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Hi Simon,

Sorry but I'm not getting this at all yet. Let's try to get the picture thing working. Years ago you could upload images but we lost that facility when we had to go to a new server. The forum has minimal funds so had to accept this. I joined imgur.com (it's free) and I upload pictures there and use the BC code to provide a link in a post.

If you take a look at one of my posts which has a picture you should be able to see the format. Once I have an idea what you're trying to make I'll have a go at sorting the geometry.

Bob

Hi Allerious101

Welcome to the forum.

Yes, it can be done. The key is the number to the left of the 4. It must be a 1 as all the boxes below it contain numbers that are bigger than something.

That means the top row is fixed. If you try a 2 in the box below the 4, you can quickly fill in a lot, but you end up with two 2s in a column. So that must be a 1.

After that all quickly falls into place. In the third row, one of the ends must be a 1 and once that is fixed the last 1 is forced. etc etc.

Hope that does it for you

Bob

Hi,

EDIT

The more I think about it, the more I think what follows is wrong. I'll leave it in place as I have proved something, just not your problem. Sorry.

If E is at 12 o'clock, A at 11.30, B at 11.00, C at 10.00 and D at 08.00, then a perpendicular from C D to X would have to be on DC extended.

My revised approach will be to assume with no loss of generality that the circle has radius 1, in which case the points can have I,j coordinates such as (cosA, sinA) for some angle A. If this leads anywhere I'll start a new post.

Old version follows

I managed this whilst on a train. Then I realised there's one element of the proof missing. Haven't got my geometry program to sort it out but you may well spot what to do so here goes with the major part.

I'll get the line joining the centroid of ABE, say X, and a point on the line CD, say Y.

I'll skip overlines to save time and use p and q as parameters.

You should already know the centroid , X, is at 1/3(a + b + e). If not, get the midpoint of BE, M, and hence the equation for AM. Centroid is one third along the line.

Direction of CD is c - d so the equation of CD is r = c + p(c - d). Say Y is a point on CD.

Equation of XY is

r = c + p(c-d) + q( 1/3(a + b + e) - c - p( c - d) )

= 1/3(a+ b+ e) + (1+p)(1-q)c -p(1-q)d.

If the concurrent point is on all five lines it must have coefficients that are equal so start by setting the c and d components equal

(1+p)(1-q) = -p(1-q). Implies p = - 1/2 so Y is the midpoint of CD and then set the a,b,e components also equal to this

1/3q = 1/2(1-q) implies q = 3/5

Substituting into the equation for XY gives

r = 1/5(a+b+e+c+d)

If you were to construct the other lines in the same way you'd get the same point each time so this is the concurrent point.

What's missing? I haven't shown that XY is perpendicular to CD. Should be easy as Y is the midpoint ... Just need to show XY goes through the centre of the circle. But I haven't found a value of q to make this happen yet. Can you?

Bob

Hi Heatherjay40,

Welcome to the forum.

When I was at school applied meant Newtonian mechanics but it could mean lots of things now. Why not join up and post something more specific.

Bob

On the line above there's 0.25 (from 0.5 X 0.5). So multiply both sides by 4 to simplify.

Bob

hi Katie,

I'll just go through the steps first and then how to solve that equation.

step 1. Split the polygon into 7 identical isosceles triangles.

step 2. Work out the top angle 360/7 = 51.4

step 3. Split one triangle in half so as to make a right angled triangle. (You need this to use trigonometry.)

step 4. Work out one bottom angle (180- 51.4)/2 = 64.29

step 5. Get an expression for the height of a triangle h = 0.5b tan(64.29) (The adjacent is 0.5 b and h is the opposite.)

step 6. Get an expression for the area of one triangle half base x height = 0.5 x b x 0.5b tan(64.29)

step 7. Work out how big one triangle actually is = 130/7 = 18.57

step 8. Make an equation to find b 0.5 x b x 0.5 x b x tan (64.29) = 18.57

step 9. Solve for b

0.5b x 0.5b tan(64.57) = 18.57

0.25 x b² tan(64.57) = 18.57

b² tan(64.57) = 4 x 18.57

b² = (4 x 18.57)/tan(64.57)

b = √ ((4 x 18.57)/tan(64.57))

Hope that helps.

Bob

OK, try this:

Looks to me like the same problem as seen in a mirror.

Bob

OK, I have it. Draw any circle and mark a point T outside the circle. Draw two lines from T; the first TPQ where P and Q are points on the circumference, and similarly, TRS.

Angles RSP and RQP are equal as the they subtended by the same chord RP so triangles TPS and TRQ are similar. Hence PT . TQ = RT . TS

Applying this to the problem: AD . AE = AF . AG therefore AE = AF . AG / AD

And so DE = AF . AG / AD - AD.

Let BC = R, then AC = 2R cos(22.5) and so AF = R(2 . cos 22.5 - 1)

This allows you to get DE in terms of R and the cosine and hence DE / BC.

The area ratio is the square of this as the triangles are similar.

Bob

hi Monox D. I-Fly

Does this look right for the problem?

The parallelogram must be a rhombus as opposite sides are equal for any parallelogram and CD = CB =radius = R for the circle.

I have drawn AC extended to cut the circle at F and G.

There's a circle theorem that should be useful here. But it's 50 years since I last used it so the exact form escapes me at present.

Choices: (1) Go up in the attic and find an old maths book. (2) Search through Euclid until I find it. Or (3) Try to re-prove it from scratch. Pride requires that I try (3) first.

Bob

ps. You wanted CD. If the shape is a rhombus then AB = BC = CD = DA.

I have made use of the following:

i x i = j x j = k x k = 0

i x j = k j x i = -k and four more like these.

CO = - OC etc.

I've just realised that using k for the multiplier was a poor use of symbol. Sorry if that is confusing. Fortunately I haven't actually used k as a unit vector in my proof, so all the ks are multipliers.

Bob

Ok, sorry for the misunderstanding. I can do that too.

To save time with typing I'm going to leave out the overlines on the vectors.

If A, B and C are colinear then AB = kBC => AO + OB = k(BO + OC) which re-arranges to OA = kOB + OB -kOC

So AxB + BxC + CxA becomes

(kOB + OB -kOC) x OB + OB x OC + OC x (kOB + OB - kOC) = -k (OC x OB) + (OB x OC) + k x (OC x OB) + (OC x OB) = 0

Hope that is finally what you want.

Bob

hi !nval!d_us3rnam3

That's much clearer. I wonder why your first post just looks like code to me. When I want maths expressions I use LaTex. Mostly that works ok. You have to start your code with square brackets math and end with square brackets /math. Does your first post look ok to you? Maybe you have some software on your machine that displays it for you. I'd like to try and get to the bottom of this for future posters.

Anyway, back to your question. Those 'x' signs must be cross products (also called vector products). You can read about them here:

and there's a calculator here:

When you use that three times for the three products you get:

(6i -2j -4k) (3i - j -2k) and (-9i +3j +6k)

These add up to (0, 0, 0) as required.

Hope that's what you wanted.

Bob

hi !nval!d_us3rnam3

Not able to follow your diagram instructions but it looks like this is just

" Prove that A = (0, 2, -1); B = (2, 0, 3); C = (3, -1, 5); are collinear. "

Sufficient to show vector AB = a multiple of vector BC as these go in the same direction and have a point in common.

Bob

hi Willie

Yes they will! **

Welcome to the forum!

Bob

** Of course, as a mathematician I have to be accurate about numbers. There are some members who aren't active and, if 'everyone' means the entire population of the world, then I've have to modify that to 'some members' but not 'everyone'. Just being realistic

Now check x + y + z = 2 + 5t + 2 - 5t - 3 = 4 - 3 = 1

So the image point is on the plane.

Bob

hi KatieWind5223

Let's say you have a regular polygon with n sides. Centre point A.

Draw lines out from A to each of the vertices of the polygon. This makes n identical isosceles triangles. Label one with points B and C. Mark the midpoint of BC as point D.

Angle BAC = 360/n, and angle BAD = (360/n)/2 = 180/n

Angle ADB = 90 so we can use trig. on the triangle ADB. AD = height of a triangle.

BD/AD = tan(BAD) , so AD = BD / tan (BAD) (note BD = half a side = s/2)

area of one triangle = 1/2 side x height = 1/2 . s . s/2 .1/tan(BAD)

area of polygon = n . 1/2 . s . s/2 / tan(180/n)

Hope that helps,

Bob

hi !nval!d_us3rnam3

Not all of your Latex was showing correctly so I have removed the command that were causing this. I think our server has less features than you.

For part (a) if you re-write the line as a column vector then you can multiply A x l. Then check the value of x + y + z.

I've come up with a way to do (b) but it's not very elegant. Here goes:

Consider (1,1,-1) and (0,1,0). These are two points in P and so may be considered as a suitable basis.

Multiply these by the second matrix, and show that both points lie on the line. If the base vectors map onto the line then all points in the plane will also.

LATER EDIT: Slight re-think. Find A where the line crosses the plane. Also pick two points in the plane, B and C. Use AB and AC as the basis. (Because these transformations are 'linear' it still comes out the same.)

Bob

hi 1conga

Welcome to the forum.

Not sure what game you are referring to.

Note to members:

Apknite has generated some controversy on-line about whether it's safe to use. This link shows mixed opinions:

Bob

hi !nval!d_us3rnam3

That's what I'm getting for D too.

Try working out D^2 and D^3. You'll spot a clear pattern and shouldn't have difficulty 'guessing' what D^n would be.

Finally consider

(PDP -¹) (PDP -¹ ) (PDP -¹ ) (PDP -¹ )…. (PDP -¹ ) = P D (P -¹P) D (P -¹P) D (P -¹P) D (P -¹P) D ….D P -¹ = P D^n P -¹

Hope that helps,

Bob

hi Monox D. I-Fly

That's correct for the first step. Then:

I'll leave you to finish this.

Bob

hi stupidblogger

New members are welcome but, if you had read my post carefully, you would realise that yours is close to unacceptable too. Please read this:

Bob

hi !nval!d_us3rnam3

I've inserted [math/math] tags to make your matrices appear properly.

If you find the inverse for P then you can do this:

That should get you started. Post back D if you need more.

Bob

hi 666 bro

Welcome to the forum.

hi Amandafuente

You seem to be using the forum to post adverts. This is against forum rules and puts the forum's existence at risk as we are sponsored by paying companies.

I have removed the link and will ban you if this is repeated.

Bob

Hi Xavier

Great to hear from you. You've come to the right place. There's loads of excellent teaching materials on the MathIsFun main site and you are welcome to ask for help here on the forum.

Bob

I'm from the UK so I don't know what a score of 233 means.

hi Jamus

Welcome to the forum.

The rules of algebra are just the rules for arithmetic so you always ought to be able to substitute numbers back to check the algebra.

Even when you don't have a number answer this can be useful.

example: Suppose I have been asked to expand a bracket and I'm wondering if I've done it correctly.

Have I done this correctly? Choose a value for x, say x = 5.

These two expressions have give the same result when x=5. There is a small chance that my choice for x gave the same results by luck alone, but it's not likely. To give myself a better chance of avoiding this I didn't choose x = 0, or x=1, or x = 3.

Recently I had some tricky algebra to simplify and I kept getting the wrong result. So I started a spreadsheet and evaluated my expressions at each step. As soon as I got a wrong number I knew I'd slipped up on that line of algebra and could find my error more easily. Ten lines later I had eliminated all my errors and I had the right result. If you want to see why I was having difficulty here's the problem:

Bob