ANOPOLIS 39

Let ABC be a triangle, and Ma, Mb, Mc the midpoints of
the altitudes AHa, BHb, CHc of ABC, resp.

Let Pa, Pb, Pc be the reflections of H in Ma, Mb, Mc, resp.

Let Kab, Kac be the circumcenters of the triangles
PaBA, PaCA, resp.


A
/|\
/ | \
Kab / H \ Kac
/ | \
/ Ma \
/ A1| \
/ Pa \
/ | \
B--------Ha-------C

Let A1 = BKac /\ CKab. Similarly B1,C1.

The Triangles ABC, A1B1C1 are perspective.
[Antreas P. Hatzipolakis, Hyacinthos, msg #5010, 19 Mar 2002]

PROOF:

Note: I will work on an acute-angled triangle.

(ABC, A1B1C1 are perspective) <==>

cotB - cot(KacBC) cotC - cot(KbaCA) cotA - cot(KcbAB)
----------------- * ----------------- * ----------------- = 1
cotC - cot(KabCB) cotA - cot(KbcAC) cotB - cot(KcaBA)

Computing cot(KacBC):


A
/|\
/ | \w
Kab / D \ Kac
/ | \
/ | E
/ Pa \
/ | \
/ | \
B--------Ha-------C-F--


Kac is the intersection of the perp. bisector DKac of APa
and the perp. bisector EKac of AC.
Let F be the orth. proj. of Kac on BC.

cot(KacBC) = BF / KacF = (BHa + HaF)/DHa = (BHa + DKac)/DHa

Computing the lengths BHa, DKac, DHa

We have:

BHa = ABcosB = ccosB = 2RsinCcosB (1)

DKac = ADtan(DAKac) = (APa/2)tan(DAC + EAKac) =

= (APa/2)tan(90 - C + w) = (APa/2)cot(C-w) =

= (APa/2)*[(cotCcotw + 1)/(cotw - cotC)] =

= (APa/2) * [(cotCcotw + 1)/(cotw - cotC)] (i)

where w: = EAKac

Computing cotw:

We have

Triangle DAKac:
cos(DAKac) = cos(90 - C + w) = sin(C-w) = AD / AKac = (APa/2) / AKac

Triangle EAKac:
cosw = AE / AKac = (AC/2) / AKac

==> sin(C-w) / cosw = APa / AC

(sinCcosw - cosCsinw)/cosw = APa / AC

1 - cotCtanw = APa / ACsinC = APa / AHa

cotw = cotC/(1 - [APa / AHa]) (ii)


(i) and (ii) ==>

DKac = (APa/2) * [(cotCcotw + 1)/(cotw - cotC)] =

(cotC[cotC/(1 - [APa / AHa])) + 1
= (APa/2) * --------------------------------- =
(cotC/(1 - [APa / AHa])) - cotC


1 + cot^2C - [APa/AHa] AHa - APasin^2C
= AHa * ---------------------- = --------------- (2)
2cotC 2sinCcosC


DHa = DPa + PaHa = (APa/2) + AHa - APa = AHa - (APa/2) (3)

(1) and (2) and (3) ==>

cot(KacBC) = (BHa + DKac)/DHa =

ccosB + [(AHa - APasin^2C)/2sinCcosC]
= ------------------------------------- =
AHa - (APa/2)


2ccosBcosCsinC + AHa - APasin^2C
= --------------------------------
2cosCsinC(AHa - (APa)/2)

Computing cotB - cot(KaBC):

cotB - cot(KaBC) =

2ccosBcosCsinC + AHa - APasin^2C
cotB - -------------------------------- =
(2AHa - APa)cosCsinC

(2AHa - APa)cosBcosCsinC - 2ccosBsinBcosCsinC - (AHa - APasin^2C)sinB
--------------------------------------------------------------------- =
(2AHa - APa)cosCsinBsinC


Similarly

cotC - cot(KabCB) =

(2AHa - APa)cosBcosCsinB - 2bcosBsinBcosCsinC - (AHa - APasin^2B)sinC
--------------------------------------------------------------------- =
(2AHa - APa)cosBsinBsinC

==>

cotB - cot(KacBC) cosB
----------------- = ---- *
cotC - cot(KabCB) cosC

(2AHa - APa)cosBcosCsinC - 2ccosBsinBcosCsinC - (AHa - APasin^2C)sinB
* ---------------------------------------------------------------------
(2AHa - APa)cosBcosCsinB - 2bcosBsinBcosCsinC - (AHa - APasin^2B)sinC

We have AHa = 2RsinBsinC, and denote APa = 2R*f

==>

cotB - cot(KacBC) cosB
----------------- = ---- *
cotC - cot(KabCB) cosC

(2sinBsinC - f)cosBcosCsinC - 2cosBsinBcosCsin^2C -
- (sinBsinC - fsin^2C)sinB
----------------------------------------------------
(2sinBsinC - f)cosBcosCsinB - 2cosBsin^2BcosCsinC -
- (sinBsinC - fsin^2B)sinC

cosB -fcosBcosCsinC - sinBsinC(sinB - fsinC)
= ----- * --------------------------------------- =
cosC -fcosBcosCsinB - sinBsinC(sinC - fsinB)

cosB sinC -f(cosBcosC + sinBsinC) - sin^2B
= ---- * ---- * -------------------------------- =
cosC sinB -f(cosBcosC + sinBsinC) - sin^2C

cosB sinC fcosA - sin^2B
= ---- * ---- * --------------
cosC sinB fcosA - sin^2C


Computing APa := 2R*f
[Pa := Reflection of H in Ma]

A
/|\
/ | \
/ H \
/ | \
/ Ma \
/ | \
/ Pa \
/ | \
B--------Ha-------C

APa = AMa + MaPa = AMa + (HMa) = AMa + (AMa - AH) = 2(AMa) - AH =

= 2(AHa/2) - AH = AHa - AH = HHa = 2RcosBcosC

==> f = cosBcosC.

Therefore

cotB - cot(KacBC) cosB sinC cosAcosBcosC - sin^2B
----------------- = ---- * ---- * --------------------- [a]
cotC - cot(KabCB) cosC sinB cosAcosBcosC - sin^2C

Similarly

cotC - cot(KbaCA) cosC sinA cosAcosBcosC - sin^2C
----------------- = ---- * ---- * ---------------------
cotA - cot(KbcAC) cosA sinC cosAcosBcosC - sin^2A

cotA - cot(KcbAB) cosA sinB cosAcosBcosC - sin^2A
----------------- = ---- * ---- * ---------------------
cotB - cot(KcaBA) cosB sinA cosAcosBcosC - sin^2B


==>

cotB - cot(KacBC) cotC - cot(KbaCA) cotA - cot(KcbAB)
----------------- * ----------------- * ----------------- = 1
cotC - cot(KabCB) cotA - cot(KbcAC) cotB - cot(KcaBA)

==> ABC, A1B1C1 are perspective.

O(/PER E)/DEI DEI=CAI

Perspector:

From [a] we get the Perspector in Barycentrics:

((1/cosA) * (sinA) * (1 / (cosAcosBcosC - sin^2A) ::)

and in Normals

((1/cosA) * (1 / (cosAcosBcosC - sin^2A) ::) =

((1/cosA) * (1 / (cosAcosBcosC - [1 - cos^2A]) ::) =

((1/cosA) * (1 / (- 1 + cosAcosBcosC + cos^2A) ::) =

((1/cosA) * (1 / (- 1 + cosA[cosBcosC + cosA]) ::) =

((1/cosA) * (1 / (- 1 + cosA[cosBcosC + cos(180 - (B + C))]) ::) =

((1/cosA) * (1 / (- 1 + cosA[cosBcosC - cos(B+C)]) ::) =

((1/cosA) * (1 / (- 1 + cosA[cosBcosC - cosBcosC + sinBsinC]) ::) =

((1/cosA) * (1 / (- 1 + cosAsinBsinC) ::) =

((1 / cosA(1 - cosAsinBsinC) ::)

I name this perspector as "CRETAN PERSPECTOR".


Antreas P. Hatzipolakis
19 March 2002