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Hyacinthos · We discuss themes on Triangle Geometry

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  • Members: 3
  • Category: Geometry
  • Founded: Dec 22, 1999
  • Language: English
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... Dear extreme trianglist, There is still a little bit not quite correct in Your solution in case (2) i.e. P and A,B are on different sides of ell. We have...
5 Mar 6, 2001
2:33 pm

Lambrou Michael
lambrou@...
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... Dear Antreas, Let P=(f,g,h) in normalized barycentrics. Then A'' = (f-cotC-cotB,g+cotC,h+cotB). By this the locus becomes ... det | f+cotC...
14 Mar 6, 2001
12:55 pm

xpolakis@...
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... ^^^^^^^ Let's replace that condition with: PA" = PA ... PB" = PB, PC" = PC ... APH...
1 Mar 5, 2001
11:13 pm

xpolakis@...
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sin(2A + B) If ------------ := k, sinB tan(A+B) compute --------- in terms of k. tanA PS: I modified the problem found in: F. Michel - M. Potron : La...
2 Mar 4, 2001
8:49 pm

Ken Pledger
Ken.Pledger@...
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... Here is it: Let ABC be the reference triangle. Consider the rectangular circumhyperbola (Hp) centered at P = (u:v:w) [a point on the NPC of ABC]. 1....
3 Mar 4, 2001
1:34 pm

xpolakis@...
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... Let's copy another problem from the above book (pp. 1-7; annee 1901): Soient Ox, Oy deux axes de coordonnees rectangulaires, soient a et a' les abscisses...
1 Mar 3, 2001
6:54 pm

xpolakis@...
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Let ABC be a triangle. Find the locus of P in 3D such that: PA^2 - a^2 = PB^2 - b^2 = PC^2 = c^2. Antreas PS: This problem (but in different wording) was given...
2 Mar 2, 2001
9:20 pm

Barry Wolk
wolkb@...
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Find the sides of a triangle if its perimeter is 16 m, "altiperimeter" 13.6 m, and area 12 sqm. [altiperimeter = sum of the three altitudes] APH...
5 Mar 2, 2001
6:35 pm

Lambrou Michael
lambrou@...
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Let AB be a given line segment and (K) a given circle. Find a point P on (K) such that ang(APB) = max. APH...
2 Mar 1, 2001
6:17 pm

Lambrou Michael
lambrou@...
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Dear friends, What is the locus of P for which the centers of the squares inscribed in triangles PBC on the side BC, PCA on CA, and PAB on AB form a triangle...
6 Feb 28, 2001
9:16 pm

wolkb@...
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... ^ - APH...
1 Feb 28, 2001
8:37 pm

xpolakis@...
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... The locus of P such that AA", BB", CC" are concurrent is the circumcubic with equation (in 3linears): x(y-z)[(y+z)cosBcosC - xcosA] + y(z-x)[(z+x)cosCcosA...
1 Feb 28, 2001
7:18 pm

xpolakis@...
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let ABC be a triangle and A'B'C' the pedal triangle of a point P. Denote PA' = x, PB' = y, PC' = z (actual 3linears of P). Let A", A* be two points on PA' (see...
1 Feb 27, 2001
1:27 pm

xpolakis@...
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1 Feb 26, 2001
8:13 pm

Paul Yiu
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Dianzhou Zhang, Professor of Mathematics, East China Normal University, Shanghai, is co-chair of a committee whose charge is to redesign high school...
1 Feb 26, 2001
7:34 pm

xpolakis@...
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Dear friends, A and B are two points on the same side of a line ell. To locate (construct if possible) a point P such that the sum of the distances from P to...
2 Feb 26, 2001
3:19 am

yiu@...
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Neuberg Theorem (1897): Let l be a line intersecting the sides BC,CA,AB of the reference triangle ABC at A1,B1,C1, respectively, and D a point on the...
2 Feb 25, 2001
9:37 pm

yiu@...
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[APH] ... In the general case: ABC = arbitrary triangle, the AA', BB', CC' concur iff M lies on our old friend (Neuberg cubic). On the other hand, if A',B',C'...
1 Feb 24, 2001
1:18 pm

xpolakis@...
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Soient A',B',C' les symetriques d'un point M par rapport aux cotes BC,CA,AB d'un triangle equilateral ABC. Demontrer que les droites AA',BB&#39;,CC' sont...
1 Feb 23, 2001
10:33 pm

xpolakis@...
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... Let A'B'C' be the pedal triangle of a point P. Let Ab, Ac be the midpoints of BA', CA' resp. Bc, Ba CB', AB' Ca, Cb...
1 Feb 19, 2001
8:53 pm

xpolakis@...
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... Solution: l C' k-l D_________________________C ... k-m | |k-m |k-m ... D'|----P--------------------|B' ... m | |...
1 Feb 19, 2001
8:43 pm

xpolakis@...
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Soient un triangle ABC et deux points D et E qui parlagent harmoniquement le cote BC. Demontrer que le cercle ADE passe par un point fixe [...] (JdME...
14 Feb 15, 2001
8:13 pm

wolkb@...
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Anallagmatic cubics, part 9. ABC are the vertices of the reference triangle. w1, w2 are conjugate points (these notes default to isotomic conjugate with ...
1 Feb 12, 2001
2:38 am

Steve Sigur
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... The locus is a quartic: We have the trigonometrical identity: q + u + v + w = 180 d. ==> tanq + tanu + tanv + tanw = tanutanvtanw + tanvtanwtanq +...
1 Feb 11, 2001
9:10 pm

xpolakis@...
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... In Nicomachus we find the terms: promekes = oblong amphimekes = odd heteromekes = even idiomekes = square. APH...
3 Feb 11, 2001
9:09 pm

xpolakis@...
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I included Hyacinthians because of the Greek connexion. While still a teen-ager I learned from Dickson the word `pronic'. E.g. in his footnote reference to ...
2 Feb 10, 2001
4:19 pm

John Conway
conway@...
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Let ABCD be a quadrilateral and P a point on its plane. Which is the locus of P such that: ang(PAB) + ang(PBC) + ang(PCD) + ang(PDA) = 180 deg. ? APH...
2 Feb 8, 2001
6:38 pm

Paul Yiu
yiu@...
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Hi, all. My conjecture is the following : Let P be in the interior of triangle ABC, and let the lines AP, BP, CP intersect the sides BC, CA, AB in L, M, N,...
4 Feb 4, 2001
1:31 am

mnas@...
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Anallagmatic cubics, part 8. ABC are the vertices of the reference triangle. w1, w2 are conjugate points (these notes default to isotomic conjugate with ...
1 Feb 3, 2001
5:22 am

Steve Sigur
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Dear all, Keith Dean and I have been discussing some constructions for reciprocal conjugacies (terminology?) lately. That is conjugacies mapping x:y:z --> u/x...
1 Jan 31, 2001
2:43 pm

Floor van Lamoen
f.v.lamoen@...
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