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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Let <math>BG = a, GE = b, AF = c, A = (0,0).</math> Then | Let <math>BG = a, GE = b, AF = c, A = (0,0).</math> Then | ||
− | <cmath>B=( | + | <cmath>B = (-a, -a), F = (0,c), BF: y = x (1 + \frac {c}{a})+c.</cmath> |
− | <cmath>E=(b, | + | <cmath>E=(b, -a), AE: y = -\frac {a}{b}x.</cmath> |
− | <cmath>D = ( | + | <cmath>D = (-a,0), C= (b,c), CD: y = c \frac {x+a}{a+b}.</cmath> |
− | <math>X = CD \cap AE \cap BF = ( | + | <math>X = CD \cap AE \cap BF = (-bk, ak),</math> where k= \frac {c} {a+b +{\frac {bc}{a}}}$ as desired. |
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 14:48, 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 k= \frac {c} {a+b +{\frac {bc}{a}}}$ as desired.
vladimir.shelomovskii@gmail.com, vvsss