259. ABC equilateral triangle ==> DEF equilateral triangle.
by Virgil Nicula, Apr 3, 2011, 4:35 PM
Proposed problem. Look at the down figure as it shows that 
, 
, 
are the midpoints of the segments ![$[BE]$](//latex.artofproblemsolving.com/f/b/0/fb061a8a7c5f9403b5f9261840de9dfea7cb68cf.png)
, ![$[CF]$](//latex.artofproblemsolving.com/2/1/b/21bdd766ec1757878aeae83d55b7ab5917af4537.png)
, ![$[AD]$](//latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
respectively and
is equilateral. Can you use the knowledge of plane geometry to prove that 
is equilateral ?
Proof 1 (metrical). Denote
. Apply the theorem of the median in the triangles :
.
In conclusion, using cyclical permutations obtain that
.
Remark. Generally, in any triangle
we have 
.
Proof 2 (vectorial). Denote
, where 
is the origin and similarly for all other vectors where the initial point is . Since , and are the
midpoints of
, 
and 
respectively, can take them as the averages of the vectors 
and , 
and , etc., so that 
. Through cyclic permutations, we have 
and
. Now we employ the use of 
(the sixth root of unity) , where 
, 
and 
.
This allows for rotation by
on the complex plane. Subtracting, the previous equations, 
and
. Note that 
for any two vectors 
and and for the equilateral triangle ,
and so on, since 
has the same magnitude as 
, but it's just rotated by . Substituting 
,
and 
have 
. Thus,
is just 
rotated by . Doing the same thing for the pairs , 
and , 
yields similar results. So 
is equilateral and our proof is complete.
Proof 3 (synthetical). Denote
, 
, 
and 
for which 
. Observe that 
is the centroid of
. Similarly prove that 
is equillateral 
a.s.o.
PP1 (SANGAKU). Let an equilateral with 
. Consider 
so that 
. Suppose that
the lengths of the inradii of the triangles
, 
, 
, 
are equally with 
. Prove that 
and in this case 
.
Proof. Let
and 
. Thus 
. Eliminate parameter 
between 
and 
and get 
.
, 
, 
are the midpoints of the segments ![$[BE]$](http://latex.artofproblemsolving.com/f/b/0/fb061a8a7c5f9403b5f9261840de9dfea7cb68cf.png)
, ![$[CF]$](http://latex.artofproblemsolving.com/2/1/b/21bdd766ec1757878aeae83d55b7ab5917af4537.png)
, ![$[AD]$](http://latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
respectively and
is equilateral. Can you use the knowledge of plane geometry to prove that 
is equilateral ?Proof 1 (metrical). Denote
. Apply the theorem of the median in the triangles :
.In conclusion, using cyclical permutations obtain that
.Remark. Generally, in any triangle
we have 
.Proof 2 (vectorial). Denote
, where 
is the origin and similarly for all other vectors where the initial point is . Since , and are themidpoints of
, 
and 
respectively, can take them as the averages of the vectors 
and , 
and , etc., so that 
. Through cyclic permutations, we have 
and
. Now we employ the use of 
(the sixth root of unity) , where 
, 
and 
.This allows for rotation by
on the complex plane. Subtracting, the previous equations, 
and
. Note that 
for any two vectors 
and and for the equilateral triangle ,
and so on, since 
has the same magnitude as 
, but it's just rotated by . Substituting 
,
and 
have 
. Thus,
is just 
rotated by . Doing the same thing for the pairs , 
and , 
yields similar results. So 
is equilateral and our proof is complete.Proof 3 (synthetical). Denote
, 
, 
and 
for which 
. Observe that 
is the centroid of
. Similarly prove that 
is equillateral 
a.s.o.PP1 (SANGAKU). Let an equilateral
with 
. Consider 
so that 
. Suppose thatthe lengths of the inradii of the triangles
, 
, 
, 
are equally with 
. Prove that 
and in this case 
.Proof. Let
and 
. Thus 
. Eliminate parameter 
between 
and 
and get 
.This post has been edited 49 times. Last edited by Virgil Nicula, Nov 22, 2015, 9:21 AM
August 2018
January 2018