58. A cevian and two incircles.
by Virgil Nicula, Jul 10, 2010, 2:20 AM
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=50559&hilit=
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=356562
P1. Let
with semiperimeter 
and incircle 
For 
denote 
of 
respectively. Prove that :
. This relation 
.
, where ![$\delta=\frac{r}{a}\cdot \left[ p-\sqrt{p(p-a)}\right]$](//latex.artofproblemsolving.com/d/6/6/d661005ea113b4346819d5da662ad51e5ffea5a7.png)
, with equality iff 
.
are tangent to 
in same point. In this case if 
are midpoints of ![$[BC]\ ,$](//latex.artofproblemsolving.com/6/5/6/6563ca77d31eb676ef646fa9c37ea181b812d864.png)
![$[AM]$](//latex.artofproblemsolving.com/1/f/9/1f9b22599237fb6240a50b5f75e8f6ced1292374.png)
and
for which 
then 
Proof. Suppose w.l.o.g. that 
Denote : the tangent points 
of 
with respectively ; the tangent points 
of with 
respectively ;
Prove easily that : 
and 
Thus, 
![$\left[(b-c)+(x-y)\right]^{2}+4r_{1}r_{2}=$](//latex.artofproblemsolving.com/1/4/4/144354563272c40f869cab6d85b127db5ff77d47.png)
![$[a-(x+y)]^{2}$](//latex.artofproblemsolving.com/e/f/1/ef1f5fad35c28aeec8b9718957e0f4017ddb4b08.png)
,
i.e.
, i.e. 
. Thus 
. But 
. In conclusion, 
Proof 1 (with derivatives). Define 
or 
Prove easily 
and 
Let 
, 
, 
The relation becomes 
and the extremum problem becomes 
i.e.
Remark 
Let 
Prove easily 
Thus,
, where 
Coming back to the initial extremum problem get 
i.e. ![$\boxed{\ \delta =\frac{r}{a}\cdot \left[ p-\sqrt{p(p-a)}\right]\ }$](//latex.artofproblemsolving.com/3/1/3/313c651e2e67d6dd58cfb242930a323cd07c92cf.png)
.
Proof 2 (without derivatives). Using the same notations from the previous proof obtain : 
Thus, 
![$k(x+y)^{2}\ge 4[(x+y)-1]$](//latex.artofproblemsolving.com/7/b/5/7b5a9829e4a3a419dd6237ed71987b08205629db.png)
and 
because 
We have equality iff 
a.s.o. In the case
![$r_{1}=r_{2}=\delta =\frac{r}{a}\cdot \left[p-\sqrt{p(p-a)}\right]$](//latex.artofproblemsolving.com/3/d/9/3d92b9c1c67a89e0855d7d36fa98c0cefd9cd277.png)
add and another relations : 
; 
;
Now I"ll show that 
. Indeed, denote 
and 
. Appling Stewart's theorem
to cevian in obtain that 
. Therefore, 
. Substitute 
in 
and obtain 
![$(x+b)\left[x^2-s(s-a)\right]=0$](//latex.artofproblemsolving.com/d/0/1/d011d980a51dc5a49cba1d0827d8b7e7ddffd267.png)
.
Proof 1. Using the above notations, the circles are tangent in the same point to the line 
.
Proof 2. 
the triangle 
is a degenerated circumscribed quadrilateral 
(from a well-known property) the incircles of the triangles
are tangent in the same point of the "diagonal" .
![$[a-(x+y)]^{2}=4r_{1}r_{2}$](//latex.artofproblemsolving.com/1/b/2/1b288ce8f4b9af44b8535c8d30f0435f5e93aa53.png)
![$\left[a-\frac{r_{1}(p-b)+r_{2}(p-c)}{r}\right]^{2}=4r_{1}r_{2}$](//latex.artofproblemsolving.com/0/6/3/063313bedb49433e40549bb4da8d47dc7c9ac1e3.png)
![$\{r[(p-b)+(p-c)]-r_{1}(p-b)-r_{2}(p-c)\}^{2}=$](//latex.artofproblemsolving.com/9/b/5/9b53359d93900bfefdbc013df7e9c91e258af88d.png)
![$[(r-r_{1})(p-b)+(r-r_{2})(p-c)]^{2}=4r^{2}r_{1}r_{2}$](//latex.artofproblemsolving.com/0/4/1/0416777ab4653a1b6cfa392e185c2b6e2be2b9e5.png)
. Otherwise, 
![$r[(p-b)-(p-c)]$](//latex.artofproblemsolving.com/c/c/c/ccc8f3b1217aedd755e255afebe853026670fcb0.png)
, i.e. 
. From 
and 
obtain that
.
Denote 
Prove easily 
, 
, 
, 
Apply the Menelaus's
theorem to
, 
. 
. From 
and 
get 
, i.e. 
and 
Apply the Menelaus' theorem to the transversal 
, i.e. 
Remark 
!
Denote the tangent points 
, 
and 
. Thus, 
, i.e. 
. Thus, 
i.e. 
. 
, 
are conjugate w.r.t. 
Thus, 
and 
is the bisector of 
.
P2. Prove similarly the following identities :
For the triangle denote the length 
of the 
-altitude and the -exincircle 
. For a point denote the -exincircles 
,
of the triangles 
, 
respectively.Prove that 
which is equivalently with 
.
For the triangle denote the length of the -altitude and the -exincircle 
For a point denote the -exincircles ,
of the triangles , respectively. Prove that 
which is equivalently with 
.
Let be a triangle with the semiperimeter and the incircle 
. For a point denote the incircle of 
and the-exincircle of the triangle . Prove that 
which is equivalently with 
.
Let 
, the incircles 
to ,
respectively and the midpoints 
of the segments ![$[BC],\ [AM]$](//latex.artofproblemsolving.com/2/6/6/2665948c3a86147486038178bb332cf6e02306b2.png)
respectively. Prove that:
Let for which exists 
so that 
. Denote 
, 
the lengths of inradii for the triangles
, 
respectively. Prove that 
, i.e. 
.
P3. Let
be a fixed line and let 
be two fixed circles so that 
is interior tangent to 
in 
. For a mobile 
denote the intersections
between the line and the tangents from to . Prove that the sum of the lengths of the radii of the incircles of the triangles 
is equal to ![$2[R-\sqrt{R(R-r)}]$](//latex.artofproblemsolving.com/2/4/c/24ce677b11ec24f58da6001506bc5e3882491136.png)
(constant).
Proof (Yetti). Assume that the line a is common tangent to the circles
. This is not stressed properly in the problem, but without it, the sum of inradii would not be constant. The circle with radius r is the incircle of 
. Let 
be the incircles of 
with inradii 
. Let 
be the tangency points of with MT. Let 
be the semiperimeters of 
and denote 
, 
. Then from with the incircle , 
. Similarly, from with the incircles , we have 
and 
. The semiperimeters 
are equal to 
and their difference to 
. Consequently, 
and P, Q are identical, i.e. touch each other on 
. By the way, this is problem 2 in chapter 1 of Honsberger's Episodes. Let 
be the common M-altitude of . Let ST be a diameter of with radius R. Denote 
. Then 
and 
, because, 
, 
are both perpendicular to 
. According to Inradii problem, we have the relation 
. Let U, V be the tangency points of with the line . The segment length UV of the common external tangent of 2 tangent circles with radii is equal to 
. Substituting these results into the inradii relation, we get 
. Let I, J be the centers of . Since they are both tangent to MT at the same point, 
. The angle between 
and 
is the same as the angle between their perpendiculars MT, ST, which is equal to 
. Hence, 
. Substituting this last result into the inradii relation, 
. Thus we obtained a quadratic equation for the sum 
in terms of the fixed radii r, R, which has the roots ![$r_x + r_y = 2R \pm \sqrt{4R^2 - 4rR} = 2\left[R \pm \sqrt{R(R - r)}\right]$](//latex.artofproblemsolving.com/0/c/e/0ce71660d68134aeb09999c20f704de2681d4c3b.png)
. The root with the plus sign is greater than 2R and not acceptable. For example, when the variable point M becomes identically with the point S diametrally opposite to the tangency point T, clearly, 
, 
.
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=356562
P1. Let
with semiperimeter 
and incircle 
For 
denote 
of 
respectively. Prove that :
. This relation 
.
, where ![$\delta=\frac{r}{a}\cdot \left[ p-\sqrt{p(p-a)}\right]$](http://latex.artofproblemsolving.com/d/6/6/d661005ea113b4346819d5da662ad51e5ffea5a7.png)
, with equality iff 
.
are tangent to 
in same point. In this case if 
are midpoints of ![$[BC]\ ,$](http://latex.artofproblemsolving.com/6/5/6/6563ca77d31eb676ef646fa9c37ea181b812d864.png)
![$[AM]$](http://latex.artofproblemsolving.com/1/f/9/1f9b22599237fb6240a50b5f75e8f6ced1292374.png)
and
for which 
then 
Proof. Suppose w.l.o.g. that 
Denote : the tangent points 
of 
with respectively ; the tangent points 
of with 
respectively ;
Prove easily that : 
and 
Thus, 
![$\left[(b-c)+(x-y)\right]^{2}+4r_{1}r_{2}=$](http://latex.artofproblemsolving.com/1/4/4/144354563272c40f869cab6d85b127db5ff77d47.png)
![$[a-(x+y)]^{2}$](http://latex.artofproblemsolving.com/e/f/1/ef1f5fad35c28aeec8b9718957e0f4017ddb4b08.png)
,i.e.
, i.e. 
. Thus 
. But 
. In conclusion, 
Proof 1 (with derivatives). Define 
or 
Prove easily 
and 
Let 
, 
, 
The relation
becomes 
and the extremum problem becomes 
i.e.
Remark 
Let 
Prove easily 
Thus,
, where 
Coming back to the initial extremum problem get 
i.e. ![$\boxed{\ \delta =\frac{r}{a}\cdot \left[ p-\sqrt{p(p-a)}\right]\ }$](http://latex.artofproblemsolving.com/3/1/3/313c651e2e67d6dd58cfb242930a323cd07c92cf.png)
.
Proof 2 (without derivatives). Using the same notations from the previous proof obtain : 
Thus, 
![$k(x+y)^{2}\ge 4[(x+y)-1]$](http://latex.artofproblemsolving.com/7/b/5/7b5a9829e4a3a419dd6237ed71987b08205629db.png)
and 
because 
We have equality iff 
a.s.o. In the case![$r_{1}=r_{2}=\delta =\frac{r}{a}\cdot \left[p-\sqrt{p(p-a)}\right]$](http://latex.artofproblemsolving.com/3/d/9/3d92b9c1c67a89e0855d7d36fa98c0cefd9cd277.png)
add and another relations : 
; 
;
Now I"ll show that 
. Indeed, denote 
and 
. Appling Stewart's theoremto cevian
in obtain that 
. Therefore, 
. Substitute 
in 
and obtain 
![$(x+b)\left[x^2-s(s-a)\right]=0$](http://latex.artofproblemsolving.com/d/0/1/d011d980a51dc5a49cba1d0827d8b7e7ddffd267.png)
. Proof 1. Using the above notations, the circles are tangent in the same point to the line 
.Proof 2.
the triangle 
is a degenerated circumscribed quadrilateral 
(from a well-known property) the incircles of the triangles
are tangent in the same point of the "diagonal" .
![$[a-(x+y)]^{2}=4r_{1}r_{2}$](http://latex.artofproblemsolving.com/1/b/2/1b288ce8f4b9af44b8535c8d30f0435f5e93aa53.png)
![$\left[a-\frac{r_{1}(p-b)+r_{2}(p-c)}{r}\right]^{2}=4r_{1}r_{2}$](http://latex.artofproblemsolving.com/0/6/3/063313bedb49433e40549bb4da8d47dc7c9ac1e3.png)
![$\{r[(p-b)+(p-c)]-r_{1}(p-b)-r_{2}(p-c)\}^{2}=$](http://latex.artofproblemsolving.com/9/b/5/9b53359d93900bfefdbc013df7e9c91e258af88d.png)
![$[(r-r_{1})(p-b)+(r-r_{2})(p-c)]^{2}=4r^{2}r_{1}r_{2}$](http://latex.artofproblemsolving.com/0/4/1/0416777ab4653a1b6cfa392e185c2b6e2be2b9e5.png)
. Otherwise, 
![$r[(p-b)-(p-c)]$](http://latex.artofproblemsolving.com/c/c/c/ccc8f3b1217aedd755e255afebe853026670fcb0.png)
, i.e. 
. From 
and 
obtain that
.
Denote 
Prove easily 
, 
, 
, 
Apply the Menelaus'stheorem to
, 
. 
. From 
and 
get 
, i.e. 
and 
Apply the Menelaus' theorem to the transversal 
, i.e. 
Remark 
!
Denote the tangent points 
, 
and 
. Thus, 
, i.e. 
. Thus, 
i.e. 
. 
, 
are conjugate w.r.t. 
Thus, 
and 
is the bisector of 
.P2. Prove similarly the following identities :
For the triangle denote the length 
of the 
-altitude and the -exincircle 
. For a point denote the -exincircles 
,
of the triangles 
, 
respectively.Prove that 
which is equivalently with 
. For the triangle denote the length of the -altitude and the -exincircle 
For a point denote the -exincircles , of the triangles , respectively. Prove that 
which is equivalently with 
. Let be a triangle with the semiperimeter and the incircle 
. For a point denote the incircle of 
and the
-exincircle of the triangle . Prove that 
which is equivalently with 
. Let 
, the incircles 
to ,
respectively and the midpoints 
of the segments ![$[BC],\ [AM]$](http://latex.artofproblemsolving.com/2/6/6/2665948c3a86147486038178bb332cf6e02306b2.png)
respectively. Prove that:
Let for which exists 
so that 
. Denote 
, 
the lengths of inradii for the triangles
, 
respectively. Prove that 
, i.e. 
.P3. Let
be a fixed line and let 
be two fixed circles so that 
is interior tangent to 
in 
. For a mobile 
denote the intersections between the line
and the tangents from to . Prove that the sum of the lengths of the radii of the incircles of the triangles 
is equal to ![$2[R-\sqrt{R(R-r)}]$](http://latex.artofproblemsolving.com/2/4/c/24ce677b11ec24f58da6001506bc5e3882491136.png)
(constant).Proof (Yetti). Assume that the line a is common tangent to the circles
. This is not stressed properly in the problem, but without it, the sum of inradii would not be constant. The circle with radius r is the incircle of 
. Let 
be the incircles of 
with inradii 
. Let 
be the tangency points of with MT. Let 
be the semiperimeters of 
and denote 
, 
. Then from with the incircle , 
. Similarly, from with the incircles , we have 
and 
. The semiperimeters 
are equal to 
and their difference to 
. Consequently, 
and P, Q are identical, i.e. touch each other on 
. By the way, this is problem 2 in chapter 1 of Honsberger's Episodes. Let 
be the common M-altitude of . Let ST be a diameter of with radius R. Denote 
. Then 
and 
, because, 
, 
are both perpendicular to 
. According to Inradii problem, we have the relation 
. Let U, V be the tangency points of with the line . The segment length UV of the common external tangent of 2 tangent circles with radii is equal to 
. Substituting these results into the inradii relation, we get 
. Let I, J be the centers of . Since they are both tangent to MT at the same point, 
. The angle between 
and 
is the same as the angle between their perpendiculars MT, ST, which is equal to 
. Hence, 
. Substituting this last result into the inradii relation, 
. Thus we obtained a quadratic equation for the sum 
in terms of the fixed radii r, R, which has the roots ![$r_x + r_y = 2R \pm \sqrt{4R^2 - 4rR} = 2\left[R \pm \sqrt{R(R - r)}\right]$](http://latex.artofproblemsolving.com/0/c/e/0ce71660d68134aeb09999c20f704de2681d4c3b.png)
. The root with the plus sign is greater than 2R and not acceptable. For example, when the variable point M becomes identically with the point S diametrally opposite to the tangency point T, clearly, 
, 
.This post has been edited 82 times. Last edited by Virgil Nicula, Dec 1, 2017, 2:59 PM
August 2018
January 2018