Tiggsy

We have ab = 20(a + b) = 20a + 20b,.......(*)

so

ab - 20a = 20b,

a(b - 20) = 20 b,

a = 20b/(b - 20).

a is to be positive so b must be greater than 20, and since the roles of a and b can be reversed a must also be greater than 20.

Let a = 20 + m, and b = 20 + n, then, substituting into (*), 

(20 + m)(20 + n) = 20(20 + m + 20 + n),

400 + 20m + 20n + mn = 800 + 20m + 20n.

mn = 400.

So we are looking for two positive integers having a product of 400.

(1)  m = 1, n = 400 from which a = 21, b = 420.

(2)  m = 2, n = 200  ...............  a = 22, b = 220.

(3)  m = 4, n = 100  ...............  a = 24, b = 120.

etc.

There are five others, (fifteen in all, counting reversals).

Think in terms of shovels.

If the father can dig, on average, f shovels per hour and it takes W shovels to dig the well, it would take the father, working alone, W/f hours to dig the well.

Similarly if the son can dig, on average, s shovels per hour, it would take him, working alone W/s hours to dig the well, (and notice in passing that this is the number to be calculated).

Now, the well can be dug if the father works for 6hrs and the son for 12 hrs, so

6f + 12s = W

and similarly

9f + 8s = W.

Multiply the first equation by 3 and the second by 2,

18f + 36s = 3W

18f + 16s + 2W,

and subtract,

20s = W,

so W/s = 20.

S = 77 + 154 + 231 + ... = 1*77 + 2*77 + 3*77 + ...

10,000/77 = 129.87

129*77 = 9933

130*77 = 10,010

S = 1*77     + 2*77     + 3*77     + ... + 129*77

S = 129*77 + 128*77 + 127*77 + ... + 1*77

Add

2S = 130*77 + 130*77 + 130*77 + ... + 130*77

     = 129*130*77

S  =  129*130*77/2 = 645645.

0.8 = y/(186 + x),

148.8 + 0.8x = y.

 y is to be an integer, so 0.8x has to be something point 2.

The smallest value of that fits is x = 4, then 148.8 + 3.2 =152.

Arrange for A to be at the origin of a co-ordinate system, B at the point (6, 0) and Cassidy at the point with co-ordinates (x, y) then 

\(\sqrt{x^{2}+y^{2}}=2\sqrt{(x-6)^{2}+y^{2}}, \)

so, squaring and tidying up,

\(x^{2}+y^{2}=4\{x^{2}-12x+36+y^{2}\}, \\ 3x^{2}+3y^{2}-48x+144=0,\\ x^{2}+y^{2}-16x+48=0, \\ (x-8)^{2}+y^{2}=16=4^{2}.\)

This is a circle with radius 4, so the area of the region is \(16\pi.\)

I think that you should check your answer Melody.

Hi Melody.

I take your point regarding simplification, and if the numbers, in this question, were different, I might agree with you.

If the thing to be simplified was \((53u^{2}v+18uv^{2})(10u+3v)\)

for example, all that you can do is remove uv from the first bracket or/and get rid of the brackets and collect up like terms. 

What we are given though is \((54u^{2}v+18uv^{2})(9u+3v),\)

and this permits

\(18uv(3u+v).3(3u+v) \\ =54uv(3u+v)^{2}.\)

My feeling is that's what they are looking for.

To get your result it would seem to be a whole lot easier to simply remove a factor 3 from the second bracket and to move that into the first bracket.

However the result would then be

\(\displaystyle (162u^{2}v+54uv^{2})(3u+v),\)

which is different from your result, and not much of an improvement on the original.

I think that they are looking for something different.

Notice that 18uv is a common factor of the two terms in the first bracket, remove that and see where that leads.

This question was posted a few days ago, but in that version CX = 4.

I'll sketch out a solution for that one. (This new version is marginally easier.)

For a triangle ABC, the sine rule says that

\(\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B} =\frac{c}{\sin C}= 2R,\)

where R is the radius of the circumscribing circle (of the triangle).

Applying that to the triangles ACX and ABX, we have,

\(\displaystyle \frac{AX}{\sin C}=2R_{1} \quad \text{ and } \quad\frac{AX}{\sin B}=2R_{2}\)

and since the two radii are equal, it follows that the angles B and C are equal meaning that the triangle ABC is isosceles.

Drop a perpendicular from A to BC, meeting BC at Y, say, then XY = 3, and calculate AY using Pythagoras.

The area of ABC will then be equal to (1/2)*BC*AY.

\(\displaystyle \frac{2x^{2}+2x-2}{x(x^{2}-1)}\equiv \frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}\)

Begin by putting the terms on the rhs over a common denominator.

\(\displaystyle \frac{2x^{2}+2x-2}{x(x^{2}-1)} \equiv \frac{A(x-1)(x+1)+Bx(x+1)+Cx(x-1)}{x(x^{2}-1)}\)

For the rhs to be identical to the lhs we need

\(\displaystyle 2x^{2}+2x-2 \equiv A(x-1)(x+1)+Bx(x+1)+Cx(x-1),\)

(the bottom lines are already identical).

There are two ways of proceeding, either collect up terms on the rhs and equate coefficients, 

\(\displaystyle 2x^{2}+2x-2 \equiv x^{2}(A+B+C)+x(B-C)+(-A) \\ 2 = A+B+C \\2 = B-C \\ -2 = -A \\ A=2, \qquad B = 1, \qquad C=-1,\)

or, quicker, substitute convenient values for x,

\(\displaystyle x=0: -2=-A , \qquad A=2,\\ x=1 : \;\;\;2=2B, \qquad \;\;\;B=1 \\ x=-1: -2=2C, \qquad C=-1.\)