196. Similar rectangles outside of a triangle.
by Virgil Nicula, Dec 23, 2010, 10:18 AM
Proposed problem. Let 
be a triangle with orthocenter 
. Construct similar rectangles 
and 
outwardly. Prove that 
.
Proof 1. Denote
. Since 
obtain that 
. Observe that
. Thus, 
. Since 
, 
, 
are altitudes in 
.
Proof 2 (synthetic). Exists
so that 
and 
. Denote 
so that 
. Construct 
so that 
and 
.
and 
.
and 
.
In conclusion, the lines
are the altitudes in the triangle 
.
Proof 3 (metric & trigonometric). Denote
and 
. Denote 
.
Observe that
and ![$\left\{\begin{array}{cccc}
\blacktriangleright & \frac {MA}{MB}=\frac {[EAC]}{[EBC]}=\frac {AE}{BE}\cdot\frac {AC}{BC}\cdot\frac {\sin\widehat{EAC}}{\sin \widehat{EBC}} & \implies & \frac {MA}{MB}=\frac {1}{\sin\phi }\cdot\frac ba\cdot\frac {\sin (A+\phi ) }{\cos B}\\\\
\blacktriangleright & \frac {NA}{NC}=\frac {[FAB]}{[FCB]}=\frac {AF}{CF}\cdot\frac {AB}{CB}\cdot\frac {\sin\widehat{FAB}}{\sin \widehat{FCB}} & \implies & \frac {NA}{NC}=\frac {1}{\sin \phi }\cdot\frac ca\cdot\frac {\sin (A+\phi ) }{\cos C}\end{array}\right\|$](//latex.artofproblemsolving.com/b/9/3/b93e1b2d929cf22237a1ddfc922c76fa7cf1a5ec.png)
![$\left(\frac cb\cdot\frac {\cos B}{\cos C}\right)\cdot \left[\sin\phi\cdot \frac ac\cdot\frac {\cos C}{\sin (A+\phi )}\right]\cdot$](//latex.artofproblemsolving.com/c/2/8/c28cd2da7173ad4875faf2a94bfffa26fdc2fa70.png)
![$\left[\frac {1}{\sin \phi}\cdot\frac ba\cdot\frac {\sin (A+\phi )}{\cos B}\right]=1$](//latex.artofproblemsolving.com/7/3/9/73989ced0dc92d921e09063a29819553634b0bbd.png)
.
See the following message, here or here
be a triangle with orthocenter 
. Construct similar rectangles 
and 
outwardly. Prove that 
.Proof 1. Denote
. Since 
obtain that 
. Observe that
. Thus, 
. Since 
, 
, 
are altitudes in 
.Quote:
An equivalent enunciation. Let be a triangle with the orthocenter . Construct the triangles 
and 
outside
of
so that 
and 
. Prove that .
be a triangle with the orthocenter . Construct the triangles 
and 
outsideof
so that 
and 
. Prove that 
.Proof 2 (synthetic). Exists
so that 
and 
. Denote 
so that 
. Construct 
so that 
and 
.
and 
.
and 
.In conclusion, the lines
are the altitudes in the triangle 
.Proof 3 (metric & trigonometric). Denote
and 
. Denote 
.Observe that
and ![$\left\{\begin{array}{cccc}
\blacktriangleright & \frac {MA}{MB}=\frac {[EAC]}{[EBC]}=\frac {AE}{BE}\cdot\frac {AC}{BC}\cdot\frac {\sin\widehat{EAC}}{\sin \widehat{EBC}} & \implies & \frac {MA}{MB}=\frac {1}{\sin\phi }\cdot\frac ba\cdot\frac {\sin (A+\phi ) }{\cos B}\\\\
\blacktriangleright & \frac {NA}{NC}=\frac {[FAB]}{[FCB]}=\frac {AF}{CF}\cdot\frac {AB}{CB}\cdot\frac {\sin\widehat{FAB}}{\sin \widehat{FCB}} & \implies & \frac {NA}{NC}=\frac {1}{\sin \phi }\cdot\frac ca\cdot\frac {\sin (A+\phi ) }{\cos C}\end{array}\right\|$](http://latex.artofproblemsolving.com/b/9/3/b93e1b2d929cf22237a1ddfc922c76fa7cf1a5ec.png)
![$\left(\frac cb\cdot\frac {\cos B}{\cos C}\right)\cdot \left[\sin\phi\cdot \frac ac\cdot\frac {\cos C}{\sin (A+\phi )}\right]\cdot$](http://latex.artofproblemsolving.com/c/2/8/c28cd2da7173ad4875faf2a94bfffa26fdc2fa70.png)
![$\left[\frac {1}{\sin \phi}\cdot\frac ba\cdot\frac {\sin (A+\phi )}{\cos B}\right]=1$](http://latex.artofproblemsolving.com/7/3/9/73989ced0dc92d921e09063a29819553634b0bbd.png)
.See the following message, here or here
This post has been edited 34 times. Last edited by Virgil Nicula, Nov 22, 2015, 5:29 PM
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