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A construction problem
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Dear Hyacinthists,
Somebody asked me the following question which is probably well known:
Given a natural number k, and three points P, Q, R on sides BC, AC,
AB of triangle ABC, such that BP/PC = CQ/QA = AR/RB = k. Construct
the triangle.
Thanks,
Barukh. 0 Attachment
On Tue, 6 Mar 2001 barukh.ziv@... wrote:
> Dear Hyacinthists,
The ratios above could be any three given (not necessarily equal).
>
> Somebody asked me the following question which is probably well known:
>
>
> Given a natural number k, and three points P, Q, R on sides BC, AC,
> AB of triangle ABC, such that BP/PC = CQ/QA = AR/RB = k. Construct
> the triangle.
>
> Thanks,
> Barukh.
>
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
>
Here is one way to proceed, using Ceva four times.
1) If RQ cuts BC at S, Ceva on ABC with RQ as transversal gives you BS:SC
and hence BP:SP.
2) If RQ cuts AP at X, Ceva on BRS with AC as transversal gives you
(using 1)), SQ:QR
3) BRS with AP as transversal gives you SX:XR. So using 2) you can find
QX:XR.
Thus X is determined.
4) ABP with RX sa transversal gives you PX:XA.
Thus A is determined. For B and C easier now to argue from the known
ratios AR:RB, AQ:QC.
That does it, although I don"t promise you it is the shortest way. This
is what came to my mind as an initial reaction.
All the best from sunny Crete,
Michael. 0 Attachment
>
<clip>
>
> On Tue, 6 Mar 2001 barukh.ziv@... wrote:
>
> > Dear Hyacinthists,
> >
> > Somebody asked me the following question which is probably well known:
> >
> >
> > Given a natural number k, and three points P, Q, R on sides BC, AC,
> > AB of triangle ABC, such that BP/PC = CQ/QA = AR/RB = k. Construct
> > the triangle.
> >
> >
>
>
> Here is one way to proceed, using Ceva four times.
>
Well, that was when I was rushing for my class. Here is a much easier
way, now that my class is over.
If X,Y divide PQ, QR into ratio 1/k respectively (i.e
PX:XQ=1/k=QY:YR) then it is easy to see that XY is parallel to AC. So
the line AC is determined. Similarly for AB,BC and we are done.
Sorry!
Michael 0 Attachment
>
I mentioned this morning the above problem to my colleague here on
> On Tue, 6 Mar 2001 barukh.ziv@... wrote:
>
> > Dear Hyacinthists,
> >
> > Somebody asked me the following question which is probably well known:
> >
> >
> > Given a natural number k, and three points P, Q, R on sides BC, AC,
> > AB of triangle ABC, such that BP/PC = CQ/QA = AR/RB = k. Construct
> > the triangle.
> >
Crete, Manolis Katsoprinakis: We were both witnesses in a court case, and
to kill time while waiting in the noisy, smoky, chaotic foyer, I explained
the problem.
He walked slowly to the end of the corridor and back, and a few
minutes later came with the following solution:
Start with any point O1, and find on the extension of O1R a point O2 such
that O1R:RO2 = 1:k. On the extension of O2Q find a point O3 such
that O2Q:QO3 = 1:k, and finally on the extension of O3P find a point O4
such that O3P:PO4 = 1:k. It is easy to see, using similarity, that the
lines BO1, AO2,CO3 and BO4 are parallel. In particular O1, B, O4 are
collinear. But O1, O4 are known, so (the uknown) B is somewhere on the
line O1O4.
Start all over again with a new point in place of O1, to find a new O4.
Where the new and old lines O1O4 meet, is B. Etc.
Cute isn't it?
Michael 0 Attachment
Dear all
Given a triangle ABC.Draw a semicircle inwardly on BC as
diameter.Then the construction problem is to construct a circle T which
touches AB,AC and the semicircle.
The circle constructed at once after finding it�s radius.
I computed the radius �r� as :
r = [P+sqrt(P^24Q)]/[2*(cotA/2)^2]
where P = 2*cotA/2*(c(a/2*cosB)) + a(1+sinB) and
Q = (cotA/2)^2*(c^2accosB)
Is this tedious expression further elegantly simplified?
Let A* be the point of tangency of circle T and semicircle on BC of
diameter.Similarly define B* and C*.Then are AA* ,BB* and CC* concurrent?
I�m sure this must have been discussed before.
Yours faithfully
Atul.A.Dixit
_________________________________________________________________
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Dear Hyacinthians,
(This isn't really a triangle geometry problem, but I thought you might find
it interesting.)
Let points A, B and circle c be given, such that A and B are both outside c.
Can we construct with ruler + compass the point P on c minimizing AP + BP?
What if A and B are inside c?
David
_________________________________________________________________
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Dear David, Antreas and other Hyacnthists
> (This isn't really a triangle geometry problem, but I thought you
might find
> it interesting.)
outside c.
>
> Let points A, B and circle c be given, such that A and B are both
> Can we construct with ruler + compass the point P on c minimizing
AP + BP?
> What if A and B are inside c?
Let O be the center of c.
We have to find the points P on c such as the line OP is a bisector
of APB (Theorem of Lagrange, for instance).
If we notice that the locus of M such as OM bisects AMB is a
strophoid with a node at A, the common points of c and the strophoid
are not rule and compass constructible.
Nevertheless, using the inversion wrt c, we can get the required
points P as the common points of c and the rectangular hyperbola,
inverse of the strophoid.
The rectangular hyperbola goes through A, the inverses B', C' of B,
C, his center is the midpoint of B'C' and the asymptots are parallel
to the bisectors of AOB.
The common points of c and the rectangular hyperbola give the maxima
and minima of AP + BP.
Friendly. JeanPierre 0 Attachment
Dear Antreas, David and other Hyacinthists
> (This isn't really a triangle geometry problem, but I thought you
might find
> it interesting.)
outside c.
>
> Let points A, B and circle c be given, such that A and B are both
> Can we construct with ruler + compass the point P on c minimizing
AP + BP?
> What if A and B are inside c?
As I didn't receive my first answer and as there were some typos, I
rewrite it.
We have to find the points of c such as the line OP is a bisector of
APB (O = center of c) : use, for example, the Lagrange theorem.
Those points maximize or minimize AP + BP.
As the locus of M such as OM bisects AMB is a strophoid with a node
at O, those points P are not rule and compass constructible.
Nevertheless, we can get those points as the common points of c and
the rectangular hyperbola, inverse of the strophoid.
The hyperbola goes through O, the inverses A', B' of A, B; the center
is the midpoint of B'C' and the asymptots are parallel to the
bisectors of AOB.
With my apologizes for the typos and the duplicaed message.
Friendly. JeanPierre 0 Attachment
Dear JeanPierre, David and other Hyacinthists
[DL]>> Let points A, B and circle c be given, such that A and B are both
[JP]
> outside c.
>> Can we construct with ruler + compass the point P on c minimizing
> AP + BP?
>> What if A and B are inside c?
> As I didn't receive my first answer and as there were some typos, I
I didn't receive your answer either !
> rewrite it.
> We have to find the points of c such as the line OP is a bisector of
This is related to "circular billiard" and optic of course.
> APB (O = center of c) : use, for example, the Lagrange theorem.
> Those points maximize or minimize AP + BP.
> As the locus of M such as OM bisects AMB is a strophoid with a node
> at O, those points P are not rule and compass constructible.
> Nevertheless, we can get those points as the common points of c and
> the rectangular hyperbola, inverse of the strophoid.
> The hyperbola goes through O, the inverses A', B' of A, B; the center
> is the midpoint of B'C' and the asymptots are parallel to the
> bisectors of AOB.
I'm not 100% happy (that's very very unusual...) with what JP wrote.
I made a drawing such that the rectangular hyperbola intersects the circle
at 4 pts : for two of them OP is internal bisector of ABP and external for
the two others. In this latter situation, there is no max. or min. for
AP+BP.
If I denote by a & b the polars of A & B in the circle intersecting at H,
the rectangular hyp. passes through H, A'=AO /\ a, B'=BO /\ b and the center
is the midpoint of A'B'.
For each common point P of the rect. hyp. and the circle, the tangent at P
to the circle meets a & b at two points whose midpoint is P.
Best regards
Bernard 0 Attachment
Dear bernard, David, Antreas and other Hyacinthists
> [DL]
of
> >> Let points A, B and circle c be given, such that A and B are both
> > outside c.
> >> Can we construct with ruler + compass the point P on c minimizing
> > AP + BP?
> >> What if A and B are inside c?
> [JPE]
> > We have to find the points of c such as the line OP is a bisector
> > APB (O = center of c) : use, for example, the Lagrange theorem.
I should have said : OP is the internal bisector of APB
> > Those points maximize or minimize AP + BP.
> > As the locus of M such as OM bisects AMB is a strophoid with a
node
> > at O, those points P are not rule and compass constructible.
This means OM is the internal or external bisector of AMB; hence, a
common point of c and the strophoid doesn't necessarily give an
extremum.
> > Nevertheless, we can get those points as the common points of c
and
> > the rectangular hyperbola, inverse of the strophoid.
center
> > The hyperbola goes through O, the inverses A', B' of A, B; the
> > is the midpoint of B'C' and the asymptots are parallel to the
Of course, there is a typo. Please read : the center is the midpoint
> > bisectors of AOB.
of A'B'
[BG]> This is related to "circular billiard" and optic of course.
circle
>
> I'm not 100% happy (that's very very unusual...) with what JP wrote.
> I made a drawing such that the rectangular hyperbola intersects the
> at 4 pts : for two of them OP is internal bisector of ABP and
external for
> the two others. In this latter situation, there is no max. or min.
for
> AP+BP.
at H,
>
> If I denote by a & b the polars of A & B in the circle intersecting
> the rectangular hyp. passes through H, A'=AO /\ a, B'=BO /\ b and
the center
> is the midpoint of A'B'.
tangent at P
> For each common point P of the rect. hyp. and the circle, the
> to the circle meets a & b at two points whose midpoint is P.
Of course, you have right : if c and the hyperbola have only two
common points, one of them gives the max and the other one the min;
if they have 4 common points, OP is the internal bisector of APB only
for two of them.
That's exactly what I thought, but my mail was not quite correct.
Note that the strophoid is the locus of the focii of the conics
tangent at A and B to OA and OB; this means that the strophoid is the
locus of M such as iMA/iMB = OA/OB, where iM is the isogonal
conjugate of P wrt OAB.
Friendly. JeanPierre 0 Attachment
Dear Hyacinthists,
I presume that the following problem is a very classical one but I
don't know. In any case, the construction is easy :
Construct a triangle A'B'C' congruent with ABC such as the line B'C'
goes through A, the line C'A' goes through B, the line A'B' goes
through C.
Friendly. JeanPierre 0 Attachment
Here is a figure of a broken triangle:
...............................................
/ \
/ \
/ \
/ \
/ \
BC
We are given the side BC, the angles B and C
but not the sides BA, CA.
Construct the Lemoine point K.
Antreas
 0 Attachment
Dear Antreas,
In Hyacinthos message #7824, you wrote:
>> Here is a figure of a broken triangle:
Nice problem, but I think (since the vertex A can be
>>
>>
>> ...............................................
>> / \
>> / \
>> / \
>> / \
>> / \
>> BC
>>
>> We are given the side BC, the angles B and C
>> but not the sides BA, CA.
>>
>> Construct the Lemoine point K.
constructed easily) you search the most elegant
solution. Here is one attempt:
Draw the altitudes of triangle ABC starting at B and
at C, and call their feet Hb, Hc. Also denote by Ma
the midpoint of BC. The circle through Ma, Hb, Hc is
the ninepoint circle of triangle ABC and meets AB
and AC (which are given as lines) at the midpoints
Mc and Mb of AB and AC, respectively. Then, if Nb,
Nc are the midpoints of the altitudes BHb, CHc, we
get the Lemoine (symmedian) point K of triangle ABC
as the intersection of the lines MbNb and McNc.
But I gvuess you meant something different, and thus
I apologize.
Sincerely,
Darij Grinberg 0 Attachment
Dear Darij
[APH]:>>> Here is a figure of a broken triangle:
[DG]:
>>>
>>>
>>> ...............................................
>>> / \
>>> / \
>>> / \
>>> / \
>>> / \
>>> BC
>>>
>>> We are given the side BC, the angles B and C
>>> but not the sides BA, CA.
>>>
>>> Construct the Lemoine point K.
>Nice problem, but I think (since the vertex A can be
It is a restriction of the problem that A shouldn't
>constructed easily) you search the most elegant
be constructed!
>solution. Here is one attempt:
Here is mine:
>
>Draw the altitudes of triangle ABC starting at B and
>at C, and call their feet Hb, Hc. Also denote by Ma
>the midpoint of BC. The circle through Ma, Hb, Hc is
>the ninepoint circle of triangle ABC and meets AB
>and AC (which are given as lines) at the midpoints
>Mc and Mb of AB and AC, respectively. Then, if Nb,
>Nc are the midpoints of the altitudes BHb, CHc, we
>get the Lemoine (symmedian) point K of triangle ABC
>as the intersection of the lines MbNb and McNc.
>
>But I gvuess you meant something different, and thus
>I apologize.
Analysis:
K is the isogonal conjugate of G.
G divides the line segment OH in a certain ratio.
O, H are isogonal points.
O is the intersection of the loci:
perpendicular bisector of BC
and the arc (BC, certainangleintermsof B,C)
Synthesis:
Draw the B,C angle bisectors meeting at I.
Draw the perpendicular bisector of BC.
Draw the arc (BC, angle 2(180(B+C)) [= 2A]), if ABC is acute
angled at A, meeting the perp. bisector of BC at O.
[if the triangle is not acute angled at A, then the angle
is 1802(180(B+C))]
Reflect BO in BI and CO in CI.
Let H be the point of their intersection.
Let G be a point on the segment OH such that GH = (2/3)OH.
Reflect BG in BI, and CG in CI.
The intersection of these reflections is the point K.
Isn't it? :)
Antreas
 0 Attachment
Dear all,
Here is a simple poser:
In a triangle ABC,
D is the midpoint of BC,
B' is the foot of the altitude from B upon AC,and
Z is the midpoint of CH.
Suppose we are given only the locations of D, B', and Z.
Construct the triangle ABC.
Vijayaprasad
[Nontext portions of this message have been removed] 0 Attachment
Dear Vijayaprasad,
I think that as H you mean the orthocenter
of ABC.
Construct the symmetric point B" of B
relative to Z.
The parallel from B" to DZ meets the
perpendicular from B' to DZ at C.
The parallel from B' to DZ meets
CD at B.
The perpendicular from B to CZ meets
CB' at A.
Best regards
Nikos Dergiades
> Dear all,
___________________________________________________________
> Here is a simple poser:
>
> In a triangle ABC,
> D is the midpoint of BC,
> B' is the foot of the altitude from B upon AC,and
> Z is the midpoint of CH.
> Suppose we are given only the locations of D, B', and
> Z.
> Construct the triangle ABC.
>
> Vijayaprasad
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Dear Nikos Dergiades,
You mean the symmetric point B'' of B' relative to Z?
Alternatively,
Let the circles D(DB') and Z(ZB') cut at C
Let B be the symmetric of C relative to D;
H be the symmetric of Z relative to Z.
Let the perpendicular from H upon BC and
the line through C, B' intersect at A.
Best Regards,
Vijayaprasad.
 In Hyacinthos@yahoogroups.com, Nikolaos Dergiades <ndergiades@...>
wrote:>
Mail
> Dear Vijayaprasad,
> I think that as H you mean the orthocenter
> of ABC.
> Construct the symmetric point B" of B
> relative to Z.
> The parallel from B" to DZ meets the
> perpendicular from B' to DZ at C.
> The parallel from B' to DZ meets
> CD at B.
> The perpendicular from B to CZ meets
> CB' at A.
> Best regards
> Nikos Dergiades
>
> > Dear all,
> > Here is a simple poser:
> >
> > In a triangle ABC,
> > D is the midpoint of BC,
> > B' is the foot of the altitude from B upon AC,and
> > Z is the midpoint of CH.
> > Suppose we are given only the locations of D, B', and
> > Z.
> > Construct the triangle ABC.
> >
> > Vijayaprasad
>
>
>
>
> ___________________________________________________________
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Dear Vijayaprasad,
>
Yes. You are right. Sorry for the typo.
> You mean the symmetric point B'' of B' relative
> to Z?
>
Best regards
Nikos Dergiades
___________________________________________________________
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