46. Outside of a quadrilateral.
by Virgil Nicula, Jun 19, 2010, 8:46 PM
PP1. Construct outside of the convex 
the triangles 
, 
. 
, 
so that
.
Denote
. Prove that 
.
Proof 1 (metric). Denote
, 
, 
, 
, 
. Thus 
, 
,
, 
, 
, 
. Apply the generalised Pitagora's theorem in the triangles 
, 
,
and 
. 
![$\boxed {MN^2=\frac {a^2}{4\cos^2x}+b^2\cos^2x-\frac 12\cdot\left(a^2+b^2-e^2\right)\cdot \cos 2x+2\sin 2x\cdot [ABC]}$](//latex.artofproblemsolving.com/6/3/5/6353b223282ed52bac22f233f30894877070077c.png)
.
![$\boxed {MR^2=\frac {a^2}{4\cos^2x}+d^2\cos^2x-\frac 12\cdot\left(a^2+d^2-f^2\right)\cdot \cos 2x+2\sin 2x\cdot [BAD]}$](//latex.artofproblemsolving.com/2/f/a/2fa0f66f4e1f38af55c253527e31b05e5d09005e.png)
.
![$\boxed {PN^2=\frac {c^2}{4\sin^2x}+b^2\sin^2x+\frac 12\cdot\left(b^2+c^2-f^2\right)\cdot \cos 2x+2\sin 2x\cdot [BCD]}$](//latex.artofproblemsolving.com/9/a/d/9ad5f261bf9260b3580d7437784a194b9c19d562.png)
.
![$\boxed {PR^2=\frac {c^2}{4\sin^2x}+d^2\sin^2x+\frac 12\cdot\left(c^2+d^2-e^2\right)\cdot \cos 2x+2\sin 2x\cdot [ACD]}$](//latex.artofproblemsolving.com/a/a/6/aa648b72688e3e578a87aa0f3cd674ec24c117b0.png)
.
Observe that
. In conclusion 
.
Remarks. For
obtain the problem from here. For 
obtain an well-known problem :
"Construct four squares outside of the convex quadrilateral on its sides. Denote the centers 
, 
, 
, 
of these squares
(in this order). Prove that
, and the midpoints of 
, 
, 
, 
are the vertices of a square".
Remark. The quadrilateral
is a "psudosquare", i.e. and . See here (<== click).
Proof 2 (with complex numbers). Define
- the point 
with the affix (complex coordinate) 
. Therefore, 
, 
, 
, 
, where 
.
Denote
and 
. Prove easily that 
, 
, 
,
, i.e. 
. Thus, 
,
and 
. Prove easily that 
.
Thus,
what means 
.
PP2. Construct outside of the convex the equilateral triangles , . , 
. Denote the
centroids
and 
of the triangles , respectively. Prove that 
and 
.
Proof (with complex numbers). Denote - the point with thw affix 
and 
, where
and 
. Therefore, 
.
Observe that
, i.e. and .
Remark. Prove easily that
and 
.
the triangles 
, 
. 
, 
so that
.Denote
. Prove that 
.Proof 1 (metric). Denote
, 
, 
, 
, 
. Thus 
, 
,
, 
, 
, 
. Apply the generalised Pitagora's theorem in the triangles 
, 
,
and 
. 
![$\boxed {MN^2=\frac {a^2}{4\cos^2x}+b^2\cos^2x-\frac 12\cdot\left(a^2+b^2-e^2\right)\cdot \cos 2x+2\sin 2x\cdot [ABC]}$](http://latex.artofproblemsolving.com/6/3/5/6353b223282ed52bac22f233f30894877070077c.png)
.
![$\boxed {MR^2=\frac {a^2}{4\cos^2x}+d^2\cos^2x-\frac 12\cdot\left(a^2+d^2-f^2\right)\cdot \cos 2x+2\sin 2x\cdot [BAD]}$](http://latex.artofproblemsolving.com/2/f/a/2fa0f66f4e1f38af55c253527e31b05e5d09005e.png)
.
![$\boxed {PN^2=\frac {c^2}{4\sin^2x}+b^2\sin^2x+\frac 12\cdot\left(b^2+c^2-f^2\right)\cdot \cos 2x+2\sin 2x\cdot [BCD]}$](http://latex.artofproblemsolving.com/9/a/d/9ad5f261bf9260b3580d7437784a194b9c19d562.png)
.
![$\boxed {PR^2=\frac {c^2}{4\sin^2x}+d^2\sin^2x+\frac 12\cdot\left(c^2+d^2-e^2\right)\cdot \cos 2x+2\sin 2x\cdot [ACD]}$](http://latex.artofproblemsolving.com/a/a/6/aa648b72688e3e578a87aa0f3cd674ec24c117b0.png)
.Observe that
. In conclusion 
.Remarks. For
obtain the problem from here. For 
obtain an well-known problem :"Construct four squares outside of the convex quadrilateral
on its sides. Denote the centers 
, 
, 
, 
of these squares(in this order). Prove that
, and the midpoints of 
, 
, 
, 
are the vertices of a square".Remark. The quadrilateral
is a "psudosquare", i.e. and . See here (<== click).Proof 2 (with complex numbers). Define
- the point 
with the affix (complex coordinate) 
. Therefore, 
, 
, 
, 
, where 
.Denote
and 
. Prove easily that 
, 
, 
,
, i.e. 
. Thus, 
,
and 
. Prove easily that 
.Thus,
what means 
.PP2. Construct outside of the convex
the equilateral triangles , . , 
. Denote thecentroids
and 
of the triangles , respectively. Prove that 
and 
.Proof (with complex numbers). Denote
- the point with thw affix 
and 
, where
and 
. Therefore, 
.Observe that
, i.e. and .Remark. Prove easily that
and 
.This post has been edited 42 times. Last edited by Virgil Nicula, Nov 23, 2015, 8:47 AM
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