Segment of length pq
I.N. Galidakis
method:
Given segments of length p and q (not necessarily integral), how does one
construct (with ruler and compass) a segment of length p*q?
I need to construct the ratio:
(p + sqrt(p^2 + 4*q))/2, given p and q.
and the construction of p^2 (inside the root) comes up. I remember how to
construct integral multiples, like 4*q, and square roots (starting with a right
triangle with sides 1,1 and proceeding in a spiral adding an orthogonal side of
1 to the found hypotenuse), but I don't know how to construct p^2, hence my
question.
Many thanks,
--
Ioannis
Patrick Coilland
One solution for constructing pq :
Choose to non parallel lines D1 and D2 with intersection O.
Put A,B on D1 such that OA=d1 and OB=d2
Put C on D2 such that OC=d3.
Build the circle owning A,B,C (if d1=d2, consider the circle tangent to
D1 at common point A=B)
Let D the second intersection of the circle with D2 and x=OD.
You get xd3=d1d2 and so x=d1d2/d3
And so you can build d1d2/d3 for any d1,d2,d3
(choose d3=1 for your problem)
I.N. Galidakis
> I.N. Galidakis a Γ©crit :
> And so you can build d1d2/d3 for any d1,d2,d3
> (choose d3=1 for your problem)
Beautiful. Thanks Patrick. Can you remind me which Theorem you are quoting in
^^^^?
Thanks again,
--
Ioannis
Philippe 92
Hi,
This a pure nonsense as you *must* be also given a unit length segment.
(the reason why is that when multiplying two length you don't get a
length but an area)
That's the reason why Patrick ended by "choose d3=1 for your problem"
That is you must be given that unit length d3.
Here you don't need to construct p^2
but merely to construct some length r with r^2 = 4*q, assuming 4 is a
length, or q is an area.
This is done by a method similar to Patricks' for instance.
(other methods exist)
On any line construct OA = q and OB = 4
draw any circle through A and B
Construct the tangent OT to that circle : OT^2 = OA.OB,
hence OT = searched r.
Then sqrt(p^2 + 4*q) is the hypothenuse of a right triangle with
sides p and r.
However it seems you want to solve a quadratic.
There is a much more direct way to construct the solutions of a
quadratic x^2 - px + q = 0
with given unit length, as above, otherwise q is an area, defined as
q1*q2 with given q1 and q2.
I assume this is the case, that is q = q1*q2 with given q1 and q2,
may be with q1 = 1, doesn't matter.
On any line draw OA = q1, OB = q2
on a perpendicular in O draw OC = p
the perpendicular bisectors of AB and OC intersect in Q
The circle centered in Q through A (and B) intersects line OC in M
and M'
We have OM.OM' = OA.OB = q
and (OM + OM')/2 = OC/2 = p/2
Hence OM and OM' are the two (real) roots of the quadratic
x^2 - px + q1*q2 = 0
You may have any sign for q and p, with the same construction,
caring of the directions of OA, OB, OC, OM and OM'.
Regards.
--
Philippe C., mail : chephip, with domain free.fr
site : http://mathafou.free.fr/ (mathematical recreations)
Patrick Coilland
>>
>> You get xd3=d1d2 and so x=d1d2/d3
> ^^^^^^^
>
>> And so you can build d1d2/d3 for any d1,d2,d3
>> (choose d3=1 for your problem)
>
> Beautiful. Thanks Patrick. Can you remind me which Theorem you are quoting in
> ^^^^?
>
It's power of point O to circle.
Tim Little
> I remember how to construct integral multiples, like 4*q, and square
> roots (starting with a right triangle with sides 1,1 and proceeding
> in a spiral adding an orthogonal side of 1 to the found hypotenuse)
A much more general method and actually simpler method of constructing
square roots is via the geometric mean with a unit interval. The
geometric mean sqrt(x y) is fairly easily constructed - draw the
circle with diameter x+y and extend a perpendicular from x to the
circumference using techniques that are likely familiar to you.
- Tim
William Elliot
> Patrick Coilland wrote:
>> I.N. Galidakis a Γ©crit :
>>>
>>> Given segments of length p and q (not necessarily integral), how does one
>>> construct (with ruler and compass) a segment of length p*q?
>>>
>>> I need to construct the ratio:
>>>
>> One solution for constructing pq :
>>
>> Choose to non parallel lines D1 and D2 with intersection O.
>>
>> Put A,B on D1 such that OA=d1 and OB=d2
>> Put C on D2 such that OC=d3.
>>
>> Build the circle owning A,B,C (if d1=d2, consider the circle tangent to
>> D1 at common point A=B)
>>
>> Let D the second intersection of the circle with D2 and x=OD.
>>
>> You get xd3=d1d2 and so x=d1d2/d3
>
>> And so you can build d1d2/d3 for any d1,d2,d3
>> (choose d3=1 for your problem)
>
Thanks Patrick. Can you remind me which Theorem you are quoting in
Theorems about similar triangles,
The similar triangles
(0,0), (1,0), (0,p)
and
(0,0), (q,0), (0,x),
where the
line (q,0), (0,x)
is constructed parallel to
line (1,0), (0,p),
gives q/1 = x/p or x = pq.
Chip Eastham
One can apply the Intersecting Chords Thm.
because it says that if chords AB and CD
of a (common) circle intersect at E, then
AE times EB equals CE times ED.
Specifically, let a line segment AB be
constructed with length p+q divided at
E into lengths p and q. Construct a
segment CE of length 1 in any direction,
and construct a circle through points
A, B, C. Extend CE to where it meets
this circle at D, so ED is of required
length pq.
As in the other constructions suggested,
a segment of unit length is necessary
for the problem to make sense.
regards, chip
I.N. Galidakis
It doesn't look to me like the unit segment is needed. Maybe it's needed
implicitly in the following. I just found a very short and elegant solution for
the construction of all "Metal Ratios":
(p +/- sqrt(p^2 + 4*q))/2 (*)
The circle which has diameter AB, with A = (0,1) and B = (p,-q) intersects the
x-axis exactly at the roots of the quadratic x^2-p*x-q=0, hence the
intersections are exactly equal to the ratios (*) I am looking for.
Thanks everyone,
> regards, chip
--
Ioannis
Chip Eastham
Hi, Ioannis:
Cartesian coordinates go beyond plane
geometry compass and ruler constructions
by requiring a reference "unit" length.
In the construction you position A = (0,1)
and conclude the roots are intersections
with the x-axis. The perpendicular from
A to the x-axis is a segment of unit length.
regards, chip
I.N. Galidakis
Hi Chip,
Would you say that the aforementioned construction is then Cartesian and not
(strictly) geometric (ruler and compass)? Is there a substantial difference
here?
Thanks,
> regards, chip
--
Ioannis
Gerry Myerson
"I.N. Galidakis" <morp...@olympus.mons> wrote:
> Would you say that the aforementioned construction is then Cartesian and not
> (strictly) geometric (ruler and compass)? Is there a substantial difference
> here?
I think the difference is what seveeral people have already mentioned;
the availability, or otherwise, of a unit length.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
I.N. Galidakis
> In article <1257879586.560072@athprx03>,
> "I.N. Galidakis" <morp...@olympus.mons> wrote:
>
>> Would you say that the aforementioned construction is then Cartesian and not
>> (strictly) geometric (ruler and compass)? Is there a substantial difference
>> here?
>
> I think the difference is what seveeral people have already mentioned;
> the availability, or otherwise, of a unit length.
I am sorry, I fail to understand your term "availability". The "unit length" is
ALWAYS available. I simply set my compass to a specific length and then mark the
ruler at this length from its starting point constructing a segment, which I
call "unit length". It can be 1 inch, or 10 stadia.
I fail to understand how geometry would be even possible without the
"availability" of a "unit length". All segments in geometry have an implicit
"length" relative to some "unit length". Absent a "unit length", geometry is
just a salad of lines and circles without meaning.
--
Ioannis
Tim Little
> I fail to understand how geometry would be even possible without the
> "availability" of a "unit length". All segments in geometry have an
> implicit "length" relative to some "unit length".
The problem is that implicit unit lengths are not at all the same as
explicit unit lengths.
Suppose you have a diagram with line segments of length x cm and y cm.
You can construct a line segment of length x+y cm, without needing a
reference as to how long a cm is.
However, if your starting diagram has nothing but those line segments,
you cannot construct a length of xy cm. You need extra marks in your
diagram to determine how long a cm is.
- Tim
Gerry Myerson
"I.N. Galidakis" <morp...@olympus.mons> wrote:
> Gerry Myerson wrote:
> > In article <1257879586.560072@athprx03>,
> > "I.N. Galidakis" <morp...@olympus.mons> wrote:
> >
> >> Would you say that the aforementioned construction is then
> >> Cartesian and not (strictly) geometric (ruler and compass)? Is
> >> there a substantial difference here?
> >
> > I think the difference is what seveeral people have already mentioned;
> > the availability, or otherwise, of a unit length.
>
> I am sorry, I fail to understand your term "availability". The "unit
> length" is ALWAYS available. I simply set my compass to a specific
> length and then mark the ruler at this length from its starting point
> constructing a segment, which I call "unit length". It can be 1 inch,
> or 10 stadia.
Yes, but. The original question was (something like) if someone hands
you line segments of lengths p and q can you construct one of length
pq. You can do this if you have a segment up your sleeve whose length
is such that the segment of length p is p times as long and the segment
of length q is q times as long. If no such segment is available to you,
you're sunk.
If I tell you this line segment: ------------------
is pi units long and this line segment: -------------
is e units long and I don't give you the unit I'm using,
you can't construct (with ruler and compass, anyway)
a line segment (pi)(e) units long.
> I fail to understand how geometry would be even possible without the
> "availability" of a "unit length". All segments in geometry have an implicit
> "length" relative to some "unit length". Absent a "unit length", geometry is
> just a salad of lines and circles without meaning.
All triangles have angles summing to pi radians, unit length or no.
The square on the hypotenuse equals the sum of the squares on
the other sides, unit length or no. There's plenty of meaningful
geometry without unit lengths. And anyway there's a difference
between "there exists a unit length" and "this line segment right
here has unit length."