Hyacinthos is a Restricted Group with 3 members.
 Hyacinthos

 Restricted Group,
 3 members
Primary Navigation
Re: [EMHL] Triangle with a median trisected by the incircle
Expand Messages
 0 Attachment
>From: Paul Yiu <yiu@...>
Dear Paul,
>
>Characterize the triangle whose incircle
>trisects one of the medians.
I think that we can solve your problem by applying Stewart Theorem:
Assume that the incircle trisects the amedian AM := m at K:
A
/\
/ \
/ K \
/ \
/ I \
/ \
BMC
By applying Stewart Theorem in the triangle AIM we get an equation with
unknown the IM (since AI = r/sin(A/2), IK = r, AK = m/3, KM = 2m/3).
Now, by applying again Stewart Theorem in the triangle BIC we get another
equation with unknown the IM
(but it is the imedian, so we can apply directly the median formula.)
By substitution we finally get a formula in terms of some of the triangle
elements, which will characterize the triangle.
This works, but I don't know how easy are the calculations!
Antreas 0 Attachment
Dear Antreas,
Thanks for your quick response. I have actually sent this out
too early by mistake. The triangle turns out to be a very simple
one: its side lengths are in the proportions of a : b : c = 5:10:13,
the bmedian 10 being trisected by the incircle, and is parallel to
I_aC.
I have a construction problem in mind: to construct an animation
picture showing a triangle with centroid on the incircle. The condition
is
5(a^2+b^2+c^2) = 6(ab+bc+ca).
I shall go home and see if Panakis has anything to say about
triangles satisfying this condition.
Can you please search if these questions have appeared before?
Best regards.
Sincerely,
Paul

From: xpolakis@...[SMTP:xpolakis@...]
Reply To: Hyacinthos@onelist.com
Sent: Thursday, March 23, 2000 3:04 PM
To: Hyacinthos@onelist.com
Subject: Re: [EMHL] Triangle with a median trisected by the incircle
From: xpolakis@...
>From: Paul Yiu <yiu@...>
Dear Paul,
>
>Characterize the triangle whose incircle
>trisects one of the medians.
I think that we can solve your problem by applying Stewart Theorem:
Assume that the incircle trisects the amedian AM := m at K:
A
/\
/ \
/ K \
/ \
/ I \
/ \
BMC
By applying Stewart Theorem in the triangle AIM we get an equation with
unknown the IM (since AI = r/sin(A/2), IK = r, AK = m/3, KM = 2m/3).
Now, by applying again Stewart Theorem in the triangle BIC we get another
equation with unknown the IM
(but it is the imedian, so we can apply directly the median formula.)
By substitution we finally get a formula in terms of some of the triangle
elements, which will characterize the triangle.
This works, but I don't know how easy are the calculations!
Antreas

GET A NEXTCARD VISA, in 30 seconds! Get rates
as low as 0.0% Intro APR and no hidden fees.
Apply NOW!
http://click.egroups.com/1/975/3/_/672428/_/953841868/
 0 Attachment
[Paul Yiu]:
>>> Characterize the triangle whose incircle
[APH]:
>>> trisects one of the medians.
>> I think that we can solve your problem by applying Stewart Theorem:
[Paul Yiu]:
>>
>> Assume that the incircle trisects the amedian AM := m at K:
>>
>> A
>> /\
>> / \
>> / K \
>> / \
>> / I \
>> / \
>> BMC
>>
>>
>> By applying Stewart Theorem in the triangle AIM we get an equation with
>> unknown the IM (since AI = r/sin(A/2), IK = r, AK = m/3, KM = 2m/3).
>>
>> Now, by applying again Stewart Theorem in the triangle BIC we get another
>> equation with unknown the IM
>> (but it is the imedian, so we can apply directly the median formula.)
>>
>> By substitution we finally get a formula in terms of some of the triangle
>> elements, which will characterize the triangle.
>>
>> This works, but I don't know how easy are the calculations!
>Thanks for your quick response. I have actually sent this out
Dear Paul:
Quick response, but theoretical ! I have to try the solution in the practice!
>too early by mistake. The triangle turns out to be a very simple
I didn't find anything related to keywords mediantrisect*incircle. Only two
>one: its side lengths are in the proportions of a : b : c = 5:10:13,
>the bmedian 10 being trisected by the incircle, and is parallel to
>I_aC.
>
>I have a construction problem in mind: to construct an animation
>picture showing a triangle with centroid on the incircle. The condition
>is
>
>5(a^2+b^2+c^2) = 6(ab+bc+ca).
>
>I shall go home and see if Panakis has anything to say about
>triangles satisfying this condition.
>
>Can you please search if these questions have appeared before?
problems related to trisect*median, which I will append below.
As for Panakis' Trigonometry: I had a quick look at Medians Section
(pp. 68 ff) but the only problem related to medianincircle I found is this:
If in a triangle ABC: amedian = a, and b = c, then the incircle divides
the amedian in extreme and mean ratio (p. 79, #12)
The problems with trisect*median keywords:
Prove that the following construction trisects an angle of 60 d. Triangle
ABC is a 30  60  90 right triangle inscribed in a circle. Median CM
is drawn to side AB and extended to M' on the circle. Using a marked
straightedge, point N on AB is located such that CN extended to N' on
the circle makes NN' = MM'. Then CN trisects the 60 d. angle ACM.
(Jack Garfunkel)
Analogous to the median, call a line from a vertex of a triangle to a
point of trisection of the opposite side a "tredian". Then, if both tredians
are drawn from each vertex, the 6 lines will intersect at 12 interior points
and divide the area into 19 subareas, each a rational part of the area
of the triangle. Find two triangles for which each subarea is an integer,
one being a Pythagorean right triangle and the other with consecutive
integers for its three sides.
(R. Robinson Rowe)
And now I am going to make the calculations, and come back later.
Antreas 0 Attachment
I wrote:
>>From: Paul Yiu <yiu@...>
^^^^^^^^^^^^^^
>>
>>Characterize the triangle whose incircle
>>trisects one of the medians.
>
>Dear Paul,
>
>I think that we can solve your problem by applying Stewart Theorem:
>
>Assume that the incircle trisects the amedian AM := m at K:
>
> A
> /\
> / \
> / K \
> / \
> / I \
> / \
> BMC
>
>
>By applying Stewart Theorem in the triangle AIM we get an equation with
>unknown the IM (since AI = r/sin(A/2), IK = r, AK = m/3, KM = 2m/3).
To avoid trigonometry let's replace this with AI^2 = r^2 + (sa)^2
Stewart Theorem in the triangle AIM:
AI^2 * MK + IM^2 * AK = IK^2 * AM + AM*AK*KM
We have: AI^2 = r^2 + (sa)^2, AM := m, MK = 2m/3, IM := x, AK = m/3
and the equality above finally becomes:
3x^2 = 3r^2 + 2m^2  6(sa)^2 (1)
>
Median Formula in the triangle BIC:
>Now, by applying again Stewart Theorem in the triangle BIC we get another
>equation with unknown the IM
>(but it is the imedian, so we can apply directly the median formula.)
2IM^2 = IB^2 +IC^2  (BC^2)/2
We have: IM := x, BI^2 = r^2 + (sb)^2, CI^2 = r^2 + (sc)^2
So, the equality finally becomes:
4x^2 = 4r^2 + 2(sb)^2 +2(sc)^2  a^2 (2)
>
Let's see:
>By substitution we finally get a formula in terms of some of the triangle
>elements, which will characterize the triangle.
(1) and (2) ==> 8m^2 + 3a^2 = 6[4(sa)^2 + (sb)^2 + (sc)^2]
Finally we get the following formula (condition) after some simple
algebraic calculations:
8a^2 + 5b^2 + 5c^2  12ab  12ac + 6bc = 0
which characterizes a triangle, whose the amedian is trisected by its
incircle.
>
Hmmm... This worked indeed, and a solution is Paul's:
>This works, but I don't know how easy are the calculations!
a = 10k, b (or c) = 5k, c (or b) = 13k
But is it unique (for nonzero possitives satisfying the triangle
inequality: bc < a < b+c ) ?
Antreas 0 Attachment
[For Munsters, the problem was to find triangles whose incircle
trisected a median.]
Given one rational point on a conic you can find them all. Writing
b/a = x, c/a = y, we have
5x^2 + 6xy + 5y^2  12x 12y + 8 = 0.
We know one point on it, namely (0.5,1.3), but the equilateral
triangle gives us the easier one, (1,1). Solving with the line
y  1 = m(x  1), with m = r/s gives
a : b : c = 5r^2 + 6rs + 5s^2 : 5r^2 + 2rs + s^2 : r^2 + 2rs + 5s^2
This gives a genuine triangle, except that it degenerates if r = s
or s + 3r = 0.
If you put r = 2, s = 1, you get (37,25,13), which isn't very attractive.
r = 1, s = 0 gives (5,5,1) and r = 3, s = 1 gives 4(17,13,5)
which immediately reminds one of the Markov equation, which is quadratic
in each of the three variables. So, put b = 13, c = 5 in the original
and you get a = 17, but also the notsolong lost 10. R.
On Fri, 24 Mar 2000 xpolakis@... wrote:
> From: xpolakis@...
>
> I wrote:
>
> >>From: Paul Yiu <yiu@...>
> >>
> >>Characterize the triangle whose incircle
> >>trisects one of the medians.
> >
> >Dear Paul,
> >
> >I think that we can solve your problem by applying Stewart Theorem:
> >
> >Assume that the incircle trisects the amedian AM := m at K:
> >
> > A
> > /\
> > / \
> > / K \
> > / \
> > / I \
> > / \
> > BMC
> >
> >
> >By applying Stewart Theorem in the triangle AIM we get an equation with
> >unknown the IM (since AI = r/sin(A/2), IK = r, AK = m/3, KM = 2m/3).
> ^^^^^^^^^^^^^^
> To avoid trigonometry let's replace this with AI^2 = r^2 + (sa)^2
>
> Stewart Theorem in the triangle AIM:
>
> AI^2 * MK + IM^2 * AK = IK^2 * AM + AM*AK*KM
>
> We have: AI^2 = r^2 + (sa)^2, AM := m, MK = 2m/3, IM := x, AK = m/3
>
> and the equality above finally becomes:
>
> 3x^2 = 3r^2 + 2m^2  6(sa)^2 (1)
>
> >
> >Now, by applying again Stewart Theorem in the triangle BIC we get another
> >equation with unknown the IM
> >(but it is the imedian, so we can apply directly the median formula.)
>
> Median Formula in the triangle BIC:
>
> 2IM^2 = IB^2 +IC^2  (BC^2)/2
>
> We have: IM := x, BI^2 = r^2 + (sb)^2, CI^2 = r^2 + (sc)^2
>
> So, the equality finally becomes:
>
> 4x^2 = 4r^2 + 2(sb)^2 +2(sc)^2  a^2 (2)
>
> >
> >By substitution we finally get a formula in terms of some of the triangle
> >elements, which will characterize the triangle.
>
> Let's see:
>
> (1) and (2) ==> 8m^2 + 3a^2 = 6[4(sa)^2 + (sb)^2 + (sc)^2]
>
> Finally we get the following formula (condition) after some simple
> algebraic calculations:
>
> 8a^2 + 5b^2 + 5c^2  12ab  12ac + 6bc = 0
>
> which characterizes a triangle, whose the amedian is trisected by its
> incircle.
>
> >
> >This works, but I don't know how easy are the calculations!
>
> Hmmm... This worked indeed, and a solution is Paul's:
>
> a = 10k, b (or c) = 5k, c (or b) = 13k
>
> But is it unique (for nonzero possitives satisfying the triangle
> inequality: bc < a < b+c ) ?
>
>
> Antreas
>
>
>
>
>
> 
> GET A NEXTCARD VISA, in 30 seconds! Get rates
> as low as 0.0% Intro APR and no hidden fees.
> Apply NOW!
> http://click.egroups.com/1/975/3/_/672428/_/953926640/
> 
>
>
> 0 Attachment
>From: Paul Yiu <yiu@...>
Dear Paul,
>
>Characterize the triangle whose incircle
>trisects one of the medians.
Let's replace the incircle with the ninepoint circle (9PC), and the median
with either a median or an altitude.
Namely, characterize the triangles whose the ninepoint circle trisects
one median or one altitude.
Since the 9PC passes through the feet of the altitudes and the medians
we can apply the Stewart Theorem for both of them with the same formula !
A
/\
/ \
/ \
/ K \
/ \
/ N \
/ \
BMC
AM : the aaltitude or the amedian
N : the 9PC center.
K : The point where the 9PC trisects AM.
Stewart Theorem in the triangle ANM:
AK*MN^2 + KM*AN^2 = AM*KN^2 + AK*KM*AM
We have: AM := d, AK = d/3, KM = 2d/3, KN = MN = R/2, AN = Rt/2
where t = sqrt(1+8cosAsinBsinC)
[Note: We get this formula for AN by applynig the median formula in the
triangle AOH [N is the midpoint of OH]. See Panakis, p. 182, #126.
The formula in Kontogiannis, p. 6, #42: 2AN = sqrt(R(R+4Rr_a)) is wrong]
So, we get after some calculations:
2R^2 * cosAsinBsinC = (d^2)/3 ==> d^2 = 3Rh_1cosA [since h_1 = 2RsinBsinC]
1. d = m_1 (amedian) : (m_1)^2 = 3Rh_1cosA
2. d = h_1 (aaltitude): (h_1)^2 = 3Rh_1cosA ==> h_1 = 3RcosA
Note that we can find the latter condition by oneline solution:
(AK = h_1/3 & AK = AH/2) ==> 2h_1 = 3AH ==> h_1 = 3RcosA
Antreas 0 Attachment
Lambrou Michael wrote:
>On Sun, 26 Mar 2000 xpolakis@... wrote:
Good memory, Michael!
>
>> Let's replace the incircle with the ninepoint circle (9PC), and the median
>> with either a median or an altitude.
>>
>> Namely, characterize the triangles whose the ninepoint circle trisects
>> one median or one altitude.
>>
>
>If I remember well the case of the 9PC tricecting a median was a proposed
>problem in CRUX (number approximately 2250: I don't have my file with me,
I took a look at CRUX in the interval [Prob. #2250, Prob.#2260], and the
problem is #2252 (October 1998):
Prove that the ninepoint circle of a triangle trisects a median if and only
if the side lengths of the triangle are proportional to its median lengths
in some order. (Proposed by K. R. S. Sastry).
The solver proves that If the ninepoint circle trisects BD [median],
then .... 2b^2 = a^2 + c^2.
By the way, in Panakis' Trigonometry (vol. II, p. 395, #21) I read this:
2b^2 = a^2 + c^2 <=> cotA + cotC = 2cotB
>and as I will be away for a few days, it might be some time before I give
Good idea to post it here.
>you the exact details). I vaguely remember that the conclusion was that
>the sides of the triangle are proportional to the medians. I might be
>wrong. I also seem to remember that I had a nice synthetic proof of this
>(but my name was mistakenly left out from the list of solvers). If my
>memory serves me right, I will give Hyacinthists my solution, upon my
>return.
>
Paul has located some errors:
>> [Note: We get this formula for AN by applynig the median formula in the
>> triangle AOH [N is the midpoint of OH]. See Panakis, p. 182, #126.
>> The formula in Kontogiannis, p. 6, #42: 2AN = sqrt(R(R+4Rr_a)) is wrong]
>
>I wish that that was ... the only mistake in Kontogiannis' otherwise
>delightfull book (on triangle inequalities).
In p. 8, #35 and #50, the E is actually C = Gergonne Point.
> Do you know the book (in Greek) of Tsaousoglou on
No, I haven't seen this book.
>Inequalities? There is a section on Triangle Inequalities.
>Tsaousoglou is a Chemical Engineer, a regular solver at CRUX, but
>unfortunately his book is not easily available. I got it from him
>directly.
Antreas 0 Attachment
Dear Antreas and Michael,
Antreas, you were one minute ahead of me. I also
checked with Crux Math, and was about to write
to praise Michael's excellent memory.
Best regards.
Sincerely,
Paul

From: xpolakis@...[SMTP:xpolakis@...]
Reply To: Hyacinthos@onelist.com
Sent: Monday, March 27, 2000 1:58 PM
To: Hyacinthos@onelist.com
Subject: Re: [EMHL] Triangle with a median trisected by the incircle
From: xpolakis@...
Lambrou Michael wrote:
>On Sun, 26 Mar 2000 xpolakis@... wrote:
Good memory, Michael!
>
>> Let's replace the incircle with the ninepoint circle (9PC), and the median
>> with either a median or an altitude.
>>
>> Namely, characterize the triangles whose the ninepoint circle trisects
>> one median or one altitude.
>>
>
>If I remember well the case of the 9PC tricecting a median was a proposed
>problem in CRUX (number approximately 2250: I don't have my file with me,
I took a look at CRUX in the interval [Prob. #2250, Prob.#2260], and the
problem is #2252 (October 1998):
Prove that the ninepoint circle of a triangle trisects a median if and only
if the side lengths of the triangle are proportional to its median lengths
in some order. (Proposed by K. R. S. Sastry).
The solver proves that If the ninepoint circle trisects BD [median],
then .... 2b^2 = a^2 + c^2.
By the way, in Panakis' Trigonometry (vol. II, p. 395, #21) I read this:
2b^2 = a^2 + c^2 <=> cotA + cotC = 2cotB
>and as I will be away for a few days, it might be some time before I give
Good idea to post it here.
>you the exact details). I vaguely remember that the conclusion was that
>the sides of the triangle are proportional to the medians. I might be
>wrong. I also seem to remember that I had a nice synthetic proof of this
>(but my name was mistakenly left out from the list of solvers). If my
>memory serves me right, I will give Hyacinthists my solution, upon my
>return.
>
Paul has located some errors:
>> [Note: We get this formula for AN by applynig the median formula in the
>> triangle AOH [N is the midpoint of OH]. See Panakis, p. 182, #126.
>> The formula in Kontogiannis, p. 6, #42: 2AN = sqrt(R(R+4Rr_a)) is wrong]
>
>I wish that that was ... the only mistake in Kontogiannis' otherwise
>delightfull book (on triangle inequalities).
In p. 8, #35 and #50, the E is actually C = Gergonne Point.
> Do you know the book (in Greek) of Tsaousoglou on
No, I haven't seen this book.
>Inequalities? There is a section on Triangle Inequalities.
>Tsaousoglou is a Chemical Engineer, a regular solver at CRUX, but
>unfortunately his book is not easily available. I got it from him
>directly.
Antreas

MAXIMIZE YOUR CARD, MINIMIZE YOUR RATE!
Get a NextCard Visa, in 30 seconds! Get rates as low as
0.0% Intro or 9.9% Fixed APR and no hidden fees.
Apply NOW!
http://click.egroups.com/1/2122/3/_/672428/_/954183518/
 0 Attachment
I wrote:
>Lambrou Michael wrote:
I had proved that (9PC intersects the amedian m) ==> m^2 = 3Rh_1cosA (1)
>
>>On Sun, 26 Mar 2000 xpolakis@... wrote:
>>
>>> Let's replace the incircle with the ninepoint circle (9PC), and the median
>>> with either a median or an altitude.
>>>
>>> Namely, characterize the triangles whose the ninepoint circle trisects
>>> one median or one altitude.
>>>
>>
>>If I remember well the case of the 9PC tricecting a median was a proposed
>>problem in CRUX (number approximately 2250: I don't have my file with me,
>
>Good memory, Michael!
>
>I took a look at CRUX in the interval [Prob. #2250, Prob.#2260], and the
>problem is #2252 (October 1998):
>
> Prove that the ninepoint circle of a triangle trisects a median if and only
> if the side lengths of the triangle are proportional to its median lengths
> in some order. (Proposed by K. R. S. Sastry).
>
>The solver proves that If the ninepoint circle trisects BD [bmedian],
>then .... 2b^2 = a^2 + c^2.
Let me see if this implies 2a^2 = b^2 + c^2.
We have: m^2 = (2(b^2 + c^2)  a^2)/4 [median formula]
Rh_1 = R*2RsinBsinC = bc/2
cosA = (b^2 + c^2  a^2)/2bc
After some computations in (1) we get indeed that 2a^2 = b^2 + c^2.
Since we are on medians: Is the following true?
Let Ga, Gb, Gc be the centroids of GBC, GCA, GAB. [G = centroid of ABC].
Draw perpendiculars to GGa, GGb, GGc at Ga, Gb, Gc, respectively, intersecting
BC, CA, AB at A', B', C', respectively.
Then the points A',B',C' are collinear.
Anyway, regardless true or not for centroids: I am wondering for which
Triangle Point(s) X_k the above is true.
Antreas 0 Attachment
The radical axis found as the difference of two Neuberg equations
(bb:00:bb1) and (cc:cc:01) is (bbcc:cc:bb0).
The vector product of this with a like:(cc:ccaa:aa0)
gives (ccaabbcc+aabb:bbcc+aabbccaa:bbccbbaa+ccaa)
for their point of intersection.
So: the radical center of the Neuberg circles is the superJunction, sJ.
The in(fra)Junction iJ = (:aabb+bbcc:) arises all over the place 
it is the midpoint of the two Brocard points, and lies on the
meridian line  but this is the most natural place I've seen the
superJunction.
John Conway 0 Attachment
On Sun, 26 Mar 2000 xpolakis@... wrote:
> Let's replace the incircle with the ninepoint circle (9PC), and the median
If I remember well the case of the 9PC tricecting a median was a proposed
> with either a median or an altitude.
>
> Namely, characterize the triangles whose the ninepoint circle trisects
> one median or one altitude.
>
problem in CRUX (number approximately 2250: I don't have my file with me,
and as I will be away for a few days, it might be some time before I give
you the exact details). I vaguely remember that the conclusion was that
the sides of the triangle are proportional to the medians. I might be
wrong. I also seem to remember that I had a nice synthetic proof of this
(but my name was mistakenly left out from the list of solvers). If my
memory serves me right, I will give Hyacinthists my solution, upon my
return.
> [Note: We get this formula for AN by applynig the median formula in the
I wish that that was ... the only mistake in Kontogiannis' otherwise
> triangle AOH [N is the midpoint of OH]. See Panakis, p. 182, #126.
> The formula in Kontogiannis, p. 6, #42: 2AN = sqrt(R(R+4Rr_a)) is wrong]
delightfull book (on triangle inequalities).
Do you know the book (in Greek) of Tsaousoglou on
Inequalities? There is a section on Triangle Inequalities.
Tsaousoglou is a Chemical Engineer, a regular solver at CRUX, but
unfortunately his book is not easily available. I got it from him
directly.
All the best,
Michael. 0 Attachment
Dear Hyacinthists,
On Sun, 26 Mar 2000 xpolakis@... wrote:
> From: xpolakis@...
>
> >From: Paul Yiu <yiu@...>
> >
> >Characterize the triangle whose incircle
> >trisects one of the medians.
>
> Dear Paul,
>
> Let's replace the incircle with the ninepoint circle (9PC), and the median
> with either a median or an altitude.
>
> Namely, characterize the triangles whose the ninepoint circle trisects
> one median or one altitude.
>
> Since the 9PC passes through the
(clip)
Thanks to Antreas and Paul for tracing the case of the median to CRUX
2252, as my memory tickled me.
Here is more or less the solution I submitted to CRUX at the time. Nothing
is not easy enough.
The following are equivalent.
a)the 9PC trisects a median
b) the Euler line is perpandicular to a median
c) b^2 + c^2 = 2a^2
d) the medians are proportinal to the sides.
Proofs. Let the 9PC cut the median AD at K with AK=KG=GD and let N be the
centre of the 9PC so that NK=ND= (the radius of 9PC)= R/2. Note that N is
on the Euler line half way between H and O, so (using OG=2HG) we have
NG=OH/6. Note NKD is isosceles so that NG, the Euler line, is
perpendicular to AD. By Pythagoras on NGK we find
(R/2)^2 = (OH/6)^2 + (AD/3)^3
Together with OH^2 = 9R^2  (a^2 + b^2 + c^2) (not so hard to prove,
but no doubt well known) this reduces to b^2 + c^2 = 2a^2. Using this
in AD = square root of ( (2(b^2 + c^2)  a^2)/4 ), we find
AD= a(root 3)/2. Similarly the other two medians are b(root3)/2,
c(root3)/2.
To reverse the argument, some reversals are done in a trivial
manner but here is is a hint for the last.
We may assume b>a>c and so, as the larger the side the smaller the median
on it, we have BE<AD<CF thus BE=pc AD=pa, CF=pb. The value of p may be
found from AD^2 + BE^2 + CF^2 = (3/4)( a^2 + b^2 + c^2).
All the best.
Michael 0 Attachment
On Mon, 27 Mar 2000 xpolakis@... wrote:
> From: xpolakis@...
I don't know if I am exceeding private versus public domains, but
>
> Lambrou Michael wrote:
>
> > Do you know the book (in Greek) of Tsaousoglou on
> >Inequalities? There is a section on Triangle Inequalities.
> >Tsaousoglou is a Chemical Engineer, a regular solver at CRUX, but
> >unfortunately his book is not easily available. I got it from him
> >directly.
>
> No, I haven't seen this book.
>
> Antreas
>
Panos Tsaousoglou is such a nice man, that feel free to give him a call
at (01) 8231081 (Athens). I am sure he will be delighted. Anyway, give him
my warmest regards.
The book is on elementary (which is not synonymous to easy) inequalities.
Michael. 0 Attachment
Michael Lambrou wrote:
>Thanks to Antreas and Paul for tracing the case of the median to CRUX
Dear Michael,
>2252, as my memory tickled me.
>Here is more or less the solution I submitted to CRUX at the time. Nothing
>is not easy enough.
>
> The following are equivalent.
> a)the 9PC trisects a median
> b) the Euler line is perpandicular to a median
> c) b^2 + c^2 = 2a^2
> d) the medians are proportinal to the sides.
I. Panakis: Plane Trigonometry. Vol II. Athens 1973, pp. 408  417,
has quite a number of properties of the triangle whose the square of a side
is the arithmetic mean of the squares of the other two.
Here is a selection:
I denote with P the equality (2a^2 = b^2 + c^2) [to not repeat it!]
1. P ==> 2  sqrt(3) < b/c < 2 + sqrt(3)
2. P ==> A <= pi/3
3. P <==> m_1 / a = m_2 / c = m_3 / b = sqrt(3) / 2 [m_i medians]
4. P <==> (m_1)^2 + (m_2)^2 + (m_3)^2 = (3a/2)^2
5. P <==> 2(m_1)^2 = (m_2)^2 + (m_3)^2
6. P <==> tanA = 3tanw [w : The Brocard angle of ABC]
7. P <==> R^2 = (R_1)^2 = (R_2)*(R_3) [R_i the circumcircles GBC, GCA, GAB]
8. P <==> KA^2 = KB^2 * KC^2
9. P <==> GK // BC
10. P ==> The circumcircles of AGB, AGC are tangent to BC at B, C, resp.
11. P ==> The apollonian circle in respect to BC passes through G
12. P ==> G_1G_2 = G_1G_3 [G_i the orth. projections of G on BC, CA, AB]
13. P ==> The sides of the triangle G_1G_2G_3 [pedal triangle of G] are
perpendiculars to AK, BK, and CK.
14. P ==> The A, O, G, K are concyclic, lying on the circle with diameter OA
15. P ==> The H, G, B, C are concyclic
and manymany more!!
Antreas 0 Attachment
Dear Antreas,
On Fri, 31 Mar 2000 xpolakis@... wrote:
> >
> > The following are equivalent.
> > a)the 9PC trisects a median
> > b) the Euler line is perpandicular to a median
> > c) b^2 + c^2 = 2a^2
> > d) the medians are proportinal to the sides.
>
> Dear Michael,
>
> I. Panakis: Plane Trigonometry. Vol II. Athens 1973, pp. 408  417,
> has quite a number of properties of the triangle whose the square of a side
> is the arithmetic mean of the squares of the other two.
>
> Here is a selection:
>
> I denote with P the equality (2a^2 = b^2 + c^2) [to not repeat it!]
>
> 1. P ==> 2  sqrt(3) < b/c < 2 + sqrt(3)
>
> 2. P ==> A <= pi/3
>
> 3. P <==> m_1 / a = m_2 / c = m_3 / b = sqrt(3) / 2 [m_i medians]
>
> 4. P <==> (m_1)^2 + (m_2)^2 + (m_3)^2 = (3a/2)^2
>
> 5. P <==> 2(m_1)^2 = (m_2)^2 + (m_3)^2
>
> 6. P <==> tanA = 3tanw [w : The Brocard angle of ABC]
>
> 7. P <==> R^2 = (R_1)^2 = (R_2)*(R_3) [R_i the circumcircles GBC, GCA, GAB]
>
> 8. P <==> KA^2 = KB^2 * KC^2
>
> 9. P <==> GK // BC
>
> 10. P ==> The circumcircles of AGB, AGC are tangent to BC at B, C, resp.
>
> 11. P ==> The apollonian circle in respect to BC passes through G
>
> 12. P ==> G_1G_2 = G_1G_3 [G_i the orth. projections of G on BC, CA, AB]
>
> 13. P ==> The sides of the triangle G_1G_2G_3 [pedal triangle of G] are
> perpendiculars to AK, BK, and CK.
>
> 14. P ==> The A, O, G, K are concyclic, lying on the circle with diameter OA
>
> 15. P ==> The H, G, B, C are concyclic
>
> and manymany more!!
>
>
Panakis never ceases to impress me. Unfortunatelly I don't have his
Geometry nor his Trigonometry, but a friend of mine does, so I will
borough them from him.
I do, however, have since my High School days his "2500 Geometrical
Locii". I must have solved, as a student, at least 30% of them! So at
least it appears when I look back.... Some arte very hard.
Where on earth did he find so much?
Thanks, Michael
Your message has been successfully submitted and would be delivered to recipients shortly.