- Dear Hyacinthists,

A problem from a local list.

The sides of a general triangle with standard points <I,G>

are in AP with common difference d.

Find dist(G,I) as a function of d.

Best regards,

Luis

_________________________________________________________________

TRANSFORME SUAS FOTOS EM EMOTICONS PARA O MESSENGER. CLIQUE AQUI PARA COMEÇAR.

http://ilm.windowslive.com.br/?ocid=ILM:Live:Hotmail:Tagline:senDimensao:TRANSFORME78:-

[Non-text portions of this message have been removed] - Dear Luis,

Let b = a + d, c = b + d, or a + c = 2b

then s = (a + b + c) / 2 = 3b/2, s - b = b/2 and since

s.r = (s - b) rb we get that rb = 3r

where r, rb are the radi of incircle and B_excircle.

The line IG has equation in barycentrics

(b - c)x + (c - a)y + (a - b)z = 0 or

-x + 2y - z = 0 and meets the line BC at D (0 : 1 : 2)

which means that BD/DC = 2 and the line IG is

parallel to AC. If M is the midpoint of AC and the

line BI meets AC at J then

CJ = ab/(a + c) = a/2

MJ = b/2 - a/2 = d/2 and

GI = 2MJ/3 = d/3.

The Nagel point Na is known that lies on line IG

and if BNa meets AC at K then

AK = s - c = 3b/2 - c and

KM = AM - AK = b/2 - (3b/2 - c) = c - b = d

Hence NaG = 2KM/3 = 2d/3

Best regards

Nikos Dergiades

> Dear Hyacinthists,

>

>

>

> A problem from a local list.

>

>

>

> The sides of a general triangle with standard points

> <I,G>

>

> are in AP with common difference d.

>

>

>

> Find dist(G,I) as a function of d.

>

>

>

> Best regards,

> Luis

>

>

> - Dear Luis and Nikos,

We can solve the problem by construction triangle ABC (b = a+d = c-d) as following:

Choose b = AC as one segment.

X as any point on segment AC and CX = d.

X1 = reflection of X in C

X2 = reflection of X in midpoint of AC

B = intersection of two circles centered at C passing X2 and centered at A passing X1

Y1 = midpoint of BX1

Y2 = midpoint of BX2

A1 = midpoint of BC

C1 = midpoint of AB

I = intesection of AY1 and CY2

G = intesection of AA1 and CC1

From this construction: GI//AC and

GI = 2/3*A1Y1 = 2/3*1/2*CX1 = 1/3*CX1 = d/3

Best regards,

Bui Quang Tuan

--- On Thu, 7/15/10, Luís Lopes <qed_texte@...> wrote:

> From: Luís Lopes <qed_texte@...>

> Subject: [EMHL] sides in arithmetic progression

> To: hyacinthos@yahoogroups.com

> Date: Thursday, July 15, 2010, 10:39 PM

>

> Dear Hyacinthists,

>

>

>

> A problem from a local list.

>

>

>

> The sides of a general triangle with standard points

> <I,G>

>

> are in AP with common difference d.

>

>

>

> Find dist(G,I) as a function of d.

>

>

>

> Best regards,

> Luis

> - Dear Tuan,

very good!

Another proof with vectors:

We have

GA + GB + GC = 0

a.IA + b.IB + c.IC = 0 or

(a + b + c).IG + a.GA + b.GB + c.GC = 0 or

3b.IG + (a - b)GA + (c - b)GC = 0 or

3b.IG = d.(GA - GC) = d.CA

and hence |IG| = |CA|.d/3b = d/3.

Best regards

Nikos Dergiades

> Dear Luis and Nikos,

>

> We can solve the problem by construction triangle ABC (b = a+d = c-d) as following:

>

> Choose b = AC as one segment.

> X as any point on segment AC and CX = d.

> X1 = reflection of X in C

> X2 = reflection of X in midpoint of AC

> B = intersection of two circles centered at C passing X2 and centered at A passing X1

> Y1 = midpoint of BX1

> Y2 = midpoint of BX2

> A1 = midpoint of BC

> C1 = midpoint of AB

> I = intesection of AY1 and CY2

> G = intesection of AA1 and CC1

>

> From this construction: GI//AC and

> GI = 2/3*A1Y1 = 2/3*1/2*CX1 = 1/3*CX1 = d/3

>

> Best regards,

> Bui Quang Tuan

>

> --- On Thu, 7/15/10, Luís Lopes <qed_texte@...> wrote:

>

> > From: Luís Lopes <qed_texte@...>

> > Subject: [EMHL] sides in arithmetic progression

> > To: hyacinthos@yahoogroups.com

> > Date: Thursday, July 15, 2010, 10:39 PM

> >

> > Dear Hyacinthists,

> >

> >

> >

> > A problem from a local list.

> >

> >

> >

> > The sides of a general triangle with standard points

> > <I,G>

> >

> > are in AP with common difference d.

> >

> >

> >

> > Find dist(G,I) as a function of d.

> >

> >

> >

> > Best regards,

> > Luis

> >

> - Sorry. there were several typos in my solution.

I repost it.

[Luis Lopes]:

> A problem from a local list.

Dear Luis

> The sides of a general triangle with standard points <I,G>

> are in AP with common difference d.

> Find dist(G,I) as a function of d.

Lemma 1:

In every triangle:

GI^2 = (bc+ca+ab)/3 - (a^2+b^2+c^2)/9 - 4Rr

Lemma 2:

If 2b = a + c [: b = a+d, c = a+2d ] ==> ac = 6Rr

L1 /\ L2 ==> 9GI^2 = b^2 - ac = (a+d)^2 - a(a+2d) = d^2

==> GI = d/3

APH