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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It is known that there is projective transformation that maps any quadrungle into square.
 
It is known that there is projective transformation that maps any quadrungle into square.
  
We use this transformation for <math>BDFG</math>. We use the claim and get the result: lines <math>AE, BF,</math> and <math>CD</math> are concurrent.
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We use this transformation for <math>BDFG</math>.
 +
We use the <i><b>Claim for square</b></i> and get the result: lines <math>AE, BF,</math> and <math>CD</math> are concurrent.
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<i><b>Claim for square</b></i>
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Let <math>ADBG</math> be the square, let <math>CEGF</math> be the rectangle, <math>A \in FG, G \in BE.</math>
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Prove that lines <math>BF, CD,</math> and <math>AE</math> are concurrent.
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<i><b>Proof</b></i>
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Let <math>BG = a, GE = b, AF = c, A = (0,0).</math> Then
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<cmath>B=(− a, − a), F = (0,c), BF: y= x (1 + \frac {c}{a})+c.</cmath>
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<cmath>E=(b, − a),  AE: y = − \frac {a}{b}x.</cmath>
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<cmath>D = (−a,0), C= (b,c), CD: y = c \frac {x+a}{a+b}.</cmath>
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<math>X = CD \cap AE \cap BF = (− b, a) \frac {c}{a+b +{\frac {bc}{a}}</math> as desired.
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'''vladimir.shelomovskii@gmail.com, vvsss'''

Revision as of 13:41, 5 December 2022

Double perspective triangles

Two triangles in double perspective are in triple perspective

Exeter B.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

\[B=(− a, − a), F = (0,c), BF: y= x (1 + \frac {c}{a})+c.\] (Error compiling LaTeX. Unknown error_msg)
\[E=(b, − a),  AE: y = − \frac {a}{b}x.\] (Error compiling LaTeX. Unknown error_msg)
\[D = (−a,0), C= (b,c), CD: y = c \frac {x+a}{a+b}.\] (Error compiling LaTeX. Unknown error_msg)

$X = CD \cap AE \cap BF = (− b, a) \frac {c}{a+b +{\frac {bc}{a}}$ (Error compiling LaTeX. Unknown error_msg) as desired.

vladimir.shelomovskii@gmail.com, vvsss