412. Geometry 5.
by Virgil Nicula, Mar 20, 2015, 1:43 PM
P13. Let 
and the 
-symmedian 
where 
Construct two similar triangles 
so that
separates 
and 
separates 
Denote 
and 
Prove that 
Proof. Denote
Apply the remarkable relations 
Thus, 
what is true. In conclusion,
P14. Let
be a point inside and let 
be the distancies from to 
respectively. Prove that 
Proof. Let
, 
and 
be the feet of the perpendiculars from to 
, 
and respectively. Note that 
is cyclic and 
is its diameter. Therefore,
Remark that
In conclusion, from 
obtain that 
Observe that 
Lemma (Gemath). Let an acute with incircle 
and circumcircle 
. The circles 
and 
are tangent internally to 
in the
same point . The circle 
is tangent externally to 
and is tangent internally to 
. Prove that 
, where 
and ![$S\equiv [ABC]$](//latex.artofproblemsolving.com/3/e/2/3e28399980b83ac5f9599575b56840d9a050dda6.png)
- the area of .
Are well known
and 
.
Proof 1. Denote
, 
. Apply the Stewart's theorem in 
for the cevian-rays 
and 
:
.
Proof 2. Apply the Pythagoras' theorem in the triangles :
Thus,
, i.e. 
. Observe that 
, where 
is the 
- exradius of 
.
Remark. Prove easily that two more interesting relations :
and 
, 
.
Observe that
![$\left\{\begin{array}{c}\boxed{\ r\le \sqrt [3]{\prod PQ}\le \frac{R}{2}\ }\\\\ \boxed{\ r\le\frac{1}{3}\cdot \sum PQ\le \frac{R^{2}}{4r}\ }\end{array}\right\|$](//latex.artofproblemsolving.com/c/f/d/cfd778ce432ccf113a47bac4192da7f28a466f37.png)
P15. Let
be a trapezoid with 
and 
. Construct outside 
and 
so that 
. Prove that 
belongs to 
.
Proof 1. Observe that
and 
. Thus,
Proof 2. Observe that 
and 
. Thus,
P16. Let
be a triangle. Ascertain 
. Answer: 
(standard notations).
Proof.
.
Particular case
.
P17. Let be a triangle. Ascertain 
(standard notations).
Proof. Prove easily that
![$4rr_a=[a-(b-c)][a+(b-c)]=a^2-(b-c)^2\implies$](//latex.artofproblemsolving.com/a/c/9/ac9f408f5cab354fe0ec8299f66cd259d585fd02.png)
. On other hand, 
.
In conclusion from the relations
and 
obtain that 
.
P18.. Let be a trapezoid so that 
so that
and 
- the semicircle with the diameter ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
and which is tangent to the line 
- the circle
which is tangent to in 
and which is tangent to 
and 
in the same point 
so that 
. Prove that 
.
Proof. Is well-known or prove easily that
, i.e. 
and 
. From the similarity 
obtain that 
. From the product of the relations , and 
obtain that 
,
i.e.
.
P19 (Miguel Ochoa Sanchez). Let with the incenter and the points 
. Prove that ![$\sqrt{[ABC]\cdot [DEF]}=r\sqrt {r(4R+r)}$](//latex.artofproblemsolving.com/f/4/b/f4bad8918e597e2404095780f79c53fcb646db1a.png)
.
Proof 1.
.
.
Otherwise. Apply an well-known relation
a.s.o.
I"ll use:![$\frac {[DEF]}{[ABC]}=$](//latex.artofproblemsolving.com/6/7/9/6792cbf40d5873df43465f27e9ccbefd1aea2345.png)
.
Proof 2..
and 
a.s.o. Thus, ![$2\cdot [EIF]=IE\cdot IF\cdot\sin\widehat{EIF}=$](//latex.artofproblemsolving.com/2/9/4/294d639c35c6a2e666971628a3917524bf669440.png)
![$\boxed{2\cdot [EIF]=\frac rs\cdot b(s-b)}$](//latex.artofproblemsolving.com/a/2/a/a2a8f3481741f48ecc1600030a31206e09cbdf79.png)
. Thus, ![$2\cdot [DEF]=2\cdot\sum [EIF]=$](//latex.artofproblemsolving.com/7/0/a/70a7e372a6f6ff23f4ddb25751fe80af34033791.png)
![$[DEF]=\frac {r^2(4R+r)}{s}\implies$](//latex.artofproblemsolving.com/9/d/6/9d65695fad230836fd3e1c6e340f60890f88203c.png)
![$[DEF]\cdot [ABC]=r^3(4R+r)\implies$](//latex.artofproblemsolving.com/0/8/c/08cdc4b8d4eb4003c319fdc2474b01c394e5d45f.png)
![$\sqrt {[DEF]\cdot [ABC]}=r\sqrt{r(4R+r)}$](//latex.artofproblemsolving.com/3/0/7/307f6ca369a87ea0cd250e3c5df9fcb4eff7a524.png)
.
P20. Let a trapezoid with 
Denote the intersection between the exterior bisectors of 
and the intersection
between the exterior bisectors of 
Prove that 
Proof. Prove easily that
where denoted 
- the distance of the point 
to the
line
Thus, , belong to the middle line of the given trapezoid, where
and 
Since 
si 
obtain that
P21 (Ruben Dario). Let a parallelogram with 
so that there is an interior point 
for which 
Prove that 
Proof (R.D). Construct the parallelogram
Therefore, 
is a parallelogram, 
and 
In conclusion,
the quadrilateral
is cyclic and 
the point 
is the circumcenter of 
and the 
-symmedian 
where 
Construct two similar triangles 
so that
separates 
and 
separates 
Denote 
and 
Prove that 
Proof. Denote
Apply the remarkable relations 
Thus, 
what is true. In conclusion, P14. Let
be a point inside and let 
be the distancies from to 
respectively. Prove that 
Proof. Let
, 
and 
be the feet of the perpendiculars from to 
, 
and respectively. Note that 
is cyclic and 
is its diameter. Therefore,
Remark that
In conclusion, from 
obtain that 
Observe that 
Lemma (Gemath). Let an acute
with incircle 
and circumcircle 
. The circles 
and 
are tangent internally to 
in thesame point
. The circle 
is tangent externally to 
and is tangent internally to 
. Prove that 
, where 
and ![$S\equiv [ABC]$](http://latex.artofproblemsolving.com/3/e/2/3e28399980b83ac5f9599575b56840d9a050dda6.png)
- the area of .Are well known
and 
.Proof 1. Denote
, 
. Apply the Stewart's theorem in 
for the cevian-rays 
and 
:
.Proof 2. Apply the Pythagoras' theorem in the triangles :
Thus,
, i.e. 
. Observe that 
, where 
is the 
- exradius of 
.Remark. Prove easily that two more interesting relations :
and 
, 
.Observe that
![$\left\{\begin{array}{c}\boxed{\ r\le \sqrt [3]{\prod PQ}\le \frac{R}{2}\ }\\\\ \boxed{\ r\le\frac{1}{3}\cdot \sum PQ\le \frac{R^{2}}{4r}\ }\end{array}\right\|$](http://latex.artofproblemsolving.com/c/f/d/cfd778ce432ccf113a47bac4192da7f28a466f37.png)
P15. Let
be a trapezoid with 
and 
. Construct outside 
and 
so that 
. Prove that 
belongs to 
.Proof 1. Observe that
and 
. Thus,
Proof 2. Observe that
and 
. Thus,
P16. Let
be a triangle. Ascertain 
. Answer: 
(standard notations).Proof.
.Particular case
.P17. Let
be a triangle. Ascertain 
(standard notations).Proof. Prove easily that
![$4rr_a=[a-(b-c)][a+(b-c)]=a^2-(b-c)^2\implies$](http://latex.artofproblemsolving.com/a/c/9/ac9f408f5cab354fe0ec8299f66cd259d585fd02.png)
. On other hand, 
.In conclusion from the relations
and 
obtain that 
.P18.. Let
be a trapezoid so that 
so that
and 
- the semicircle with the diameter ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
and which is tangent to the line 
- the circlewhich is tangent to
in 
and which is tangent to 
and 
in the same point 
so that 
. Prove that 
.Proof. Is well-known or prove easily that
, i.e. 
and 
. From the similarity 
obtain that 
. From the product of the relations , and 
obtain that 
,i.e.
.P19 (Miguel Ochoa Sanchez). Let
with the incenter and the points 
. Prove that ![$\sqrt{[ABC]\cdot [DEF]}=r\sqrt {r(4R+r)}$](http://latex.artofproblemsolving.com/f/4/b/f4bad8918e597e2404095780f79c53fcb646db1a.png)
.Proof 1.
.
.Otherwise. Apply an well-known relation
a.s.o.I"ll use:
![$\frac {[DEF]}{[ABC]}=$](http://latex.artofproblemsolving.com/6/7/9/6792cbf40d5873df43465f27e9ccbefd1aea2345.png)
.Proof 2..
and 
a.s.o. Thus, ![$2\cdot [EIF]=IE\cdot IF\cdot\sin\widehat{EIF}=$](http://latex.artofproblemsolving.com/2/9/4/294d639c35c6a2e666971628a3917524bf669440.png)
![$\boxed{2\cdot [EIF]=\frac rs\cdot b(s-b)}$](http://latex.artofproblemsolving.com/a/2/a/a2a8f3481741f48ecc1600030a31206e09cbdf79.png)
. Thus, ![$2\cdot [DEF]=2\cdot\sum [EIF]=$](http://latex.artofproblemsolving.com/7/0/a/70a7e372a6f6ff23f4ddb25751fe80af34033791.png)
![$[DEF]=\frac {r^2(4R+r)}{s}\implies$](http://latex.artofproblemsolving.com/9/d/6/9d65695fad230836fd3e1c6e340f60890f88203c.png)
![$[DEF]\cdot [ABC]=r^3(4R+r)\implies$](http://latex.artofproblemsolving.com/0/8/c/08cdc4b8d4eb4003c319fdc2474b01c394e5d45f.png)
![$\sqrt {[DEF]\cdot [ABC]}=r\sqrt{r(4R+r)}$](http://latex.artofproblemsolving.com/3/0/7/307f6ca369a87ea0cd250e3c5df9fcb4eff7a524.png)
.P20. Let a trapezoid
with 
Denote the intersection between the exterior bisectors of 
and the intersection
between the exterior bisectors of 
Prove that 
Proof. Prove easily that
where denoted 
- the distance of the point 
to theline
Thus, , belong to the middle line of the given trapezoid, where
and 
Since 
si 
obtain that
P21 (Ruben Dario). Let a parallelogram
with 
so that there is an interior point 
for which 
Prove that 
Proof (R.D). Construct the parallelogram
Therefore, 
is a parallelogram, 
and 
In conclusion,the quadrilateral
is cyclic and 
the point 
is the circumcenter of 
This post has been edited 83 times. Last edited by Virgil Nicula, Nov 12, 2016, 11:37 AM
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