geometry – Singapore Maths Tuition

Varignon’s Theorem (Surprising Geometry Theorem)

I first learnt it from Quora:

“Take any quadrilateral. It doesn’t have to be any special kind of a quadrilateral. Then connect the midpoints of its sides.

Surprisingly, you will always get a parallelogram!”

Quite a nice result. Some googling revealed that the name of this theorem is called Varignon’s Theorem.

An illustration is found here:

Math Riddle (Proof that all triangles are equilateral)

Can you spot the subtle mistake in this video?

Very interesting and a good exercise in geometry proving! 🙂

Featured book:

Tutor in a Book’s Geometry

Need help with Geometry? Designed to replicate the services of a skilled private tutor, the new and improved Tutor in a Book’s Geometry is at your service! TIB’s Geometry is an extremely thorough, teen tested and effective geometry tutorial.

TIB’s Geometry includes more than 500 of the right, well-illustrated, carefully worked out and explained proofs and problems. Throughout TIB’s Geometry, there is ongoing, specific guidance as to the most effective solution and test taking strategies. Recurring patterns, which provide solutions to proofs, are pointed out, explained and illustrated using the visual aids that students find so helpful. Also included are dozens of graphic organizers, which help students understand, remember and recognize the connections between concepts.

TIB’s author Jo Greig intended this book to level the playing field between the students who have tutors and those that don’t. As a long time, very successful private mathematics tutor and teacher, Jo Greig knew exactly how best to accomplish this! TIB’s Geometry 294 pages are packed with every explanation, drawing, hint and memory tool possible! Not only does it have examples of the right proofs and problems, it also manages to impart every bit of the enthusiasm that great tutors impart to their private tutoring students. Ms. Greig holds a bachelors’ degree in mathematics. Dr. J. Shiletto, the book’s mathematics editor, holds a Ph.D in mathematics.

The Three Square Geometry Problem – Numberphile

Watch this interesting video about the “Three Square Geometry Problem”!

Theoretically, a fifth-grader or P5/PSLE student can solve it! The featured solution is truly brilliant and requires one to “think out of the box”.

Featured book:

Tutor in a Book’s Geometry

Tips on attempting Geometrical Proof questions (E Maths Tuition)

Tips on attempting Geometrical Proof questions (O Levels E Maths/A Maths)

1) Draw extended lines and additional lines. (using pencil)

Drawing extended lines, especially parallel lines, will enable you to see alternate angles much easier (look for the “Z” shape). Also, some of the more challenging questions can only be solved if you draw an extra line.

2) Use pencil to draw lines, not pen

Many students draw lines with pen on the diagram. If there is any error, it will be hard to remove it.

3) Rotate the page.

Sometimes, rotating the page around will give you a fresh impression of the question. This may help you “see” the way to answer the question.

4) Do not assume angles are right angles, or lines are straight, or lines are parallel unless the question says so, or you have proved it.

For a rigorous proof, we are not allowed to assume anything unless the question explicitly says so. Often, exam setters may set a trap regarding this, making the angle look like a right angle when it is not.

5) Look at the marks of the question

If it is a 1 mark question, look for a short way to solve the problem. If the method is too long, you may be on the wrong track.

6) Be familiar with the basic theorems

The basic theorems are your tools to solve the question! Being familiar with them will help you a lot in solving the problems.

Hope it helps! And all the best for your journey in learning Geometry! Hope you have fun.

“There is no royal road to Geometry.” – Euclid

Animation of a geometrical proof of Phytagoras... — Animation of a geometrical proof of Pythagoras theorem (Photo credit: Wikipedia)

O Level 2007 E Maths Paper 2 Q3 Solution (Geometry Question)

(a)
$\angle DAB=\angle DCB$
(opposite angles of parallelogram)

$\displaystyle\angle PAD=\frac{1}{2}\angle DAB=\frac{1}{2}\angle DCB=\angle RCB$
(shown)

(b)
$AD=BC$

$\angle PAD=\angle RCB$
(from part a)

$\angle ADC=\angle ABC$
(opposite angles of parallelogram)

$\displaystyle\angle ADP=\frac{1}{2}\angle ADC=\frac{1}{2} ABC=\angle RBC$

Thus, triangles ADP and CBR are congruent (ASA).

(c)(i)
$\angle ADC+\angle DAB=180^\circ$
$\angle DAP+\angle ADP=90^\circ$

Considering the triangle ADP,
$\angle DPA=180^\circ-90^\circ=90^\circ$
(shown)

(ii)
$\angle DAB+\angle ABC=180^\circ$

$\angle BAQ+\angle ABQ=90^\circ$

Considering the triangle ABQ,
$\angle PQR=180^\circ-90^\circ=90^\circ$
(shown)

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)

[Answer Key] (b) (i) 1:3

(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