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• Dear Hyacinthists, A problem from a local list. The sides of a general triangle with standard points are in AP with common difference d. Find dist(G,I)
Message 1 of 5 , Jul 15, 2010
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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

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• 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
Message 1 of 5 , Jul 15, 2010
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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

> 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
Message 1 of 5 , Jul 15, 2010
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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 -
Message 1 of 5 , Jul 15, 2010
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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

> 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. ... Dear Luis Lemma 1: In every triangle: GI^2 = (bc+ca+ab)/3 - (a^2+b^2+c^2)/9 - 4Rr Lemma 2: If
Message 1 of 5 , Jul 16, 2010
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Sorry. there were several typos in my solution.
I repost it.

[Luis Lopes]:

> 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.

Dear Luis

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
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