Maths Olympiad Questions
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(1) It is given that sq rt a - sq rt b = 20 where a and b are real numbers.
Find the maximum possible value of a - 5b
(2) The expression 1000 sin10 cos 20 cos 30 cos 40 can be simplified into
a sin b where a and b are integers with 0 < b < 90 degrees. Find the value of
100a + b. All anagles are acute and in degrees.
(3) Without the use of a calculator, evaluate
(sin 80) / (sin 20) - sq rt 3 / (2 sin 80)
All angles are in degrees.
(4) Without the use of a calculator, evaluate
[ 25 + 10 x sq rt 5 ]^(1/3) + [ 25 - 10 x sq rt 5 ]^(1/3)
(5) Let a = [ 1 + sq rt 2009 ] / 2.
Without the use of a calculator, evaluate [ a^3 - 503a - 500 ] ^ 10.
These are the "easier questions" and good ''O" level add maths students should be able to solve them.
Have fun in maths.
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I hate math
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Originally posted by Lee012lee:
(1) It is given that sq rt a - sq rt b = 20 where a and b are real numbers.
Find the maximum possible value of a - 5b
(2) The expression 1000 sin10 cos 20 cos 30 cos 40 can be simplified into
a sin b where a and b are integers with 0 < b < 90. Find the value of
100a + b
(3) Without the use of a calculator, evaluate
(sin 80) / (sin 20) - sq rt 3 / (2 sin 80)
(4) Without the use of a calculator, evaluate
[ 25 + 10 x sq rt 5 ]^(1/3) + [ 25 - 10 x sq rt 5 ]^(1/3)
(5) Let a = [ 1 + sq rt 2009 ] / 2.
Without the use of a calculator, evaluate [ a^3 - 503a - 500 ] ^ 10.
These are the "easier questions" and good ''O" level add maths students should be able to solve them.
Have fun in maths.
Thanks for sharing. Hope you don't mind if I give my 2 cents about something minor.
If you meant for all the angles in your trigo ratios to be in degrees (judging from 0<b<90), please state so explicitly in the questions (with 'deg' or a sentence saying all angles in degrees), otherwise they'll be assumed to be in radians. I used to make my students write the little degree symbol on every single angle throughout their working and I think it's a good habit everyone should have (just like squiggling under every vector that should be squiggled).
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5.
a = (1 + sqrt(2009)) / 2
(a^3 - 503a - 500)^10
= [(1^3 + 3 sqrt(2009) + 3 * 2009 + 2009 sqrt(2009)) / 2^3 - (503 + 503 sqrt(2009)) / 2 - 500]^10 (Cubic expansion)
= [(1 + 2012 sqrt(2009) + 6027) / 8 - (503 * 4 + 503 * 4 sqrt(2009)) / 8 - 4000 / 8]^10 (Common denominator)
= { [1 + 2012 sqrt(2009) + 6027 - 2012 - 2012 sqrt (2009) - 4000] / 8}^10
= (16 / 8) ^ 10
= 1024 -
For those interested:
Q1
sq rt a = 20 + sq rt b
so a -5b = (20 + sq rt b)^2 - 5b
= 400 + 40sq rt b - 4b
= -4(b - sq rt b - 100)
= -4 ( [sq rt b - 5]^2 - 125)Min value of ( [sq rt b + 5]^2 - 125) = -125
So max value of -4 ( [sq rt b + 5]^2 + 125) = -4 * -125 = 500 (note negative sign)Q2 requires factor formulae in Trigo, and the whole trigo chapter has sadly been taken out of A level syllabus... so it's only tested a little in topics here and there.... most students become not that strong in trigo because of this...
1000 sin10 cos 20 cos 30 cos 40
= 500 cos 30 sin 10 (2cos 40 cos 20)
= 500 cos 30 sin 10 (cos 60 + cos 20)
= 500 cos 30 sin 10 (0.5 + cos 20)
= 500 cos 30 (0.5sin 10 + cos 20 sin 10 )
= 500 cos 30 (0.5sin 10 + [0.5 sin 30 - 0.5 sin10] )
= 500 cos 30 (0.25)
= 125 cos 30
= 125 sin 60
Hence a = 125, b = 60
100a + b = 12560Have fun!
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Originally posted by eagle:
For those interested:
Q1
sq rt a = 20 + sq rt b
so a -5b = (20 + sq rt b)^2 - 5b
= 400 - 40sq rt b - 4b
= -4(b + sq rt b + 100)
= -4 ( [sq rt b + 5]^2 + 75)Min value of ( [sq rt b + 5]^2 + 75) = 75
So max value of -4 ( [sq rt b + 5]^2 + 75) = -4 * 75 = -300 (note negative sign)Q2 requires factor formulae in Trigo, and the whole trigo chapter has sadly been taken out of A level syllabus... so it's only tested a little in topics here and there.... most students become not that strong in trigo because of this...
1000 sin10 cos 20 cos 30 cos 40
= 500 cos 30 sin 10 (2cos 40 cos 20)
= 500 cos 30 sin 10 (cos 60 + cos 20)
= 500 cos 30 sin 10 (0.5 + cos 20)
= 500 cos 30 (0.5sin 10 + cos 20 sin 10 )
= 500 cos 30 (0.5sin 10 + [0.5 sin 30 - 0.5 sin10] )
= 500 cos 30 (0.25)
= 125 cos 30
= 125 sin 60
Hence a = 125, b = 60
100a + b = 12560Have fun!
Hi Eagle,
You have made a mistake in the expansion of (20 + sq rt b)^2 - 5b.
It should be 400 + 40sq rt b - 4b.
The answer for question one ie the maximum possible value of a - 5b is 500.
Question 2 can be solved without the use of factor formulae.
It can be solved by the use of double angle formulae and the relationship between acute angles of sin and cos.
Let's have fun in maths.
PS : Recently, I have just come across an online IQ-test formulated by a Singapore Mensa Chairman. The url link is www.test-iq.com. Have fun in maths.
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oh yup, made a mistake :D
Typed it out straight without doing it on paper first...
I go change it
Add: But I got -500 as the max value, not 500. Anything wrong?
Ok fixed one more careless mistake
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Question 1
It is given that sq rt a - sq rt b = 20 where a and b are real numbers.
Find the maximum possible value of a - 5b
Steps :
(1) Bring sq rt b to the other side
sq rt a - sq rt b = 20
sqrt a = 20 + sq rt b
(2) Square both sides
(sq rt a)^2 = (20 + sq rt b)^2
a = 400 + 40 sq rt b + b
(3) Minus 5b on both sides
a - 5b = 400 + 40 sq rt b + b - 5b
a - 5b = 400 + 40 sq rt b - 4b
(4) Using the completing the squares method
a - 5b = 400 - 4 (b - 10 sq rt b)
= 400 - 4 (b - 10 sq rt b + (10/2)^2 - (10/2)^2 )
= 400 - 4 (sq rt b - 5)^2 + 100
= 500 - 4 (sq rt b - 5)^2
(5) When - 4 (sq rt b - 5)^2 = 0,
The maximum possible value of a - 5b = 500
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Question 2
The expression 1000 sin10 cos 20 cos 30 cos 40 can be simplified into
a sin b where a and b are integers with 0 < b < 90 degrees. Find the value of
100a + b. All anagles are acute and in degrees.
Steps :
(1) 2 cos 10 sin10 cos 20 cos 30 cos 40
sin10 cos 20 cos 30 cos 40 = ------------------------------------------------
2 cos 10
sin 2(10) cos 20 cos 30 cos 40
= ----------------------------------------
2 cos 10
sin 20 cos 20 cos 30 cos 40
= ------------------------------------
2 cos 10
sin 2(20) cos 40 cos 30
= -------------------------------
2 x 2 cos 10
sin 40 cos 40 cos 30
= -------------------------------
4 cos 10
sin 2(40) cos 30
= --------------------
2 x4 cos 10
sin 80 cos 30
= ---------------------
8 cos 10
cos 10 sin 60
= ----------------------
8 cos 10
Hence 1000 1000 sin10 cos 20 cos 30 cos 40 = 1000 x (1/8) sin 60
= 125 sin 60
(2) Since a sin b = 125 sin 60, a = 125 b = 60
So, it is 100a + b = 100 (125) + 60 = 12560
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A little mistake on the last part of solution for Qns 1 by Lee012lee. It should be "400 - 4 (sq rt b - 5) ^2 + 100", not "400 - 4 (b - 5 sq rt b) ^2 + 100".
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Originally posted by normalguy87:
A little mistake on the last part of solution for Qns 1 by Lee012lee. It should be "400 - 4 (sq rt b - 5) ^2 + 100", not "400 - 4 (b - 5 sq rt b) ^2 + 100".
Hi Normalguy87,
Yes, I have made a mistake.
Thanks for pointing out the mistake.
I have made the changes in the answer.
Regards.