All About Math Olympiad Training & Questions
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Absolutely Bo Chap:
Keban Baru CC is organising Maths Olympiad Training for upper primary / lower secondary at $50 for 2 x half-day sessions next week, just after Christmas. Some problems with numerical answers were given in the posters in the CC.
I cannot see how any of these problems can be solved using upper primary / lower secondary mathematics, especially the last problem. And the posters/banner claimed that only simple algebra is needed.
I had tried giving these problems to some JC students. They can solve Problems 1 to 3, but the methods are quite complicated and involve messy algebra that are beyond any primary school kids. And none of them can solve Problem 4, though it is also true that none of them were Maths Olympiad students.
Can anyone help?
1. Let ABCD be a square with area 100 cm square. Let points P, Q, R and S be the midpoints of sides CD, DA, AB and BC respectively. Determine the area of the square formed by lines AP, BQ, CR and DS.
Ans: \t20 cm square
2. Road Runner is walking over a railway-bridge. When he is 10 m away from the middle of the railway-bridge, Wile E. Coyote approaches him from behind on a rocket. The rocket travels at a speed of 100 km/h. The distance of the rocket from the railway-bridge is the same as the length of the railway-bridge. If the Road Runner runs towards the rocket to reach the beginning of the railway-bridge, he will miss the rocket by 4 m. If Road Runner runs in the other direction, the rocket will reach him 8 m before the end of the bridge. What is the length of the railway-bridge?
Ans: \t44 m
3. Two boats are docked at the different sides of a river, which we labelled as side A and side B. They start to travel to the opposite sides at the same time. When they reach the opposite sides, they immediately reverse and travel back to their original positions. Both boats travel at constant but different speeds. The first time they meet, they are at 21 m away from side A. When they meet on their way back, they meet 28 m away from side B. What is the width of the river?
Ans: \t35 m
4. Let the area of a 1 cm x 1 cm x 1 cm triangle be A. Find the area of a 2 cm x 2 cm x 2 cm x 1 cm x 1 cm x 1 cm hexagon that is inscribed in a circle in terms of A.
(This was the problem on the banner, but the banner was taken down. Is it because the problem was too difficult?)
Ans: \t13 A
given answers are correct.
For Q1, you need to move the areas around to form a cross with 5 squares. Each square is 1/5 of the area of the original square = 20 cm2
For Q2 and Q3, not necessary to use algebra. Concepts of speed ratio/distance ratio can be applied to solve these.
For Q4, secondary school students should be able to solve it using trigo and pythagoras theorem.
cheers. -
kiasuMathsOlympiadtrainer:
Keban Baru CC is organising Maths Olympiad Training for upper primary / lower secondary at $50 for 2 x half-day sessions next week, just after Christmas. Some problems with numerical answers were given in the posters in the CC.
I cannot see how any of these problems can be solved using upper primary / lower secondary mathematics, especially the last problem. And the posters/banner claimed that only simple algebra is needed.
I had tried giving these problems to some JC students. They can solve Problems 1 to 3, but the methods are quite complicated and involve messy algebra that are beyond any primary school kids. And none of them can solve Problem 4, though it is also true that none of them were Maths Olympiad students.
Can anyone help?
1. Let ABCD be a square with area 100 cm square. Let points P, Q, R and S be the midpoints of sides CD, DA, AB and BC respectively. Determine the area of the square formed by lines AP, BQ, CR and DS.
Ans: \t20 cm square
Answer is 20 square cm. Using cut and paste, we will form a cross with 5 squares and each square will be 1/5. My apologies for missing out this qn earlier.
2. Road Runner is walking over a railway-bridge. When he is 10 m away from the middle of the railway-bridge, Wile E. Coyote approaches him from behind on a rocket. The rocket travels at a speed of 100 km/h. The distance of the rocket from the railway-bridge is the same as the length of the railway-bridge. If the Road Runner runs towards the rocket to reach the beginning of the railway-bridge, he will miss the rocket by 4 m. If Road Runner runs in the other direction, the rocket will reach him 8 m before the end of the bridge. What is the length of the railway-bridge?
Ans: \t44 m
Speed of rocket = 100km/h
Speed of Road runner = v km/h
Let the length of half the bridge be L km.
Scenario 1: Run towards each other
we form a equation using Distance moved by rocket:
100 X (L-10)/v = 2L-4
Scenario 2: Rocket chase after road runner
100 X (L+10-8)/v = 4L-8
Take scenario 2 divide by scenario 1 equation (simulataneous eqn):
(L+2)/(L-10)=2
L+2=2L-20
L=22
2L=44m
3. Two boats are docked at the different sides of a river, which we labelled as side A and side B. They start to travel to the opposite sides at the same time. When they reach the opposite sides, they immediately reverse and travel back to their original positions. Both boats travel at constant but different speeds. The first time they meet, they are at 21 m away from side A. When they meet on their way back, they meet 28 m away from side B. What is the width of the river?
Ans: \t35 m
Let the distance of the river be 21+X m, with first boat travelling 21 m and second boat travel X m. When they meet the second time, a total of 3 times the length is travelled. You need to plot out the path way of each boat to be able to see that altogether 3 times the length of the river is covered.
Looking at the second time they meet, first boat will travelled 63 m, while second boat has travelled 3X m. When they meet, first boat has covered 1 length of the river and 28 m.
1 Length of river + 28 m = 63 m
1 Length of river = 35 m
4. Let the area of a 1 cm x 1 cm x 1 cm triangle be A. Find the area of a 2 cm x 2 cm x 2 cm x 1 cm x 1 cm x 1 cm hexagon that is inscribed in a circle in terms of A.
(This was the problem on the banner, but the banner was taken down. Is it because the problem was too difficult?)
Ans: \t13 A
The answer is wrong as well. If you try to draw out the hexagon, it will comprise of 3 1X1X1 triangles and 1 2X2 square.
Area of 2X2 squares= 4 X Area of 1X1 square
The part in blue below is slightly wrong previously:
Using pythagoras theorem, we can see that the ratio of a 1X1X1 triangle to 1X1 square will be sqrt3/4 : 1=1 : 4/sqrt 3 = A : 4/sqrt3 A
Thus total area will be (4/sqrt3X4+3)A=(3+ 16/sqrt3)A
I think they realised the mistakes that's why they take that down. Nowadays there are alot of \"Maths Olympiad Trainer\" or \"best tuition teacher\" (as they claim to be) around. Just be careful of what you sign up for.
Have a MERRY XMAS! -
kiasuMathsOlympiadtrainer:
nothing wrong with Q4. Radius of circle is sqrt(7/3).kiasuMathsOlympiadtrainer:
Keban Baru CC is organising Maths Olympiad Training for upper primary / lower secondary at $50 for 2 x half-day sessions next week, just after Christmas. Some problems with numerical answers were given in the posters in the CC.
I cannot see how any of these problems can be solved using upper primary / lower secondary mathematics, especially the last problem. And the posters/banner claimed that only simple algebra is needed.
I had tried giving these problems to some JC students. They can solve Problems 1 to 3, but the methods are quite complicated and involve messy algebra that are beyond any primary school kids. And none of them can solve Problem 4, though it is also true that none of them were Maths Olympiad students.
Can anyone help?
1. Let ABCD be a square with area 100 cm square. Let points P, Q, R and S be the midpoints of sides CD, DA, AB and BC respectively. Determine the area of the square formed by lines AP, BQ, CR and DS.
Ans: \t20 cm square
Answer is 20 square cm. Using cut and paste, we will form a cross with 5 squares and each square will be 1/5. My apologies for missing out this qn earlier.
2. Road Runner is walking over a railway-bridge. When he is 10 m away from the middle of the railway-bridge, Wile E. Coyote approaches him from behind on a rocket. The rocket travels at a speed of 100 km/h. The distance of the rocket from the railway-bridge is the same as the length of the railway-bridge. If the Road Runner runs towards the rocket to reach the beginning of the railway-bridge, he will miss the rocket by 4 m. If Road Runner runs in the other direction, the rocket will reach him 8 m before the end of the bridge. What is the length of the railway-bridge?
Ans: \t44 m
Speed of rocket = 100km/h
Speed of Road runner = v km/h
Let the length of half the bridge be L km.
Scenario 1: Run towards each other
we form a equation using Distance moved by rocket:
100 X (L-10)/v = 2L-4
Scenario 2: Rocket chase after road runner
100 X (L+10-8)/v = 4L-8
Take scenario 2 divide by scenario 1 equation (simulataneous eqn):
(L+2)/(L-10)=2
L+2=2L-20
L=22
2L=44m
3. Two boats are docked at the different sides of a river, which we labelled as side A and side B. They start to travel to the opposite sides at the same time. When they reach the opposite sides, they immediately reverse and travel back to their original positions. Both boats travel at constant but different speeds. The first time they meet, they are at 21 m away from side A. When they meet on their way back, they meet 28 m away from side B. What is the width of the river?
Ans: \t35 m
Let the distance of the river be 21+X m, with first boat travelling 21 m and second boat travel X m. When they meet the second time, a total of 3 times the length is travelled. You need to plot out the path way of each boat to be able to see that altogether 3 times the length of the river is covered.
Looking at the second time they meet, first boat will travelled 63 m, while second boat has travelled 3X m. When they meet, first boat has covered 1 length of the river and 28 m.
1 Length of river + 28 m = 63 m
1 Length of river = 35 m
4. Let the area of a 1 cm x 1 cm x 1 cm triangle be A. Find the area of a 2 cm x 2 cm x 2 cm x 1 cm x 1 cm x 1 cm hexagon that is inscribed in a circle in terms of A.
(This was the problem on the banner, but the banner was taken down. Is it because the problem was too difficult?)
Ans: \t13 A
The answer is wrong as well. If you try to draw out the hexagon, it will comprise of 3 1X1X1 triangles and 1 2X2 square.
Area of 2X2 squares= 4 X Area of 1X1 square
The part in blue below is slightly wrong previously:
Using pythagoras theorem, we can see that the ratio of a 1X1X1 triangle to 1X1 square will be sqrt3/4 : 1=1 : 4/sqrt 3 = A : 4/sqrt3 A
Thus total area will be (4/sqrt3X4+3)A=(3+ 16/sqrt3)A
I think they realised the mistakes that's why they take that down. Nowadays there are alot of \"Maths Olympiad Trainer\" or \"best tuition teacher\" (as they claim to be) around. Just be careful of what you sign up for.
Have a MERRY XMAS!
Area of triangle of 1 cm side = sqrt(3)/4 cm2 = A
Area of hexagon = 13 * sqrt(3)/4 cm2 = 13 A
cheers. -
MathIzzzFun:
nothing wrong with Q4. Radius of circle is sqrt(7/3).kiasuMathsOlympiadtrainer:
[quote=\"kiasuMathsOlympiadtrainer\"]Keban Baru CC is organising Maths Olympiad Training for upper primary / lower secondary at $50 for 2 x half-day sessions next week, just after Christmas. Some problems with numerical answers were given in the posters in the CC.
I cannot see how any of these problems can be solved using upper primary / lower secondary mathematics, especially the last problem. And the posters/banner claimed that only simple algebra is needed.
I had tried giving these problems to some JC students. They can solve Problems 1 to 3, but the methods are quite complicated and involve messy algebra that are beyond any primary school kids. And none of them can solve Problem 4, though it is also true that none of them were Maths Olympiad students.
Can anyone help?
1. Let ABCD be a square with area 100 cm square. Let points P, Q, R and S be the midpoints of sides CD, DA, AB and BC respectively. Determine the area of the square formed by lines AP, BQ, CR and DS.
Ans: \t20 cm square
Answer is 20 square cm. Using cut and paste, we will form a cross with 5 squares and each square will be 1/5. My apologies for missing out this qn earlier.
2. Road Runner is walking over a railway-bridge. When he is 10 m away from the middle of the railway-bridge, Wile E. Coyote approaches him from behind on a rocket. The rocket travels at a speed of 100 km/h. The distance of the rocket from the railway-bridge is the same as the length of the railway-bridge. If the Road Runner runs towards the rocket to reach the beginning of the railway-bridge, he will miss the rocket by 4 m. If Road Runner runs in the other direction, the rocket will reach him 8 m before the end of the bridge. What is the length of the railway-bridge?
Ans: \t44 m
Speed of rocket = 100km/h
Speed of Road runner = v km/h
Let the length of half the bridge be L km.
Scenario 1: Run towards each other
we form a equation using Distance moved by rocket:
100 X (L-10)/v = 2L-4
Scenario 2: Rocket chase after road runner
100 X (L+10-8)/v = 4L-8
Take scenario 2 divide by scenario 1 equation (simulataneous eqn):
(L+2)/(L-10)=2
L+2=2L-20
L=22
2L=44m
3. Two boats are docked at the different sides of a river, which we labelled as side A and side B. They start to travel to the opposite sides at the same time. When they reach the opposite sides, they immediately reverse and travel back to their original positions. Both boats travel at constant but different speeds. The first time they meet, they are at 21 m away from side A. When they meet on their way back, they meet 28 m away from side B. What is the width of the river?
Ans: \t35 m
Let the distance of the river be 21+X m, with first boat travelling 21 m and second boat travel X m. When they meet the second time, a total of 3 times the length is travelled. You need to plot out the path way of each boat to be able to see that altogether 3 times the length of the river is covered.
Looking at the second time they meet, first boat will travelled 63 m, while second boat has travelled 3X m. When they meet, first boat has covered 1 length of the river and 28 m.
1 Length of river + 28 m = 63 m
1 Length of river = 35 m
4. Let the area of a 1 cm x 1 cm x 1 cm triangle be A. Find the area of a 2 cm x 2 cm x 2 cm x 1 cm x 1 cm x 1 cm hexagon that is inscribed in a circle in terms of A.
(This was the problem on the banner, but the banner was taken down. Is it because the problem was too difficult?)
Ans: \t13 A
The answer is wrong as well. If you try to draw out the hexagon, it will comprise of 3 1X1X1 triangles and 1 2X2 square.
Area of 2X2 squares= 4 X Area of 1X1 square
The part in blue below is slightly wrong previously:
Using pythagoras theorem, we can see that the ratio of a 1X1X1 triangle to 1X1 square will be sqrt3/4 : 1=1 : 4/sqrt 3 = A : 4/sqrt3 A
Thus total area will be (4/sqrt3X4+3)A=(3+ 16/sqrt3)A
I think they realised the mistakes that's why they take that down. Nowadays there are alot of \"Maths Olympiad Trainer\" or \"best tuition teacher\" (as they claim to be) around. Just be careful of what you sign up for.
Have a MERRY XMAS!
Area of triangle of 1 cm side = sqrt(3)/4 cm2 = A
Area of hexagon = 13 * sqrt(3)/4 cm2 = 13 A
cheers.[/quote]Yes you are right. My bad. I missed out on the \"inscribed in circle part\"=) -
Hi,
thanks for sharing the MO questions.
are these questions real MO competition questions?
thanks -
HyperKiasu:
Real? But I can say for sure that the first 3 are very SIMILAR to past MO questions. So yes they are \"real\".Hi,
thanks for sharing the MO questions.
are these questions real MO competition questions?
thanks -
Thanks!
Can Problem 4 be solved using primary school mathematics as claimed? -
Apparently, the trainer was a Silver Medallist in the International Mathematical Olympiad, and he claim that Problem 4 can be solved using primary school mathematics ā¦ that is, no Pythagoras Theorem or Trigonometry.
-
Absolutely Bo Chap:
Apparently, the trainer was a Silver Medallist in the International Mathematical Olympiad, and he claim that Problem 4 can be solved using primary school mathematics ... that is, no Pythagoras Theorem or Trigonometry.
Maybe MathizzzFun might know since he can tell how to decompose that hexagon into 13 triangles.
Else we have to wait for that trainer to perform magic. Its possible that primary maths is all that is needed.. but it will probably took an IMO brain to figure out how
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Absolutely Bo Chap:
Visualize the figure as a hexagon with alternate 2cm and 1 cm sides... suppose the hexagon is ABCDEF
4. Let the area of a 1 cm x 1 cm x 1 cm triangle be A. Find the area of a 2 cm x 2 cm x 2 cm x 1 cm x 1 cm x 1 cm hexagon that is inscribed in a circle in terms of A.
(This was the problem on the banner, but the banner was taken down. Is it because the problem was too difficult?)
Ans: \t13 A
AB=2cm
BC=1cm
CD=2cm
DE=1cm
EF=2cm
FA=1cm
Extend AB, DC to meet at X
Extend BA, EF to meet at Y
Extend FE CD to meet at Z
Now XYZ is an equilateral triangle with sides 4 cm. If you divide each side into 4 x 1cm sections. Subdivide this triangle into 16 x 1 cm side equilateral triangle.
The hexagon will consist of 13 such triangles.![](\"http://i50.tinypic.com/2a6ui2w.png\")
http://i50.tinypic.com/2a6ui2w.png\">
cheers.
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