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I realise in your diagram that triangle PQT is right-angled;

Where did you get that idea from. There's no right angles in my diagram.

My diagram wasn't meant to be accurately drawn anyway but is like phrontister's.

Other problem:

Let the width be w, then L = 2x + 2w and xw = 7.5

So you can eliminate w and make L the subject.

Bob

Triangles PQT, TSP and STR are congruent, so that result follows whatever the other answers come to.

I've improved my diagram by making it accurately to scale and my calculated angle seems to be right. If you show your working, I'll be able to see what is going wrong.

Bob

hi math9maniac

Part c exactly!

Part a. I don't get this. Would you post your working out please.

Part b. Would be correct if part a was correct, as it depends on it.

Bob

hi math9maniac

The key to many geometry questions is to get the right diagram.

I drew PS and QR parallel to it. Then SR and PT parallel to it. PXRS is a parallelogram where X lies on QR. As SR = PT one interpretation of the information is that T = X. The alternative is that T is on the other side of XP. I have shown this possibility as T'

But the usual convention when describing parallels is to preserve the directional sense which makes T where I have put it. Also later the question talks about PTRS. Again the convention is to take the points in the given order which confirms this interpretation. If T is at T' then the shape PTRS crosses over itself and finding its area would be silly.

The dotted line shows you how to get the ratio of areas very simply. You can get the angle using the cosine rule and the height with simple trig.

Bob

below number block wrote:

Question: do you have to continue marking the whole chart, or can you stop at some point?

I think it's a good teaching point to allow students to think about this themselves.

Bob

hi MathsIsFun

That's a useful little activity that could be a helpful part of a lesson on primes.

Bob

hi coolbee

There is a method using the remainder theorem but, in case you've not heard of that. here's a way that just uses factors.

There must be a third factor. As I want 24x^3 I'll assume this factor is (4x + r)

By comparing coefficients with the original expression you can find r and hence p.

Bob

Oh that's great!!!

I chose 2 x 10^23 and 6 x 10^17 for a Sun and Earth system (scaled down mass-wise) and managed this straight away. Think I need to adjust the velocity so that we don't all burn up on closest approach!

Nice demonstration of Kepler's laws too.

Oh joy. I've just added Mars. Shows opposition and all that. This is better than 'sliced bread' !

Bob

hi phanthanhtom,

Using Geometer's Sketchpad, I choose points for B and C, and then M slightly closer to B than C. Then I constructed an angle BAC and its bisector and moved A until the bisector went through M. I recorded that point and then tried to find another A. Here is the result:

There were no points to the right of M and above C, and similarly none below the lowest point on the left where I have shown the construction. No idea what these curves are. And they'd be different if I moved my initial points.

I suppose you could specify coordinates (Bx,By), (Mx,My) and Cx,Cy) and then find Ax and Ay in terms of these. That would give parametric equations for the locus. But I think it's a horrible calculation.

Sorry I cannot help more

Bob

hi MathsIsFun

Good page. I found the animation tricky to use but I think this has great potential. Mostly, new particles either zoomed off or just crashed into the growing central mass. After some practice I managed to get one particle to orbit the central mass. The 'ellipse' had an axis progression (Mercury type orbit?). Have you built in some Einsteinian relativity element or is this just an accuracy issue? Would it be possible to provide greater control of each particle so it's easier to set up orbits? eg create a new particle, then specify its direction of motion, then the magnitude of this motion. That would allow controllable experimentation.

I made the central mass about 10 x 10^25 and the satellite 1 x 10^23. I say 'about' because most of my attempted satellites just crashed into the 'planet'. It took a while before I managed to get this one with the right tangential velocity.

Bob

hi coolbee,

At first sight, I thought geometric series

http://www.mathsisfun.com/algebra/seque … etric.html,

but then the third term would need to be 2^3 not 2^7.

So I don't follow what the general rule could be here. Could that be a misprint?

Bob

hi coolbee

I think the following might work:

The expression you are asked to evaluate looks like it is the coefficient in the above for the term containing x^{3p}

You can determine this coefficient independently using the binomial expansion.

Bob

hi Monox D. I-Fly

I have put a full analysis for the icosahedron here:

http://www.mathisfunforum.com/viewtopic.php?id=22966

The dodecahedron should follow quickly using duality but I won't have a chance to type it up until next week.

Bob

hi coolbee

I did it like this:

so

and

So now it's just a matter of substituting into the given expression in x. You'll find there's lots of cancelling.

Bob

MathsIsFun has also made this page:

http://www.mathsisfun.com/calculus/slop … point.html

that allows you to do the shrinking interactively. I suggest this: Move point B onto the curve. Move A onto the curve and note the gradient between the two points. Move A closer to B and note the new gradient. You can zoom in for greater accuracy and if you create a box around an area the image will focus in on that area.

There's also a page that shows a graph and its derivative here:

http://www.mathsisfun.com/calculus/deri … otter.html

Bob

see Maths Teaching Resources.

Bob

**bob bundy**- Replies: 2

The regular icosahedron is one of the five Platonic solids (http://www.mathsisfun.com/geometry/plat … -five.html

It consists of 20 identical equilateral triangles joined 5 at each vertex. It has 20 faces, 30 edges and 12 vertices. If you put a dot at the centre of each face and join together adjacent face dots the resulting solid is a dodecahedron, having 12 faces, 30 edges and 20 vertices. These two are 'duals' of each other; a fact I shall make use of later.

If you join opposite vertices, all these lines go through the centre so you can make a sphere that goes through all the vertices. I shall call the centre to vertex distance the radius of the icosahedron. In another thread, Monox D. I-Fly asked about this distance, as it is key to working out the volume of the solid. Here is his diagram:

I shall assume a side length of 1, and show how to work out this distance. Then I'll work out the volume. I also plan to make a separate thread that shows how to use this to get the volume of a dodecahedron.

Outline of my method. (1) Obtain expressions for tan(36), cos(36) and sin(36). I have done that here: http://www.mathisfunforum.com/viewtopic.php?id=22964

(2) Calculate E'C'. (3) prove that E'C'CA is a rectangle. (4) Calculate AC' and hence (5) the radius. (6) Calculate the area of any face. (7) Calculate the volume of a pyramid with such a triangle as its base and O as its vertex. (8) Hence calculate the volume of the solid.

(2) Triangle E'D'C' is isosceles so it can be split in two down its line of symmetry, creating a triangle that is 36-54-90. Thus E'C' is given by

(3) E'C'CA is a quadrilateral; it has E'C' = AC and E'A = C'C so it is a parallelogram. But E'C = AC' = 2.radius so it is a parallelogram with equal diagonals, ie. a rectangle.

(4) So, by Pythagoras

(5)

(6) Area of any equilateral triangle:

(7) I need the perpendicular distance from O to a triangle.

Let M be the midpoint of B'C' and G the centre of triangle PB'C' (third way up median)

(8) every triangle has a pyramid like this and there is no space between them where they meet at O so 20 times the above should give the volume of the icosahedron:

I don't think that is right. So I've got to check to find where I've slipped up.

All errors found and corrected.

Note:

Thus the above simplifies like this:

which is the value given by Monox D. I-Fly here http://www.mathisfunforum.com/viewtopic.php?id=22950

I think I'll rest with that for now.

Bob

**bob bundy**- Replies: 0

I seem to have overloaded the servers maximum number of links so I'm continuing the list here.

tan(36)

http://www.mathisfunforum.com/viewtopic … 87#p378587

The Regular Icosahedron

http://www.mathisfunforum.com/viewtopic.php?id=22966

Note: I was having difficulty making my answer match the one given by Monox D. I-Fly. I have worked out the two versions are the same but I cannot get the final proof to post; why I don't know. So I'll try again later

Bob

**bob bundy**- Replies: 2

Start with an isosceles triangle ABC with angle C = 36, and A = B = 72.

Bisect A and continue this line to cut BC at D. E is the 'foot' of the perpendicular from D to AC.

There are two more isosceles triangles in this diagram; ADC is 36,36, 108; and ABD is 36,72,72 again. I have marked the angles of 36 with a dot to avoid too much labelling on the diagram.

Let AC = BC = x and AB = AD = 1.

As ABC and ABD are similar, the ratios of corresponding sides are equal:

This quadratic has two solutions but as we know x is positive we can discard the negative solution:

So

As x has a square root in its form and the above also I will simplify by working with tan squared.

Now errors can occur if you square and then square root an expression. But in this case we can safely take the positive root as we know tan(36) will be positive.

If a right angled triangle has an opposite of \sqrt{5 - 2\sqrt{5}} and an adjacent of 1, then the smallest angle will be 36 degrees.

By Pythagoras:

Therefore

and

Bob

hi Hannibal lecter

Every line will have a slope. It is also called the gradient. The number tells you how much the line goes up for every 1 across.

Start anywhere on the line and then count across one unit. How far up must you go before you reach the line again. That number is the gradient or slope. In this question the slope is 3.

If the line is steeper you have to go up a bigger amount so the number gives an idea of how steep the line is. A slope of 2 means the line is not so steep. A slope that is negative means the line slope downwards as you go from left to right.

The interactive graph on this page:

http://www.mathsisfun.com/data/straight_line_graph.html

may help with this.

Bob

hi greendragon

Your initial 'guess' was right. The chain rule on that gives minus and 4 so you need an improved guess of

Your last line is not correct ... it should be cosine.

You said why you need the minus yourself.

Bob

hi bingbong

Start with a right angled triangle and 'O', 'A' and 'H' having their usual meanings.

Write out the usual Pythagoras formula and divide by O squared.

That proves the formula for angles 0 up to 90.

Outside that range the trig values are defined by symmetry so eg. sin^2(120) = sin^2(60). Therefore the result continues to hold for all angles.

Bob

Got a fresh idea. The icosahedron and dodecahedron are duals. So once you know about one the other should be relatively easy

I think the icosahedron is the easier one to tackle so here's a way to do it:

That rectangle E'C'CA is the thing to use after all! You can calculate the length of E'C' by splitting the triangle D'E'C' to make a 54-36-90 triangle and using trig. E'A is the side length for the solid. The diagonals E'C and C'A will cross at the centre of the solid so you can use Pythagoras to calculate E'C and hence calculate the perpendicular height of a pyramid. The base area of a pyramid should be easy as they are all equilateral triangles.

Once you've got that sorted (let me know when you have) I'll explain how to transfer those measurements to a dual dodecahedron.

Bob

hi MathsIsFun

I like this page a lot and also the calculations page.

Bob

I can do the green using trig on a pentagon. So far I haven't worked out the connection between the radius and the side length. But I'm working on it

Bob