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Difference between revisions of "Double perspective triangles"

(→‎Two triangles in double perspective are in triple perspective)
(→‎Two triangles in double perspective are in triple perspective)
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==Two triangles in double perspective are in triple perspective==
 
==Two triangles in double perspective are in triple perspective==
 
[[File:Exeter B.png|500px|right]]
 
[[File:Exeter B.png|500px|right]]
 +
[[File:Exeter C.png|500px|right]]
 
Let <math>\triangle ABC</math> and <math>\triangle DEF</math> be in double perspective, which means that triples of lines <math>AF, BD, CE</math> and <math>AD, BE, CF</math> are concurrent. Prove that lines <math>AE, BF,</math> and <math>CD</math> are concurrent (the triangles are in triple perspective).
 
Let <math>\triangle ABC</math> and <math>\triangle DEF</math> be in double perspective, which means that triples of lines <math>AF, BD, CE</math> and <math>AD, BE, CF</math> are concurrent. Prove that lines <math>AE, BF,</math> and <math>CD</math> are concurrent (the triangles are in triple perspective).
  
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<cmath>E=(b, -a),  AE: y = -\frac {a}{b}x.</cmath>
 
<cmath>E=(b, -a),  AE: y = -\frac {a}{b}x.</cmath>
 
<cmath>D = (-a,0), C= (b,c), CD: y = c \frac {x+a}{a+b}.</cmath>
 
<cmath>D = (-a,0), C= (b,c), CD: y = c \frac {x+a}{a+b}.</cmath>
−
<math>X = CD \cap AE \cap BF = (-bk, ak),</math> where k=  \frac {c} {a+b +{\frac {bc}{a}}}$ as desired.
+
<cmath>X = CD \cap AE \cap BF = (-bk, ak),</cmath>
 +
where <math>k=  \frac {c} {a+b +{\frac {bc}{a}}}</math> as desired.
  
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''

Revision as of 14:50, 5 December 2022

Double perspective triangles

Two triangles in double perspective are in triple perspective

Exeter B.png
Exeter C.png

Let $\triangle ABC$ and $\triangle DEF$ be in double perspective, which means that triples of lines $AF, BD, CE$ and $AD, BE, CF$ are concurrent. Prove that lines $AE, BF,$ and $CD$ are concurrent (the triangles are in triple perspective).

Proof

Denote $G = AF \cap BE.$

It is known that there is projective transformation that maps any quadrungle into square.

We use this transformation for $BDFG$. We use the Claim for square and get the result: lines $AE, BF,$ and $CD$ are concurrent.

Claim for square Let $ADBG$ be the square, let $CEGF$ be the rectangle, $A \in FG, G \in BE.$ Prove that lines $BF, CD,$ and $AE$ are concurrent.

Proof

Let $BG = a, GE = b, AF = c, A = (0,0).$ Then \[B = (-a, -a), F = (0,c), BF: y = x (1 + \frac {c}{a})+c.\] \[E=(b, -a), AE: y = -\frac {a}{b}x.\] \[D = (-a,0), C= (b,c), CD: y = c \frac {x+a}{a+b}.\] \[X = CD \cap AE \cap BF = (-bk, ak),\] where $k= \frac {c} {a+b +{\frac {bc}{a}}}$ as desired.

vladimir.shelomovskii@gmail.com, vvsss

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