I will be going to China for the China Western Math Olympiad (CWMO) 2010, from 26 to 1 November. There are 2 teams, total 8 of us from the National Team. The CWMO 2010 will be held in Taiyuan, Shanxi of China.
As my year end exams are from 29 Oct to 2 Nov, I will not be able to take 3 papers - Chinese, Biology (29 Nov) and English (1 Nov morning). Not sure of the make-up for the missed papers ...
Wednesday, 13 October 2010
Wednesday, 6 October 2010
Math Qn on Red and Blue Marbles
This is posted by LL ☺
Someone posted the following question on the cbox:
Container A contains 250 red marbles and 200 blue marbles. Container B contains 600 red marbles and 150 blue marbles.How many red and blue marbles must be moved from Container A to Container B such that 25% of the marbles in Container A are red and 75% of the marbles in Container B are red?
LL's solution:
before............after
A red 250.........1 unit(u)
..blue 200........3 units(u)
B red 600.........3 parts(p)
..blue 150........1 part(p)
4u+4p=1200 (total of red and blue marbles in containers A and B)
->1u+1p=300
3p+u=850 (total red marbles)
(3p+u)-(1u+1p)=850-300
2p=550, p=275
At first, container B has 150 blue balls.
Now, container B has 1 part=275 which is increased by 275-150=125 blue marbles
At first, container B has 600 red balls.
Now, container B has 3 parts=275x3=825 which is increased by 825-600=225 red marbles
Therefore, 225 red and 125 blue marbles must be moved from Container A to Container B.
Someone posted the following question on the cbox:
Container A contains 250 red marbles and 200 blue marbles. Container B contains 600 red marbles and 150 blue marbles.How many red and blue marbles must be moved from Container A to Container B such that 25% of the marbles in Container A are red and 75% of the marbles in Container B are red?
LL's solution:
before............after
A red 250.........1 unit(u)
..blue 200........3 units(u)
B red 600.........3 parts(p)
..blue 150........1 part(p)
4u+4p=1200 (total of red and blue marbles in containers A and B)
->1u+1p=300
3p+u=850 (total red marbles)
(3p+u)-(1u+1p)=850-300
2p=550, p=275
At first, container B has 150 blue balls.
Now, container B has 1 part=275 which is increased by 275-150=125 blue marbles
At first, container B has 600 red balls.
Now, container B has 3 parts=275x3=825 which is increased by 825-600=225 red marbles
Therefore, 225 red and 125 blue marbles must be moved from Container A to Container B.
Thursday, 30 September 2010
Random Math Question
Find the shaded area.
Let the midpoint of line AD be E. (Centre of semi-circle). Let the arc of AC and AD cut at point F.
AE = EF = 3.5 (radius of circle)
BA = BF = 7 (radius of circle)
EB is common. Then,
Triangle ABE is congruent to Triangle FBE. Hence,
angle ABE = angle FBE = tan-1(AE/AB) = tan-1(1/2) (Note: tan-1 means tangent inverse)
angle ABF = angle ABE + angle FBE = 2 tan-1(1/2) (get this angle in degree)
Area of sector ABF = (2 tan-1 (1/2))/360 x 7 x 7 x pi = (49 pi tan-1(1/2))/180
angle AEB = angle FEB = tan-1(AB/AE) = tan-1(2)
angle AEF = angle AEB + angle FEB = 2 tan-1(2) (get this angle in degree too)
Area of sector AEF = (2 tan-1(2))/360 x 3.5 x 3.5 x pi = (49 pi tan-1(2))/720
Area of quadrilateral ABFE = 3.5x7 = 24.5
Area of shaded region = Area of sector ABF + Area of sector AEF - Area of quadrilateral ABFE
= (49 pi tan-1(1/2))/180 + (49 pi tan-1(2))/720 - 24.5
= 11.8cm2 (round off to 1 decimal point)
(exact pi is used, not 22/7)
Let the midpoint of line AD be E. (Centre of semi-circle). Let the arc of AC and AD cut at point F.
AE = EF = 3.5 (radius of circle)
BA = BF = 7 (radius of circle)
EB is common. Then,
Triangle ABE is congruent to Triangle FBE. Hence,
angle ABE = angle FBE = tan-1(AE/AB) = tan-1(1/2) (Note: tan-1 means tangent inverse)
angle ABF = angle ABE + angle FBE = 2 tan-1(1/2) (get this angle in degree)
Area of sector ABF = (2 tan-1 (1/2))/360 x 7 x 7 x pi = (49 pi tan-1(1/2))/180
angle AEB = angle FEB = tan-1(AB/AE) = tan-1(2)
angle AEF = angle AEB + angle FEB = 2 tan-1(2) (get this angle in degree too)
Area of sector AEF = (2 tan-1(2))/360 x 3.5 x 3.5 x pi = (49 pi tan-1(2))/720
Area of quadrilateral ABFE = 3.5x7 = 24.5
Area of shaded region = Area of sector ABF + Area of sector AEF - Area of quadrilateral ABFE
= (49 pi tan-1(1/2))/180 + (49 pi tan-1(2))/720 - 24.5
= 11.8cm2 (round off to 1 decimal point)
(exact pi is used, not 22/7)
Friday, 24 September 2010
Games on iPad
This is posted by Lim Li.
In Angry Birds, you are supposed to kill the green thingy by shooting the birds using a catapult. I do not play it often as I am stuck on level 3-12.
In Plants VS Zombies, you are supposed to kill the zombies by planting plants to attack the zombies and to prevent them from eating your brains. Sometimes, I play this game with my cousin, Liu Qin, who just completed the game like me and my sister and brother (we completed the game long ago).
My mother keeps saying that this is our favourite game because she always sees us playing it. However, we play it everyday because we need to get the gems (one gem can be obtained per day from the word game) to buy stuff which help you play the game better. My high score is 278686, but the highest in the world is 30 something millions!
In Angry Birds, you are supposed to kill the green thingy by shooting the birds using a catapult. I do not play it often as I am stuck on level 3-12.
In Plants VS Zombies, you are supposed to kill the zombies by planting plants to attack the zombies and to prevent them from eating your brains. Sometimes, I play this game with my cousin, Liu Qin, who just completed the game like me and my sister and brother (we completed the game long ago).
My mother keeps saying that this is our favourite game because she always sees us playing it. However, we play it everyday because we need to get the gems (one gem can be obtained per day from the word game) to buy stuff which help you play the game better. My high score is 278686, but the highest in the world is 30 something millions!Thursday, 16 September 2010
A Great Educational Toy
This is posted by LJ's mum.
My kids practically grow up with Tomy Megasketcher, a magnetic drawing board. It is their favourite toy. It has a unique screen that allows one to draw clear and crisp pictures, unlike the rough pictures drawn on honeycomb type magnetic board made by Fisher. I bought the first one when Lim Min was 4 or 5 years old and to-date I must have bought over 10 of them (about 1 every year) as the impression left on the drawing surface become quite faint after some months of heavy usage.
When my kids were young (before schooling age), they loved to to draw random pictures and complicated mazes (Lim Jeck's favourite), and now they use it to practise spelling/ tingxie, solve maths problems or just doodle and scribble.
It is a great learning companion for them all these years!
Lim Jeck solving a question that someone posted on Mathlinks.
Lim Li creating different ways of writing the word 'BAD'.
My kids practically grow up with Tomy Megasketcher, a magnetic drawing board. It is their favourite toy. It has a unique screen that allows one to draw clear and crisp pictures, unlike the rough pictures drawn on honeycomb type magnetic board made by Fisher. I bought the first one when Lim Min was 4 or 5 years old and to-date I must have bought over 10 of them (about 1 every year) as the impression left on the drawing surface become quite faint after some months of heavy usage.When my kids were young (before schooling age), they loved to to draw random pictures and complicated mazes (Lim Jeck's favourite), and now they use it to practise spelling/ tingxie, solve maths problems or just doodle and scribble.
It is a great learning companion for them all these years!
Lim Jeck solving a question that someone posted on Mathlinks.
Lim Li creating different ways of writing the word 'BAD'.
Sunday, 12 September 2010
Math Qn
Someone posted the following question on the cbox:
Let AG be x cm.
Then GB = 12-x
Triangle AGF is similar to Triangle ABC, hence AG = GF = x
Tiangle BGF is similar to Triangle BAE, hence
BG/GF = BA/AE
=> (12-x)/x = 12/6 (since AB = 12, AE = 6)
=> 12-x = 2x
=> x=4
Area of Triangle DEF = 1/2 x ED x AG = 1/2 x 6 x 4 = 12
Area of Triangle DFC = 1/2 x DC x (BC - GF) = 1/2 x 12 x (12-4) = 48
Hence, Area of shaded region = 12 + 48 = 60 cm2.
Let AG be x cm.Then GB = 12-x
Triangle AGF is similar to Triangle ABC, hence AG = GF = x
Tiangle BGF is similar to Triangle BAE, hence
BG/GF = BA/AE
=> (12-x)/x = 12/6 (since AB = 12, AE = 6)
=> 12-x = 2x
=> x=4
Area of Triangle DEF = 1/2 x ED x AG = 1/2 x 6 x 4 = 12
Area of Triangle DFC = 1/2 x DC x (BC - GF) = 1/2 x 12 x (12-4) = 48
Hence, Area of shaded region = 12 + 48 = 60 cm2.
Sunday, 5 September 2010
911
SMO (Senior) - Individual 8th 120
SMO (Senior) - Team 3rd 88 (I think, 350/4)
SMO (Open) - Individual 8th 120
SMO (Open) - Team 4th 83 (250/3)
IMO Silver - 500
Total - 911 :)
SMO (Senior) - Team 3rd 88 (I think, 350/4)
SMO (Open) - Individual 8th 120
SMO (Open) - Team 4th 83 (250/3)
IMO Silver - 500
Total - 911 :)
Sunday, 29 August 2010
Yay!
I went for the Hwa Chong Math Quest 2010 yesterday. We (my mother, Lim Li and I) took 3 buses (186, 153, 174) from my house and alighted at HCI's 1st bus stop (there are 3 bus stops that lead to 3 different HCI entrance gates). There was some construction going on and the first gate was closed. So we walked to the 2nd bus stop. Still no open gate in sight. We then took a bus (luckily it's distance-based fare now, so it was free or cost only a few cents) to the 3rd bus stop, and entered HCI through the 3rd gate. The whole HCI was like a ghost town (it's Saturday at 8 am) and there was not a single soul to guide us - no signs on the Math Quest and no Hwa Chong students around to provide direction. It took us quite long to find LT3 (competition venue at 2nd level) as the direction signs to the LTs are not very clear. Fortunately I was not late. My mother and Lim Li then left for the RGS Speech Day and Prize Presentation Ceremony, as Lim Min has won an award.Yay, NUS High has won the Individual 1st and Team 1st in the Hwa Chong Math Quest. I got a ridiculously huge trophy, a TI Graphic Calculator (my 3rd one, I had 2 from the NJC Math Challenge 2010), a 2 GB TI Thumb Drive and $50 Popular Book Vouchers.
Monday, 23 August 2010
More Photos
These photos were taken by our teachers, Mrs Ng and Ms Lim.
Group photo with our buddies and teachers (Mrs Ng and Ms Lim).
Group photo with our buddies and teachers (Mrs Ng and Ms Lim).Monday, 16 August 2010
CGMO Photos
CGMO Opening Ceremony.
Our quad-sharing hostel room in Shijiazhuang No. 2 Middle High School. With the exception of Zhang Aidi who climbed up the upper deck to sleep, the rest of us brought the mattresses down and slept on the floor.
Uniform provided by CGMO organiser, which we had to wear for opening and closing ceremonies.
This building is the library of Shijiazhuang No. 2 Middle High School.Friday, 13 August 2010
CGMO Results Out!
This is posted by LJ's mum.
This year's CGMO officially closes today. About 200 students participated, of which 20 are awarded Gold medals, 40 Silver medals and 60 Bronze medals.
The two Singapore teams comprising 8 girls won a total of 5 medals, as follows:
Silver - Aidi
Bronze - Catharine, Lim Min, Jazlene, Pin Lin
Congrats to the girls! It is a marked improvement over last year's haul of 1 Bronze medal.
This year's CGMO officially closes today. About 200 students participated, of which 20 are awarded Gold medals, 40 Silver medals and 60 Bronze medals.
The two Singapore teams comprising 8 girls won a total of 5 medals, as follows:
Silver - Aidi
Bronze - Catharine, Lim Min, Jazlene, Pin Lin
Congrats to the girls! It is a marked improvement over last year's haul of 1 Bronze medal.
Sunday, 8 August 2010
CGMO 2010
Lim Min has flown to Shanghai early this morning (1.15 am) on SQ826, and she will be taking another flight at 12.50 pm to Shijiazhuang Daguocun. She is part of the RI/RGS team (comprising Catharine, Xin Xuan, Aidi and Lim Min) which is taking part in the China Girls Maths Olympiad held at Shijiazhuang, the capital of Hebei Province, about 40 miles southwest of Beijing. There are 2 teams from Singapore; NUS High is sending the other team of 4 girls.
The programme/itinerary is as follows:
8 Aug (Sun) - Arrival, check-in to Shijiazhuang No. 2 Middle School Student Dormitory
9 Aug (Mon) - Opening ceremony
10 Aug (Tue) - Competition Day 1
11 Aug (Wed) - Competition Day 2
12 Aug (Thu) - Excursion / Closing dinner
13 Aug (Fri) - Closing ceremony / departure to Singapore (arrival at Changi Airport at 12.25 am, 14 Aug)
Wishing both Singapore teams all the best!
The programme/itinerary is as follows:
8 Aug (Sun) - Arrival, check-in to Shijiazhuang No. 2 Middle School Student Dormitory
9 Aug (Mon) - Opening ceremony
10 Aug (Tue) - Competition Day 1
11 Aug (Wed) - Competition Day 2
12 Aug (Thu) - Excursion / Closing dinner
13 Aug (Fri) - Closing ceremony / departure to Singapore (arrival at Changi Airport at 12.25 am, 14 Aug)
Wishing both Singapore teams all the best!
Monday, 2 August 2010
NMOS 2010 Special Round Qn 5
Question 5: A company could use any of the 3 builder (Anthony, Bob and Charles) to build a wall. Anthony would take A days to build the wall. Likewise, Bob and Charles would take B days and C days to build the wall respectively.
Anthony and Bob take 16.8 days to complete a wall.
Anthony and Charles take 21 days to complete a wall.
Bob and Charles take 28 days to complete a wall.
Solution:
Anthony builds 1 wall in A days. Hence, Anthony builds 1/A wall in 1 day.
Similarly, Bob builds 1/B wall in 1 day, and Charles builds 1/C wall in 1 day.
In 1 day, Anthony and Bob together will build (1/A + 1/B) wall. In 16.8 days, Anthony and Bob will build [16.8x(1/A + 1/B)] wall which is 1 wall since Anthony and Bob complete the wall in 16.8 days. Hence, we get the following equations:
16.8(1/A + 1/B) = 1
21(1/A + 1/C) = 1
28(1/B + 1/C) = 1
Implying..
1/A + 1/B = 1/16.8 -------- (1)
1/A + 1/C = 1/21 ---------- (2)
1/B + 1/C = 1/28 ---------- (3)
(1)+(2)-(3)
2(1/A) = 1/16.8 + 1/21 - 1/28 = 1/14
Therefore, 1/A = 1/28, hence, A = 28.
From (1), 1/B = 1/16.8 - 1/28 = 1/42. Hence, B = 42.
From (2), 1/C = 1/21 - 1/28 = 1/84. Hence, C = 84.
Answer: A+B+C=154
Anthony and Bob take 16.8 days to complete a wall.
Anthony and Charles take 21 days to complete a wall.
Bob and Charles take 28 days to complete a wall.
Solution:
Anthony builds 1 wall in A days. Hence, Anthony builds 1/A wall in 1 day.
Similarly, Bob builds 1/B wall in 1 day, and Charles builds 1/C wall in 1 day.
In 1 day, Anthony and Bob together will build (1/A + 1/B) wall. In 16.8 days, Anthony and Bob will build [16.8x(1/A + 1/B)] wall which is 1 wall since Anthony and Bob complete the wall in 16.8 days. Hence, we get the following equations:
16.8(1/A + 1/B) = 1
21(1/A + 1/C) = 1
28(1/B + 1/C) = 1
Implying..
1/A + 1/B = 1/16.8 -------- (1)
1/A + 1/C = 1/21 ---------- (2)
1/B + 1/C = 1/28 ---------- (3)
(1)+(2)-(3)
2(1/A) = 1/16.8 + 1/21 - 1/28 = 1/14
Therefore, 1/A = 1/28, hence, A = 28.
From (1), 1/B = 1/16.8 - 1/28 = 1/42. Hence, B = 42.
From (2), 1/C = 1/21 - 1/28 = 1/84. Hence, C = 84.
Answer: A+B+C=154
Sunday, 1 August 2010
NMOS 2010 Special Round Answers
1. 20
2. 2
3. 15
4. 216
5. 154
6. 54
7. 25
8. 12
9. 38
10. 75
11. 542
12. 48
13. 108
14. 12
15. 65
16. 24
17. 40
18. 25
19. 4
20. 62
Note: many answers were contributed by Lim Min and Lim Jeck.
Lim Li - if you have different answers and you are sure that you are correct, please let me know.
2. 2
3. 15
4. 216
5. 154
6. 54
7. 25
8. 12
9. 38
10. 75
11. 542
12. 48
13. 108
14. 12
15. 65
16. 24
17. 40
18. 25
19. 4
20. 62
Note: many answers were contributed by Lim Min and Lim Jeck.
Lim Li - if you have different answers and you are sure that you are correct, please let me know.
Friday, 23 July 2010
More Photos
Friday, 16 July 2010
Photos and Random Comments on 51st IMO
Below are some photos taken (a few are from Mr Lu's and Gar Goei's facebook).
Group departing from Terminal 3 on 4 July. Participants are in the back row, observers and deputy leader in front row.
Transit at Beijing. Group photo at Birdnest stadium.
Hotel in Astana.
The Baiterek monument and the United Buddy Bears at Astana city.
Ivan and Gabriel with the China Buddy Bear.
Singapore Team participants with observers and Simon (bear).
Group departing from Terminal 3 on 4 July. Participants are in the back row, observers and deputy leader in front row.
Transit at Beijing. Group photo at Birdnest stadium.
Hotel in Astana.
The Baiterek monument and the United Buddy Bears at Astana city.
Ivan and Gabriel with the China Buddy Bear.
Singapore Team participants with observers and Simon (bear).
- organisation of IMO failed, a lot of delays, from hotel in Astana we took 6-hr coach ride (!) to the competition venue
- China is the only country that wins 6 Golds; Nie Zipei from China is the only participant with a perfect score of 42
- China is the only country that wins 6 Golds; Nie Zipei from China is the only participant with a perfect score of 42
- Belgium beats China in Qn 1 by scoring 42 (i.e. all 6 participants got complete solution for Qn 1) as China scores 'only' 41 in Qn 1, however all 6 Belgium contestants get Honourable Mention (all because of Qn 1, contestants who do not qualify for a medal but score full mark in at least 1 qn will get Honourable Mention)
- Singapore also beats China, Russia, USA and Japan in Qn 1 :)
- Highest team score for Qn 5 is 24. This must be the hardest Qn 5 in the history of IMO.
- North Korea got disqualified. Heard it was because all the contestants use the same lemma (i.e. a proven statement) for Qn 3. You can read more about this incident here in Mathlinks.
- This year's Gold cut-off point of 27 is the lowest in the history of IMO
- Initial cut-off points for Silver and Bronze were 22 and 16 respectively, but Jury voted for 21 and 15. If Bronze cut-off is 16, 43% of contestants are awarded medals (Gold, Silver and Bronze) and if it is 15, 51% are awarded medals.
Monday, 12 July 2010
IMO 2010 Results Out!
This is posted by LJ's mum.
The IMO 2010 results were released about half an hour ago. Here is Singapore's results: 4 Silvers, 1 Bronze and 1 Honourable Mention.

Singapore is ranked 22 out of 96 countries (last year, she was ranked 30 out of 104 countries). Top 5 countries are: China, Russia, USA, Korea and Kazakhstan.
Full results are here. This year the cut-off points for Gold, Silver and Bronze medals are 27, 21 and 15 respectively. Looking at the Singapore Team's individual contestant's results for each question, I think we could have won 4 Golds!! How?? Typically Qn 3 and Qn 6 of the IMO are the hardest questions, so it is understandable if none of them could solve the questions or manage to secure even 1 point. However, IF only Ivan could solve Qn 2 (Geometry) completely, and IF only Jie Jun, Lim Jeck and Yan Sheng could solve Qn 5 (Combinatorics) completely, then this time Singapore would have won 4 Golds, and she would also be ranked among the top 10 countries :D
But sadly the fact is, Singapore has participated 23 times in IMO since 1988, and has only won a grand total of 1 Gold :-( .. nevertheless, I do have the confidence that Singapore can win more Golds within the next years as we see more and more younger students being inducted into the national training team!!
The IMO 2010 results were released about half an hour ago. Here is Singapore's results: 4 Silvers, 1 Bronze and 1 Honourable Mention.

Singapore is ranked 22 out of 96 countries (last year, she was ranked 30 out of 104 countries). Top 5 countries are: China, Russia, USA, Korea and Kazakhstan.
Full results are here. This year the cut-off points for Gold, Silver and Bronze medals are 27, 21 and 15 respectively. Looking at the Singapore Team's individual contestant's results for each question, I think we could have won 4 Golds!! How?? Typically Qn 3 and Qn 6 of the IMO are the hardest questions, so it is understandable if none of them could solve the questions or manage to secure even 1 point. However, IF only Ivan could solve Qn 2 (Geometry) completely, and IF only Jie Jun, Lim Jeck and Yan Sheng could solve Qn 5 (Combinatorics) completely, then this time Singapore would have won 4 Golds, and she would also be ranked among the top 10 countries :D
But sadly the fact is, Singapore has participated 23 times in IMO since 1988, and has only won a grand total of 1 Gold :-( .. nevertheless, I do have the confidence that Singapore can win more Golds within the next years as we see more and more younger students being inducted into the national training team!!
Sunday, 4 July 2010
Off to Astana, Kazakhstan
Will be flying in a few hours' time.
Programme of the 51st IMO
5 July - Arrival
6 July - Opening ceremony
7-8 July - 1st and 2nd day of contest
9-12 July - Excursion
13 July - Closing ceremony / farewell dinner
14 July - Departure (reaching Singapore on 15 July)
Questions are usually uploaded quite soon, a few hours after the contest. They can be checked here, at Mathlinks.
I think final results should come in on 11 or 12 July, and can probably be checked from here.
Programme of the 51st IMO
5 July - Arrival
6 July - Opening ceremony
7-8 July - 1st and 2nd day of contest
9-12 July - Excursion
13 July - Closing ceremony / farewell dinner
14 July - Departure (reaching Singapore on 15 July)
Questions are usually uploaded quite soon, a few hours after the contest. They can be checked here, at Mathlinks.
I think final results should come in on 11 or 12 July, and can probably be checked from here.
Saturday, 3 July 2010
Very Moving Thai Video Clip
If you haven't watched this before, you will be surprised at the ending. It will bring a smile to your face :)
The Thais are really creative people.
The Thais are really creative people.
Thursday, 1 July 2010
Counting Down to 51st IMO
We have been training for IMO at NUS since before the June holidays and the training will officially end this evening.
Official website for 51st IMO: http://www.imo2010org.kz/
Team Singapore:
We will be flying to Beijing from Singapore on 4 July morning, then from Beijing to Almaty (Kazakhstan), and then from Almaty to Astana (capital of Kazakhstan, competition venue).
We were told that the Kazakh food is mainly meat (mostly horse meat!) and milk products, and the most popular drink is tea with milk. In case the food does not appeal to us, we are encouraged to bring instant cup noodles.
National currency is Tenge, 1 tenge = 100 tiyns. 1 SGD is about 100 Tenge.
Official website for 51st IMO: http://www.imo2010org.kz/
Team Singapore:
We will be flying to Beijing from Singapore on 4 July morning, then from Beijing to Almaty (Kazakhstan), and then from Almaty to Astana (capital of Kazakhstan, competition venue). We were told that the Kazakh food is mainly meat (mostly horse meat!) and milk products, and the most popular drink is tea with milk. In case the food does not appeal to us, we are encouraged to bring instant cup noodles.
National currency is Tenge, 1 tenge = 100 tiyns. 1 SGD is about 100 Tenge.
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