# Math Blog

## Challenging Geometry E Maths Question — St Andrew’s Sec 3 Maths Tuition Question

Question:

ABCD is a rectangle. M and N are points on AB and DC respectively. MC and BN meet at X. M is the midpoint of AB.

(a) Prove that $\Delta CXN$ and $\Delta MXB$ are similar.

(b) Given that area of $\triangle CXN$: area of $\triangle MXB$=9:4, find the ratio of,

(i) DN: NC

(ii) area of rectangle ABCD: area of $\triangle XBC$. (Challenging)

(ii) 20:3

Suggested Solutions:

(a)
$\angle MXB=\angle NXC$ (vert. opp. angles)

$\angle MBX = \angle XNC$ (alt. angles)

$\angle BMX = \angle XCN$ (alt. angles)

Therefore, $\Delta CXN$ and $\Delta MXB$ are similar (AAA).

(b) (i) $\displaystyle\frac{NC}{BM}=\sqrt{\frac{9}{4}}=\frac{3}{2}$

Let $BM=2u$ and $NC=3u$

Then $DC=2\times 2u=4u$

So $DN=4u-3u=u$

Thus, $DN:NC=1u:3u=1:3$

(ii)

We now have a shorter solution, thanks to a visitor to our site! (see comments below)

From part (a), since $\Delta CXN$ and $\Delta MXB$ are similar, we have $MX:XC=2:3$

This means  that $MC:XC=5:3$

Thus $\triangle MBC:\triangle XBC=5:3$ (the two triangles share a common height)

Now, note that $\displaystyle\frac{\text{area of }ABCD}{\triangle MBC}=\frac{BC\times AB}{0.5 \times BC \times MB}=\frac{AB}{0.5MB}=\frac{2MB}{0.5MB}=4$

Hence area of $ABCD=4\times\triangle MBC$

We conclude that area of rectangle ABCD: area of $\triangle XBC=4(5):3=20:3$

Here is a longer solution, for those who are interested:

Let area of $\triangle XBC =S$

Let area of $\triangle MXB=4u$

Let area of $\triangle CXN=9u$

We have $\displaystyle\frac{S+9u}{S+4u}=\frac{3}{2}$ since $\triangle NCB$ and $\triangle CMB$ have the same base BC and their heights have ratio 3:2.

Cross-multiplying, we get $2S+18u=3S+12u$

So $\boxed{S=6u}$

$\displaystyle\frac{\triangle BCN}{\triangle BDC}=\frac{3}{4}$ since $\triangle BCN$ and $\triangle BDC$ have the same base BC and their heights have ratio 3:4.

Hence,

$\begin{array}{rcl} \triangle BDC &=& \frac{4}{3} \triangle BCN\\ &=& \frac{4}{3} (9u+6u)\\ &=& 20u \end{array}$

Thus, area of $ABCD=2 \triangle BDC=40u$

area of rectangle ABCD: area of $\triangle XBC$=40:6=20:3

## O Level Logarithm Question (Challenging)

Question:

Given $\displaystyle\log_9{a} = \log_{12}{b} =\log_{16}{(a+b)}$, find the value of  $\displaystyle\frac{a}{b}$.

Solution:

Working with logarithm is tricky, we try to transform the question to an exponential question.

Let $\displaystyle y=\log_9{a} = \log_{12}{b} =\log_{16}{(a+b)}$

Then, we have $a=9^y=3^{2y}$

$b=12^y=3^y\cdot 2^{2y}$

$a+b=16^y=2^{4y}$.

Here comes the critical observation:

Observe that $\boxed{a(a+b)=b^2}$.

Divide throughout by $b^2$, we get $\displaystyle (\frac{a}{b})^2+\frac{a}{b}=1$.

Hence, $\displaystyle (\frac{a}{b})^2+\frac{a}{b}-1=0$.

Solving using quadratic formula (and reject the negative value since $a$ and $b$ has to be positive for their logarithm to exist),

We get $\displaystyle\frac{a}{b}=\frac{-1+\sqrt{5}}{2}$.

If you have any questions, please feel free to ask me by posting a comment, or emailing me.

(I will usually explain in much more detail if I teach in person, than when I type the solution)

## Hwa Chong IP Sec 2 Maths Question – Equation of Parabola

Question:

Given that a parabola intersects the x-axis at x=-4 and x=2, and intersects the y-axis at y=-16, find the equation of the parabola.

Solution:

Sketch of graph:

Now, there is a fast and slow method to this question. The slower method is to let $y=ax^2+bx+c$, and solve 3 simultaneous equations.

The faster method is to let $y=k(x+4)(x-2)$.

Why? We know that x=4 is a root of the polynomial, so it has a factor of (x-4). Similarly, the polynomial has a factor of (x-2). The constant k (to be determined) is added to scale the graph, so that the graph will satisfy y=-16 when x=0.

So, we just substitute in y=-16, x=0 into our new equation.

$-16=k(4)(-2)$.

$-16=-8k$.

So $k=2$.

In conclusion, the equation of the parabola is $y=2(x+4)(x-2)$.

## H2 Maths Tuition: Foot of Perpendicular (from point to plane) (Part II)

This is a continuation from H2 Maths Tuition: Foot of Perpendicular (from point to line) (Part I).

## Foot of Perpendicular (from point to plane)

From point (B) to Plane ( $p$)

## Equation (I):

Where does F lie?

F lies on the plane  $p$.

$\overrightarrow{\mathit{OF}}\cdot \mathbf{n}=d$

## Equation (II):

Perpendicular

$\overrightarrow{\mathit{BF}}=k\mathbf{n}$

$\overrightarrow{\mathit{OF}}-\overrightarrow{\mathit{OB}}=k\mathbf{n}$

$\overrightarrow{\mathit{OF}}=k\mathbf{n}+\overrightarrow{OB}$

## Final Step

Substitute Equation (II) into Equation (I) and solve for k.

## Example

[VJC 2010 P1Q8i]

The planes $\Pi _{1}$ and $\Pi _{2}$ have equations $\mathbf{r\cdot(i+j-k)}=6$ and $\mathbf{r\cdot(2i-4j+k)}=-12$ respectively. The point $A$  has position vector  $\mathbf{{9i-7j+5k}}$ .

(i) Find the position vector of the foot of perpendicular from  $A$ to $\Pi _{2}$ .

## Solution

Let the foot of perpendicular be F.

### Equation (I)

$\overrightarrow{\mathit{OF}}\cdot \left(\begin{matrix}2\\-4\\1\end{matrix}\right)=-12$

### Equation (II)

$\overrightarrow{\mathit{OF}}=k\left(\begin{matrix}2\\-4\\1\end{matrix}\right)+\left(\begin{matrix}9\\-7\\5\end{matrix}\right)=\left(\begin{matrix}2k+9\\-4k-7\\k+5\end{matrix}\right)$

Subst. (II) into (I)

$2(2k+9)-4(-4k-7)+(k+5)=-12$

Solve for k,  $k=-3$ .

$\overrightarrow{\mathit{OF}}=\left(\begin{matrix}3\\5\\2\end{matrix}\right)$

## H2 Maths Tuition

If you are looking for Maths Tuition, contact Mr Wu at:

Email: mathtuition88@gmail.com

## Foot of Perpendicular is a hot topic for H2 Prelims and A Levels. It comes out almost every year.

There are two versions of Foot of Perpendicular, from point to line, and from point to plane. However, the two are highly similar, and the following article will teach how to understand and remember them.

## H2: Vectors (Foot of perpendicular)

From point (B) to Line ( $l$)

(Picture)

### Equation (I):

Where does F lie? F lies on the line  $l$.

$\overrightarrow{\mathit{OF}}=\mathbf{a}+\lambda \mathbf{m}$

### Equation (II):

Perpendicular:

$\overrightarrow{\mathit{BF}}\cdot \mathbf{m}=0$

$(\overrightarrow{\mathit{OF}}-\overrightarrow{\mathit{OB}})\cdot \mathbf{m}=0$

### Final Step

Substitute Equation (I) into Equation (II) and solve for  $\lambda$.

## Example:

[CJC 2010 P1Q7iii]

Relative to the origin $O$ , the points $A$ , $B$ and $C$  have position vectors  $\left(\begin{matrix}1\\2\\1\end{matrix}\right)$ , $\left(\begin{matrix}2\\1\\3\end{matrix}\right)$ and $\left(\begin{matrix}-1\\2\\3\end{matrix}\right)$ Find the shortest distance from  $C$ to $\mathit{AB}$ . Hence or otherwise, find the area of triangle $\mathit{ABC}$ .

[Note: There is a 2nd method to this question. (cross product method)]

## Solution:

Let the foot of perpendicular from C to AB be F.

Equation (I):

$\overrightarrow{\mathit{OF}}=\overrightarrow{\mathit{OA}}+\lambda \overrightarrow{\mathit{AB}}=\left(\begin{matrix}1+\lambda \\2-\lambda \\1+2\lambda \end{matrix}\right)$

Equation (II):

$(\overrightarrow{\mathit{OF}}-\overrightarrow{\mathit{OC}})\cdot \overrightarrow{\mathit{AB}}=0$

$\left(\begin{matrix}2+\lambda \\-\lambda \\-2+2\lambda \end{matrix}\right)\cdot \left(\begin{matrix}1\\-1\\2\end{matrix}\right)=0$

$\lambda =\frac{1}{3}$

$\overrightarrow{\mathit{CF}}=\overrightarrow{\mathit{OF}}-\overrightarrow{\mathit{OC}}=\left(\begin{matrix}2\frac{1}{3}\\-{\frac{1}{3}}\\-1\frac{1}{3}\end{matrix}\right)$

$\left|{\overrightarrow{{\mathit{CF}}}}\right|=\sqrt{\frac{22}{3}}$

Area of  $\Delta \mathit{ABC}=\frac{1}{2}\left|{\overrightarrow{\mathit{AB}}}\right|\left|{\overrightarrow{\mathit{CF}}}\right|=\sqrt{11}$

For the next part, please read our article on Foot of Perpendicular (from point to plane).

## H2 Maths Tuition

If you are looking for Maths Tuition, contact Mr Wu at:

Email: mathtuition88@gmail.com

## Sec 2 IP (HCI) Revision 1: Expansion and Factorisation

This is a nice worksheet on Expansion and Factorisation by Hwa Chong Institution (HCI).

There are no solutions, but if you have any questions you are welcome to ask me, by leaving a comment, or by email.

Hope you enjoy practising Expansion and Factorisation.

Expansion and Factorisation by Hwa Chong Institution (HCI)

## Is d/dx (a^x)=x a^{x-1}? (a conceptual error in O/A Level Math)

In O Level, students are taught that $\boxed{\frac{d}{dx}(x^{n})=nx^{n-1}}$

So naturally, students may think that $\displaystyle\frac{d}{dx}(a^{x})=xa^{x-1}??$ (a is a constant)

Well, actually that is good pattern spotting, but unfortunately it is incorrect. Do not be too disheartened if you make this mistake, it is a common mistake.

The above is a conceptual error as  $\boxed{\frac{d}{dx}(x^{n})=nx^{n-1}}$ only holds when n is a constant.

Fortunately, this question is rarely tested, though it is quite possible that it can come up in A Levels.

To fully understand the following steps, it would help read my other post (Why is e^(ln x)=x?) first.

First, we write $\displaystyle a^x=e^{\ln a^x}=e^{x\ln a}$.

Hence
$\displaystyle \begin{array}{rcl} \frac{d}{dx} (a^x)&=&\frac{d}{dx} (e^{x\ln a})\\ &=&e^{x\ln a}(\ln a)\\ &=&e^{\ln a^x}(\ln a)\\ &=&a^x(\ln a) \end{array}$

After fully understanding the above steps, you may memorize the formula if you wish:

$\boxed{\frac{d}{dx} (a^x)=a^x(\ln a)}$

Memory Tip: If you let a=e, you should get $\boxed{\frac{d}{dx} (e^x)=e^x(\ln e)=e^x}$

The above steps involve the chain rule, which I will cover in a subsequent post.

## How to avoid Careless Mistakes for Maths?

Many parents have feedback to me that their child often makes careless mistakes in Maths, at all levels, from Primary, Secondary, to JC Level. I truly empathize with them, as it often leads to marks being lost unnecessarily. Not to mention, it is discouraging for the child.

Also, making careless mistakes is most common in the subject of mathematics, it is rare to hear of students making careless mistakes in say, History or English.
Fortunately, it is possible to prevent careless mistakes for mathematics, or at least reduce the rates of careless mistakes.

From experience, the ways to prevent careless mistakes for mathematics can be classified into 3 categories, Common Sense, Psychological, and Math Tips.

Common Sense

1. Firstly, write as neatly as possible. Often, students write their “5” like “6”. Mathematics in Singapore is highly computational in nature, such errors may lead to loss of marks. Also, for Algebra, it is recommended that students write l (for length) in a cursive manner, like $\ell$ to prevent confusion with 1. Also, in Complex Numbers in H2 Math, write z with a line in the middle, like Ƶ, to avoid confusion with 2.
2. Leave some time for checking. This is easier said than done, as speed requires practice. But leaving some time, at least 5-10 minutes to check the entire paper is a good strategy. It can spot obvious errors, like leaving out an entire question.

Psychological

1. Look at the number of marks. If the question is 5 marks, and your solution is very short, something may be wrong. Also if the question is just 1 mark, and it took a long time to solve it, that may ring a bell.
2. See if the final answer is a “nice number“. For questions that are about whole numbers, like number of apples, the answer should clearly be a whole number. At higher levels, especially with questions that require answers in 3 significant figures, the number may not be so nice though. However, from experience, some questions even in A Levels, like vectors where one is suppose to solve for a constant $\lambda$, it turns out that the constant is a “nice number”.

Mathematical Tips

Mathematical Tips are harder to apply, unlike the above which are straightforward. Usually students will have to be taught and guided by a teacher or tutor.

1. Substitute back the final answer into the equations. For example, when solving simultaneous equations like x+y=3, x+2y=4, after getting the solution x=2, y=1, you should substitute back into the original two equations to check it.
2. Substitute in certain values. For example, after finding the partial fraction $\displaystyle\frac{1}{x^2-1} = \frac{1}{2 (x-1)}-\frac{1}{2 (x+1)}$, you should substitute back a certain value for x, like x=2. Then check if both the left-hand-side and right-hand-side gives the same answer. (LHS=1/3, RHS=1/2-1/6=1/3) This usually gives a very high chance that you are correct.

Thanks for reading this long article! Hope it helps! 🙂

I will add more tips in the future.

## Recommended Maths Book:Math Doesn’t Suck: How to Survive Middle School Math Without Losing Your Mind or Breaking a Nail

This book is a New York Times Bestseller by actress Danica McKellar, who is also an internationally recognized mathematician and advocate for math education. It should be available in the library. Hope it can inspire all to like Maths!

## Why is e^(ln x)=x? (O Level Math/ A Level Math Tuition)

Why is $\boxed{e^{\ln x}=x}$?

This formula will be useful for some questions in O Level Additional Maths, or A Level H2 Maths.

There are two ways to show or prove this, first we can let

$y=e^{\ln x}$

Taking natural logarithm (ln) on both sides, we get

$\ln y=\ln x\ln e=\ln x$

So $y=x$. Substitute the very first equation and we get $e^{\ln x}=x$. 🙂

Alternatively, we can view $e^x$ and $\ln x$ as inverse functions of each other. So, we can let $f(x)=e^x$ and $f^{-1}(x)=\ln x$. Then, $e^{\ln x}=f(f^{-1})(x)=x$ by definition of inverse functions. This may be a better way to remember the result. 🙂

The above method of inverse functions can be used to remember $\ln (e^x)=x$ too.

## The mass of particles of a certain radioactive chemical element (O Level Math Tuition Question)

The mass of particles of a certain radioactive chemical element is halved every 10 months.  During a chemical experiment, the initial mass of particles of the chemical element is 3mg.

(i) write down an expression, in terms of t, for the mass of particles after t years.
(ii) Hence, find the value of t, if the mass is reduced to 0.046875 mg after t years.

Solution:

(i)

$1\: \text{year} = 12\: \text{months}$

Therefore, $t\: \text{years}=12t\: \text{months}$.

How many 10 months are there in $t\: \text{years}$? (Ans: $\frac{12t}{10}=1.2t$)

Hence, the mass of particles after $t$ years is $3\times(0.5)^{1.2t}$ mg.

(ii)
We need to solve $3(0.5)^{1.2t}=0.046875$.

Dividing by 3, we have $(0.5)^{1.2t}=0.015625$.

Ln both sides, we have $1.2t\ln{0.5}=\ln{0.015625}$.

Hence, $t=\frac{\ln{0.015625}}{1.2\ln{0.5}}=5$.

If you liked our solution above, please consider signing up for Maths Tuition with us! 🙂