379. Geometrical extremity I.
by Virgil Nicula, Jun 29, 2013, 1:47 PM
PE12. We have given a circle 
and two nonparallel tangents 
, 
at points 
. For a point mobile 
of the smaller
arc
the tangent to 
at meets at 
and at 
. How must be chosen so that the triangle 
has maximum area ?
Proof. I'll use the standard notations for the
with the 
-exincircle 
. Thus, 
.
Thus,![$ S = [ABC] = r_a(p - a)$](//latex.artofproblemsolving.com/8/2/2/822626bf5040809d71e018e7d778827ad5e76648.png)
is 
is 
is 
is 
, where 
(constant) 
is 
is 
is 
. Remark that can use (or can prove easily)
the well-known
, i.e. the function 
is convex. Have the equality iff 
.
PE13. Given are
and a point 
in the interior of a fixed angle with vertex 
. A line through cut the
sides of the angle at
and 
. Find the position of the line for which 
is minimum.
Lemma. Given are
and the variables 
for which 
. Then the product 
is maximum if and only if 
.
Proof of the lemma. 
. Then the product is maximum if and only if the product 
is maximum .
Proof of the proposed problem. Denote
. Observe that the points and are fixed, i.e. 
and 
are constant and 
. Apply the upper lemma for 
and 
. Thus, the product is minimum 
is maximum is maximum 
. In conclusion, construct the points 
. Then the points 
, 
are fixed and 
.
Here is the construct of the required points , for which is minimum.
PE14. Given are
and 
Let 
for which 
Project 
, 
on a mobile line 
, 
in 
, 
and let 
Prove that
the product
is maximum if and only if , are (harmonic) conjugate w.r.t. 
, 
i.e. the ray 
is the bisector of 
Construct the 
-maximum position of 
Proof. Let the foot
of the 
-bisector in 
and is imilarly to itself for any position of the line Thus, the ratio (homogeneous) 
is constant. From the next problem obtain
and 
is maximum iff 
i.e. the ray 
is the - bisector in the triangle 
Construct. Let a fixed circles
, 
and 
, 
Let 
, 
so that 
. Let 
which belongs to the -bisector of 
The required-maximum position of is perpendicular 
raised from on 
Indeed, if denote 
so that 
then 
i.e. 
PE15. Let
be a fixed triangle and let ![$M\in [BC]$](//latex.artofproblemsolving.com/6/f/1/6f11082ff5ed53afefce1cb91c180d8a7342b4c1.png)
be a mobile point. Prove that the ratio 
is maximum if and only if the point 
is the foot of the - bisector.
Proof. Apply the Stewart's theorem :
Therefore, the ratio 
It is achieved (see the next problem) when 
i.e. the point is the foot of the - bisector.
PE16. (commonplace). Given are the positive constants , 
, 
If 
then the sum 
is minimum if and only if 
obtain 
Have the equality 
i.e. 
PE17. Let
be a fixed point which is interior to a given circle 
and for which 
. Let 
be a variable quadrilateral which
is inscrbed in the circle and 
, 
Find the range of the value of the perimeter of in terms of 
and 
Proof. Denote the perimeter
I"ll use the well-known property (or prove easily) in a convex quadrilateral what is inscribed in the circle 
Thus, if
then 
Particularly, 
Denote
Thus, 
(constant). Therefore, 
![$ \boxed {\{x,y\}\subset\left(\ 0\ ,\ 2R^2\ \right]}$](//latex.artofproblemsolving.com/2/5/6/25616c98de80de79a0dfd63a9ea6c84a70cb3fd9.png)
and
The problem is to find the range of 
,
where![$ \left\|\begin{array}{c} \{x,y\}\subset \left(\ 0\ ,\ 2R^2\ \right] \\
\\
xy = 4R^2(R^2 - \delta^2)\end{array}\right\|$](//latex.artofproblemsolving.com/6/a/6/6a6f04869b4bc9a08e9eb2db5a7ebd92d6b182a7.png)
. The minimum is touched iff is harmonically, i.e. 
or inversely, i.e.
is a deltoid. The maximum is touched iff 
, i.e. is a trapezoid.
In conclusion,![$ \boxed {2\sqrt 2\left[\sqrt {R(R + \delta )} + \sqrt {R(R - \delta )}\right]\le \pi\le 2\left[R\sqrt 2 + \sqrt {2R^2 - \delta^2}\right]}\ .$](//latex.artofproblemsolving.com/e/c/3/ec3971654172bf60ed2624404cbe7f30f5f7f4bf.png)
PE18. Let with 
and circumcircle . Let 
so that 
separates , . Find the position of for which 
is maximum.
Proof. Let
, 
, 
and 
. Thus, ![$ PA + PB + PC = y + z + \frac {by + cz}{a} = \frac 1a\cdot [(a + b)y + (a + c)z]$](//latex.artofproblemsolving.com/f/e/4/fe47751a27fd59e45a4ee568aaefbcf7a69ecd86.png)
.
![$ [(a + b)y + (a + c)z]^2\ \stackrel{C.B.S.}{\le}\ \left[(a + b)^2 + (a + c)^2\right]\cdot \left(y^2 + z^2\right)$](//latex.artofproblemsolving.com/9/0/e/90eee2ce50cba9d52efea559a3eb60ded6660ed7.png)
, i.e. 
with the equality iff 
.
Remark. Observe that
and 
, i.e. 
. We can construct easily now the maximum position of the point .
PE19. Let
be a quadrilateral and 
. Find and construct the positions of 
and 
, 
so that the sum 
is minimum.
Proof. Let the minimum position
of the mobil point and 
. 
. (constant).
Then the given sum is minimum when
. Let 
. Apply the Menelaus' theorem to 
, i.e. 
. Let the reflection 
of w.r.t. 
. Then , are harmonical conjugate w.r.t. , .
Construct with the line and compasses the position of the minimum for which 
. It is easily !
PE20. Let and a mobile point 
for which denote 
. Find the position of the point so that the length of the segment ![$[XY]$](//latex.artofproblemsolving.com/b/d/5/bd5db5e85aa6daea3eebecaea5d26721edd15203.png)
is minimum.
Proof 1. Let
. Thus, 
i.e. 
. Let 
, i.e.
. The generalized Pythagoras' theorem to 
. So 
. I"ll look for the range of 
, i.e. the set of 
so that the equation
has at least one positive real root 
has at least one positive real root 
.
In conclusion the range of the function is 
. Prove easily that 
and
which is the distance from or to the -median of .
Proof 2. Let the midpoint of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and 
. So 
, i.e. 
. Appear two cases: 
is a parallelogram 
. Hence 
is minimum iff 
is minimum, i.e. 
and in this case
. Since the area ![$\sigma [ABC]=\frac S2=\frac 12\cdot m_a\cdot\delta_{AD}(B)$](//latex.artofproblemsolving.com/2/f/5/2f5bde1212529ccf2b2b56346cf9a7825bd839f0.png)
obtain that the minimum value of is 
, where 
and ![$S=\sigma[ABC]$](//latex.artofproblemsolving.com/2/f/4/2f45db55ab04b4386cbc8528db2babc0344608ed.png)
.
Remarks.
If is the foot of the -symmedian, then 
, i.e. 
and 
. Using the relations 
and
obtain that 
. From the upper extremum problem get 
, i.e. 
.
If 
then prove easily that 
. Indeed, if denote 
then 
and 
. But 
. In the first case when 
prove analogously.
If is the foot of the -bisector, then 
and 
, i.e. 
, where 
is the length of the -bisector.
PE21. Find the maximum value of the product
, where 
and 
.
Proof 1.
. We have equality if and only if 
.
Proof 2. Denote
and 
. Thus, 
.
In conclusion,
with equality iff 
and .
PE22. Find the maximum area of the triangle with 
, where 
is a real positive variable.
Proof 1. Let the Apollonius' circle with the diameter ![$[MN]$](//latex.artofproblemsolving.com/1/6/2/162e03b9cf481dcfb3d5bdf078be84feab5d2f6e.png)
, where , 
so that 
and 
. Since 
then 
.
Thus,
. Thus, the area ![$[ABC]$](//latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
is
is 
the maximum area of is 
, i.e. 
.
Proof 2. Prove easily that
, where 
. Thus, 
and 
is
is 
is 
is 
. Since 
(constant),
then is 
, 
and 
and 
.
PE23. Let
be side lengths of a right triangle with 
. Find the minimum value for 
and for 
.
Proof 1. Let
. Then 
, i.e. becomes 
. Let ![$S=\sin x+\cos x=t\in\left(1,\sqrt 2\right]$](//latex.artofproblemsolving.com/a/b/9/ab98d6a4411e5bfd90157fb3c3fd41031f0f60e8.png)
for any 
.
Then
and 
becomes 
, where
![$t\in\left(1,\sqrt 2\right]$](//latex.artofproblemsolving.com/7/3/0/730188fa3fd4143fa848773d480938130806d18a.png)
. Let ![$t-1=u\in \left(0,\sqrt 2-1\right]$](//latex.artofproblemsolving.com/7/3/0/730676f42dd8227651ba15c105524d5db3130dea.png)
and 
becomes 
which is evidently strict decreasing 
. The minimum of 
is touched in 
the minimum of is touched in 
the minimum of is touched in 
and in this case 
, i.e. 
with the equality iff 
Proof 2.
Remark.
.
PP24. Find the side length of a square, which has one side along an edge of the sector, that can be inscribed in a sector with central angle
and radius
, where 
. Determine whether a square is the largest possible rectangle (area) that can be inscribed in such a manner.
Proof.. Call the center of the sector and let the endpoints of its arc be and , such that the side of the rectangle rests on 
. Let the rectangle be 
with
on , 
on 
and on 
. Suppose w.l.o.g. that 
. Denote 
. Then 
, 
and the area of the rectangle is
. Prove easily that 
. Observe that ![$ \left[\sigma '(x) - \sigma '\left(\frac {\theta}{2}\right)\right]$](//latex.artofproblemsolving.com/8/c/c/8cc844adb6a54b68447e648b83bd085aee6f2649.png)
and
have same sign for any 
. Therefore, 
for any and maximum value is 
.
Remark. The rectangle is a square iff 
and in this case
, i.e. 
.
and two nonparallel tangents 
, 
at points 
. For a point mobile 
of the smallerarc
the tangent to 
at meets at 
and at 
. How must be chosen so that the triangle 
has maximum area ?Proof. I'll use the standard notations for the
with the 
-exincircle 
. Thus, 
.Thus,
![$ S = [ABC] = r_a(p - a)$](http://latex.artofproblemsolving.com/8/2/2/822626bf5040809d71e018e7d778827ad5e76648.png)
is 
is 
is 
is 
, where 
(constant) 
is 
is 
is 
. Remark that can use (or can prove easily)the well-known
, i.e. the function 
is convex. Have the equality iff 
.PE13. Given are
and a point 
in the interior of a fixed angle with vertex 
. A line through cut thesides of the angle at
and 
. Find the position of the line for which 
is minimum.Lemma. Given are
and the variables 
for which 
. Then the product 
is maximum if and only if 
.Proof of the lemma.
. Then the product is maximum if and only if the product 
is maximum .Proof of the proposed problem. Denote
. Observe that the points and are fixed, i.e. 
and 
are constant and 
. Apply the upper lemma for 
and 
. Thus, the product is minimum 
is maximum is maximum 
. In conclusion, construct the points 
. Then the points 
, 
are fixed and 
.Here is the construct of the required points
, for which is minimum.PE14. Given are
and 
Let 
for which 
Project 
, 
on a mobile line 
, 
in 
, 
and let 
Prove thatthe product
is maximum if and only if , are (harmonic) conjugate w.r.t. 
, 
i.e. the ray 
is the bisector of 
Construct the 
-maximum position of 
Proof. Let the foot
of the 
-bisector in 
and is imilarly to itself for any position of the line Thus, the ratio (homogeneous) 
is constant. From the next problem obtain
and 
is maximum iff 
i.e. the ray 
is the - bisector in the triangle 
Construct. Let a fixed circles
, 
and 
, 
Let 
, 
so that 
. Let 
which belongs to the -bisector of 
The required
-maximum position of is perpendicular 
raised from on 
Indeed, if denote 
so that 
then 
i.e. 
PE15. Let
be a fixed triangle and let ![$M\in [BC]$](http://latex.artofproblemsolving.com/6/f/1/6f11082ff5ed53afefce1cb91c180d8a7342b4c1.png)
be a mobile point. Prove that the ratio 
is maximum if and only if the point 
is the foot of the - bisector.Proof. Apply the Stewart's theorem :
Therefore, the ratio 
It is achieved (see the next problem) when 
i.e. the point is the foot of the - bisector.PE16. (commonplace). Given are the positive constants
, 
, 
If 
then the sum 
is minimum if and only if 
obtain 
Have the equality 
i.e. 
PE17. Let
be a fixed point which is interior to a given circle 
and for which 
. Let 
be a variable quadrilateral whichis inscrbed in the circle
and 
, 
Find the range of the value of the perimeter of in terms of 
and 
Proof. Denote the perimeter
I"ll use the well-known property (or prove easily) in a convex quadrilateral what is inscribed in the circle 
Thus, if
then 
Particularly, 
Denote
Thus, 
(constant). Therefore, 
![$ \boxed {\{x,y\}\subset\left(\ 0\ ,\ 2R^2\ \right]}$](http://latex.artofproblemsolving.com/2/5/6/25616c98de80de79a0dfd63a9ea6c84a70cb3fd9.png)
and
The problem is to find the range of 
,where
![$ \left\|\begin{array}{c} \{x,y\}\subset \left(\ 0\ ,\ 2R^2\ \right] \\
\\
xy = 4R^2(R^2 - \delta^2)\end{array}\right\|$](http://latex.artofproblemsolving.com/6/a/6/6a6f04869b4bc9a08e9eb2db5a7ebd92d6b182a7.png)
. The minimum is touched iff is harmonically, i.e. 
or inversely, i.e. is a deltoid. The maximum is touched iff 
, i.e. is a trapezoid.In conclusion,
![$ \boxed {2\sqrt 2\left[\sqrt {R(R + \delta )} + \sqrt {R(R - \delta )}\right]\le \pi\le 2\left[R\sqrt 2 + \sqrt {2R^2 - \delta^2}\right]}\ .$](http://latex.artofproblemsolving.com/e/c/3/ec3971654172bf60ed2624404cbe7f30f5f7f4bf.png)
PE18. Let
with 
and circumcircle . Let 
so that 
separates , . Find the position of for which 
is maximum.Proof. Let
, 
, 
and 
. Thus, ![$ PA + PB + PC = y + z + \frac {by + cz}{a} = \frac 1a\cdot [(a + b)y + (a + c)z]$](http://latex.artofproblemsolving.com/f/e/4/fe47751a27fd59e45a4ee568aaefbcf7a69ecd86.png)
.![$ [(a + b)y + (a + c)z]^2\ \stackrel{C.B.S.}{\le}\ \left[(a + b)^2 + (a + c)^2\right]\cdot \left(y^2 + z^2\right)$](http://latex.artofproblemsolving.com/9/0/e/90eee2ce50cba9d52efea559a3eb60ded6660ed7.png)
, i.e. 
with the equality iff 
.Remark. Observe that
and 
, i.e. 
. We can construct easily now the maximum position of the point .PE19. Let
be a quadrilateral and 
. Find and construct the positions of 
and 
, 
so that the sum 
is minimum.Proof. Let the minimum position
of the mobil point and 
. 
. (constant).Then the given sum is minimum when
. Let 
. Apply the Menelaus' theorem to 
, i.e. 
. Let the reflection 
of w.r.t. 
. Then , are harmonical conjugate w.r.t. , .Construct with the line and compasses the position of the minimum
for which 
. It is easily !PE20. Let
and a mobile point 
for which denote 
. Find the position of the point so that the length of the segment ![$[XY]$](http://latex.artofproblemsolving.com/b/d/5/bd5db5e85aa6daea3eebecaea5d26721edd15203.png)
is minimum.Proof 1. Let
. Thus, 
i.e. 
. Let 
, i.e.
. The generalized Pythagoras' theorem to 
. So 
. I"ll look for the range of 
, i.e. the set of 
so that the equation
has at least one positive real root 
has at least one positive real root 
.In conclusion the range of the function
is 
. Prove easily that 
and
which is the distance from or to the -median of .Proof 2. Let the midpoint
of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and 
. So 
, i.e. 
. Appear two cases: 
is a parallelogram 
. Hence 
is minimum iff 
is minimum, i.e. 
and in this case
. Since the area ![$\sigma [ABC]=\frac S2=\frac 12\cdot m_a\cdot\delta_{AD}(B)$](http://latex.artofproblemsolving.com/2/f/5/2f5bde1212529ccf2b2b56346cf9a7825bd839f0.png)
obtain that the minimum value of is 
, where 
and ![$S=\sigma[ABC]$](http://latex.artofproblemsolving.com/2/f/4/2f45db55ab04b4386cbc8528db2babc0344608ed.png)
.Remarks.
If is the foot of the -symmedian, then 
, i.e. 
and 
. Using the relations 
and
obtain that 
. From the upper extremum problem get 
, i.e. 
.
If 
then prove easily that 
. Indeed, if denote 
then 
and 
. But 
. In the first case when 
prove analogously.
If is the foot of the -bisector, then 
and 
, i.e. 
, where 
is the length of the -bisector.PE21. Find the maximum value of the product
, where 
and 
.Proof 1.
. We have equality if and only if 
.Proof 2. Denote
and 
. Thus, 
.In conclusion,
with equality iff 
and .PE22. Find the maximum area of the triangle
with 
, where 
is a real positive variable.Proof 1. Let the Apollonius' circle
with the diameter ![$[MN]$](http://latex.artofproblemsolving.com/1/6/2/162e03b9cf481dcfb3d5bdf078be84feab5d2f6e.png)
, where , 
so that 
and 
. Since 
then 
.Thus,
. Thus, the area ![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
is 
is 
the maximum area of is 
, i.e. 
.Proof 2. Prove easily that
, where 
. Thus, 
and 
is 
is 
is 
is 
. Since 
(constant),then
is 
, 
and 
and 
.PE23. Let
be side lengths of a right triangle with 
. Find the minimum value for 
and for 
.Proof 1. Let
. Then 
, i.e. becomes 
. Let ![$S=\sin x+\cos x=t\in\left(1,\sqrt 2\right]$](http://latex.artofproblemsolving.com/a/b/9/ab98d6a4411e5bfd90157fb3c3fd41031f0f60e8.png)
for any 
.Then
and 
becomes 
, where![$t\in\left(1,\sqrt 2\right]$](http://latex.artofproblemsolving.com/7/3/0/730188fa3fd4143fa848773d480938130806d18a.png)
. Let ![$t-1=u\in \left(0,\sqrt 2-1\right]$](http://latex.artofproblemsolving.com/7/3/0/730676f42dd8227651ba15c105524d5db3130dea.png)
and 
becomes 
which is evidently strict decreasing 
. The minimum of 
is touched in 
the minimum of
is touched in 
the minimum of is touched in 
and in this case 
, i.e. 
with the equality iff 
Proof 2.
Remark.
.PP24. Find the side length of a square, which has one side along an edge of the sector, that can be inscribed in a sector with central angle
and radius
, where 
. Determine whether a square is the largest possible rectangle (area) that can be inscribed in such a manner.Proof.. Call the center of the sector
and let the endpoints of its arc be and , such that the side of the rectangle rests on 
. Let the rectangle be 
with
on , 
on 
and on 
. Suppose w.l.o.g. that 
. Denote 
. Then 
, 
and the area of the rectangle is
. Prove easily that 
. Observe that ![$ \left[\sigma '(x) - \sigma '\left(\frac {\theta}{2}\right)\right]$](http://latex.artofproblemsolving.com/8/c/c/8cc844adb6a54b68447e648b83bd085aee6f2649.png)
and
have same sign for any 
. Therefore, 
for any and maximum value is 
.Remark. The rectangle
is a square iff 
and in this case
, i.e. 
.This post has been edited 173 times. Last edited by Virgil Nicula, Feb 18, 2017, 3:39 PM
August 2018
January 2018