ADGEOM 2497 - ADGEOM 2509

 

#2497

[Paris Pamfilos]:

Dear friends,

what is the biggest rectangle circumscribing a triangle ?

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#2498

By "circumscribing" you mean each side of the rectangle contains at least a vertex of the triangle, right?
(otherwise, the rectangle can be arbitrarily large)

Assuming this, the triangle and rectangle share a vertex A. Let ADEF be the rectangle. Then B, C must be on DE, EF, respectivley. Let x be the angle BAD. Then ADEF has area bc*cos(x)*sin(x+A), which achieves maximum (1/2)bc(1+sinA) at x=pi/4-A/2. Hence the answer is (1/2)bc+(area of ABC) if b,c are the two longest sides of the triangle.

Is this correct?

Kind regards,

Li

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#2499

Dear Li,

you are right, this is correct. Thank you.

Best regards

Paris Pamfilos

 

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#2500

 

Dear Paris,

Thanks. Note that this rectangle can be constructed by ruler and compass.
Also, if we interpret "maximal" as for perimeter, then the maximal perimeter is 2sqrt{b^2+c^2+4(area of ABC)}, where b,c are the longest sides, achieved at x=arctan(bcosA/(c+bsinA)), also constructible by ruler and compass.

Best regards,

Li

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#2502

Dear Paris, Li and all friends:

 

Let me surf this wave and add some algebraic results.

 

Let`s rename {ra}=A-Ab-Aa-Ac the A-rectangle with maximum area and build {rb}=B-Bc-Bb-Ba, {rc}=C-Ca-Cc-Cb cyclically.

 

We need make clear that {ra}, {rb}, {rc} are not necessarily with same area. "{ra} is the rectangle with maximum area having a vertex on A and non-adjacent sides  to A (or produced) containing B and C, respectively". So, if B or C are outside the segment-sides of {ra}, rectangle may not exactly circumscribe the triangle: it just satisfies the quoted properties .  Similarly {rb} and {rc}.  

 

Trilinear coordinates of vertices of {ra}=A-Ab-Aa-Ac are:

Aa = -(S^2+SB*SC)/a : c*S+b*SB : b*S+c*SC

Ab =   (c*SC-b*S)/a : S+b*c : -SA

Ac =   (b*SB-c*S)/a : -SA   : S+b*c

 

Let A'B'C' be the triangle bounded by lines (Ab-Ac), (Bc-Ba), (Ca-Cb).  

A'B'C' and ABC are perspective at:

  Z' = SA/(S*a^2+SW*b*c) : :    (trilinears)

      =  (5.433748995839291, 3.87270111163872, -1.548320439614409)

 

A'B'C' is also perspective to these triangles:

CIRCUMPERP1,  CIRCUMPERP2, CIRCUMSYMMEDIAL, EXCENTRAL, INTOUCH, LUCAS CENTRAL, LUCAS TANGENTS, YFF CENTRAL

 

Let A"B"C" be the triangle bounded by lines (Bc-Cb), (Ca-Ac), (Ab-Ba). 

A"B"C" and ABC are perspective at:

  Z" =  1/(a*((SA-S)*s+S*a)) : :

       = On {Feuerbach hyperbola} and line (8,637) (they intersect at X(8) and Z")

       = (152.850459312073800, 180.66525185037790, -191.981644943157400)

 

ABC and A"B"C" are orthologic with centers Q=X(1)=Incenter and

  Q" = (((b+c)*a^3+(b-c)^2*(a^2-2*(b+c)*s))*s-S*(2*a^3-(b+c)*(a^2+(b-c)^2)))/a : :  (trilinears)

        =  (3,142)/\ (482,1565)

        = ( -0.933631641210944, -1.41645962554592, 5.052197287844295 )

 

A"B"C" is also orthologic to next triangles:

ANTICOMPLEMENTARY,CIRCUMPERP1,CIRCUMPERP2,EULER,EXCENTRAL,EXTOUCH,FUHRMANN,GREBE INNER,GREBE OUTER,HEXYL,INCENTRAL,INTOUCH,JOHNSON,MEDIAL,MIDARC,MIXTILINEAR

 

Best regards

César Lozada

 

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#2508

Hi César,

Unless I am missing something, when I construct Aa, Ab, Ac from your trilinears, I do not get a rectangle.  Should these have been barycentrics?

Best regards,
Randy Hutson

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#2509

Hi, Randy:

 

You are right. I copied and pasted incorrectly and a minus is missing in front of  first coordinates of Ab and Ac.

 

Trilinear coordinates of vertices of {ra}=A-Ab-Aa-Ac are:

Aa = -(S^2+SB*SC)/a : c*S+b*SB : b*S+c*SC

Ab = -(c*SC-b*S)/a : S+b*c : -SA

Ac = -(b*SB-c*S)/a : -SA   : S+b*c

Could you please check it again? Thank you in advance

 

Note: Given perspectors and orthologic centers were checked again. They seem to be ok.

 

Best regards

César Lozada