## [EMHL] Re: Distance between incenter and orthocenter

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• between incenter and orthocenter Dear Barry and Walther , I post again my last message because the formulas are badly formatted. Sorry ! ... thank you for your
Message 1 of 5 , Sep 30, 2001
between incenter and orthocenter

Dear Barry and Walther ,

I post again my last message because the formulas are badly formatted. Sorry
!

>I wrote:
>> Dear friends,
>>
>> Can you help me to prove the formula :
>>
>> IH^2 = 4R^2 + 4Rr + 3r^2 -p^2

I am not an expert with barycentrics (can you recommend some references ?)

I have find a vectorial proof (similar to barycentric solution). The use of
vectors in geometry problems is explained , for example, in the book :

(*) R. Rusczyk, S. Lehoczky -The art of problem solving, vol. 2
Greater Testing Concepts, P.O. 5014
New York, NY 10185-5014

Choosing for origin the circumcenter of the triangle, we have [see (*)] :

--> --> --> -->
OH = OA + OB + OC

--> 1 --> --> -->
OI = ------- [a OA + b OB + c OC ]
a+b+c

Conseguently:

--> --> --> 1 --> --> -->
IH = OH - OI = ------- [(b+c)OA + (a+c)OB + (a+b)OC ]
a+b+c

and

-->
IH^2 = (Norm( IH ))^2 =

1
= ----[(b+c)^2R^2 +(a+c)^2R^2 +(a+b)^2R^2 +2R^2(b+c)(a+c)cos(2C) + ... ]=
4p^2

R^2
= ----[2(a^2 +b^2 +c^2) +2(ab +ac +bc) +2(b+c)(a+c)(1 -2(sinC)^2) +...] =
4p^2

R^2 c^2
= ----[2(a^2 +b^2 +c^2) +2(ab +ac +bc) +2(b+c)(a+c)(1 - 2------ ) +...] =
4p^2 4R^2

R^2 a^4+b^4+c^4
=----[2(a^2+b^2+c^2)+2(ab+ac+bc)+2(a^+b^2+c^2+3ab+3ac+3bc)- -----------]=
4p^2 R^2

R^2 a^4+b^4+c^4
=----[2(a^2+b^2+c^2)+2(ab+ac+bc)+2(a^+b^2+c^2+3ab+3ac+3bc)- -----------]=
4p^2 R^2

R^2 (a^2+b^2+c^2)^2-2(a^2b^2+a^2c^2+b^2c^2)
=----[4(a^2+b^2+c^2)+5(ab+ac+bc)- --------------------------------------]
4p^2 R^2

By substituing the well-known formulas

(ab + ac + bc) = p^2 + 4Rr + r^2

(a^2 + b^2 + c^2) = 2p^2 - 8Rr -2r^2

(a^2b^2 + a^2c^2+ b^2c^2) = (ab + ac + bc)^2 - 2abc(a + b + c) =

= (p^2 + 4Rr + r^2)^2 - 16Rrp^2

we obtain the result:

IH^2 = 4R^2 + 4Rr + 3r^2 - p^2

Best regards

Ercole Suppa

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