for the second factor. The solutions for x are 1 and 3±√21/2. b) Determine all solutions (x,y) for these equations: $\begin{array}{rcl} y &=& \log_{10} (x^4) \\ y &=& (\log_{10} x)^3 \end{array}$ This first equation can be written like this: $y = 4 \log_{10} x$ Substitute g for $\log{10}x$ and combine the two equations: $\begin{array}{rcl} 4g &=& g^3 \\ g^3 - 4g &=& 0 \\ g (g+2) (g-2) &=& 0 \end{array}$ The solutions for g are 0, -2, and 2. Because $x=10^g$, replace the g values with their respective x values to get 100, 1, and $\frac{1}{100}$. Now calculate the y values by substituting back the x values, and the solutions are (1,0), (100,8), and (1/100, -8). ### Question 8 8. a) Oi-Lam tosses 3 coins, and removes those that come up heads (if any). He then tosses the remaining coins (if any). Determine the probability he tosses exactly one head (on the second toss). Using the pascal triangle, there is a $\frac{1}{8}$ chance of having no coins to toss on the second round, $\frac{3}{8}$ chance of one coin, $\frac{3}{8}$ chance of having two coins, and $\frac{1}{8}$ chance of having no coins. For each case the probability of coming up with exactly one head is given by the first diagonal: Namely the probability of having exactly one head when tossing 3 coins is 3 divided by the sum of the row, giving $\frac{3}{8}$. So adding it up, we have this expression: $\begin{array}{l} \frac{1}{8} \cdot (1 \cdot 0 + 3 \cdot \frac{1}{2} + 3 \cdot \frac{1}{2} + 1 \cdot \frac{3}{8}) \\ = \frac{27}{64} \end{array}$ The total probability is 27/64. This fraction is the same as the answer to 6a, reversed. Coincidence? b) Here P is the center of the larger circle with AC as diameter; Q is the center of the smaller circle with BD as diameter. $\angle PRQ$ is 40. Determine $\angle ARD$. It’s best to solve this problem by first drawing a few additional lines: We also give the variables a, b, c, and d to several of the angles as shown above. Because of isosceles triangles, $\angle RAP = \angle ARP = a+b$. Similarly, $\angle RBQ = \angle BRQ = b+40$. The two must be equal so $a+b = b+40$. From that, a=20. With the exact same logic, d=20. AC is a diameter so ARC is a right angle. We know d so ARD = 110°. ### Question 9 9. a) i) Prove that: $\cot \theta - \cot 2\theta = \frac{1}{\sin 2\theta}$ If you know the trigonometric identities, this question shouldn’t be too difficult: $\begin{array}{l} \cot \theta - \cot 2 \theta \\ = \frac{\cos \theta}{\sin \theta} - \frac{\cos 2 \theta}{\sin 2 \theta} \\ = \frac{\cos \theta}{\sin \theta} - \frac{\cos^2 \theta - \sin^2 \theta}{2 \sin \theta \cos \theta} \\ = \frac{2 \cos^2 \theta - \cos^2 \theta + \sin^2 \theta}{2 \sin \theta \cos \theta} \\ = \frac{1}{2 \sin \theta \cos \theta} \\ = \frac{1}{\sin 2 \theta} \end{array}$ ii) S is given by this formula: $S = \frac{1}{\sin 8^\circ} + \frac{1}{\sin 16^\circ} + \cdots + \frac{1}{\sin 4096^\circ} + \frac{1}{\sin 8192^\circ}$ Determine, without a calculator, the value of $\alpha$: $S = \frac{1}{\sin \alpha}$ It’s important to know to use the formula given in part (i) to solve this problem. Other than that, it’s not too hard: $\begin{array}{l}\frac{1}{\sin 8} + \frac{1}{\sin 16} + \cdots + \frac{1}{\sin 4096} + \frac{1}{\sin 8192} \\ = \cot 4 - \cot 8 + \cot 8 - \cdots + \cot 4096 - \cot 8192 \\ = \cot 4 - \cot 8192 \\ = \cot 4 - \cot 92 \\ = \frac{\cos 4}{ \sin 4} - \frac{\cos 92}{\sin 92} \\ = \frac{\cos 4}{\sin 4} + \frac{\sin 2}{\cos 2} \\ = \frac{\cos 4 \cos 2 + \sin 4 \sin 2}{\sin 4 \cos 2} \\ = \frac{\cos (4-2)}{\sin 4 \cos 2} \\ = \frac{1}{\sin 4} \end{array}$ So the value of $\alpha$ is . b) In $\triangle ABC$, $a < \frac{1}{2} (b+c)$ (where a would be the side opposite of angle A). Prove that $\angle A < \frac{1}{2} (\angle B + \angle C)$. The Sine law states the following: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$ From this we can multiply to find the relationship between sides, their angles, and an unknown constant k: $\begin{array}{l} a = k \sin A \\ b = k \sin B \\ c = k \sin C \end{array}$ We now substitute and simplify: $\begin{array}{rcl} k \sin A &<& \frac{1}{2} (k \sin B + k \sin C) \\ \sin A &<& \frac{1}{2} (\sin B + \sin C) \\ \sin A &<& \frac{1}{2} (2 \sin \frac{B+C}{2} \cos \frac{B-C}{2}) \\ \sin A &<& \sin \frac{B+C}{2} \cos \frac{B-C}{2}\end{array}$ It’s obvious that $\cos \frac{B-C}{2} \leq 1$. So if we divide both sides of the equation by $\cos \frac{B-C}{2}$, the right hand side is what we desire, and the left hand side is smaller. Thus, $\begin{array}{rcl} \sin A &<& \sin \frac{B+C}{2} \\ A &<& \frac{1}{2} (B+C) \end{array}$ This is what we set out to prove. ### Question 10 10. Let $T(n)$ be the number of triangles with integer sides and perimeter n (where n is a positive integer). For example $T(6) = 1$ because only (2,2,2) has a perimeter of 6 and nothing else. a) Determine $T(10)$, $T(11)$, and $T(12)$. Triangles with perimeter 10 are (2,4,4) and (3,3,4). So $T(10)$ is 2. Triangles with perimeter 11 are (1,5,5), (2,4,5), (3,3,5), and (3,4,4). So $T(11)$ is 4. Triangles with perimeter 12 are (2,5,5), (3,4,5), and (4,4,4). So $T(12)$ is 3. The answers are 2, 4, and 3 respectively. b) If m is a positive integer and $m \geq 3$, prove that $T(2m) = T(2m-3)$. Let a, b, and c be the sides of the triangle so that $a \leq b \leq c$. Let U be the triangle with perimeter 2m, and V be the triangle with perimeter 2m-3. In order for the triangle to be non-degenerate, $a+b > c$. Because otherwise the ‘triangle’ would be a line, or something (I’m going to refer to it as a degenerate triangle). For every triangle V, there must be at least one triangle U. If we represent the sides of V by (a,b,c), we can add one to each side giving (a+1,b+1,c+1). This triangle can’t be degenerate. So $T(2m) \geq T(2m-3)$. Now we want to prove that the converse is true: for any triangle U with side lengths (a,b,c), a triangle V with side lengths (a-1,b-1,c-1) is also non-degenerate. The only case where this might not be true is when c is exactly one more than a+b, in which case the resulting triangle would be degenerate. In this case, $a+b = c+1$. But this cannot be true. The perimeter of U has to be even, so a+b+c is even. But in the equation, if a+b is even then c is odd, and if a+b is odd then c is even. Therefore it’s impossible for a+b+c to be even and at the same time for the equation $a+b = c+1$ to be true. Thus (a-1,b-1,c-1) is never degenerate. For every triangle U, there must be at least one triangle V, so $T(2m-3) \geq T(2m)$. But we’ve already proved that $T(2m) \geq T(2m-3)$. Combined with the other inequality, $T(2m) = T(2m-3)$ which is what we need. c) Determine the smallest n where $T(n) > 2010$. The formula for $T(n)$ is fairly well known. It’s given by: $\begin{array}{rcll} T(n) &=& round(\frac{n^2}{48}) & \textrm{when n is even} \\ && round(\frac{(n+3)^2}{48}) & \textrm{when n is odd} \end{array}$ Let’s use the even one just because. Solving the inequality $\frac{n^2}{48} > 2010$, we get $n > 310.6$. Since the equation is only relevant for even integers, the first even integer greater than 310.6 is 312. But since we proved that $T(n) = T(n-3)$ for even integers n, $T(309) = T(312)$. There is no smaller n meeting the requirements, so n is 309. This solution uses an external formula and would probably not be considered correct. I don’t know how to do it without this formula. ### Notes Because I took the contest several days later than everyone else, I had a considerable and unfair advantage because some of the questions were available already. So I was able to research the questions and figure out how to solve them before writing it. Nevertheless I wasn’t able to solve all of the questions before the contest. But perhaps this unfair advantage balances my lack of experience (this contest is for grade 12 students). This blog post itself may be useful to anyone taking the euclid after today. Notice that ‘today’ in this blog post is inconsistent because I wrote this blog post over the course of two days. \frac{\cost \theta}{\sin \theta} – \frac{\cos^2 \theta – \sin^2 \theta}{2 \sin \theta \cos \theta} ## 2010 Galois Contest April 9, 2010 I took the Galois contest this morning, it was a 75 minute test, with calculators allowed. It has four questions, each worth ten points, and a full written response is required. I’m going to post my answers here. These are my own answers, so no guarantee that they’re correct. Although some schools haven’t took it yet, it should still be fine. Nobody would care if I leak answers. Yay me. Anyways, first question. ### Problem 1 1. Emily has an old showerhead using 18L of water per minute and a new showerhead using 13L a minute. a) How long does it take to use 260L of water using the new showerhead? 13L per minute and 260L of water, so 20 minutes. b) How much less water is used in 10 minutes with the new showerhead than the old one? In 10 minutes the old showerhead uses 180L of water and the new one only 130L. So the new one uses 50L less water. c) Water costs 8 cents per 100L. How much money is saved by using the new showerhead instead of the old one, in a 15 minute shower? In 15 minutes the old showerhead uses 270L of water and the new one uses 195. So the difference is 75L which costs 6 cents. d) How long (in minutes) does it take to save$30 by using the new showerhead?

If m is the time showered in minutes then the cost of the old shower per minute (in cents) is $0.08 \cdot 18m$ or $1.44m$ and the new shower $0.08 \cdot 13m$ or $1.04m$, and so the difference is $0.4m$.

If we save $30 then we save 3000 cents, and $0.4m = 3000$ and m = 7500 minutes. In the contest I somehow thought$30 was 30000 cents and not 3000, so I was off by an order of magnitude. I wonder how much I’ll still get for that question.

### Problem 2

2 a) Determine the area of this quadrilateral:

The area of a trapezoid is $A=\frac{1}{2}h(s_1+s_2)$ where $s_1$ and $s_2$ are the two parallel sides.

Using this formula we get $\frac{1}{2} \cdot 12 \cdot (12+2)$ or 84.

b)

Determine an expression for the area of $\triangle COB$ in terms of p.

The area of a triangle is $\frac{bh}{2}$, and by substitution the area of COB is $\frac{12p}{2}$ or 6p.

From now on I will use the notation [ABC] to describe the area of triangle ABC.

c) Determine an expression for [QCA] in terms of p.

Using QC as the height, QC can be represented by $12-p$. Substituting into the area formula for a triangle again, the expression should be 12-p.

d) If [ABC], determine the value of p.

Use what we got before. Notice that [ABC] = [QABO] – [COB] – [QCA], which we determined to be $84 - 6p - (12-p)$ or $72-5p$.

We know that $[ABC]=27$ so we solve the equation $72-5p=27$ for p, and get p=9.

### Problem 3

3 a) Solve this system for $(x,y)$:

$\begin{array}{l} x+y=42 \\ x-y=10 \end{array}$

Using elimination or substitution, you should get (x,y) = (26,16).

b) If p is an even integer and q is an odd integer, prove that the following system of equations,

$\begin{array}{l} x+y=p \\ x-y=q \end{array}$

has no positive integer solutions.

By subtraction, we can get $2y = p-q$. The left hand side is even so the right hand side must be even. But clearly the right hand side is odd (even integer minus odd integer). Contradiction.

c) Determine all solutions to the equation $x^2 - y^2 = 420$ for (x,y) being positive integers.

We can rewrite the equation as $(x+y) (x-y) = 420$, and let p=x+y and q=x-y. From what we’ve seen in (b), p-q has to be even so p and q must be both even, or both odd. Because pq=420, p and q are both even (odd * odd != even).

There are four pairs of even integer pairs (p,q) satisfying pq=420: (2,210), (6,70), (10,42), and (14,30).

The four pairs each give the solutions (104,102), (38,32), (26,16), and (22,8) for (x,y).

### Problem 4

4 a)

Explain why [PQT]:[PTR] = 3:5.

The two triangles have different bases but share the same altitude, or height. Thus the ratio of their areas is $\frac{6h}{2} : \frac{10h}{2}$ or 3:5.

b)

Here BD=DC; AE:EC=1:2; AF:FD=3:1. If [DEF]=17 determine [ABC].

Following a similar logic as in part (a), we know that [ABD]=[ADC], [DAE]:[DEC]=1:2, and [EAF]:[EFD]=3:1.

Since [EFD]=17, [EAF]+[EFD]=4*17 or 68. So [DAE]=68, and [DAE]+[DEC]=3*68 or 204. Finally [ADC]=204, so [ABC]=2*204 = 408.

This is just the exact same logic repeated three times.

c)

Here, VY:YW = 4:3, [PYW]=30, and [PZW]=35.

Because VY:YW = 4:3, we can find [VPY] to be 40 the same way that we did it in part (b).

In $\triangle VZW$, $\triangle VPW= 70$ and $\triangle PZW=35$, so the ratio is 2:1. Therefore the ratio VP:PZ = 2:1.

In order to simplify things a little bit, let $\triangle VWP = x$, $\triangle UXP = y$, and $\triangle UPZ = z$.

Here’s a diagram, with new labels:

In the whole triangle, [UVY]:[UYW] = 4:3. We can write this algebraically:

$x+y+40 : z+65 = 4:3$

Now using the line XW as a base, the ratios of the triangles on either side is equal. Namely, [XVP]:[PVW] = [UXP]:[UPW]. Writing this algebraically:

$x:70 = y:z+35$

Finally we use VZ as a base. We know the ratio to be 2:1, so writing it algebraically would be this:

$x+y:z = 2$

Solving the three equations for the three variables, we get (x,y,z) = (56,84,70). [UXP]=y, so [UXP]=84.

I was unable to get this far on the contest, and this problem is by far the hardest in the set. Looking back, I was actually on the right track, but gave up my approach thinking it was completely wrong.

If I’m lucky, I might get some part marks for my work here.

I think I did better on this contest than I did on the Cayley. Probably still won’t win any prizes, though.

### In other news

I’ve placed in the first group in the Canadian Computing Competition, along with Wen Li Looi from my school.