[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 341
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Еще раз: прикол в том, что АВ != АС. Точка А не является центром окружности, связанной с этой дугой.
I understand that, but it doesn't change anything.if AB != AC then A is doomed to fall on the drawn line
если АВ != АС то А обречена попасть на нарисованную прямую
Why is that, dare I ask?
Anyway, the idea of a line dividing the perimeter in half is this. You have to divide the arc and the sum of the two segments separately.
The arc is easy to divide. Just find the centre of the circle (it can be anywhere), and then knowing the centre, divide the arc in half.
Dividing the sum of two segments in half is also technically easy. I haven't come up with an elegant construction yet.
If a is a smaller segment, b is larger, then (a+b)/2 = a/2 + b/2. Divide both segments in half, and from the middle of the larger one towards point A, you draw a half of the smaller one.
The problem is that this is not quite correct. There seems to be no concept of "more/smaller" in computations with a compass and ruler. OK, we'll figure it out.
P.S. You can also do this: If a is a smaller segment and b is larger, then (a+b)/2 = a + (b-a)/2. That is, from point A towards the end of the larger segment, we draw half the difference of the segments. Slightly more elegant, but again not quite correct.
Это почему же, смею спросить?
OK, how about this?We erase AB and AC in the drawing.
that leaves only arc BC.
we make two circles with centre B and centre C and the same radius = BC
we get a line from the two intersection points of these circles.
This line divides the arc in half.
we need to draw a line that's erased at the beginning.
however long AB and AC are, if they are equal then A is doomed to be on the line
Perimeter in half
2 points:
the first is the centre of the arc
(circle B and C represent two identical intersecting circles.
a line passing through the intersection points divides the arc in half).
second:
Draw two circles at centre B with radius AC, with centre C with radius AB.
Find the intersection (D) of one of the circles with AC or AB.
Divide AD in half - we get the second point.
какими бы не были по длине АВ и АС, если они равны,то А обречена оказаться на прямой
If they are equal, yes, of course, where can she go? But the general case is just that, when not equal. In the general case A will not be on this line.
You can draw an arc through two points at any time. Consequently, its centre can be almost anywhere.
The perimeter problem is simple and straightforward - I've already solved it. It's more difficult with the area.
The first point D is the midpoint of the arc
S(dce)=S(abd)+S(aed)
S(adc)-S(aed)=S(abd)+S(aed)
1/2*AD*hc-1/2*AD*he =1/2*AD*hb+1/2*AD*he
hc -he =hb+he
By projecting on the BC we obtain
BF=FC
The second point E:
Point of intersection of AC and the line (EF) parallel to AD
and passing through the middle of BC.
Hello!
Some time ago I had to solve the following geometrical problem: there is a pipe or a sleeve with diameter D in which it is necessary to lay cables with diameter d in number of n pieces, and the clearance (delta) between the pipe (sleeve) and the nearest cable should be observed. I can't figure out a formula or a series where I write d, n, delta in the input data, but the output is D
So that the diameter of the pipe (sleeve) is minimal.
qwerty1235813, what brand of cable are we talking about, if not a secret, which pipes (steel, PVC, HDPE, ABC)? Are the cables the same diameter? Range of variation n?