133. Some problems with projections/perpendiculars (1).
by Virgil Nicula, Sep 28, 2010, 10:10 AM
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=369056
Proof 1 of PP1 (synthetic). Suppose w.l.o.g. that the tangent
in a point 
to 
cut the side ![$[CD]$](//latex.artofproblemsolving.com/e/7/0/e70960e9e5738a46ad23f794e796ef3cb4ad7e2c.png)
in 
and the sideline 
in 
.
Denote the midpoints
, 
of ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
, . Observe that 
, i.e. 
and 
, i.e. 
.
Proof 2 of PP1 (metric). Suppose w.l.o.g. that the tangent separates 
, 
and 
. Thus, 
. Denote 
and 
.
Prove easily that
. Thus, 
.
Proof 3 of PP1 (analytic). Choose the origin in . Thus, the circle 
has the equation 
and 
, 
, 
, 
. A point 
verifies the relation 
and the equation of the tangent in to is 
. Therefore, 
, what is truly.
Remark (Toshio Seimyia, Japan). If separates , and 
, 
, 
, then 
.
Proof. Suppose w.l.o.g. . Denote 
, 
, 
. Prove easily that 
. Observe that 
, 
and 
. Therefore, 
.
Proof of the "Generalization" (analytic).
Proof of the "Another particular case". Apply the relation
to the degenerated tangential quadrilaterals 
, 
, 
,
and 
. From the product of these relations obtain 
.
Proof of PP2 (analytic). Choose the origin in
and 
, 
, 
, where . Therefore,
. Thus, the length 
of the projection ![$\mathrm{pr}_{BC}[MN]$](//latex.artofproblemsolving.com/5/f/2/5f20f567bce4ba6e9de324c23f650bbc8305942d.png)
is given by 
![$\left|v\cdot\frac {(1-u)\cdot m^2+2v\cdot m+(u+1)}{v(1-u)\cdot m^2+\left[v^2+(1-u)(1+u)\right]\cdot m+v(u+1)}\right|=1$](//latex.artofproblemsolving.com/3/9/8/39815b41c8f2923a1960d90d5f78cc7c0153167f.png)
(constant).
Remark. If
and 
, then can prove easily synthetically. Denote ![$\{U,V\}\subset [AD]$](//latex.artofproblemsolving.com/0/9/7/0977a9fc3848f0ee8c56ebf5ea1f2006059c8bf7.png)
so that 
,
. Prove easily that 
and 
, i.e. ![$\mathrm{pr}_{AD}([AM]\cup [AN])=AD$](//latex.artofproblemsolving.com/9/1/e/91ea96142204ca0b57c3a19dd45c624f0cc5c718.png)
. Since
![$\mathrm{pr}_{BC}[MN]=\mathrm{pr}_{BC}([AM]\cup[AN])=\mathrm{pr}_{AD}([AM]\cup[AN])$](//latex.artofproblemsolving.com/1/d/2/1d2ecb04a25a0bbeb68ffe32430280450f85e7ec.png)
obtain ![$\mathrm{pr}_{BC}[MN]=AD=\frac a2$](//latex.artofproblemsolving.com/d/4/c/d4c24752b48c3f204046e0543eb4c51d951581cc.png)
.
Quote:
PP1. (M.D. Vasiliou, Greece & W. Pompe, Polonia) Let be a mobile tangent line to the incircle 
of the square
. Denote the projections 
, 
, 
, 
of , 
, 
, on respectively. Prove that .
Generalization (Toshio Seimyia, Japan) Let be a mobile tangent line to the incircle of the convex quadrilateral .
Denote the projections , , , of , , , on respectively. Prove that 
.
Another particular case. Denote the tangent points
, 
, 
of the incircle of 
with the sides. Let be a mobile tangent line
to the incircle of the triangle 
. Denote the distance 
of a point 
to the line . Prove that 
.
PP2. (M. Fianu, Romania) Let be a -right angle triangle. Denote the midpoint of the ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
. Consider two mobile points , which belong
to the sidelines , 
respectively so that 
. Prove that the projection of the segment ![$[MN]$](//latex.artofproblemsolving.com/1/6/2/162e03b9cf481dcfb3d5bdf078be84feab5d2f6e.png)
on 
is constant. Generalization.
be a mobile tangent line to the incircle 
of the square
. Denote the projections 
, 
, 
, 
of 
, 
, 
, 
on respectively. Prove that 
.Generalization (Toshio Seimyia, Japan) Let
be a mobile tangent line to the incircle of the convex quadrilateral .Denote the projections
, , , of , , , on respectively. Prove that 
.Another particular case. Denote the tangent points
, 
, 
of the incircle of 
with the sides. Let be a mobile tangent lineto the incircle
of the triangle 
. Denote the distance 
of a point 
to the line . Prove that 
.PP2. (M. Fianu, Romania) Let
be a -right angle triangle. Denote the midpoint of the ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
. Consider two mobile points 
, 
which belongto the sidelines
, 
respectively so that 
. Prove that the projection of the segment ![$[MN]$](http://latex.artofproblemsolving.com/1/6/2/162e03b9cf481dcfb3d5bdf078be84feab5d2f6e.png)
on 
is constant. Generalization. Proof 1 of PP1 (synthetic). Suppose w.l.o.g. that the tangent
in a point 
to 
cut the side ![$[CD]$](http://latex.artofproblemsolving.com/e/7/0/e70960e9e5738a46ad23f794e796ef3cb4ad7e2c.png)
in and the sideline in .Denote the midpoints
, 
of ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
, . Observe that 
, i.e. 
and 
, i.e. 
.Proof 2 of PP1 (metric). Suppose w.l.o.g. that the tangent
separates , 
and 
. Thus, 
. Denote 
and 
.Prove easily that
. Thus, 
.Proof 3 of PP1 (analytic). Choose the origin in
. Thus, the circle 
has the equation 
and 
, 
, 
, 
. A point 
verifies the relation 
and the equation of the tangent in to is 
. Therefore, 
, what is truly.Remark (Toshio Seimyia, Japan). If
separates , and 
, 
, 
, then 
.Proof. Suppose w.l.o.g.
. Denote 
, 
, 
. Prove easily that 
. Observe that 
, 
and 
. Therefore, 
.Proof of the "Generalization" (analytic).
Proof of the "Another particular case". Apply the relation
to the degenerated tangential quadrilaterals 
, 
, 
,
and 
. From the product of these relations obtain 
.Proof of PP2 (analytic). Choose the origin in
and 
, 
, 
, where . Therefore,
. Thus, the length 
of the projection ![$\mathrm{pr}_{BC}[MN]$](http://latex.artofproblemsolving.com/5/f/2/5f20f567bce4ba6e9de324c23f650bbc8305942d.png)
is given by 
![$\left|v\cdot\frac {(1-u)\cdot m^2+2v\cdot m+(u+1)}{v(1-u)\cdot m^2+\left[v^2+(1-u)(1+u)\right]\cdot m+v(u+1)}\right|=1$](http://latex.artofproblemsolving.com/3/9/8/39815b41c8f2923a1960d90d5f78cc7c0153167f.png)
(constant).Remark. If
and 
, then can prove easily synthetically. Denote ![$\{U,V\}\subset [AD]$](http://latex.artofproblemsolving.com/0/9/7/0977a9fc3848f0ee8c56ebf5ea1f2006059c8bf7.png)
so that 
,
. Prove easily that 
and 
, i.e. ![$\mathrm{pr}_{AD}([AM]\cup [AN])=AD$](http://latex.artofproblemsolving.com/9/1/e/91ea96142204ca0b57c3a19dd45c624f0cc5c718.png)
. Since![$\mathrm{pr}_{BC}[MN]=\mathrm{pr}_{BC}([AM]\cup[AN])=\mathrm{pr}_{AD}([AM]\cup[AN])$](http://latex.artofproblemsolving.com/1/d/2/1d2ecb04a25a0bbeb68ffe32430280450f85e7ec.png)
obtain ![$\mathrm{pr}_{BC}[MN]=AD=\frac a2$](http://latex.artofproblemsolving.com/d/4/c/d4c24752b48c3f204046e0543eb4c51d951581cc.png)
.This post has been edited 76 times. Last edited by Virgil Nicula, Nov 23, 2015, 7:22 AM
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