Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 205
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All right, you've got me, here's a little harder:
Prove that the ratio AB / CB = 5
In other words, that point C cuts off from the segment AB exactly one fifth.
// If you're really smart, work out an algorithm for dividing the base of a trapezoid into an arbitrary number of equal parts using a "ruler without divisions".
--
Those who wish may join the clever club. ;)
Consecutive division of a segment. The principle of construction is the same as in the previous case.
This is clear to me and understandable, but there is no desire to bother with the proof. Although the proof is, again, not beyond school geometry.
Sequential division of a segment. The principle of construction is the same as in the previous case.
This is clear to me and understandable, but there is no desire to bother with the proof. Although the proof is, again, not beyond the limits of school geometry.
Yes, it's beautiful. But I still don't understand why it's an exact algorithm.
Thinking over the proof.
I think I've proved it for division by three. Watch your hands:
Let's draw a line through points K and L . For now just assume it's parallel to the bases (I'll prove it later).
Now compare triangles AFH and HFB. Their bases are equal and lie on the same line, hence the line parallel to the bases that intersects them (in this case IL) will be divided into equal segments.
I.e. [if the line IL is parallel to the base of the trapezium] we have proved that IK = KL
Similarly, by consideration of triangles AGH and HGB we prove that the segment KL = LJ
But then we have the pairwise equality of two segments to the third one, i.e. IK = KL = LJ, which proves that the line IKLJ is divided into three equal parts by the mentioned points.
If now we prove that it is parallel to the trapezium bases, then we also prove that all lines parallel to it (in particular to the bases of our trapezium) are also divided by triples by rays emanating from the intersection point of side-side extensions of trapezium N.
It remains to prove that the line IJ is parallel to AB (and, of course, to FG). I shall now proceed with this.
Thus we obtain pairs of triangles AKH-FKG and HLB-FLG. They are pairwise similar, and their similarity coefficients coincide by construction (I can describe them but it is too complicated for you), so it follows that the areas S(AKH) = S(HLB) and, correspondingly, S(FKG ) = S(FLG) are equal.It suffices to consider one pair, e.g. AKH-HLB. They have equal bases and areas, hence equal heights, which is exactly what is required to prove parallelism of line IJ to bases of a trapezium.
Ъ.
// Crooked, verbose, but all seems to be correct. Check.
It remains to prove that the line IJ is parallel to AB (and of course to FG). I will do that now.
If we consider the quadrilateral FGLK, the intersection point of its diagonals, the midpoint of base FG and point H lie on one line. Sincethe intersection point of the diagonals of the trapezium, the intersection point of the extensions of its sides and the midpoint of its base lie on one line, it means that FGLK is a trapezium... QTD :)
If we consider the quadrilateral FGLK - the intersection point of its diagonals, the midpoint of its base FG and the point H lie on the same line. Sincethe intersection point of the diagonals of the trapezium, the intersection point of the extensions of its sides and the midpoint of its base lie on one line, it means that FGLK is a trapezium... WTD :)
Yes, it is beautiful and seems to be correct too. // As long as the necessary conditions with sufficient ones are not overdrawn.
PS. I did some more tinkering, everything seems to be clear.
@Mathemat : But! Although this proof is suitable for sending to your moderators, it certainly does not particularly advance in proving the correctness of the whole generator.
It would be nice to somehow attach matte induction to this case. I think // Probably linear algebra will have to be unpacked after all. Though it would be better, of course, to do without... :)
Since this is "Pure maths, physics, logic and brain games in general" ;) I propose a problem worthy of attention:
Everybody knows Newton's laws. Suppose that the motion chart of some (any) financial instrument (the price) is a trajectory of a body of mass m=1.
Determine the force acting on this body.
Since this is "Pure maths, physics, logic and brain games in general" ;) I propose a problem worthy of attention:
Everybody knows Newton's laws. Suppose that the motion chart of some (any) financial instrument (the price) is a trajectory of a body of mass m=1.
Determine the force acting on this body.
It's in humour.
It shows the extent of your understanding.