All About Math Olympiad Training & Questions
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TanJ:
Both Platinum and Gold medalist will definitely qualify for invitation round since there are less than 150 people for P and G altogether. They might include some \"better\" Silvers too.
Does gettng a gold qualify for invitation round? How will my kid know?Maths Hub:
Congrats to all students who did well in the SMOPS 2013! :rahrah: :rahrah: :rahrah:
http://www.hci.sg/aphelion/apmops/smops2013results.htm
But then.... second round heavyweights.... chances as good as nil....
Cheers. -
MathsOlympiadtrainer:
Thanks.
Both Platinum and Gold medalist will definitely qualify for invitation round since there are less than 150 people for P and G altogether. They might include some \"better\" Silvers too.TanJ:
Does gettng a gold qualify for invitation round? How will my kid know?
But then.... second round heavyweights.... chances as good as nil....
Cheers. -
Hi, need help again on the flwg qns, thx...
1. In the diagram below, ABCD is a square. The points A, B & G are collinear. The line segments AC & DG intersect at E, and the line segments DG & BC intersect at F. Suppose that DE = 15 cm, EF = 9 cm & FG = x cm, find the value of x.
http://i42.tinypic.com/21l7uwz.png\">
2. Find the ratio of the area of the larger equilateral triangle PQR to that of the smaller equilateral triangle LMN.
http://i42.tinypic.com/21mc76b.jpg\">
http://i44.tinypic.com/24g0g36.png\">
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pinky88:
Let the length of the square be 1 unit. We try to make use of similar triangle. (shall skip the proving of similar triangles here)Hi, need help again on the flwg qns, thx...
1. In the diagram below, ABCD is a square. The points A, B & G are collinear. The line segments AC & DG intersect at E, and the line segments DG & BC intersect at F. Suppose that DE = 15 cm, EF = 9 cm & FG = x cm, find the value of x.
http://i42.tinypic.com/21l7uwz.png\">
Triangle ADE~Triangle CFE
So AD/CF=DE/FE
CF--> 3/5 unit
FB--> 2/5 unit
Next,
Triangle GDA~Triangle GFB
So DA/FB=DG/FG=5/2
5 units--> x + 15 + 9
2 units--> x
Then we have 3 units--> 24 and thus, 2 units-->16 -
pinky88:
First of all, we need to apply fraction of figure knowledge to this question. For example, if we cut a triangle A into ratio of 1:2, the 2 smaller triangles B and C will be 1/3 and 2/3 of A respectively. Now if we cut the smallest triangle B into the ratio of 1:3, the new smallest triangle D will be 1/4 of B, which is equivalent to 1/4 X 1/3 = 1/12 of A.Hi, need help again on the flwg qns, thx...
2. Find the ratio of the area of the larger equilateral triangle PQR to that of the smaller equilateral triangle LMN.
http://i42.tinypic.com/21mc76b.jpg\">
Now let's look at the question,
Triangle PML-->1/6 X 5/6 = 5/36 of Triangle PQR
Triangle MRN-->1/6 X 5/6 = 5/36 of Triangle PQR
Triangle NQL-->1/6 X 5/6 = 5/36 of Triangle PQR
Triangle LMN--> 1 - 5/36 X 3 = 21/36 = 7/12 of Triangle PQR
PQR:LMN=12:7 -
pinky88:
3. This is a not-obvious qn on Pythagoras. We make use of pythagorean triplets to find them. Some common pythagorean triplets include (3,4,5), (5,12,13), (7,24,25), (8,15,17).Hi, need help again on the flwg qns, thx...
http://i44.tinypic.com/24g0g36.png\">
870:464=15:8
615:855=41:57
123:477=41:159
326:614=163:307
536:462=268:231
Since only in the first pair, it is part of the triplets (8,15,17), and none of the other 4 are, therefore only A is the answer. -
pinky88:
4a) You need to read up more on modulo operations. In summary,Hi, need help again on the flwg qns, thx...
http://i44.tinypic.com/24g0g36.png\">
If we take 59 / 29 = 2 R 1. We just need to look at the remainders for powers.
59/29 = 2R1
87/29 = 3R0
115/29 = 3R28 = 4R(-1)
So remainder in 59^59 + 87^87 - 115^115 = 1^59 + 0^87 + (-1)^115 = 1 + 0 -1 = 0
4b) n^6039<2013^2013
((n^3)^2013)<2013^2013
n^3<2013
Since 12*12*12=1728
and 13*13*13=2197
Therefore the largest n will be 12. -
Thank u v much for ur help, MathsOlympiadTrainer.
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pinky88:
Thank u v much for ur help, MathsOlympiadTrainer.
You are welcome!:lol: :lol:
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pinky88:
A square number must ends with digit 00, 1, 4, 5, 6, 9.
http://i44.tinypic.com/24g0g36.png\">
In addition, if it ends with 5, then it must end with 25. this is because for eg a number (m5)^2
(10m +5)^2 = 100m^2 + 100 m + 25 will end with 25
It can be calculated that (b), (c), (d) and (e) will not work just by looking at their last 2 digits when calculating.