http://math.stackexchange.com/questions … 72#1418072

For the equation:

Use the standard approach of using Pell equations. If you use solutions of this equation.

Decisions can be recorded.

If this equation Pell.

Then decisions can be recorded.

Or.

For this.

Number

can have any sign.An interesting case of when.

The formulas can be seen that while the

can be anything. And ]]>where R is the radius of the circumcircle.

If you are not yet familiar with the construction of a circumcircle and the angle properties of a circle, you are advised to look these up first.

The circumcircle

http://www.mathisfunforum.com/viewtopic … 42#p365742

Angle properties of a circle.

http://www.mathisfunforum.com/viewtopic.php?id=17799

Posts 6 and 7

Proof

angle AOF = half angle AOB = ACB (using the angle props)

Therefore

The rest follows by a similar argument.

Bob

]]>Theorem.

There is a circle that goes through all of the points D, E, F; G, H, I; J, K, L.

Proof.

Construct the circumcircle of DEF. (I have shown the centre, M, but left out the construction lines to avoid cluttering the diagram.)

As F and D are the midpoints of AB and BC, FD//AC.

As F and J are the midpoints of AB and AN, FJ//BH

But AHB is 90. => JFD is also 90.

We already know that D and F lie on a circle; now we know that J also lies on that circle as JD subtends an angle of 90 at the circumference of the circle. (In fact JD is a diameter.) This follows from one of the angle properties of a circle which you will find here: http://www.mathisfunforum.com/viewtopic.php?id=17799

Similarly K and L will lie on this circle.

JGD is 90. => G is also on this circle.

Similarly H and I will lie on this circle.

Bob

]]>Proof.

In triangle ABC, mark the midpoints D, E, F of the sides.

Consider triangles AFE and ABC. They have two lines and the included angle in common, so they are similar, with a length scale factor of x2.

So EF is parallel to BC. Similarly for FD//AC and ED//AB.

I have already shown that the perpendicular bisectors of the sides of ABC are concurrent (meet at a point) . See http://www.mathisfunforum.com/viewtopic.php?id=22442

But these lines are the altitudes of DEF. So the altitudes of DEF are concurrent, and O is the orthocentre of DEF.

Will this be true for any triangle?

Label the sides of the triangle D, E and F. Draw lines parallel to each side going through the vertices to produce triangle ABC.

Because of the parallel lines and the common sides AFE, DEF, EDC and FBD are all congruent, so D, E and F are the midpoints of ABC. The result follows.

Bob

]]>Find the midpoints D, E and F of the sides.

Construct the perpendicular bisectors of the sides (the line that goes through the midpoint and is at 90 to the side).

Call the point where the bisector from D and from E meet, point O.

In triangles OBD and OCD, one side is common (OD), BD = DC, and ODB = ODC = 90. So the triangles are congruent (SAS).

So OB = OC.

Similarly OC = OA.

So in triangles OAF and OBF, OA = OB, OF is common and AF = BF, so the triangles are congruent (SSS). So angle OAF = OBF = 90.

So the third bisector goes through O as well.

A circle, centre O, with radius OA will also go through B and C.

Bob

]]>The purpose of this thread is to provide a link to posts that have geometric proofs. I shall gradually add to the list. From time to time I may move items about in the list to make it easier to find a particular item.

If you have requests, please reply in this thread and I’ll try to find (or create) what you want. Use the usual Help Me section if you need help with a particular problem. Normally I wouldn’t delete someone’s post, but I reserve the right to do so in this thread if it is becoming hard to keep track of new requests.

Two tangents to a circle from the same point are equal.

http://www.mathisfunforum.com/viewtopic … 09#p365709

post 6

Angle properties of a circle.

http://www.mathisfunforum.com/viewtopic.php?id=17799

Posts 6 and 7

The circumcircle

http://www.mathisfunforum.com/viewtopic … 42#p365742

The orthocentre

http://www.mathisfunforum.com/viewtopic … 50#p365750

Nine point circle

http://www.mathisfunforum.com/viewtopic … 52#p365752

The Sine Rule

http://www.mathisfunforum.com/viewtopic … 08#p365808

Bob

]]>By taking the mean and the standard deviation a couple of guessers could get together and do better than a single one. How much better? That would depend on the sd. But would that increase the expectation of the guessers? Probably not.

]]>There are no more primes after prime, z.

A=3*5*7*11*13*17*19*23*29*31............................................*z.

p= No. not factorable by any primes in A or 2.

p+/-1= otherwise A - 2p and A- 2p +/-2= twin primes.

Some gaps must be >4 as p+1 and p+4-1=.

No room for .

These gaps must be >6 as p+1 and p+3 and p+6-1=.

Again no room for .

And so on until gaps that must be >2 become gaps that must be >infinity.]]>

0/. 6 - 0 = 006

1/. 5 - 2 = ....3

2/. 6 - 4 = 2....

3/. 7 - 5 = ....52

4/. 7 - 6 = ....

5/. 9 - 6 = ....

6/. 10 - 6 = 6....

7/. 9 - 8 = ....21

8/. 11 - 8 = ....83

9/. 13 - 7 = 9....

I made a big mistake:

Primes minus No.'s with remainder e would = (not >1) using my method..........which is not true.

Each block can be 1D, 2D, 3D, 4D ....

This is to get the position of an element within a block.

The sets of blocks consist of: rows, columns, levels and blocks.

To record a position within a block rows are counted first, then columns, then the levels are bigger and blocks.

The blocks are always counted from the top level to the bottom, from left to right, just as the left and right columns, rows up and down and praise blocks also from left to right.

For example, one formed by 3 rows, 3 columns and 3 levels block, this block comprises 27 elements (3 * 3 * 3).

If we want to know which position is an element located at row 3, column 2 and Level 3, we say that this at position 26, counting as we mentioned earlier.

As we see it is very simple.

Now things get complicated a bit if we say for example you have 27 blocks, each with 27 elements (3 * 3 * 3), in total there are 729 elements in the set of all blocks and want to know which position is an element that is in row 3, column 3, Level 2 and block 27. His position is 720.

We can also reverse the problem, we know the composition of a block and know your position, determine which row, column, and block level is.

With this work I show these calculations and their possible applications in various fields.

Regards

]]>