206. Four problems with extremum in a quadrilateral.
by Virgil Nicula, Jan 9, 2011, 9:51 AM
PP1. Let 
be a convex quadrilateral. Denote 
. Ascertain and construct the positions of
the points
and 
, where 
so that the sum 
is minimum.
PP2. Let be a cyclical convex quadrilateral with circumcircle 
and 
be a mobile point for which 
separates 
, 
. Denote 
,
and 
. Ascertain the position of the point for which the sum 
is minimum and prove that in this case
the tangent
in the point to the circle and the lines 
, are concurrently.
PP3. Let be a parallelogram. For two mobile points and denote 
, 
and
. Prove that there is the relation 
and the area ![$\sigma [NPMQ]$](//latex.artofproblemsolving.com/b/b/4/bb4050d8127131a7b1990eddbd63f05d285a5bf0.png)
is maximum if and only if 
.
PP4. Let be a rectangle with the circumcircle 
. For a mobile point 
so that the line separates , denote
and . Construct the position of the point for which the sum 
is minimum.
.
be a convex quadrilateral. Denote 
. Ascertain and construct the positions ofthe points
and 
, where 
so that the sum 
is minimum.
of the mobil point 
. Thus, 
(constant). Then the minimum of the given sum is achieved when 
.
.
,
. Denote the reflection 
of the point 
.
.
. It is easily !PP2. Let
be a cyclical convex quadrilateral with circumcircle 
and 
be a mobile point for which 
separates 
, 
. Denote 
,
and 
. Ascertain the position of the point for which the sum 
is minimum and prove that in this casethe tangent
in the point to the circle and the lines 
, are concurrently.
. Since the product 
is constant obtain that the sum 
, i.e. the line 
is angle-bisector in 
respectively. Therefore, in this case the point 
. Denote 
and 
. Apply the Menelaus' theorem to the transversal 
and 
. From 
obtain 
. Thus, 
. But 
. Hence 
, i.e. 
.PP3. Let
be a parallelogram. For two mobile points and denote 
, 
and
. Prove that there is the relation 
and the area ![$\sigma [NPMQ]$](http://latex.artofproblemsolving.com/b/b/4/bb4050d8127131a7b1990eddbd63f05d285a5bf0.png)
is maximum if and only if 
.
obtain that 
, i.e. 
.![$\sigma [ABCD]=S$](http://latex.artofproblemsolving.com/0/f/3/0f34c7b01e86c3da9a64b65acf1ea18ac9403d06.png)
and ![$\left\{\begin{array}{ccccccc}
\frac {MA}m=\frac {MB}1=\frac {AB}{m+1} & ; & \sigma [PMN]=\sigma [PAD]=x & ; & \sigma [PND]=\alpha & ; & \sigma [PMA]=\beta\\\\
\frac {ND}n=\frac {NC}1=\frac {CD}{n+1} & ; & \sigma [QBC]=\sigma [QMN]=y & ; & \sigma [QNC]=u & ; & \sigma [QMB]=v\end{array}\right\|$](http://latex.artofproblemsolving.com/a/c/7/ac7fc17f40d906c1357b902c76ea548959835a78.png)
.
.![$\left\|\begin{array}{ccc}
\left[\frac {mS}{2(m+1)}-x\right]\cdot\left[\frac {nS}{2(n+1)}-x\right]=x^2 & \implies & x=\frac {mnS}{2(m+n+2mn)}\\\\
\left[\frac {S}{2(m+1)}-y\right]\cdot\left[\frac {S}{2(n+1)}-y\right]=y^2 & \implies & y=\frac {S}{2(m+n+2)}\end{array}\right\|$](http://latex.artofproblemsolving.com/b/0/8/b08d7347ba3090cda71cbb745f5dd6a6a4383607.png)
. Therefore, ![$\sigma [NPMQ]=x+y=\frac 12\cdot f(m,n)\cdot S$](http://latex.artofproblemsolving.com/e/6/d/e6d949fa28381afc01cbe76ef90e68227cb5ee64.png)
, where
, 
, 
. Observe that 
and 
for any 
. Remark that 
is paralelogram iff 
Denote 
and 
. Thus, 
.
obtain that 
.PP4. Let
be a rectangle with the circumcircle 
. For a mobile point 
so that the line separates , denote and . Construct the position of the point for which the sum 
is minimum.
. Observe that 
, 
, 
and 
,
. Therefore, 
(constant). Hence the sum 
is minimum 
is minimum 
, i.e. the position of the point .
This post has been edited 55 times. Last edited by Virgil Nicula, Nov 22, 2015, 4:38 PM
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