Difference between revisions of "Double perspective triangles"
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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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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 | + | 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. | ||
+ | |||
+ | <i><b>Claim for square</b></i> | ||
+ | |||
+ | Let <math>ADBG</math> be the square, let <math>CEGF</math> be the rectangle, <math>A \in FG, G \in BE.</math> | ||
+ | |||
+ | Prove that lines <math>BF, CD,</math> and <math>AE</math> are concurrent. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>BG = a, GE = b, AF = c, A = (0,0).</math> Then | ||
+ | <cmath>B = (-a, -a), F = (0,c), BF: y = x (1 + \frac {c}{a})+c.</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>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''' |
Latest revision as of 14:51, 5 December 2022
Double perspective triangles
Two triangles in double perspective are in triple perspective
Let and
be in double perspective, which means that triples of lines
and
are concurrent. Prove that lines
and
are concurrent (the triangles are in triple perspective).
Proof
Denote
It is known that there is projective transformation that maps any quadrungle into square.
We use this transformation for .
We use the Claim for square and get the result: lines
and
are concurrent.
Claim for square
Let be the square, let
be the rectangle,
Prove that lines and
are concurrent.
Proof
Let Then
where
as desired.
vladimir.shelomovskii@gmail.com, vvsss