Guys I need some help ---

1. If p & q be the intercepts on the coordinate axes by the tangent to (x/a)ⁿ + (y/b)ⁿ = 1, then show that (a/p)^n/n-1 + (b/q)^n/n-1 = 1.

2. A point is moving among the vertical parabola 12y = x³ , which of the coordinates changes faster ?

3. Can you tell me any general formula for finding all the primes?

4. Find the minimum possible least common multiple of twenty (not necessarily distinct) natural numbers whose sum is 801.

5. Let ABCD be a convex quadrilateral in which angle BAC = 50 degrees , angle CAD = 60 degrees , angle CBD = 30 degrees and angle BDC = 25 degrees. If E is the point of intersection of AC and BD, find angle AEB.

6. Let ABC be a triangle with AB = AC and angle BAC = 30 degrees. Let A' be the reflection of A on line BC ;B' be the reflection of B on line AC ; C' be the reflection of C on line AB. Show that A' , B' , C' form the vertices of an equilateral triangle.

7. Could you prove that

I think Dan has already answered Q7.

Q1 is just mere algebra once you obtain the slope of the line.

Q4 By pigeonhole, one of the number must be > 40. So LCM must be at least 41. Now 41 is a prime and 801-19 x 41 =/= 1 or 41, so 41 is unattainable. However, 42 is. (801=19 x 42+3)

Kerwin

2) This curve is parametrised by (t,t³ /12). Then we have:

dx/dt = 1, dy/dt = t² /4

So the x co-ordinate changes faster than the y co-ordinate if and only if:

t² /4 < 1

i.e.

-2 < t < 2.

For t = 2 or t = -2, dx/dt = dy/dt = 1 so the co-ordinates change at an equal rate here. For |t| > 2 we have t² /4 > 1, i.e. the y co-ordinate changes faster than the x co-ordinate.

For 3) there is no known formula.

To say that there is no known formula for the primes is not entirely true. What is true is that there is no known useful formula. But you can come up with very contrived and useless formulae. For example, if we define a constant k by

k = (Sum from n=1 to infinity) 10^-n! p_n

where p_n is the n^th prime, (so the decimal expansion of k is just lots of zeros with the prime numbers embedded in them at increasingly large distances apart, in fact k = 0.230005000000000000000007000... ) then you can fairly easily write down a formula to give you the n^th prime in terms of k. But this obviously isn't the slightest bit useful for calculating the primes.

James.

In fact, there is even a slightly less contrived example, but equally useless. There is a polynomial of degree 25 in 26 variables, p(x,...,z) such that if p(x,...,z)> 0 (and x,...,z are integers) then p(x,...,z) is prime. Unfortunately, p(x,...,z) is negative for almost all values of x,...,z. Here is the polynomial:

p(x,...,z) =

(k+2) { 1 - ( [wz+h+j-q]^2 + [(gk+2g+k+1)(h+j)+h-z]^2 +
[16(k+1)^3 (k+2) (n+1)^2 +1-f^2]^2 + [ 2n+p+q+z-e]^2 +
[ e^3 (e+2)(a+1)^2 + 1 - o^2]^2 + [(a^2-1)y^2 + 1 - x^2]^2 +
[16r^2 y^4 (a^2-1) + 1-u^2]^2 +
[ ( (a+u^2 (u^2-a))^2 - 1 ) (n+4dy)^2 + 1 - (x+cu)^2]^2 +
[(a^2-1)l^2 + 1 - m^2]^2 + [ai+k+1-l-i]^2
+ [n+l+v-y]^2 +
[p+l(a-n-1)+b(2an+2a-n^2-2n-2)-m]^2 +
[q+y(a-p-1)+s(2ap+2a-p^2-2p-2)-x]^2 +
[z+pl(a-p)+t(2ap-p^2-1)-pm]^2
)
}

Pretty nice, eh?

Cool. I haven't seen that before.

When you write x,...,z, don't you really mean a,...,z?

James.

Yes.

Also, it's quite straightforward to prove that there is no polynomial that will generate all the primes (or even generate prime-values-only for integral x).

Olof

Olof,

I think you mean polynomial of one variable.

Kerwin

Yupp, sure did. My phrasing has been way off lately.

Thanks,
Olof

Also, the way I said it, it isn't quite correct. P(x) = x will obviously generate all the primes - what I meant was generate all the primes only.

Olof

In addition to that (Hehe J ), I should have mentioned in the bracket of the first post: "as long as the polynomial isn't a constant", as P(x) = 5 will generate a prime for all values of x.

Kerwin,
for 1st one,
i found the slope but still could noy get the solution.I think i messed up some where.

please can you give me its solution.

Also,

I have heard that fermat had developed a formula which gives the primes but is limited to a certain extent though.

Does anyone have any info on that?

love arun.

kerwin,
I don't quite understand your solution to question no.4.

please explain.

love arun

Arun,

For 1, we know the equation of the curve is
(x/a)ⁿ +(y/b)ⁿ =1. Differentiating wrt x, we get y'=-(b/a)ⁿ (x/y)^n-1 . So the equation of tangent at (x₁ ,y₁ ) is

(y-y₁ )=-(b/a)ⁿ (x₁ /y₁ )^n-1 (x-x₁ )

i.e. bⁿ x₁ ^n-1 x+aⁿ y₁ ^n-1 y=bⁿ x₁ ⁿ +aⁿ y₁ ⁿ =aⁿ bⁿ

Hence the x-intercept p is aⁿ /x₁ ^n-1 and the y-intercept q is bⁿ /y₁ ^n-1

Now rearrange slightly to get the required result [remember that (x₁ ,y₁ ) lies on the curve (x/a)ⁿ +(y/b)ⁿ =1]

For question 4, pigeonhole principle states that if there are n+1 objects in n pigeonholes, there must be a pigeonhole with at least two objects. Applying the principle here, we see that at least one of the number must be > 40 (as 801=20 x 40+1). Now the LCM of the twenty number is bounded below by the largest of the twenty, so the LCM is at least 41. However, 41 is a prime, so in order to attain 41, we need to have all our numbers either 41 or 1, but there can be no such combination (by playing about with all possible combinations) Hence we must try 42, which is attainable by splitting 801 as 19 fourty-twos and 1 three.

Kerwin

Yup i gotcha dude!!!Thanks kerwin!

Any thoughts on 5th and 6th,Guys??
love arun

Number 6 should follow from drawing the figure described (instead of just reflecting the point, reflect the whole triangle) noting the lengths of BC and then A'C; then using the law of sines on the triangle formed with an angle of 135 degrees on it.

I didn't quite understand BRAD.
Could you please explain it?

love arun

Here's a picture; see if you can see what I am describing. (Hint: we want to prove that the red triangle is a 30-60-90 triangle; and we also can use the law of sines on the green triangle for this. Hint 2: see if you can use trig to find what length x is.)

BRAD,
Please make yourself more clear.I cannot follow you.You see i am a slow starter.

love
arun

Sorry, I've probably been poor in explaining my ideas. Anyways, we can see using the defintion of cosine (and the fact that the red line bisects the base of our original figure to find what the green line labeled x equals.(degrees will be used in these calculations)

cos(75)=(¹ /₂ )/x

so

x=(1/2) x 1/cos(75)
=sqrt(6)/[3-sqrt(3)]

I found this by looking at a table; I can prove it to you if you'd like.

Also, by the law of sines; we are given the requirement

sin(A)/1=sin(B)/(sqrt(6)/[3-sqrt(3)])

and also A+B=45.

It is sufficient to note that A=15; and B=30 will satisfy this. Once we know these dimensions, we can just fill in the triangle and note that indeed, an equilateral triangle is formed.

Sorry for any earlier confusion, (and especially sorry if there still is any),

Brad

Brad,
sorry i still can't follow you.

a labelled diagram and a complete solution will be appreciated.

please,i am really sorry i am troubling you.

love arun

Is it just telling what I'm talking about in the diagram that's giving you problems. If there's nothing else, I'll try to post an easier to follow diagram with a solution using standard notation tomorrow.

Brad

BRAD,
Your words,
"I'll try to post an easier to follow diagram with a solution using standard notation tomorrow."
are appreciated.

Yes it is what i would like to have!!

Thanks!!!!! BRAD for taking so much trouble.

love arun

[Editor: Brad put a diagram here, but unfortunately it has since disappeared.]

M is the midpoint of AC.
if we let AC=1, then MC=1/2. Also < MB'C=15 (I am using < to represent "angle"). So

sin(< MB'C)=sin(15)=(1/2)/B'C

B'C=1/(2sin(15))=sqrt(6)/[3-sqrt(3)]

Also, A'C=1. Lastly, < B'CA'=135. Therefore, as all angles in a triangle add up to 180,

< A'B'C+< CA'B'=180-135=45. (1)

And, by the law of sines,

sin(< C'B'A)/A'C=sin(< CA'B')/B'C

sin(< C'B'A)/1=sin(< CA'B')/(sqrt(6)/[3-sqrt(3)]) (2)

We then note that the solution to this [(1) and (2)] is < CB'A'=15 , < CA'B'=30. (which can be shown be substituting values). Also, triangle C'B'A' is symmetrical about line BB', so we can easily see that it is both isosceles and that < C'B'A' is 60. Keeping in mind that it is isoscleles, we see that < B'C'A'=< B'A'C'=60. As all the angles of the triangle are 60, we know the triangle is equilateral.

Brad

BRAD,
Shouldn't A' lie on line BC and B' on line AC and C' on line AB?

I can understand the rest of the proof but the diagram produces doubts in my mind.

can you explain the diagram please ?

love arun

What I think the question means is A' is the reflection of A across line BC. Similar of B' and C'.

Brad