229. Concurs online MATEFBC, ed. 5 -subiect 2.
by Virgil Nicula, Feb 23, 2011, 1:41 PM
Quote:
Let 
be a 
-right-angled triangle. Denote 
.
Prove that the line
is common tangent line of the circles with the diameters ![$[MA]$](//latex.artofproblemsolving.com/d/d/f/ddf91ec100609c83be5c154de4e74c7649687ba6.png)
, ![$[MB]$](//latex.artofproblemsolving.com/6/5/c/65cd8f377128d62f97243a68e9ad7956a303d43d.png)
.
be a 
-right-angled triangle. Denote 
.Prove that the line
is common tangent line of the circles with the diameters ![$[MA]$](http://latex.artofproblemsolving.com/d/d/f/ddf91ec100609c83be5c154de4e74c7649687ba6.png)
, ![$[MB]$](http://latex.artofproblemsolving.com/6/5/c/65cd8f377128d62f97243a68e9ad7956a303d43d.png)
.Method 1. For to show a line
is tangent to the circle 
in 
choose conveniently a secant through 
which cut the circles in
and 
, where 
. Remain to prove that 
or (metrically) 
. Come back to the proposed problem.![$\begin{array}{ccccc}
\widehat{MAQ}\equiv\widehat{PMQ}\equiv\widehat{RQM} & \Longrightarrow & \widehat{MAQ}\equiv\widehat{RQM} & \Longrightarrow & RQ\ \mathrm{is\ tangent\ to\ the\ circle\ with\ diameter}\ [MA]\\\\
\widehat{MBR}\equiv\widehat{PMR}\equiv\widehat{QRM} & \Longrightarrow & \widehat{MBR}\equiv\widehat{QRM} & \Longrightarrow & RQ\ \mathrm{is\ tangent\ to\ the\ circle\ with\ diameter}\ [MB]\end{array}$](http://latex.artofproblemsolving.com/0/e/a/0eac138e381508c34108afcbaeae7640dd8d875d.png)
.Method 2. Denote the midpoints
, 
, 
of the segments ![$[PM]$](http://latex.artofproblemsolving.com/e/8/3/e83377b926603aeb0a97093c89affe949af4af7a.png)
, ![$[AM]$](http://latex.artofproblemsolving.com/1/f/9/1f9b22599237fb6240a50b5f75e8f6ced1292374.png)
, ![$[BM]$](http://latex.artofproblemsolving.com/b/8/d/b8df50365ac4274a59b15564d2715bbc99693b79.png)
respectively. For to show that is commontangent of the circles with the diameters
, must prove that is perpedicularly on the lines 
si 
. Prove easily thatby simmetry that the quadrilaterals
and 
are cyclical deltoids with symmetry (diameters) 
, 
respectively.In conclusion,
, 
. Remark that the quadrilateral 
is cyclically because
(I used the theorem of cathetus).Method 3. For to show that
is perpendicularly on the lines si find a line which is parallel with these on which is perpendicularly.Denote the midpoint
of ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
and prove easly that the required line is 
, i.e. 
and 
. Indeed, 
and are izogonallines in angle
and since the quadrilateral 
is cyclically obtain that the line is antiparallel to the sideline 
, i.e. 
.This post has been edited 13 times. Last edited by Virgil Nicula, Nov 22, 2015, 2:50 PM
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