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Hi Agnishom

Question 6 means probably that triangles AED and BEC are similar . We have to proof that:

if EB/EA=EC/ED, then AD=BC.

EB/EA=EC/ED -->EB*ED=EA*EC --> A,B,C,D are concyclic --> AD=BC

**hammana**- Replies: 2

If a, b, c, d are 4 positive numbers, show that

a/b+b/c+c/d+d/a>=4

Hi!

It seems to me that some data is missing, because I can draw a figure satisfying the conditions for DE/EF, EK/GK, CD/AC, with an arbitrary value for CD/BC.

Hi anonimnystefy

that is right, your problem inspired this one, their solutions are quite similar.

**hammana**- Replies: 3

On 2012-04-05 at 13:38 in the "Help me" section the following was proposed

"Given an arbitrary triangle find the equilateral triangle circumscribed around the original one such that the area of that equilateral triangle is maximized"

I am proposing the dual problem, by replacing "circumscribed around" by "inscribed inside" and " maximized" by "minimized"

Hi

Hereis another way to see things:

There are 8^3 arrangements with repetition of the digits 1 to 8 , 3 by 3

The arrangements 3 by 3 of the digits 3,4,5,6 containing at least once the digits 3 and 6 are:

`3-6-x; 6-3-x; 3-x-6; 6-x-3; x-3-6; x-6-3`

x being any digit 3 to 6 included. The total makes 24; the answer would be 24/8^3=3/64

Who knows where is the truth ?

Hi bobbym

Can you try this site. It looks very professional

and goes up to 50 000 000

http://primes.utm.edu/lists/small/millions/

Hi mathmatiKs

If you program in Liberty Basic, I think there is a way to obtain 100 000 000 prime numbers, but they would be written in a file an the program will certainly take a long time to run

Hi phrontister

http://en.wikipedia.org/wiki/Sudoku_algorithms

The above site discusses the various algorithms to solve Sudoku. I learned that I am using the "Brute Force Algorithm" which is the least elegant one. The site gives the worst case grid to show the disadvantage of this algorithm. In fact it ooks me 9 hours to solve it, but the same grid rotated 90 degrees is solved in 18 minutes. I tried tho see if the compiled version runs faster, but the .exe file i obtain does not work.

Hi bobbym

I think I made a wrong manipulation. Is it all right now?

**hammana**- Replies: 3

Hi bobbym

I am proposing 2 exercises, associating theory with practice. They assume a basic knowledge of mechanics. Feel free to delete what you think is inappropriate in this forum.

It is observed that a ladder, uniform in design, length=6 m, mass=10 Kg, slips down whenever its angle with the floor exceeds 60°. We assume there is no friction on the wall.

a - If the ladder makes an angle of 65° with the floor, how far can a man of 60 Kg climb along the ladder before the ladder slips

b - What is the minimum angle of the ladder with the floor that allows the man to climb safely to the top of the ladder.

Consider a cylinder of wood, density=0.7 Kg/dm^3, length=100 cm, section square 10x10 cm, floating in a basin of water.

a - Why the cylinder can not remain floating vertically

b - A square slice of copper 10x10 cm, density = 8.92 Kg/dm^3, is fixed at the bottom of the cylinder. What is the minimum thickness of the copper slice, calculated to +/- .1 mm, that will allow the cylinder to float vertically (It is easier to use a trial and error method, or Excel or a computer for the calculation.)

c - A square slice of copper, 10x10 cm, 2.5 cm thick, is fixed at the bottoom of the cylinder, What is the height of the emerging part of the cylinder when it is at rest

d - The cylinder is pushed down until its top is at the water level, then left to oscillate freely. Neglecting the friction forces , and assuming g=9.8 m/sec^2, What is the period of its oscillations, Find the function y(t) giving the emerging part as a functio of time.

Hi anonimnystefy

Thanks for the advice. I still have some progress to make

I assume you know these basic formula of trigonometry

Consider point M1 of polar coordonates r ,b its cartesian coordinates are

If we rotate it by an angle a we obtain point M2 polar coordinates, its cartesian coordinates would be

Develop cos(a+b) and sin(a+b) as indicated above, replace r*cos(b) by , r*sin(b) by you get

equivalent to:

(Note that you go from M2 to M1 by changing a to -a)

In equation , replace by their values as functions of you get:

Hi anonimnystify

I will try to make myself clear. I refer to your figure in post #24 (I do'nt know yet how to post a figure). If C1 is the center of circle ABD, Its projection H1 on CD is the middle of AD. The projection H2 of C2, center of the circle ABC, is the middle of AC. H1 H2 = C1 C2 = CD/2. the length of CD is twice the distance of the centers C1 C2. If we draw any other line passing through A, making an angle x with C1 C2, and intersecting the circles at P and Q, the same reasoning shows that the distance PQ is 2*C1 C2*cos(x) . The factor cos(x) makes the difference.

Hi anonoimnystefy

the proof

Hi bobbeam

This hint may help:

If we build three equilateral triangles on sides AB, BC , CA, and the three circles circumscribed to these tiangles, the vertexes of any equilateral triangle circumscribed to ABC should lie on these circles. The problem becomes the following: If M is an intersection of any 2 circles, a line passing by M intersects thes circles in 2 points, P and Q. How to choose this line to maximize PQ

**hammana**- Replies: 28

Hi

Progrmming a Sudoku solver has been a challenge for me. The difficulty was to imagine the appropriate algorithm. I finally wrote the following in LB (Liberty Basic). I found no bug in it so far.

It is based on the following algorithm.

Cells are numbered from left to right, up to down. their number r goes from 1 to 81.

l1, c1 are the line and column of the cell. l2, c2 are the line and column of the upper left cell of the 3x3 square containing the cell r.

the core of the algorithm is the table tbv(81,10). each line correspond to a cell. Columns 1 to 9 indicate the digit allowed for that cell. I+f cell r is allowed the value v, tbv(r,v)=0, otherwise it contains the number of the cell that forbids that value. as explained below)

All possible values are tested for all cells, in ascending order. After assigning a value v to a cell r we should block that value in all empty cells of the same line, column and square, (see sub block). then check that no cell after r is blocked (all values forbidden). Otherwise we should erase everything done (see sub unblock) and test the next value. If we have reached the last possible value for cell r we go back to cell r-1 and repeat the same procedure.

tbv(r,10) is given any number other than zero if r is a cell which has a preassigned value, these lines are always skipped in the process.

If you like to write such program in LB and have any question, do'nt hesitate to call

REM enter here your values of a and b

a=50109:b=10

e=1

while (a/b)>(b^e)

e=e+1

fl=fl+1

wend

e1=e

'print e1

'stop

for e=e1 to 1 step-1

if a>b^e then

n=int(a/b^e)

a=a-n*b^e

print n;

else

print 0;

end if

next e

print a

Hi bobbym

Since I have plenty of free time, as a retired Telecom engineer, I can afford the luxury of looking for elegant solutions. My first and natural approach to this exercise was to introduce 3 variables: s=sidelength of the triangle, x1=DR, x2=AQ. I ended with 3 equations on which I spent much time without success. Then I thought to introduce 2 varaibles an the problem could be solved easisly

Yesterday I found that the solution is evident and perhaps the problem is not worth to be proposed as an exercise

I know that the running time of a program (or the solution of a problem) is inversely proportional to the time you spend for it, and I thank you for the time you spent for me.

I wanted to learn something about Latex to improve my presentation, but where the tutorial which appeared usually on top of the page has disappeared?

I wanted to slove the 3 equations you wrote in your reply in the litteral form, replacing "10" by "a" the side of the square and "3" by "d", the distance AP, so that the values of x1 and x2 are obtained as functions of "a" and "d". I found by other approaches of the problem that x1 and x2 have very simple values:

x1=(2*a-d)/sqr(3) and x2=(a+d)/sqr(3)

which I could not deduce from the previously mentioned equations.

I was also surprised to see that the sum x1+x2 is independant of d, which means that when then vertex P moves along the side AD, the middle of the opposite side QR remains fixed

Hi Bobbym

I see that you calculated first the position of Q and R. I failed to do it that way. Iwas interested in this problem after seeing the same topic recently posted in the "Help me" section where I lefted post #4. If you have time i would appreciate to have your comments on my post there

I am sorry again. Answers given above are the square of the sidelength

Hi Bobbym

Hi I am sorry. I choose the wrong value. I take this opportunity to reedit this exercise and make it more interersting.

An equilateral triangle P Q R is inscribed in a square A B C D . P is on AD, Q on AB and R on CD.

What is the side length of the triangle if the side length of the square = 10 cm and the distance AP=3 cm.

You can start working on numerical values and use numerical method to find the answer. It is more rewarding

to start with AD=a, Ap=d, calculate the litteral value of the side lenth of the triangle. I hope I am right this time

By the way, my professor of mathematics defined a mathematician as one who thiks a, writes b, says c , while it sould be d.

**hammana**- Replies: 10

An equilateral triangle ** P Q R ** is inscribed in a square ** A B C D **. P is on AD, Q on AB and R on CD.

What is the side length of the triangle if the side length of the square = 10 cm and the distance AP=3 cm.