## Suggested Solution to Question 24 and 25 of the Fermat Math Contest 2018

Posted: March 20, 2018 in Mathematics

Q24.

Denote $G_i, R_i, B_i, Y_i$ as the i-th bucket that contains the puck.

If all 4 pucks are distributed into the same Green bucket, then we don’t need to worry about the Red, Blue, and Yellow buckets since they won’t have 4 pucks in them. So,

$\displaystyle P(G_a, G_a, G_a, G_a) = 3 \left(\frac13\right)^4 = \frac{1}{27}$

Now, 3 pucks are distributed into the same Green bucket, then we only need to exclude the case where 3 pucks are put into the same Red bucket. There are 4 ways to permute aaab and 3 numbers to pick for a and b which is 6 ways, so

$\displaystyle P(G_a, G_a, G_a, G_b) \cdot (1 - P(R_a, R_a, R_a)) = 4 \cdot 6 \cdot \left(\frac13\right)^4 \cdot \left(1-3 \left(\frac13\right)^3\right) = \frac{64}{243}$

If 2 pucks are distributed into the same Green bucket, then we only need to consider the case where all 3 pucks are distributed evenly into three different Red buckets and all 2 pucks are distributed evenly into two different Blue buckets. The most difficult part of the problem is in this case here. We need to find the number of ways to permute n repeated objects, aabc, which is $\dfrac{4!}{2!1!1!}$, and since the permutation of b and c has already counted, we need to only consider how many numbers go into the choice of a which is 3.

$\displaystyle P(G_a, G_a, G_b, G_c) \cdot P(R_a, R_b, R_c) \cdot P(B_a, B_b) \\ = \frac{4!}{2!1!1!}\cdot 3 \cdot \left(\frac13\right)^4 \cdot 3!\left(\frac13\right)^3 \cdot (3P2) \left(\frac13\right)^2 \\ = \frac{4}{9}\cdot\frac{2}{9}\cdot\frac{2}{3} = \frac{16}{243}$

Final step, add the three results and we get

$\displaystyle \frac{1}{27} + \frac{64}{243} + \frac{16}{243} = \frac{89}{243}$

Q25.

For $1 \leq k \leq 2018$ and $P, Q, R \in \{1,2,\ldots,9\}$

$P_{2k} - Q_k = (R_k)^2$

When k = 1,

$11P - Q = R^2$. The 8 quadruples are easy to find, they are

$(1,7,2,1), (1,2,3,1), (2,6,4,1), (3,8,5,1), (4,8,6,1), (5,6,7,1), (6,2,8,1), (8,7,9,1)$

When k = 2,

$1111P - 11Q = 11^2R^2 \Rightarrow 101P - Q = 11R^2$

When k = 3,

$111111P - 111Q = 111^2R^2 \Rightarrow 1001P - Q = 111R^2$

$\therefore (10^k+1)P-Q=\dfrac{10^k-1}{10-1}\cdot R^2 \Rightarrow (10^k+1)P-Q=\dfrac{10^k-1}{9} \cdot R^2$

or

$P-Q \equiv 11R^2 \pmod{100} \text{ for } k \geq 2$

$\underline{\quad R \qquad R^2 \qquad 11R^2 \qquad 11R^2 \mod{100} \quad} \\ \text{ } \quad 1 \qquad 1 \quad \qquad 11 \qquad\qquad\qquad 11 \\ \text{ } \quad 2 \qquad 4 \quad \qquad 44 \qquad\qquad\qquad 44 \\ \text{ } \quad 3 \qquad 9 \quad \qquad 99 \qquad\qquad\qquad 99 \equiv -1 \\ \text{ } \quad 4 \qquad 16 \quad \quad 176 \qquad\qquad\qquad 76 \\ \text{ } \quad 5 \qquad 25 \quad \quad 275 \qquad\qquad\qquad 75 \\ \text{ } \quad 6 \qquad 36 \quad \quad 396 \qquad\qquad\qquad 96 \equiv -4 \\ \text{ } \quad 7 \qquad 49 \quad \quad 539 \qquad\qquad\qquad 39 \\ \text{ } \quad 8 \qquad 64 \quad \quad 704 \qquad\qquad\qquad 4 \\ \text{ } \quad 9 \qquad 81 \quad \quad 891 \qquad\qquad\qquad 91$

Therefore, R can only be 3, 6, or 8.

When R = 3,

$(10^k + 1)P - Q = 10^k - 1$

$10^k P + P - Q = 10^k - 1$

$\text{Since } P - Q \equiv -1 \mod{100}$

$10^k P - 1 = 10^k - 1$

$\Rightarrow P = 1, Q = 2 \quad \therefore (1,2,3,k) \text{ for } 1 < k \leq 2018$

When R = 6,

$(10^k + 1)P - Q = \dfrac{10^k-1}{9} \cdot 36$

$10^k P - 4 = 4\cdot 10^k - 4$

$\Rightarrow P = 4, Q = 8 \quad \therefore (4,8,6,k) \text{ for } 1 < k \leq 2018$

When R = 8,

$(10^k + 1)P - Q =\dfrac{10k-1}{9} \cdot 64$

$9(10^k P + 4) = 64 \cdot 10^k - 64$

$9 \cdot 10^k P + 36 = 64 \cdot 10^k - 64$

$100 = (64 - 9P) 10^k$

Now, obviously $k > 2$ has no solution.

If k = 2,

$1 = 64 -9P$

$\Rightarrow P = 7, Q = 3$

$\therefore (7,3,8,2)$ is a solution.

In conclusion, N = 8 + 2017 + 2017 + 1 = 4043.

Therefore the sum of digits of N is 11.

## Expressing Pi in an interesting way

Posted: March 18, 2018 in Mathematics

What is $\pi$? $\pi$ is a constant. It is a ratio of the circumference to the diameter of a circle. It’s value is approximately 3.14 or 3.14159 if one cares about accuracy, however 3.14159 is only correct to 6 significant figures. If you pull out your everyday scientific calculator and press the $\pi$ button, it will show 3.141592654 on the display. Although this value is good to 10 significant figures, it still contains an error. I know it has error because I have $\pi$ memorized to 3.14159265358979323846… You may wonder why I memorize $\pi$ to 21 digits? It could be the password to one of my dozen email addresses. You can try hacking my email with it 🙂

No matter how many digits of $\pi$ you write down on a piece of paper, it is still just an approximation. The tail of $\pi$ never terminates. It is an irrational number. Most mathematicians don’t really like approximation, so you rarely see one write 3.14 as $\pi$. It’s just not correct and not cool to write $\pi = 3.14$ (other than PIE=314). So today I am gonna show you a cool way to write $\pi$.

$\pi = \left(\left(-.5\right)!\right)^2 \quad \text{ or } \quad \pi = \Gamma^2(\frac12)$

The symbol “!” means factorial. By definition,

$n! = n(n-1)(n-2)\cdots 2 \cdot 1$ for all positive integers n. So

$1! =1$,

$2! = 2\cdot 1 = 2$,

$3! =3\cdot2\cdot1 = 6$

and so on.

What about $(-0.5)!$? The above definition didn’t include non-positive integer factorial. So we need to find a way to extend the factorials to a larger set of number that includes decimal numbers. One way to do it is by using Gamma function.

$\displaystyle \Gamma(s) = \int_0^\infty t^{s-1} e^{-t} \text{ d}t$

First of all, we need to verify that this Gamma function indeed satisfies

$f(1)=1 \text{ and } f(n+1) = nf(n)$

$\displaystyle \Gamma(1) = \int_0^\infty e^{-t} \text{ d}t = -e^{-t}\bigg|_0^\infty = 1$

and by using integration by parts, we get

$\displaystyle \Gamma(s+1) = \int_0^\infty t^s e^{-t} \text{ d}t = -t^s e^{-t}\bigg|_0^\infty + s\int_0^\infty t^{s-1} e^{-t} \text{ d}t = s\Gamma(s)$

And therefore the Gamma function does indeed extend the factorials to the real numbers, and in fact also the complex numbers as well.

$\Gamma(1) = 1$

$\Gamma(2) = 1\cdot\Gamma(1) = 1$

$\Gamma(3) = 2\cdot\Gamma(2) = 2\cdot 1$

$\Gamma(4) = 3\cdot\Gamma(3) = 3\cdot 2 \cdot 1$

Hence, we see that

$n! = \Gamma(n+1)$

Now, let’s compute $\Gamma\left(\frac12\right)$.

By using substitution of $w=\sqrt{t}$, we get

$\displaystyle\Gamma\left(\frac12\right) = \int_0^\infty \frac{e^{-t}}{\sqrt{t}} \text{ d}t = 2\int_0^\infty e^{-w^2} \text{ d}w$

Now let $\displaystyle I=\int_0^\infty e^{-w^2} \text{ d}w$,

$\displaystyle I^2 = \int_0^\infty e^{-x^2}\text{ d}x \int_0^\infty e^{-y^2}\text{ d}y =\int_0^\infty \int_0^\infty e^{-(x^2+y^2)} \text{ d}x\text{ d}y$

then by using polar coordinates,

$\displaystyle I^2=\int_0^{\pi/2} \int_0^\infty e^{-r^2} r\text{ d}r\text{ d}\theta = \frac{\pi}{2} \int_0^\infty re^{-r^2} \text{ d}r$

By substituting $u =r^2$ we get

$\displaystyle I^2 = \frac{\pi}{4} \int_0^\infty e^{-u} \text{ d}u = \frac{\pi}{4}(-e^{-u})\bigg|_0^\infty = \frac{\pi}{4}$

$\therefore I = \dfrac{\sqrt\pi}{2}$

Therefore,

$\displaystyle \Gamma\left(\frac12\right) = \sqrt\pi$

or

$\pi = \Gamma^2\left(\dfrac12\right)$

or

$\pi = \left(\left(-\dfrac12\right)!\right)^2$

if you accept the factorial notation 🙂

## How to Draw a Parallelogram with One Ruler

Posted: March 8, 2018 in Mathematics

Usually in school, you learn how to construct a parallelogram with compass and straightedge. But today, I am gonna show you how to construct a parallelogram with only one ruler. Of course, you need a piece of paper, a pencil, a hand, and your brain as well, etc …

Actually the method is very simple, even an eight year old can do it I promise. So here it is:

1. Draw a quadrilateral (4-sided shape), any quadrilateral will do even the non-convex one.
2. Find the midpoint of all four sides with your ruler.
3. Connect those four midpoints together and you are done.

Here are some examples.

The proof is very simple.

Let the four points $A(a_1, a_2)$$B(b_1, b_2)$$C(c_1, c_2)$$D(d_1, d_2)$ be the vertices of a quadrilateral.

Then the four midpoints are

$\displaystyle P\left(\frac{a_1+b_1}{2}, \frac{a_2+b_2}{2}\right), Q\left(\frac{b_1+c_1}{2}, \frac{b_2+c_2}{2}\right), R\left(\frac{c_1+d_1}{2}, \frac{c_2+d_2}{2}\right), S\left(\frac{d_1+a_1}{2}, \frac{d_2+a_2}{2}\right)$

Find the slopes of the sides and show that the opposite sides are parallel:

$m_{PQ} = \dfrac{c_2-a_2}{c_1-a_1} = m_{RS} \Rightarrow PQ \;//\; RS$

$m_{QR} = \dfrac{d_2-b_2}{d_1-b_1} = m_{SP} \Rightarrow QR \;//\; SP$

Therefore PQRS is a parallelogram.

## The Two Special Sets of Numbers

Posted: May 3, 2017 in Mathematics

Sum of Zeroth Powers

$1^0 + 6^0 + 7^0 + 23^0 + 24^0 + 30^0 + 38^0 + 47^0 + 54^0 + 55^0 = 10$

$2^0 + 3^0 + 10^0 + 19^0 + 27^0 + 33^0 + 34^0 + 50^0 + 51^0 + 56^0 = 10$

Sum of First Powers

$1^1 + 6^1 + 7^1 + 23^1 + 24^1 + 30^1 + 38^1 + 47^1 + 54^1 + 55^1 = 285$

$2^1 + 3^1 + 10^1 + 19^1 + 27^1 + 33^1 + 34^1 + 50^1 + 51^1 + 56^1 = 285$

Sum of Second Powers

$1^2 + 6^2 + 7^2 + 23^2 + 24^2 + 30^2 + 38^2 + 47^2 + 54^2 + 55^2 = 11685$

$2^2 + 3^2 + 10^2 + 19^2 + 27^2 + 33^2 + 34^2 + 50^2 + 51^2 + 56^2 = 11685$

Sum of Third Powers

$1^3 + 6^3 + 7^3 + 23^3 + 24^3 + 30^3 + 38^3 + 47^3 + 54^3 + 55^3 = 536085$

$2^3 + 3^3 + 10^3 + 19^3 + 27^3 + 33^3 + 34^3 + 50^3 + 51^3 + 56^3 = 536085$

Sum of Fourth Powers

$1^4 + 6^4 + 7^4 + 23^4 + 24^4 + 30^4 + 38^4 + 47^4 + 54^4 + 55^4 = 26043813$

$2^4 + 3^4 + 10^4 + 19^4 + 27^4 + 33^4 + 34^4 + 50^4 + 51^4 + 56^4 = 26043813$

Sum of Fifth Powers

$1^5 + 6^5 + 7^5 + 23^5 + 24^5 + 30^5 + 38^5 + 47^5 + 54^5 + 55^5 = 1309753125$

$2^5 + 3^5 + 10^5 + 19^5 + 27^5 + 33^5 + 34^5 + 50^5 + 51^5 + 56^5 = 1309753125$

Sum of Sixth Powers

$1^6 + 6^6 + 7^6 + 23^6 + 24^6 + 30^6 + 38^6 + 47^6 + 54^6 + 55^6 = 67334006805$

$2^6 + 3^6 + 10^6 + 19^6 + 27^6 + 33^6 + 34^6 + 50^6 + 51^6 + 56^6 = 67334006805$

Sum of Seventh Powers

$1^7 + 6^7 + 7^7 + 23^7 + 24^7 + 30^7 + 38^7 + 47^7 + 54^7 + 55^7 = 3512261547765$

$2^7 + 3^7 + 10^7 + 19^7 + 27^7 + 33^7 + 34^7 + 50^7 + 51^7 + 56^7 = 3512261547765$

Sum of Eighth Powers

$1^8 + 6^8 + 7^8 + 23^8 + 24^8 + 30^8 + 38^8 + 47^8 + 54^8 + 55^8 = 185039471773893$

$2^8 + 3^8 + 10^8 + 19^8 + 27^8 + 33^8 + 34^8 + 50^8 + 51^8 + 56^8 = 185039471773893$

## 2187 in Star Wars

Posted: April 15, 2017 in Mathematics

Last night the Pearl was playing the Star Wars movie episode 4. When Princess Leia was captured in the Death Star, she was held in the cell number 2187. Cell “2187” sounds very familiar. The black dude who is casting as Finn in the Force Awaken also has the code name “FN-2187”. I don’t know if the number 2187 has any special meaning to the Star Wars story or the screenwriter, but 2187 reminds me of tutoring full time 10 years ago on topic like Geometric Sequence because $3^7=2187$.

There are also some interesting facts about the number 2187. If you write the number in backward, you get 7812. When you add the two numbers, you get 9999. When you split the number 2187 into two 2-digit numbers 21 and 87 and then multiple them, you get 1827. And also notice that $2187 = 3^7 = 3^3 \cdot 3^4 = 27 \cdot 81$.

## Stanford Summer Institute Math Problems

Posted: April 9, 2017 in Mathematics

(1)  Suppose $x, y \in\mathbb{Z}$.

$x^2-9y^2=(x-3y)(x+3y)=2017=1\cdot2017=2017\cdot1=(-1)(-2017)=(-2017)(-1)$

Case $(x-3y)(x+3y) = 1\cdot2017 \Rightarrow (x,y)=(1009, 336)$

Case $(x-3y)(x+3y) = 2017\cdot1 \Rightarrow (x,y)=(1009, -336)$

Case $(x-3y)(x+3y) = (-1)(-2017) \Rightarrow (x,y)=(-1009, -336)$

Case $(x-3y)(x+3y) = (-2017)(-1) \Rightarrow (x,y)=(-1009, 336)$

(2)  Since there are more 2’s than 5’s in 2017! and $2\cdot5=10$, all we need to do is to count the number of 5’s in the prime factorization of 2017!.

$\displaystyle \left\lfloor\frac{2017}{5}\right\rfloor + \left\lfloor\frac{2017}{25}\right\rfloor + \left\lfloor\frac{2017}{125}\right\rfloor + \left\lfloor\frac{2017}{625}\right\rfloor = 403 + 80 + 16 + 3 = 502$

(3)  Suppose that $n \in\mathbb{Z}$.

$n^2+9n+14 = m^2$, for some $m\in\mathbb{Z}$

By completing the square,

$\displaystyle \left(n+\frac{9}{2}\right)^2 - \frac{25}{4} = m^2$

$\displaystyle \left(n+\frac{9}{2}-m\right)\left(n+\frac{9}{2}+m\right) = \frac{25}{4}$

$(2n+9-2m)(2n+9+2m) = 25 = 1\cdot25 = 25\cdot1 = 5\cdot5 \\ = (-1)(-25) = (-25)(-1) = (-5)(-5)$

Case $1\cdot25 \text{ and } 25\cdot1 \Rightarrow 2n+9=\dfrac{1+25}{2}=13 \Rightarrow n=2$

Case $5 \cdot 5 \Rightarrow 2n+9=\dfrac{5+5}{2}=5 \Rightarrow n=-2$

Case $(-1)(-25) \text{ and } (-25)(-1) \Rightarrow 2n+9=-\dfrac{1+25}{2}=-13 \Rightarrow n=-11$

Case $(-5)(-5) \Rightarrow 2n+9=-\dfrac{5+5}{2}=-5 \Rightarrow n=-7$

If we define zero being a perfect square, then $n=2, -2, -11, -7$. Otherwise $n=2, -11$.

(4)  $ax^4+bx^3+cx^2+dx+e=0$

$x=\sqrt3+\sqrt5$

$x^2=8-2\sqrt{15}$

$x^3=18\sqrt3-14\sqrt5$

$x^4=124-32\sqrt{15}$

$a(124-32\sqrt{15}) + b(18\sqrt3-14\sqrt5) + c(8-2\sqrt{15}) + d(\sqrt3-\sqrt5) + e = 0$

$\sqrt{15}: \quad -32a-2c=0 \Rightarrow c=-16a \Rightarrow \dfrac{c}{a} = -16$

$\sqrt5: \quad -14b-d=0$

$\sqrt3: \quad 18b+d=0$

$\therefore b=d=0$

constant term: $\displaystyle 124a+8c+e=0 \Rightarrow 124+\frac{8b}{a}+\frac{c}{a}=0$

$\displaystyle \therefore \frac{e}{a} = -\frac{8c}{a}-124 = 128-124=4$

In particular,

$\displaystyle x^4+\frac{c}{a}x^2+\frac{e}{a}=0 \Rightarrow x^4-16x^2+4=0$

In general, for any integer n,

$a=n, b=0, c=-16n, d=0, e=4n$

(5)  Let $r$ be the radius of the smallest semi-circle.

$2\sqrt2 r + 4\sqrt2 r = 6\sqrt2 \quad \therefore \; r=1$

Note that the perimeter of semi-circle is $\pi r$,

$\text{perimeter of the curve} = \pi + 2\pi + 4\pi = 7\pi$

(6)  Let the side length of B, C, D, E, F, G, H, and I be b, c, d, e, f, g, h, and i respectively.

$g = f+1$

$b = f+2$

$c = f+3$

$d = f+3+e$

$i = f+3+2e$

$h = f+g = 2f+1$

$b+c+d=i+h \Rightarrow 3f+8+e = 3f+4+2e \Rightarrow e=4$

$d+i=b+g+h \Rightarrow 2f+18=4f+4 \Rightarrow f=7$

$\therefore \text{Area} = (d+i)(i+h) = 32\cdot33 = 1056$

(7)  By Fermat’s little theorem, $3^6 \equiv 1 \mod 7$

$\displaystyle 8\sum_{k=51}^{100}10^k + 10^{50}a + 9\sum_{k=0}^{49}10^k \pmod{7} \equiv \sum_{k=0}^{49}3^{k+51} + 3^{2}a + 2\sum_{k=0}^{49}3^k$

$\displaystyle \equiv -\sum_{k=0}^{49}3^k + 2a + 2\sum_{k=0}^{49}3^k \equiv 2a+\sum_{k=0}^{49}3^k \equiv 2a+\frac{3^{50}-1}{3-1} \equiv 0$

$\displaystyle 4a + 3^{50}-1 \equiv 0 \pmod{7}$

$\displaystyle 4a + 1 \equiv 0 \pmod7$

$\displaystyle a \equiv 5 \pmod7$

$\displaystyle \therefore a = 5$

(8)  By Pythagoras theorem,

$x^2+13 = 4^2 = 16 \therefore x=\sqrt3$

(9)  By Euclidean Algorithm,

$11n+7 = (7n+1)(1)+4n+6$

$7n+1 = (4n+6)(1) + 3n-5$

$4n+6 = (3n-5)(1) + n+11$

$3n-5 = (n+11)(3) - 38$

hence $(11n+7, 7n+1) = (n+11, -38)$

$(n+11, -38) > 1 \quad\text{if}\quad 2|n+11 \text{ or } 19|n+11$

$n+11\equiv 0 \mod 2 \Rightarrow n\equiv1 \mod 2 \quad\therefore n=2k+1$

$n+11\equiv 0 \mod 19 \Rightarrow n \equiv 8 \mod 19 \quad\therefore n=19k+8$

(10)  Divide the numbers from 1 to 100 into congruence classes mod 11.

[1] : 1, 12, … , 100; There are 10 members so we can choose at most 5 members, since we don’t want any pair of numbers with difference of 11. (Any adjacent members in the class have difference of 11)

[2] : 2, 13, … , 90; There are 9 members so we can choose at most 5 members.

[3] : 3, 14, … , 91; There are 9 members so we can choose at most 5 members.

… … , similarly for classes such as [4] to [11], each of them have 9 members. Therefore we can at most pick $5\times 11 = 55$ numbers without any pair in the set having difference of 11.

## DSE Math 2016 Paper 1 ANS

Posted: March 27, 2017 in Mathematics

Exam week’s coming, so here it is.

###### Section A (35 marks)

(1)  Simplify $\displaystyle\frac{(x^8 y^7)^2}{x^5 y^{-6}}$ and express your answer with positive indices.  (3 marks)

Solution

$\displaystyle\frac{(x^8 y^7)^2}{x^5 y^{-6}} = \frac{x^{16}y^{14}}{x^5 y^{-6}} = x^{11}y^{20}$

(2)  Make $x$ the subject of the formula $Ax=(4x+B)C$.  (3 marks)

Solution

$Ax=4Cx+BC$

$Ax-4Cx=BC$

$\therefore x = \dfrac{BC}{A-4C}$

(3)  Simplify $\dfrac{2}{4x-5}+\dfrac{3}{1-6x}$.  (3 marks)

Solution

$\displaystyle \frac{2(1-6x)+3(4x-5)}{(4x-5)(1-6x)} = \frac{2-12x+12x-15}{(4x-5)(1-6x)} = \frac{13}{(4x-5)(6x-1)}$

(4)  Factorize

(a)  $5m-10n$.

(b)  $m^2+mn-6n^2$.

(c)  $m^2+mn-6n^2-5m+10n$.  (4 marks)

Solution

(a)  $5(m-2n)$

(b)  $(m+3n)(m-2n)$

(c)  $(m+3n)(m-2n)-5(m-2n) = (m-2n)(m+3n-5)$

(5)  In a recreation club, there are 180 members and the number of male members is 40% more than the number of female members. Find the difference of the number of male members and the number of female members.  (4 marks)

Solution

Let $f$ be the number of female members.

$1.4f + f = 180 \quad \Rightarrow \quad f = \dfrac{180}{2.4} = 75$

$\therefore \text{difference} = 1.4f-f=0.4f=0.4(75)=30$

(6)  Consider the compound inequality

$x+6 < 6(x+11) \text{ or } x\leq -5 \qquad (*)$

(a)  Solve (*).

(b)  Write down the greatest negative integer satisfying (*).  (4 marks)

Solution

(a)  $x+6<6x+66 \text{ or } x\leq -5 \\ \Leftrightarrow 5x>-60 \text{ or } x\leq -5 \\ \Leftrightarrow x>-12 \text{ or } x\leq -5 \\ \Leftrightarrow x\in\mathbb{R}$

(b)  $-1$

(7)  In a polar coordinate system, $O$ is the pole. The polar coordinate of the points $A$ and $B$ are $(12, 75^\circ)$ and $(12, 135^\circ)$ respectively.

(a)  Find $\angle AOB$.

(b)  Find the perimeter of $\triangle AOB$.

(c)  Write down the number of folds of rotational symmetry of $\triangle AOB$.  (4 marks)

Solution

(a)  $\angle AOB = 135^\circ - 75^\circ = 60^\circ$

(b)  Obviously the triangle is isosceles and hence $\alpha = \beta = 60^\circ$ and hence $\triangle AOB$ is equilateral. Therefore the perimeter is 36.

(c)  3

(8)  It is given that $f(x)$ is the sum of two parts, one part varies as $x$ and the other part varies as $x^2$. Suppose that $f(3)=48$ and $f(9)=198$.

(a)  Find $f(x)$.

(b)  Solve the equation $f(x)=90$.  (5 marks)

Solution

(a)  Let $f(x) = ax^2+bx$ for some non-zero real numbers $a$ and $b$.

$f(3) = 9a+3b = 48 \Leftrightarrow 3a+b=16 \qquad (1)$

$f(9) = 81a+9b = 198 \Leftrightarrow 9a+b=22 \qquad (2)$

(2) – (1) gives $6a=6 \Rightarrow a=1 \Rightarrow b=13$

$\therefore f(x) = x^2+13x$

(b)  $x^2+13x-90=0 \Rightarrow (x+18)(x-5)=0 \Rightarrow x=-18 \text{ or } x=5$

(9)  The frequency distribution table and the cumulative frequency distribution table below show the distribution of the heights of the plants in a garden.

(a)  Find $x$, $y$, and $z$.

(b)  If a plant is randomly selected from the garden, find the probability that the height of the selected plant is less than 1.25 m but not less than 0.65 m.

Solution

(a)  $a=2, \therefore x=2+4=6, y=37-15=22, z=37+3=40$

(b)  $\dfrac{22-6}{40}=\dfrac{16}{40}=\dfrac{2}{5}$

(10)  The coordinates of the points $A$ and $B$ are $(5, 7)$ and $(13,1)$ respectively. Let $P$ be a moving point in the rectangular coordinate plane such that $P$ is equidistant from $A$ and $B$. Denote the locus of $P$ by $\Gamma$.

(a)  Find the equation of $\Gamma$.  (2 marks)

(b)  $\Gamma$ intersects the x-axis and y-axis at $H$ and $K$ respectively. Denote the origin by $O$. Let $C$ be the circle which passes through $O$, $H$, and $K$. Someone claims that the circumference of $C$ exceeds 30. Is the claim correct? Explain your answer.  (3 marks)

Solution

(a)  $(x-5)^2+(y-7)^2=(x-13)^2+(y-1)^2$

$\Rightarrow -10x+25-14y+49=-26x+169-2y+1$

$\Rightarrow 16x=12y+96$

$\Rightarrow 4x-3y-24=0$

(b)  Let $y=0$, then $x=6$, hence $H(6, 0)$

Let $x=0$, then $y=-8$, hence $K(0, -8)$

Since $\triangle HKO$ is a right triangle where $\angle O = 90^\circ$, hence $HK$ is the diameter of the circle $C$.

$\text{Circumference} = \pi d = \pi\sqrt{6^2+8^2} = 10\pi > 30$

Therefore, the claim is correct.

(11)  An inverted right circular conical vessel contains some milk. The vessel is held vertically. The depth of milk in the vessel is $12 \text{ cm}$. Peter pours $444\pi\text{ cm}^2$ of milk into the vessel without overflowing. He now finds that the depth of milk in the vessel is $16\text{ cm}$.

(a)  Express the final volume of milk in the vessel in terms of $\pi$.  (3 marks)

(b)  Peter claims that the final area of the wet curved surface of the vessel is at least $800\text{ cm}^2$. Do you agree? Explain your answer.  (3 marks)

Solution

(a)  Let $x$ be the original volume of milk.

$\displaystyle\frac{444\pi+x}{x}=\left(\frac{16}{12}\right)^3=\left(\frac{4}{3}\right)^3$

$27(444\pi+x)=64x$

$27\cdot444\pi=37x$

$x=27\cdot4\cdot3\pi = 81\cdot4\pi=324\pi$

$\therefore \text{Final volume} = 444\pi+324\pi = 768\pi\text{ cm}^3$

(b)  $\dfrac{\pi r^2 (16)}{3} = 768 \pi \Rightarrow r^2=144 \Rightarrow r=12\text{ cm}$

$\text{Curved surface area} = \pi r s = \pi(12)\sqrt{12^2+16^2} = 240\pi \approx 754 \text{ cm}^2 < 800\text{ cm}^2$

Therefore, disagree.

(12)  The bar chart below shows the distribution of the ages of the children in a group, where $a>11$ and $4. The median of the ages of the children in the group is 7.5.

(a)  Find $a$ and $b$.  (3 marks)

(b)  Four more children now join the group. It is found that the ages of these four children are all different and the range of the ages of the children in the group remains unchanged. Find

(i)  the greatest possible median of the ages of the children in the group.

(ii)  the least possible mean of the ages of the children in the group.  (4 marks)

Solution

(a)  Since the median age is 7.5, hence $11+a=11+b+4 \Rightarrow a=b+4$

Now, $a>11 \text{ and } 47 \text{ and } 4.

Since $a,b\in\mathbb{Z}^+ \therefore b=8\text{ or } 9$

$\therefore (a,b) = (12,8)\text{ or }(13,9)$

(b) (i)  To maximize the median, we add age 7, 8, 9, 10. Therefore the new median will be 8.

(b) (ii)  To minimize the mean, we add age 6, 7, 8, 9.

Case $(a, b)=(12,8)$,

$\text{mean} = \dfrac{12\cdot6+13\cdot7+12\cdot8+9\cdot9+4\cdot10}{12+13+12+9+4} = \dfrac{380}{50}=\dfrac{38}{5}=7.6$

Case $(a, b)=(13,9)$,

$\text{mean} = \dfrac{12\cdot6+14\cdot7+12\cdot8+10\cdot9+4\cdot10}{12+14+12+10+4} = \dfrac{396}{52}=\dfrac{99}{13}=7.\overline{615384}$

(13)  In figure 1, $ABC$ is a triangle. $D$, $E$, and $M$ are points lying on $BC$ such that $BD=CE$, $\angle ADC = \angle AEB$ and $DM=EM$.

(a)  Prove that $\triangle ACD\cong\triangle ABE$.  (2 marks)

(b) Suppose that $AD=15\text{ cm}$, $BD=7\text{ cm}$ and $DE=18\text{ cm}$.

(i)  Find $AM$.

(ii)  Is $\triangle ABE$ a right-angled triangle? Explain your answer.  (5 marks)

Solution

(a)  $\angle ADC = \angle AEB \quad \text{(given)}$

$AD=AE \quad \text{(sides opp. equal }\angle\text{s)}$

$BD=CE \quad\text{(given)}$

$BE = BD+DE = CE+DE = CD\quad (BD=CE)$

$\therefore \triangle ACD\cong\triangle ABE \quad\text{(SAS)}$

(b) (i)  $AD=AE=15 cm\quad \text{(given)}$

Hence, $\triangle ADE$ is isosceles.

$DM=EM\quad\text{(given)}$

$\therefore AM \perp DE\quad\text{(property of isos }\triangle \text{)}$

$\therefore AM=\sqrt{15^2-9^2} = 12\text{ cm} \quad\text{(Pyth. thm)}$

(b) (ii) $DM=EM=9\text{ cm}$

By Pyth. thm, $AB=\sqrt{12^2+(7+9)^2} = 20\text{ cm}$

$AB^2+AE^2 = 20^2+15^2 = 25^2 = AE^2$

Therefore, $\triangle ABE$ is right-angled.

(14)  Let $p(x)=6x^4+7x^3+ax^2+bx+c$, where $a$, $b$, and $c$ are constants. When $p(x)$ is divided by $x+2$ and when $p(x)$ is divided by $x-2$, the two remainders are equal. It is given that $p(x)=(lx^2+5x+8)(2x^2+mx+n)$, where $l$, $m$, and $n$ are constants.

(a)  Find $l$, $m$, and $n$.  (5 marks)

(b)  How many real roots does the equation $p(x)=0$ have? Explain your answer.  (5 marks)

Solution

(a)  $p(-2)=p(2)$

$6(-2)^4+7(-2)^3+a(-2)^2+b(-2)+c = 6(2)^4+7(2)^3+a(2)^2+b(2)+c$

$-56-2b = 56+2b$

$4b = -112$

$b=-28$

$\therefore p(x)=6x^4+7x^3+ax^2-28x+c$

$(lx^2+5x+8)(2x^2+mx+n) = (2l)x^4+(lm+10)x^3+(ln+5m+16)x^2+(5n+8m)x+8n$

By matching the coefficient,

$2l=6 \Rightarrow l=3$

$lm+10=7 \Rightarrow m=-1$

$5n+8m=-28 \Rightarrow n=-4$

(b)  $p(x)=(lx^2+5x+8)(2x^2+mx+n)=(3x^2+5x+8)(2x^2-x-4)=0$

$\therefore (3x^2+5x+8)=0\text{ or }(2x^2-x-4)=0$

For $3x^2+5x+8=0, \quad \Delta=25-4\cdot3\cdot8 = -71\quad \therefore$ no real root.

For $2x^2-x-4=0, \quad \Delta=1+4\cdot2\cdot4 = 33\quad \therefore$ two real roots.

Therefore, $p(x)=0$ has two real roots.

###### Section B (35 marks)

(15)  If 4 boys and 5 girls randomly form a queue, find the probability that no boys are next to each other in the queue.  (3 marks)

Solution

Consider this : _ G _ G _ G _ G _ G _

There are 6 places to put 4 B’s so there are 6P4 ways to place the boys. Moreover, there are 5! ways to permute those 5 G’s. Therefore,

$\displaystyle \text{P(no boys are next to each other)}=\frac{_6P_4 \cdot 5!}{9!} = \frac{6\cdot5\cdot4\cdot3}{9\cdot8\cdot7\cdot6} = \frac{5}{42}$

(16)  In a test, the mean of the distribution of the scores of a class of students is 61 marks. The standard scores of Albert and Mary are -2.6 and 1.4 respectively. Albert gets 22 marks. A student claims that the range of the distribution is at most 59 marks. Is the claim correct? Explain your answer.  (3 marks)

Solution

$Z=\dfrac{X-\mu}{\sigma} \quad \Rightarrow\quad -2.6 = \dfrac{22-61}{\sigma} \quad\Rightarrow\quad \sigma = 15$

So, the standard deviation is 15. Now, let’s calculate what Mary gets on the test.

$X=Z\sigma+\mu = 1.4\cdot 15 + 61 = 82$

$82-22 = 60$, which is greater than 59. Therefore the claim is false.

(17)  The 1st term and the 38th term of an arithmetic sequence are 666 and 555 respectively. Find

(a)  the common difference of the sequence.  (2 marks)

(b)  the greatest value of $n$ such that the sum of the first $n$ term of the sequence is positive.  (3 marks)

Solution

(a)  $t_{38} =t_1 + 37d$

$\therefore d=\dfrac{t_{38}-t_1}{37} = \dfrac{555-666}{37} = \dfrac{-111}{37}=-3$

(b)  $S_n = \dfrac{(2a+(n-1)d)n}{2} > 0$

$\displaystyle \frac{(2(666)+(n-1)(-3))n}{2}>0$

Since $n$ is positive integer, we can divide both side by $n$ without switching the sign.

$2(666)-3(n-1)>0$

$n-1 < 2(222)$

$n < 445$

Therefore, the greatest such $n$ is 444.

(18)  Let $f(x) = \dfrac{-1}{3}x^2 + 12x - 121$.

(a)  Using the method of completing the square, find the coordinates of the vertex of the graph of $y = f(x)$.  (2 marks)

(b)  The graph of $y=g(x)$ is obtained by translating the graph of $y=f(x)$ vertically. If the graph of $y=g(x)$ touches the x-axis, find $g(x)$.  (2 marks)

(c)  Under a transformation, $f(x)$ is changed to $latex \dfrac{-1}{3}x^2 – 12x – 121$. Describe the geometric meaning of the transformation.  (2 marks)

Solution

(a)  $\displaystyle f(x) = \frac{-1}{3}x^2 + 12x - 121$

$= -\frac13 (x^2-36x+18^2-18^2) - 121$

$= -\frac13(x-18)^2+\frac{18^2}{3}-121$

$= -\frac13(x-18)^2 -13$

Therefore, the vertex is $(18, -13)$.

(b)  Basically translate the graph of $f(x)$ 13 units upward.

$\therefore g(x) = -\frac13(x-18)^2$

(c)  Since $\displaystyle\frac{-1}{3}x^2-12x-121=\frac{-1}{3}(-x)^2+12(-x)-121 = f(-x)$

Therefore, the transformation is reflecting the graph of f(x) horizontally along the y-axis.

(19)  Figure 2 shows a geometric model $ABCD$ in the form of tetrahedron. It is given that $\angle BAD=86^\circ, \angle CBD=43^\circ, AB=10\text{ cm}, AC=6\text{ cm}, BC=8\text{ cm} \text{ and } BD=15\text{ cm}$.

(a)  Find $\angle ABD$ and $CD$.  (4 marks)

(b)  A craftsman claims that the angle between $AB$ and the face $BCD$ is $\angle ABC$. Do you agree? Explain your answer.  (2 marks)

Solution

(a)  By Sine Law, $\displaystyle \frac{\sin\angle ADB}{AB}=\frac{\sin\angle BAD}{BD}$

$\displaystyle \frac{\sin\angle ADB}{10}=\frac{\sin 86^\circ}{15}$

$\therefore \angle ADB = \sin^{-1} \dfrac{10\sin 86^\circ}{15}$

$\therefore \angle ABD = 180^\circ - 86^\circ - \sin^{-1} \dfrac{2\sin 86^\circ}{3} = 52.31439868 \approx 52.3^\circ \text{ (3 s.f.)}$

By Cosine Law, $CD^2=AC^2+AD^2-2AC\cdot AD \cos\angle CBD$.

$\therefore CD = \sqrt{8^2+15^2-2\cdot8\cdot15\cdot\cos 43^\circ} = 10.65246974 \approx 10.7 \text{ cm (3 s.f.)}$

(b)  By Sine law, $\displaystyle \frac{AD}{\sin\angle ABD}=\frac{BD}{\sin\angle BAD}$

$\therefore AD=\dfrac{15\sin\angle ABD}{\sin 86^\circ} \approx 11.8996447\ldots$

$AC^2+CD^2=6^2+10.65246974^2=149.4751116\neq AD^2=141.6015440$

Therefore, do not agree.

(20)  $\triangle OPQ$ is an obtuse-angled triangle. Denote the in-centre and the circumcentre of $\triangle OPQ$ by $I$ and $J$ respectively. It is given that $P$, $I$, and $J$ are collinear.

(a)  Prove that $OP=PQ$.  (3 marks)

(b)  A rectangular coordinate system is introduced so that the coordinates of $O$ and $Q$ are $(0,0)$ and $(40,30)$ respectively while the y-coordinate of $P$ is 19. Let $C$ be the circle which passes through $O$, $P$, and $Q$.

(i)  Find the equation of $C$.

(ii)  Let $L_1$ and $L_2$ be two tangents to $C$ such that the slope of each tangent is $\dfrac34$ and the y-intercept of $L_1$ is greater than that of $L_2$. $L_1$ cuts the x-axis and y-axis at $S$ and $T$ respectively while $L_2$ cuts the x-axis and the y-axis at $U$ and $V$ respectively. Someone claims that the area of the trapezium $STUV$ exceeds 17000. Is the claim correct? Explain your answer.  (9 marks)

Solution

(a)  Since $P$, $I$, and $J$ are collinear, we can draw a line passing through all three points and meet the line segment $OQ$ at $K$. Since $I$ is incenter, $PI$ is angle bisector of $\angle OPQ$, hence $\angle OPK = \angle QPK$. Since $J$ is circumcenter, then $PK$ is perpendicular bisector of $OQ$ and so $\angle PKO = \angle PKQ = 90^\circ$. $PK = PK$ obviously by common sides. Hence, $\triangle PKO \cong \triangle PKQ$ by ASA. Therefore the corresponding sides of the two congruent triangles are congruent and so $OP=PQ$.

(b) (i)  Let $P=(x, 19)$.  Since $OP = PQ$,

$\sqrt{x^2+19^2} = \sqrt{(x-40)^2+(19-30)^2}$

$x^2+361=x^2-80x+1600+121$

$80x=1360$

$x=17$

Let circle $C: x^2+y^2+Dx+Ey+F=0$

Substitute $O(0,0)$ into circle C yields $F=0$. Hence,

$P(17, 19): 17^2+19^2+17D+19E=0 \Rightarrow 650+17D+19E \qquad (1)$

$Q(40, 30): 40^2+30^2+40D+30E=0 \Rightarrow 250+4D+3E=0 \qquad (2)$

3(1)-19(2):

$1950+51D-4750-76D=0$

$\therefore D=-112$

$\therefore E=66$

Therefore, the equation of circle C is $x^2+y^2-112x+66y=0$

(b) (ii)  Since $L_1$ and $L_2$ have slope of $\dfrac34$, They both have the equation of the form: $y=\dfrac34x+b$

They are both tangents to the circle hence:

$x^2+y^2-112x+66y =0 \quad\text{and}\quad y=\dfrac34x+b$

$x^2+\left(\frac34x+b\right)^2-112x+66\left(\frac34x+b\right)=0$

$x^2+\frac{9}{16}x^2+\frac{3b}{2}x+b^2-112x+\frac{99}{2}x+66b=0$

$16x^2+9x^2+24bx+16b^2-1792x+792x+1056b=0$

$25x^2+(24b-1000)x+16b^2+1056b=0$

Since they are tangents to the circle,

$\Delta = (24b-1000)^2-4(25)(16b^2+1056b)=0$

$576b^2-48000b+1000000-1600b^2-105600b=0$

$1024b^2+153600b-1000000=0$

$16b^2+2400b-15625=0$

$(4b+625)(4b-25)=0$

$b=-\dfrac{625}{4} \text{ or } b=\dfrac{25}{4}$

$\displaystyle\therefore \quad L_1: y=\frac34x+\frac{25}{4} \quad\text{and}\quad L_2: y =\frac34x-\frac{625}{4}$

$L_1:$ Let $y=0 \Rightarrow x=-\dfrac{25}{3} \Rightarrow S\left(-\dfrac{25}{3},0\right) \Rightarrow T\left(0,\dfrac{25}{4}\right)$

$L_2:$ Let $y=0 \Rightarrow x=\dfrac{625}{3} \Rightarrow U\left(\dfrac{625}{3},0\right) \Rightarrow V\left(0,-\dfrac{625}{4}\right)$

$\displaystyle \therefore ST=\sqrt{\left(\frac{25}{3}\right)^2+\left(\frac{25}{4}\right)^2}=25\sqrt{\frac{1}{9}+\frac{1}{16}}=\frac{25}{12}$

$\displaystyle \therefore UV=\sqrt{\left(\frac{625}{3}\right)^2+\left(\frac{625}{4}\right)^2}=625\sqrt{\frac{1}{9}+\frac{1}{16}}=\frac{3125}{12}$

$C: x^2+y^2-112x+66y=0$

$\Rightarrow x^2-112x+56^2+y^2+66y+33^2=56^2+33^2=4225$

$\Rightarrow (x-56)^2+(y+33)^2=65^2 \Rightarrow r=65 \Rightarrow d=130$

$\displaystyle\therefore \text{Area} = \frac{(ST+UV)d}{2} = \left(\frac{125}{12}+\frac{3125}{12}\right)65 =17604.1\overline{6}>17000$

Therefore, the claim is correct.