Welcome to the forum.

If you 'google' it you'll find several sites that have calculators for this. You just have to enter the data.

Bob

]]>Suppose a group of

people are to play exactly matches with available tables.Each match has following restrictions:

Exactly two players are playing.

None of the players have met in a match before.

None of the players have played at that table before.

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I hope I mentioned everything needed. So for uneven amount of tables (and therefore matches) it is pretty clear that you could just divide the number of people into two sets, ie. {1,3,5...} and {2,4,6..}. Then you can place {1,2}, {3,4}, {5,6}... at each of their own table in the beginning and then just rotate all uneven numbers one place to the right and all the even numbers one place to the left.

It is not possible to solve the problem with

, but it is if . For I have found a solution as well, but is rather complicated compared to and uneven since I can't just permute two disjoint groups.I suspect that this problem is solvable for all

, but can anyone proves this? And can anyone come up with an algorithm that is not bruteforce to solve this for each ?]]>My daughter's maths teacher apparently showed them a solution without solving equation. My daughter said she understood his explanation (kind of short in fact) but she couldn't reproduce it for me. In my opinion, trying to solve this problem without equation is against the nature. But I cannot say I know everything. So I agree 100% with you Jane, and thanks for reply.

Probably her teacher just did what janetpc2008 suggested: TAKE A PIECE OF GRAPH PAPER. As a math person myself, I am also glad whenever my students solve problems that way. It shows that they understand the initiate concept rather than merely memorizing exhausting formulas.

]]>I have extended BA to D, BQR to S and constructed CSD parallel to PA.

As BP = PC the triangle BCD is similar and twice the size of BPA.

Therefore if AQ = QP = x, then SC = 2x

In triangles AQR and CSR, AP is parallel to CS, AR and RC are part of the same line, and similarly QR and RS, so these triangles are similar. As CS = twice AQ then RS is twice RQ.

Let BQ = 3y = QS. R divides QS in the ratio 1:2 then QR = y and RS = 2y.

thus BQ = 3.QR

Bob

]]>First, determine the length of the side GH, using the cosine law. Then, you should be able to work out the other angles from there, since you already know one of them.

]]>i like math because it is a "language" which is internationally available.

]]>Do you know the ellipse? The parabola sits on the border between the ellipses and the hyperbolas.

An ellipse is a single closed curve and the hyperbola is always open and has two disjoint parts.

This page: http://www.mathsisfun.com/geometry/conic-sections.html tells about the conic sections.

Start with a double cone (that is a cone and another upside-down cone sitting on its vertex.

If you cut (make a section) with a horizontal plane the shape of the section is a circle.

Incline the plane a little and you get an ellipse. As there are many angles for the cut there are many different ellipses you can make in this way.

Tilt the plane beyond the angle of slope of the cone and you get the 'family' of hyperbolas. The planes cut the lower cone and the upper cone so you get two sections. Again there are many angles that will result in hyperbolas.

When the angle of the cut equals the angle of the slope of the cone you get a parabola. The cut can slice the lower (say) cone but will just never cut again on the top cone. This is why I say it sits at the border between the hyperbolas and the ellipses.

Copernicus discovered that the planets orbit the Sun in elliptical orbits. Newton showed mathematically why this is so. It is possible for a planet to orbit a star in a circular orbit but in practice the orbit will always be at least slightly elliptical.

Comets that return also have elliptical orbits. But it is possible for an object to enter the solar system and pass close to the Sun before swinging back out and never returning. Such an orbit is a hyperbola. Again is it possible for the orbit to be a parabola but it's not likely.

Search lights use reflecting mirrors that are parabolic. If the bulb is at the focus of the parabola, then the light beam will produced parallel rays. For the same reason astronomical reflecting telescopes have a parabolic surface so that all the light gathered is concentrated at the focus.

Bob

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