one billion two hundred and thirty four million five hundred and sixty seven thousand eight hundred and ninety.

Correct?]]>

There are a couple of ways to approach this type of problem. The best ones are computational but I suppose it can be done the classical way too.

Looks like you can break the shape up into 4 right triangles. Please try that and let me know how you do.

]]>Here is how to prove it computationally, you see that M knows it is impossible.

This may take a while to finish.

```
FindInstance[{a1*b1 == 1/11, (a2*b1 + a1*b2) == 1/
11, (a3*b1 + a2*b2 + a1*b3) == 1/
11 , (a4*b1 + a3*b2 + a2*b3 + a1*b4) == 1/
11, (a5*b1 + a4*b2 + a3*b3 + a2*b4 + a1*b5) == 1/
11, (a6*b1 + a5*b2 + a4*b3 + a3*b4 + a2*b5 + a1*b6) == 1/
11, (a6*b2 + a5*b3 + a4*b4 + a3*b5 + a2*b6) == 1/
11, (a6*b3 + a5*b4 + a4*b5 + a3*b6) == 1/
11, (a6*b4 + a5*b5 + a4*b6) == 1/11, (a6*b5 + a5*b6) == 1/11,
a6*b6 == 1/11, 1 > a1 >= a2 >= a3 >= a4 >= a5 >= a6 >= 0,
1 > b1 >= b2 >= b3 >= b4 >= b5 >= b6 > 0,
a1 + a2 + a3 + a4 + a5 + a6 == 1,
b1 + b2 + b3 + b4 + b5 + b6 == 1}, {a1, a2, a3, a4, a5, a6, b1, b2,
b3, b4, b5, b6}, Reals]
```

2) Is there a function with uncountably many strict extremal points?

No, there cannot be. The following is a proof, which I found on the internet, is for maximas. We could produce a similar proof for minimas and then union them to get the same result.

I cannot believe that I actually found a solution to this problem that I understood. I thoroughly remember having this conversation here on this forum two years ago, it still feels like yesterday. I remember the sequences of events that followed.

And now, here I am fully (well, at least much better) being able to comprehend this. It does turn out I have learnt *something*.

Welcome to the forum.

Whenever I teach fractions, I start with diagrams.

example: http://www.mathisfunforum.com/viewtopic … =17631&p=6

Bob

]]>I've never heard it called that but, as you will see from this link, I did know this:

http://www.mathisfunforum.com/viewtopic … 92#p368592

Bob

]]>LATER EDIT: Had a brain-wave about how to search for it and here it is (many solutions)

http://www.mathisfunforum.com/viewtopic.php?id=20628

Bob

]]>I suggest you to read through my thread " An algorithm to generate primes " .

You will find our similarity and difference .

A soccer ball is constructed using 32 regular polygons with equal side lengths. Twelve of the polygons are pentagons, and the rest are hexagons. A seam is sewn wherever two edges meet. What is the number of seams in the soccer ball?

Number of hexagons = 32 - 12

Each hexagon has 6 sides and each pentagon has 5. Before the ball is sewn together work out how many sides in total. To make the ball pairs of sides are sewn together so halve this total.

If you post your working I'll check it over.

Bob

]]>Given what was said above and the similarities that Bob gave, we get MX as 6, and we know that XY is 3. This shows that MY = 9, thus PQ = 18.

]]>Loosely speaking, an algorithm is a step by step recipe of solving a problem.

Algorithms in mathematics are ubiquitous. Can you calculate 123 + 23? Yes, you can do so because you learnt an algorithm to do so in primary school.

Algorithms are a lot of fun. An interesting (and very approachable) book on algorithms is "Algorithms Unlocked" by Thomas Cormen.

]]>2 circles A and B both with radius 1 unit rotate freely and randomly

outside a circle E also with radius 1 unit on its circumference .

Find the expected value of the overlapping area of A and B .