Some More Questions to Think About (1 Viewer)

CM_Tutor

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1. Consider the sequence a<sub>0</sub>, a<sub>1</sub>, a<sub>2</sub>, ... which is an increasing sequence of real numbers (ie a<sub>n</sub> < a<sub>n+1</sub> for all non-negative integers n), defined by a<sub>n+1</sub> = 2<sup>n</sup> - 3a<sub>n</sub>, for n = 0, 1, 2, ...

(a) If the first term of the sequence is a<sub>0</sub> > 0, find expressions for a<sub>1</sub>, a<sub>2</sub> and a<sub>3</sub>, in terms of a<sub>0</sub>, and show that
a<sub>4</sub> = 2<sup>3</sup> - 3 * 2<sup>2</sup> + 3<sup>2</sup> * 2 - 3<sup>3</sup> + 3<sup>4</sup>a<sub>0</sub>

(b) By examining the pattern, show that a<sub>n</sub> = 2<sup>n</sup> / 5 + (-1)<sup>n</sup>(a<sub>0</sub> - 1 / 5)3<sup>n</sup>

(c) Show that a<sub>0</sub> = 1 / 5 produces such a sequence.

(d) By considering large values of n, with A = a<sub>0</sub> - 1 / 5 <> 0, show that a<sub>0</sub> = 1 / 5 is the only value of a<sub>0</sub> for which this sequence increases.

2. Consider the arc of the parabola y = x<sup>2</sup> for 0 <= x <= b, where b is a positive constant. Let the points A and P be, respectively, (0, a) and (t, t<sup>2</sup>), where a => 0 and 0 <= t <= b.

(a) Find AP<sup>2</sup> in terms of t.

(b) We seek to find the position of P such that the distance AP is minimised. Discuss the value of AP<sub>min</sub>, and the value of t for which it occurs, and how they vary depending on the values of a and b. (Hint: there are three cases to consider).

[I can make this question more directed if people need more help.]

3. Following an unfortunate accident involving a football in a Chemistry Laboratory at a certain co-educational school, each boy in year 12 was fined 30 cents, and each girl 2 cents. There were a total of 120 students in year 12, and the boys outnumber the girls. The money raised by the fines was enough to buy 12 new test tubes (costing 7 cents each), and some number of beakers (costing 90 cents each), with no money left over. We seek to find out how many boys there are in year 12, and how many new beakers were purchased.

---Note: For those who want a real challenge, try and solve this problem without looking at the parts that follow---

(a) If b, g and x are, respectively, the number of boys, girls and beakers, show that 1758 = 14g + 45x, g < 60.

(b) Explain why the number of new beakers purchased must have been an even number.

(c) If X = x / 2, and G = g / 3, discuss the range of possible values of X and G.

(d) Show that X - 6 is divisible by 7.

(e) Find b, g and x.

4. The circumcircle of triangle ABC is the circle that passes through all of its vertices.

(a) Without using the sine rule, show that the radius R of the circumcircle is given by
R = a / (2sin A) = b / (2sin B) = c / (2sin C), where a, b and c are the lengths of BC, CA and AB, respectively.

(b) Prove that sin<sup>2</sup>A + sin<sup>2</sup>B + sin<sup>2</sup>C <= 9 / 4

(c) Show that if the circumcircle of a triangle ABC has a radius of at most 1 cm, then the sum of the squares of the sides of the triangle does not exceed 9 cm<sup>2</sup>

5. If a, b, c, d, e and f are six positive real numbers, such that

a<sup>2</sup> + b<sup>2</sup> + c<sup>2</sup> = 16
d<sup>2</sup> + e<sup>2</sup> + f<sup>2</sup> = 49
ad + be + cf = 28

show that (a + b + c) / (d + e + f) = 4 / 7
[Hint available]
 

CM_Tutor

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Nike33, I didn't say that A was the focus. Certainly, if a = 1 / 4, then AP<sub>min</sub> = a and t = 0, but it isn't true to say that a must be 1 / 4.
 

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Happy birthday CM_Tutor :p (sorry to offtopic your thread)
 

nike33

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that was abit harsh :p ill reattempt it later

..after doing 15 parametrics qns then getting a point (o,a) that isnt the focus..hehe
 
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CM_Tutor

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Originally posted by nike33
that was abit harsh :p ill reattempt it later
Sorry, didn't mean to be harsh, and your answer is on the track to one of the three cases... :)
 
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Originally posted by CM_Tutor

3. Following an unfortunate accident involving a football in a Chemistry Laboratory at a certain co-educational school, each boy in year 12 was fined 30 cents, and each girl 2 cents. There were a total of 120 students in year 12, and the boys outnumber the girls. The money raised by the fines was enough to buy 12 new test tubes (costing 7 cents each), and some number of beakers (costing 90 cents each), with no money left over. We seek to find out how many boys there are in year 12, and how many new beakers were purchased.

---Note: For those who want a real challenge, try and solve this problem without looking at the parts that follow---

(a) If b, g and x are, respectively, the number of boys, girls and beakers, show that 1758 = 14g + 45x, g < 60.
b+g = 120
b= 120-g
b>g, therefore g<= 59

No money is left over so:
30b + 2g = 12.7 + 90x
3600-28g = 84 + 90x
14g + 45x = 1758 (skipping some manipulation) (g<= 59)

(b) Explain why the number of new beakers purchased must have been an even number.
g is an integer, so 14g is an even integer
therefore 1758 - 14g is an even integer
therefore x is an even integer, as an even integer divided by an odd integer (45) is an even integer

(c) If X = x / 2, and G = g / 3, discuss the range of possible values of X and G.
Maximum possible money raised with b=120, money = 3600 cents
Therefore maximum x is 40, max X is 20.
Minimum possible money with b=61, g=59 money = 1888 cents
therefore minimum x is 20 (rounding down), min X is 10.
10<= X <= 20

g<=59, therefore 0<= g<= 59
0<= G<= 19 2/3

(d) Show that X - 6 is divisible by 7.
X is an integer as x is even, X= x/2
1758 = 42g + 90x
Rearranging, 1218 - 42G = 90X - 540
g is an integer, so 3G is an integer (G = g/3)
7 ( 87 - 3G) = 45(X-6)
as 87-3G and X-6 are both integers, X-6 is divisible by 7.
(e) Find b, g and x.
[/B]
Therefore, X = 13 (10<=X<=20)
Substituting, G=14
B=106

Unlucky boys indeed:(. Will look at other questions later.
 

Xayma

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Originally posted by CM_Tutor
4. The circumcircle of triangle ABC is the circle that passes through all of its vertices.

(a) Without using the sine rule, show that the radius R of the circumcircle is given by
R = a / (2sin A) = b / (2sin B) = c / (2sin C), where a, b and c are the lengths of BC, CA and AB, respectively.
Ill make a start on 4 and do the rest later.

The Centre of the circle (O) is at the intersection of the perpendicular bisectors.

Now < BOC=2A (Angle at centre of the circle= 2*Angle at circumfrence)
Let the midpoint of BC=M (ie the perpendicular bisector leaves M)
Now OMC=A (As OB=OC (both radii, OAC is an isos triangle, therefore the perpendicular line to the vertex (O) from the base bisects the angle at the vertex))
MC=a/2 (M is the midpoint of BC and BC=a)

Now sin A=MC/OC (basic trig)
sin A=(a/2)/R
R=a/2sinA

Similarly R=b/2sinB
R=c/2sinC.
 
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Gotta run out right now, but for 4a it's just much quicker to prove d = a/sinA, i'd feel.

This is done by constructing a line though the diameter meeting line BC (of length a) on the other side of the circle, and constructing another line from A to meet the other end of BC. Now, angle BCA (or ABC depending on how you construct triangle) is 90 as your new triangle is a right angle, so R = a/sinA (using trig definition). Similary R= b/sinB, r= c/sinC.
 

Xayma

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That could work as well, oh well I probably wont encounter one this hard, grr I dont like b) I know that it will be a max when A=B=C (ie it is 9/4). but proving it.
 

Archman

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Well there's jensen's inequality which says
sin^2((A+B+C)/3) >= (sin^2A+sin^2B+sin^2C)/3 for 0<=A,B,C<=2pi

But that's not in the syllabus.
Can't seem to find a nice hsc way though.
I doubt you solve it through proving (a^2+b^2+c^2)/R^2 <= 9 because that's part c
 

nike33

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here a hint. (i think) ill post my ans tommorow

r = a/2sinA = b/2sinB = c/2sinC
1/(2r) = sinA/a
1/(4r^2) = sin^2A/a^2 = sin^2B/b^2 = sin^2C/c^3=2
3/(4r^2) = sin^2A/a^2 + sin^2B/b^2 + sin^2C/c^2
 
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CM_Tutor

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Originally posted by George W. Bush
Therefore, X = 13 (10<=X<=20)
Substituting, G=14
B=106

Unlucky boys indeed:(. Will look at other questions later.
Very close. Unfortunately, since G = 14, g = 42, and thus b = 78, and x = 26 :)
 

Lexicographer

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This is really quite intriguing for my first foray into the 4u maths sector. Too bad my stuff at uni falls between 3 and 4, I've got nowhere to fit in. :(
 

Xayma

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Originally posted by Lexicographer
This is really quite intriguing for my first foray into the 4u maths sector. Too bad my stuff at uni falls between 3 and 4, I've got nowhere to fit in. :(
A fair bit of the questions you can do without 4U knowledge, I just stay out of the topics I cant do ;)
 

nike33

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think about this..

h = sin^2A + sin^2B + sin^2C
given A + B + C = 180..A,B,C > 0 and h is a maximum...find h
then you have proven it (h is 4/9)
 

Xayma

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Thats the way I was thinking it, but I cant think of how to prove it, hmm so I went of thinking about how to prove it would be a max when A=B=C as the rate of change decreases as A gets bigger
 
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nike33

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5. If a, b, c, d, e and f are six positive real numbers, such that

a2 + b2 + c2 = 16
d2 + e2 + f2 = 49
ad + be + cf = 28

show that (a + b + c) / (d + e + f) = 4 / 7
[Hint available]


if you havent done the last one...(did this thismorning on the trian..sad i know) heres hint..

ie adding them (timsing last one by 2)

(a+d)^2 + (b+e)^2 + (c+f)^2 = 121
then (a+b+c+d+e+f)^2 -2(ab+.....) = 121
now the ab+...bit = (a+b+c)(d+e+f) - 28

edit: yes archman i had left out 6 terms from that :(
 
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