Tiggsy

As a sort of lead in, consider the simpler situation S = 1!.2!.3!.4!.5!.6!.7!.8! = (1).(1.2).(1.2.3). ... .(1.2.3.4.5.6.7.8).

It's easier to see the product if it's written as,

1! = 1

2! = 1.2

3! = 1.2.3

4! = 1.2.3.4

5! = 1.2.3.4.5

6! = 1.2.3.4.5.6

7! = 1.2.3.4.5.6.7

8! = 1.2.3.4.5.6.7.8

S will equal the product of all of the numbers on the rhs of the equals signs.

Notice that all of the even numbers 2, 4, 6, 8 occur an odd  number of times, while the odd numbers 3, 5, 7 each occur an even number of times, (we can ignore the 1's).

\(\displaystyle S = 2^{7}.3^{6}.4^{5}.5^{4}.6^{3}.7^{2}.8^{1}, \\ = (2^{7}.4^{5}.6^{3}.8^{1}).(3^{6},5^{4}.7^{2}) ,\\ = (2^{6}.4^{4}.6^{2}).(2.4.6.8).(3^{6}.5^{4}.7^{2}), \\ =(2^{6}.4^{4}.6^{2}).2^{4}.(1.2.3.4).(3^{6}. 5^{4}.7^{2}), \\ = (2^{6}.4^{4}.6^{2}).2^{4}.4!.(3^{6}.5^{4}7^{2}).\)

Apart from the 4! in the middle everything else is a perfect square, so if we divide S by 4! the result will be a perfect square.

The same routine can be used for S = 1!.2!.3!. ... 100! .

The factorial in the middle will be 50!.

This only works if the final factorial is a multiple of 4. Try this for S = 1!.2!.3!.4!.5!.6! for example, and it doesn't work because the power of 2 that appears in the middle will be 3, so not a perfect square.

Add together the terms on the rhs.

\(\displaystyle \frac{A(x-3)^{2}+B(x+5)(x-3)+C(x+5)}{(x+5)(x-3)^{2}}.\)

This is to be identical with the fraction on the lhs. The denominators are identical and the numerators must also be identical.

That is,

\(\displaystyle A(x-3)^{2}+B(x+5)(x-3)+C(x+5)\equiv1\;\;\;\textbf{ for all values of x}.\)

\(\displaystyle \text{If }\quad x=3, \quad 8C=1, \quad \text{so}\quad C=1/8.\)

\(\displaystyle \text{If}\quad x=-5 \quad 64A=1, \quad \text{so}\quad A=1/64.\)

\(\displaystyle \text{If}\quad x = 0 \quad 9A-15B+5C=1, \quad \text{so}\quad B = -1/64.\)

For #1, extend QRS across the circle to meet the other side at T and drop a perpendicular from the centre O onto this line meeting it at N.

TQ.RQ = PQ^2, so (TR + 3).3 = 36, so TR = 9.

NS = 9/2 - 3 = 3/2.

From the triangle ONS, use Pythagoras to calculate ON^2, and then from the triangle ONT use Pythagoras again to calculate OT, the radius of the circle.

There's a trig method for #2.

Join M to B and A.

Since angle C is a right angle, BA, (length 10) will be a diameter of the circle and in which case angle BMA is also a right angle.

Angles ABM and BAM are both equal to 45 deg, (angles on the same arc), so BM = AM.

Now use Pythagoras in the triangle ABM, BM^2 + AM^2 = 2BM^2 = 10^2 = 100, BM = sqrt(50) = 5sqrt(2).

Call angle CBA theta, then using the sine rule in triangle CBM, 

5sqrt(2)/ sin(45) = CM/ sin(theta + 45),

so CM = 5sqrt(2)(sin(theta).cos(45) + cos(theta).sin(45))/sin(45).

Simplify that to get 7sqrt(2), (same as geno).

The field is conservative if curl F is zero. 

The first two components turn out to be zero, the third is 2Bx - Ax and will be zero if A = 2B.

For the gradient, integrate the first compont of F partially wrt x, the second partially wrt y and the third partially wrt z.

The results are Ax^2y/2 +f(y,z), Bx^2y + Cy^/2 - 3yz^2 + g(x,z) and -3yz^2 - z^2 + h(x,y), respectively.

All that you need to do after that is to deduce expressions for f(y,z), g(x,z) and h(x,y) so that the three are identical.

\(\displaystyle x^{4}=y^{4}-176\sqrt{7},\\ \text{so}\\ 176\sqrt{7}=y^{4}-x^{4}=(y^{2}+x^{2})(y^{2}-x^{2})=22(y+x)(y-x)=88(y-x),\\ \text{so} \\ y-x=2\sqrt{7}, \\ x-y = -2\sqrt{7}.\)

From Alan #3,

\(\displaystyle a=9^{1/8}=(3^{2})^{1/8}=3^{1/4}=\sqrt[4]{3},\\ m+n=7.\)

This is known as Van Aubel's Theorem.

Sorry Alan, that doesn't work. The identity has to be true for all x, not just x = 1/a.

Also, there's no reason why a should equal b.

Start by moving the first term to the rhs, and then take the tangent of both sides.

\(\displaystyle \tan(\tan^{-1}(ax)+\tan^{-1}(bx))=\tan\left(k-\tan^{-1}\left(\frac{1}{x}-\frac{x}{8}\right)\right)\)

(k for the moment, rather than pi/2).

Expand both sides using the tan(A + B) identity. Also, on the rhs, then divide top and bottom by tan(k).

\(\displaystyle \frac{ax+bx}{1-abx^{2}}=\frac{1-(1/x-x/8)/\tan(k)}{1/\tan(k)+(1/x-x/8)}.\)

Now replace k by pi/2 and this becomes

\(\displaystyle \frac{ax+bx}{1-abx^{2}}=\frac{1}{(1/x)-(x/8)}.\)

Cross multiply, multiply out and collect up terms,

\(\displaystyle a+b-\frac{x^{2}}{8}(a+b-8ab)=1.\)

For this to be an identity, true for all values of x, it's necessary that a + b - 8ab = 0, and also that a + b = 1.

Solving those simultaneously leads to

\(\displaystyle a =\frac{2\pm \sqrt{2}}{4}\quad \text{and} \quad b=\frac{2 \mp \sqrt{2}}{4},\)

after which

\(\displaystyle a^{2}+b^{2}=\frac{3}{4}.\)

\(\displaystyle 6xyz +30xy+21xz+2yz+105x+10y+7z=812.\)

Factorise.

\(\displaystyle (3x+1)(2y+7)(z+5)=812 = 2\times 2\times7\times29.\)

The only x, y, z that fit are x =1, y = 11 and z = 2, so x + y + z = 14.

Hi OldTimer

Been sitting on this for a couple of days, but better late than never.

I'm assuming that you would prefer a leg-up rather than a complete solution, but if that's not the case, or if you need more help, then post again.

For part (i), drop a perpendicular from A onto BC, meeting BC at D say, work out what the angle LAD is in terms of angles of the triangle and then write down a trig identity for this angle.

For part (ii), make the obvious substitutions from part (i) and arrive at (via the sine rule) the requirement that

\(\displaystyle \cot B\cot C+ \cot C \cot A + \cot A \cot B =1.\)

To prove this, make use of the identity

\(\displaystyle \cot (A+B)=\frac{\cot A \cot B - 1}{\cot A + \cot B}.\)

That's not one that you tend to remember but it comes easily from the identity for \(\displaystyle \tan (A+B), \text{(which you should remember)}.\)

(You'll still need to use \(\displaystyle \cot (A+B)=1/\tan(A+B)\)

on the LHS, and remember that A + B + C = 180 deg. )

Tiggsy