235. Proofs of some geometry problems with complex numbers.
by Virgil Nicula, Mar 3, 2011, 10:41 AM
Lemma 1. If 
is the point 
with affix 
, Then 
.
Lemma 2. The triangle
is equilateral if and only if 
.
Proof 1. Denote
. Observe that 
and 
. Thus, is an equilateral triangle 
or 
or 
.
Proof 2. is an equilateral triangle 
.
P1. Triangles , 
are equilateral and denote the midpoints 
, 
, 
of ![$[AD]$](//latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
, ![$[BE]$](//latex.artofproblemsolving.com/f/b/0/fb061a8a7c5f9403b5f9261840de9dfea7cb68cf.png)
, ![$[CF]$](//latex.artofproblemsolving.com/2/1/b/21bdd766ec1757878aeae83d55b7ab5917af4537.png)
respectively. Show that 
is equilateral.
Proof. I"ll show that
. Indeed, 
. In conclusion, 
.
P2. Let
be a quadrilateral such that 
and 
. Construct the equilateral triangles
, 
, 
such that the points 
lie on the same side of 
. Prove that are collinear points.
Proof 1. Let with the affix 
and the rotation 
, i.e. 
, 
and 
. Construct 
for which the quadrilateral 
is a
parallelogram. Thus,
, i.e. 
and 
the triangle 
is equilateral 
.
Therefore,
![$p+r-2q=(a-d)+[(b-c)-(a-d)]\cdot\omega =$](//latex.artofproblemsolving.com/b/2/a/b2a72f893e5775c67f6fc375ee36220a682d079f.png)
![$(a-d)+[(a-d)\cdot\omega-(a-d)]\cdot\omega =$](//latex.artofproblemsolving.com/0/e/3/0e36eebfa201f0949b2b92340358e80134f85e46.png)
, i.e. 
.
Proof 2. Denote:
. Prove easily that 
is an equilateral. So 
and 
is cyclic, so
. From 
and 
we have: 
. Similarly, we also have: 
. Hence 
are collinear. We also have is midpoint of 
P3. Let be a parallelogram and let 
be two squares what belong to outside of the parallelogram. Prove that 
Proof. Denote the affix of the point from our plane and denote 
Suppose w.l.o.g. 
and 
Therefore,
![$\left\{\begin{array}{ccccc}
m-q & = & [b+i(b-a)]-[-a-i(a+b)] & = & (a+b)+2ib\\\\
p-d & = & [b-i(a+b)]-(-b) & = & 2b-i(a+b)\end{array}\right\|\ .$](//latex.artofproblemsolving.com/a/e/d/aed4709127ad663d135c547208b3ca8ae9cbf8f5.png)
Observe that
and 
Prove similarly that 
and 
is the point 
with affix 
, Then 
.Lemma 2. The triangle
is equilateral if and only if 
.Proof 1. Denote
. Observe that 
and 
. Thus, is an equilateral triangle 
or 
or 
.Proof 2.
is an equilateral triangle 
. P1. Triangles
, 
are equilateral and denote the midpoints 
, 
, 
of ![$[AD]$](http://latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
, ![$[BE]$](http://latex.artofproblemsolving.com/f/b/0/fb061a8a7c5f9403b5f9261840de9dfea7cb68cf.png)
, ![$[CF]$](http://latex.artofproblemsolving.com/2/1/b/21bdd766ec1757878aeae83d55b7ab5917af4537.png)
respectively. Show that 
is equilateral.Proof. I"ll show that
. Indeed, 
. In conclusion, 
.P2. Let
be a quadrilateral such that 
and 
. Construct the equilateral triangles
, 
, 
such that the points 
lie on the same side of 
. Prove that are collinear points.Proof 1. Let
with the affix 
and the rotation 
, i.e. 
, 
and 
. Construct 
for which the quadrilateral 
is aparallelogram. Thus,
, i.e. 
and 
the triangle 
is equilateral 
.Therefore,
![$p+r-2q=(a-d)+[(b-c)-(a-d)]\cdot\omega =$](http://latex.artofproblemsolving.com/b/2/a/b2a72f893e5775c67f6fc375ee36220a682d079f.png)
![$(a-d)+[(a-d)\cdot\omega-(a-d)]\cdot\omega =$](http://latex.artofproblemsolving.com/0/e/3/0e36eebfa201f0949b2b92340358e80134f85e46.png)
, i.e. 
.Proof 2. Denote:
. Prove easily that 
is an equilateral. So 
and 
is cyclic, so
. From 
and 
we have: 
. Similarly, we also have: 
. Hence 
are collinear. We also have is midpoint of 
P3. Let
be a parallelogram and let 
be two squares what belong to outside of the parallelogram. Prove that 
Proof. Denote the affix
of the point from our plane and denote 
Suppose w.l.o.g. 
and 
Therefore,
![$\left\{\begin{array}{ccccc}
m-q & = & [b+i(b-a)]-[-a-i(a+b)] & = & (a+b)+2ib\\\\
p-d & = & [b-i(a+b)]-(-b) & = & 2b-i(a+b)\end{array}\right\|\ .$](http://latex.artofproblemsolving.com/a/e/d/aed4709127ad663d135c547208b3ca8ae9cbf8f5.png)
Observe that
and 
Prove similarly that 
and 
This post has been edited 49 times. Last edited by Virgil Nicula, Sep 24, 2016, 1:08 AM
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