Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

hi Krash,

Welcome to the forum.

And Q2 needs the sine rule:

Q3. Because MN is half way between SR and PQ there are lots of similarity properties you can use here:

MX = ½ SR = YN

MN = ½ (SR + PQ)

MY = ½ PQ = XN

I think that's enough to work out PQ.

Bob

hi hussam

Thanks. This has got me thinking.

The starting point for a mathematical theory will be a set of axioms; from which other properties can be worked out.

is not part of the axioms, so it has to be checked. Just because something appears to be true doesn't make it true. Here's an example.

generates primes numbers for n = 0, 1, 2, 3, 4, ......

so you might conclude that it is a **prime number generator**. If you keep checking values of n, mostly it works. But consider n = 41. Now the 'generator' has produced a number that isn't prime. It's a useful lesson in why mathematicians spend so long proving the apparently obvious.

So back to square roots. Why should rule (1) apply?

Clearly we can find examples where it does work:

It is conventional to assume that the square root symbol means take the positive square root. But supposing we take a more general meaning for square root such that

And let's choose √ 4 = -2 ; √ 9 = -3 and √36 = -6. Now we find √ 4 x √ 9 ≠ √ 36.

When it comes to complex numbers, you cannot say 'by square root I mean the positive square root' because complex numbers lie somewhere in the Argand diagram not necessarily on a line. The fundamental theorem of complex numbers tells us that every complex number will have two square roots (except zero) so rule (1) won't work.

What you have done is to produce a counter example (like n = 41) that demonstrates that the rule fails. It just shows how careful you must be not to assume a rule holds without careful checking.

You have probably heard that you can 'prove' that 2 = 1 with some simple algebra. It's another example of the need for careful checking. If you haven't met this one have a look here:

http://en.wikipedia.org/wiki/Mathematic … on_by_zero

Bob

hi SaltnPepper

Welcome to the forum.

The word 'product' is used to mean multiply numbers together such as:

eg. What is the product of 5 and 16? Answer 5 x 16 = 80

So maybe this means work out 147 x 11.

Bob

hi Niharika,

When I get set a question, I assume that the questioner has devised one that does have a sensible solution. If you're not told much that should mean that if you do whatever is possible, it ought to lead to an answer. So these questions may look daunting but let's see what is possible and hope my faith is upheld.

Q2. I had to remind myself what the equation of a normal is, based on a parameter m. My version isn't quite the same as yours. So I looked it up in a book and mine is correct. I think you had the gradient of a normal as m. It should be -m.

So your starting point was good but needs a small adjustment.

If you sketch the parabola, you'll see that typical normals have negative gradients.

So if you're given x and y, you get a cubic in m, which you'd expect to have three solutions ie values of m.

I couldn't see that it would be very easy to factorise the cubic, especially with x and y in it, so I thought maybe I could start from the other end and assume the cubic has the form:

That has two coincident roots (p twice) and one separate one (q), so it fits the question.

You can probably do the rest yourself. Expand the cubic and compare coefficients with (1).

The algebra isn't that nice but, remember: "it ought to lead to an answer"

So eliminate p and q and you'll be left with an equation containing x, y and a. That will be the required locus.

Q3. I had no idea where the vertex would be for that equation. I know you cannot do this in a test but it's worth having a look at the graph. You'll find it here:

http://www.mathsisfun.com/data/function … +a^2x/2-2a

The function grapher has a slider that lets you alter the value of 'a', so you'll be able to see how the vertex changes for yourself. Of course, that's no answer to the question, but hopefully it will make it seem more manageable.

If the graph was simpler (eg. y = x^2) then you would know all about the vertex, so what I did next was to change the coordinates to bring the vertex to the origin. This is a handy technique that may help with other questions. You try to make the equation simpler by changing to a new x and y coordinate system. Here's my method:

One simplification that suggested itself to me was to change the x coordinate to X = ax. the equation then becomes:

Already, it doesn't look so frightening. Now if I 'complete the square' for that quadratic I can set it up for the next coordinate change.

So now I can make a second change of coordinates thus:

ie.

Now that parabola is very simple. We know it has a vertex at (0,0) for all a.

So put

and that's nearly done. You want the locus of 'a' so eliminate it from the above and the resulting equation in the old y and x will be the locus.

LATER EDIT: and you can put this locus into the function grapher as the second function to see if this works. (I found I needed to change a to -a to get the second half of the solution. I didn't find a way to make a negative on the slider.)

I suggest you have a try at these to finish them off and then see if you can do another yourself. If not, you know where to come.

Bob

hi Stefy,

In my head I said that too. Just forgot to put it down. Thanks. On the graphs it is obvious.

Bob

hi denis_gylaev

Q1. replace x by k-x to get

Simplify and set the x coefficients equal and the constant terms equal, to get k.

Q2. There's a lot of code here that doesn't display with this forum's Latex. At least, not for me. Did it for you?

Here's what I think you meant:

f can take any values except between 0 and -1. g can only take values above 1.

Check out at http://www.mathsisfun.com/data/function-grapher.php

Bob

hi Niharika,

Oh!

What I'm lacking here is information. Here's a suggestion:

Post three problems: (1) One you have done and got right; (2) One you are struggling to complete (say what you have tried); and (3) One you don't know where to start.

Bob

hi hussam,

But you are also a contributor. Your trig / triangle exercises are really interesting. So thank you for those.

Bob

ps. I've done Q1-Q5. I'm taking a temporary break from them because I wasn't getting anything else done!

A={x} and B={{x}}

Are they the same? No. A ={x} and B = {A}. These are not the same any more than sin(sin(x)) is the same as sin(x).

As far as I am aware, set theory was only thought to create a difficulty because of this paradox:

Let A = {the set of all sets that do not contain themselves}

Is A ∈ A ?

Well, A is a set so that's a good start.

Let's say it does contain itself.

Then it is in the set of all sets that don't contain themselves. =><= ie. a contradiction.

So, alternatively, let's say that A doesn't contain itself.

Then it should be placed in A, by the definition of A. And we have a contradiction again!

The paradox is avoided if you simply require that sets must be 'well defined'. That means, you must be able to decide whether an item is a member or not. So sets like A are excluded from set theory ... paradox avoided.

http://en.wikipedia.org/wiki/Russell's_paradox

Bob

hi Niharika,

Imagine two identical cones one up-side-down with their vertices touching. The cones extend to infinity. Assume they are hollow.

Any plane will cut through the cone(s) and the resulting 2D shape is called a conic section.

There are five possibilities:

A circle: this occurs when the cut is horizontal.

An ellipse: this occurs when the cut is sloping at an angle less than the angle of the cone.

A parabola: this occurs when the cut is exactly parallel with the angle of the cone.

A hyperbola: this occurs when the cut is sloping at an angle greater than the angle of the cone. The section will be in two parts; one from the top cone; and one from the up-side-down cone.

Two straight lines: this occurs when the cut is vertical through the common vertex.

The general formula for any conic section takes the form:

[This arises from the general equation for the cones intersecting with the equation for a plane in 3D.]

If E = 0 and A = B the conic will be a circle.

If A ≠ B but they have the same sign, then the conic will be an ellipse.

If either A or B = 0 (but not both) then the conic will be a parabola.

If A and B have opposite signs, then the conic will be a hyperbola.

If the equation is factorable as

then the conic is two straight lines.

The ellipse, hyperbola and two straight lines have two axes of symmetry. The parabola has one. The circle has many.

But the issue is complicated because the format of the equations can be changed by a change of axes.

eg. With the format above, a hyperbola will have two axes of symmetry and these might be the x and y axes by suitable choice of the coefficients. By changing the axes so that the asymptotes are the axes, the equation becomes:

Does that help?

Bob

Give me a chance to try these first.

Bob

hi Hussam,

OK. Thanks.

Bob

Are you giving us these as challenges or because you cannot do them and want help?

Bob

Look here first:

https://www.mathsisfun.com/geometry/conic-sections.html

then come back to the forum with a more specific query.

Bob

hi Niharika,

Do you mean cones / pyramids / that sort of solid or, rather, conic sections like circle, ellipse, parabola, hyperbola?

Bob

hi hussam,

Bob

M is the centre of the 9 point circle. (green)

O is the centre of the inscribed circle. (red)

P is the point common to both (ie. where they meet tangentially)

Now what?

Bob

hi SuperLynx

I don't think you could get the volume of any of those with any precision.

eg. For a hair dryer, do you want to count the air inside the dryer or just the solid part.

One way to get the volume of an irregular solid is to immerse it in a container that has graduation marks so you can see how much the level of water goes up. I do NOT recommend you do this for any electrical item! Even with an object that can be safely immersed, the meniscus (http://en.wikipedia.org/wiki/Meniscus will lead to errors in measurement.

Bob

hi classof2020

As Jeff says, trig is the quick way to do this. But you can do it just using the area of a triangle formula (=0.5 x base x height)

In this diagram, I've introduced the lines ZU and CV, perpendicular to AB.

As they are both perpendicular => ZU is parallel to CV so AZU and ACV are similar triangles. => If ZU = 2h, then CV = 5h.

So, if you say AX = x, you can work out all these areas in terms of x and h:

AZX, AZY, AYC, ABC and hence the areas ZYC and YCB by subtraction.

Bob

hi Jeff,

Maybe this will help.

The triangle is ABC,

D,E and F are the midpoints of the sides and the three red lines are the perpendicular bisectors leading to the circumcentre. (not labelled and circle left out for clarity)

G, H and I are the 'feet' of the perpendiculars from the vertices to the opposite sides and N is the orthocentre. (circle left out)

J, K and L are the midpoints of AN, BN and CN respectively.

M is the centre of the circle that goes through these nine points.

Bob

hi Jeff,

Thanks for the link. Hhmmm. He didn't make it easy did he? The diagram isn't next to the text so you have to keep flitting back and forth, many points aren't labelled and there's no triangle shown. So I'm really confused about this and don't know if I'll ever be able to follow what you want, let alone help with an answer.

It would help if you could reproduce the diagram with some letters to identify what is what, and then maybe I can follow what you are seeking.

Thanks,

Bob

hi SuperLynx

There's no such thing as 'absolute' precision so what makes precise measurements will depend on the context. For example, if you are making a tank to hold one gallon and the finished tank only holds 0.9 gallons, then the measurement wasn't precise enough in the making.

If you're trying to land an unmanned spacecraft on a comet, then the volume of the spacecraft isn't as important as measuring exactly where the comet will be and making small adjustments to the position of the craft.

Sometimes you know the length, width and height of something but would still have a lot of trouble getting the volume. eg. You are going to cut down a tree and transport the timber away. You want to know the weight of wood so you can organise what size truck(s) to use.

So I don't think there is a single answer to your question. It all depends on context.

Bob

I have joined imgur.com and so I can upload images to that site. If you do the same, you can copy the BBcode here.

Bob