153. Geometrical minimum/maximum.
by Virgil Nicula, Oct 11, 2010, 6:13 PM
PP1. Consider 
and for a mobile point 
construct the parallelogram 
, where 
. Find ![$\max\ [AXMY]$](//latex.artofproblemsolving.com/b/b/9/bb9142951acc81e5d59a1269cf7c6ecab7c3130a.png)
and prove that
, where ![$S=[ABC]$](//latex.artofproblemsolving.com/b/3/a/b3ae3d445111e4dd28be75922309d3270079368c.png)
and 
is the length of the 
-median. Deduce 
, where 
is the length of the -angle bisector.
Proof. Denote the midpoint
of the side ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and 
. Observe that 
, i.e. 
. Thus,
the area![$[AXMY]=MX\cdot MY\cdot\sin A$](//latex.artofproblemsolving.com/4/3/6/436589b871b7dd12bf437c5ba81cdabe3a006651.png)
is maximum 
is maximum 
, i.e. 
and in this case ![$[AXMY]=\frac S2$](//latex.artofproblemsolving.com/f/9/5/f95d49ea7f79cce9269edca3e6f58d4ce44001b3.png)
.
IF
.
IF
.
In conclusion, for any
have 
, i.e. the quadrilateral 
is a parallelogram. Therefore, 
is minimum 
is minimum
and in this case ![$S=2\cdot [ABD]=m_a\cdot XY$](//latex.artofproblemsolving.com/7/d/f/7df8a09822a245a6971ff87fb68d649398b1a2c3.png)
.
Particular cases.
If is the foot of the -symmedian, then 
and 
, i.e. 
and 
because 
.
If is the foot of the -angle bisector, then 
and 
.
Remark. In the first case (from the proof)
and in the second case 
("mean and extreme ratio").
PP2. Prove that the symmedian center (Lemoine's point)
of a triangle 
is the point for which 
is minimum (Grebe).
Proof. Denote the distancies from to the sidelines 
, 
, 
respectively. From the evident relation 
, where and the C.B.S. - inequality obtain 
with equality iff 
. Prove easily
that the point which verifies the relations 
is the Lemoine's point 
. Indeed, if 
, then for any point of the -symmedian have 
![$\frac bc\cdot \frac {[ASB]}{[ASC]}=$](//latex.artofproblemsolving.com/0/4/6/046c32ebe12634401c2677cedf4070fa3ac12407.png)
, i.e. 
a.s.o.
PP3. Consider an angle
and a fixed point 
. Find the positions of 
, 
so that 
and the area ![$[MON]$](//latex.artofproblemsolving.com/5/0/5/5050d989919c2b014d10271c0aee5919df527c39.png)
is minimum.
Proof. Denote
and ![$[OAU]=[OAV]=a$](//latex.artofproblemsolving.com/b/7/5/b75eeaec47a6a2f7afea7dfa0946fe69a3dfa3d7.png)
, ![$[MAU]=x$](//latex.artofproblemsolving.com/f/4/e/f4ecf75721766e558e629705098527e904156022.png)
and ![$[NAV]=y$](//latex.artofproblemsolving.com/9/6/c/96c3ed456c81970358f4fec01ae68fb23510435c.png)
. Observe that 
. Area ![$[MON]=(2a+x+y)$](//latex.artofproblemsolving.com/8/8/9/8891d50831afbee6ac94c41a296879883c9c2304.png)
is minimum 
is minimum 
because 
, i.e. 
.
Remark. For finding the positions of , for which construct the symmetric 
of 
w.r.t. and afterwards
construct , 
for which 
is a parallelogram. Otherwise, construct the parallel line through the point (fixed) to the line 
(fixed).
Proof (analytic). Consider 
, 
, line 
and 
, 
where 
. If 
, then 
. Observe that
, 
and 
. Thus 
. See CLIV
PP4. Prove that in any triangle there is the inequality 
.
Proof. Denote the centroid
and the midpoints , 
, 
of the sides , ![$[CA]$](//latex.artofproblemsolving.com/4/5/c/45c1acd47628de406680d04c09fe6314c3847acf.png)
, ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
respectively. Apply the Ptolemeu's inequality for the
quadrilaterals
, 
, 
and obtain 
.
PP5. Let be a triangle. Consider the points , 
, 
so that 
, 
, 
and 
.
The cevians
, 
, 
meet again the circumcircle of in the points 
, 
, 
respectively. Prove that 
![$ -3+2\cdot \left[\sqrt {\frac {(m+1)(p+1)}{m}}+\sqrt {\frac {(m+1)(n+1)}{n}}+\sqrt {\frac {(n+1)(p+1)}{p}}\right]\ge$](//latex.artofproblemsolving.com/9/6/9/969c81e4c4def45b45dc39ff9f64464d7040a38e.png)
![$-3+4\cdot \left[\sqrt {\frac {p+1}{m+1}}+\sqrt {\frac {m+1}{n+1}}+\sqrt {\frac {n+1}{p+1}}\right]\ge 9$](//latex.artofproblemsolving.com/3/3/4/33462cdd3d1bf58446d8842c44c4b08e4f11b650.png)
.
and for a mobile point 
construct the parallelogram 
, where 
. Find ![$\max\ [AXMY]$](http://latex.artofproblemsolving.com/b/b/9/bb9142951acc81e5d59a1269cf7c6ecab7c3130a.png)
and prove that
, where ![$S=[ABC]$](http://latex.artofproblemsolving.com/b/3/a/b3ae3d445111e4dd28be75922309d3270079368c.png)
and 
is the length of the 
-median. Deduce 
, where 
is the length of the -angle bisector.Proof. Denote the midpoint
of the side ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and 
. Observe that 
, i.e. 
. Thus,the area
![$[AXMY]=MX\cdot MY\cdot\sin A$](http://latex.artofproblemsolving.com/4/3/6/436589b871b7dd12bf437c5ba81cdabe3a006651.png)
is maximum 
is maximum 
, i.e. 
and in this case ![$[AXMY]=\frac S2$](http://latex.artofproblemsolving.com/f/9/5/f95d49ea7f79cce9269edca3e6f58d4ce44001b3.png)
.IF
.IF
.In conclusion, for any
have 
, i.e. the quadrilateral 
is a parallelogram. Therefore, 
is minimum 
is minimum 
and in this case ![$S=2\cdot [ABD]=m_a\cdot XY$](http://latex.artofproblemsolving.com/7/d/f/7df8a09822a245a6971ff87fb68d649398b1a2c3.png)
.Particular cases.
If is the foot of the -symmedian, then 
and 
, i.e. 
and 
because 
. If is the foot of the -angle bisector, then 
and 
.Remark. In the first case (from the proof)
and in the second case 
("mean and extreme ratio").PP2. Prove that the symmedian center (Lemoine's point)
of a triangle 
is the point for which 
is minimum (Grebe). Proof. Denote the distancies from
to the sidelines 
, 
, 
respectively. From the evident relation 
, where and the C.B.S. - inequality obtain 
with equality iff 
. Prove easilythat the point
which verifies the relations 
is the Lemoine's point 
. Indeed, if 
, then for any point of the -symmedian have 
![$\frac bc\cdot \frac {[ASB]}{[ASC]}=$](http://latex.artofproblemsolving.com/0/4/6/046c32ebe12634401c2677cedf4070fa3ac12407.png)
, i.e. 
a.s.o. PP3. Consider an angle
and a fixed point 
. Find the positions of 
, 
so that 
and the area ![$[MON]$](http://latex.artofproblemsolving.com/5/0/5/5050d989919c2b014d10271c0aee5919df527c39.png)
is minimum.Proof. Denote
and ![$[OAU]=[OAV]=a$](http://latex.artofproblemsolving.com/b/7/5/b75eeaec47a6a2f7afea7dfa0946fe69a3dfa3d7.png)
, ![$[MAU]=x$](http://latex.artofproblemsolving.com/f/4/e/f4ecf75721766e558e629705098527e904156022.png)
and ![$[NAV]=y$](http://latex.artofproblemsolving.com/9/6/c/96c3ed456c81970358f4fec01ae68fb23510435c.png)
. Observe that 
. Area ![$[MON]=(2a+x+y)$](http://latex.artofproblemsolving.com/8/8/9/8891d50831afbee6ac94c41a296879883c9c2304.png)
is minimum 
is minimum 
because 
, i.e. 
.Remark. For finding the positions of
, for which construct the symmetric 
of 
w.r.t. and afterwardsconstruct
, 
for which 
is a parallelogram. Otherwise, construct the parallel line through the point (fixed) to the line 
(fixed). Quote:
Proposed problem. Let an angle and a fixed inner point of . Find the positions of , so that and ![$[MN]$](//latex.artofproblemsolving.com/1/6/2/162e03b9cf481dcfb3d5bdf078be84feab5d2f6e.png)
is minimum.
and a fixed inner point of . Find the positions of , so that and ![$[MN]$](http://latex.artofproblemsolving.com/1/6/2/162e03b9cf481dcfb3d5bdf078be84feab5d2f6e.png)
is minimum.
, 
, line 
and 
, 
where 
. If 
, then 
. Observe that
, 
and 
. Thus 
. See CLIVPP4. Prove that in any triangle
there is the inequality 
. Proof. Denote the centroid
and the midpoints , 
, 
of the sides , ![$[CA]$](http://latex.artofproblemsolving.com/4/5/c/45c1acd47628de406680d04c09fe6314c3847acf.png)
, ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
respectively. Apply the Ptolemeu's inequality for thequadrilaterals
, 
, 
and obtain 
. PP5. Let
be a triangle. Consider the points , 
, 
so that 
, 
, 
and 
.The cevians
, 
, 
meet again the circumcircle of in the points 
, 
, 
respectively. Prove that 
![$ -3+2\cdot \left[\sqrt {\frac {(m+1)(p+1)}{m}}+\sqrt {\frac {(m+1)(n+1)}{n}}+\sqrt {\frac {(n+1)(p+1)}{p}}\right]\ge$](http://latex.artofproblemsolving.com/9/6/9/969c81e4c4def45b45dc39ff9f64464d7040a38e.png)
![$-3+4\cdot \left[\sqrt {\frac {p+1}{m+1}}+\sqrt {\frac {m+1}{n+1}}+\sqrt {\frac {n+1}{p+1}}\right]\ge 9$](http://latex.artofproblemsolving.com/3/3/4/33462cdd3d1bf58446d8842c44c4b08e4f11b650.png)
.This post has been edited 67 times. Last edited by Virgil Nicula, Nov 22, 2015, 9:14 PM
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