394. A cevian in a triangle and two incircles.
by Virgil Nicula, Apr 12, 2014, 1:13 AM
PP1. Let 
with semiperimeter 
and incircle 
For 
let the incircles 
of 
respectively. Prove that :
Prove easily that the 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 
The circles 
are tangent to the line 
in the same point. In this case denote the middlepoints 
of the side ![$[BC]\ ,$](//latex.artofproblemsolving.com/6/5/6/6563ca77d31eb676ef646fa9c37ea181b812d864.png)
the segment![$[AM]$](//latex.artofproblemsolving.com/1/f/9/1f9b22599237fb6240a50b5f75e8f6ced1292374.png)
respectively ; the point 
for which 
Prove easily that 
Proof. Suppose w.l.o.g. that 
Denote : the tangent points 
of the circles with the line respectively ; the tangent points 
of the circles
with the line
respectively ; 
Prove easily that : 
and 
Thus,
![$\left[(b-c)+(x-y)\right]^{2}+4r_{1}r_{2}=[a-(x+y)]^{2}$](//latex.artofproblemsolving.com/3/1/4/314ea7977a02f33789e74435e3cccdfdf935e0b7.png)
Proof 1 (with derivatives). Let 
or 
Prove easily that 
Let 
The relation becomes 
and the problem 
i.e. 
Remark
Let 
Prove easily that 
Thus, for any 
, where
Coming back to the initial problem get 
i.e. ![$\boxed{\ \delta =\frac{r}{a}\cdot \left[ p-\sqrt{p(p-a)}\right]\ }\ .$](//latex.artofproblemsolving.com/2/3/5/2354e1652fbc38a7fb56d2bd2c225d7966fb9f37.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.
Remark 1. 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 : 
From the Stewart's theorem applying for obtain 
Remark 2. In same case![$r_1=r_2=\delta=\frac{r}{a}\cdot \left[ p-\sqrt{p(p-a)}\right]$](//latex.artofproblemsolving.com/0/a/9/0a9d7bd303eecab9d93a6dd6db908f12c9752977.png)
. Let 
. Thus, 
![$\frac {\frac{r}{a}\cdot \left[ p-\sqrt{p(p-a)}\right]}{\frac {(r-\delta)(p-c)}r}=$](//latex.artofproblemsolving.com/a/7/c/a7cbd7319cd87cf6b9746d52dde6cfecf231e345.png)
![$\frac {r^2\left[p-\sqrt{p(p-a)}\right]}{(p-c)\left[ar-pr+r\sqrt{p(p-a)}\right]}=$](//latex.artofproblemsolving.com/4/1/8/418eef6fdcb1973c874c4e46f99bdee0ea3c6934.png)
![$\frac {r\left[p-\sqrt{p(p-a)}\right]}{(p-c)\left[\sqrt{p(p-a)}-(p-a)\right]}=$](//latex.artofproblemsolving.com/e/2/1/e21798fc72d171c12ca0d839330eae018bda233a.png)
. On other hand, 
.
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)
![$\{r[(p-b)+(p-c)]-r_{1}(p-b)-$](//latex.artofproblemsolving.com/9/5/0/950e8ac7d65c3ca95fd760e14e857ebe7e98f1f6.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.
![$(p-b)r_{1}-(p-c)r_{2}=r[(p-b)-(p-c)]$](//latex.artofproblemsolving.com/9/3/3/933825ca2f65cc6e3f3254b1e15390ccdd2173c6.png)
,
i.e.
From the relations 
and 
obtain 
Denote the intersections 
Prove easily that 
, 
, 
, 
Apply the Menelaus's theorem to the transversals
.
From the relations
and 
obtain 
, i.e. 
and 
Apply the Menelaus' theorem to the transversal 
in 
, i.e. 
Remark 
Denote the tangent points 
, 
of the incircle 
and the intersection 
Thus, 
i.e. 
Thus, 
i.e. 
The points 
, 
are conjugate
w.r.t.
Thus, 
and 
the ray 
is the bisector of the angle 
I hope for no mistakes in the my proof !
A similar identity. For the triangle denote the length 
of the 
- altitude and the - exincircle 
For a point denote the -excircles 
,
of the triangles 
, 
respectively. Prove that exists the identity 
, which is equivalently with 
PP2. Let an equilateral and for 
denote the incircles 
. Find the length of the side .
Proof. Suppose w.l.o.g.
. Denote 
. Prove easily that 
and 
. Observe that 
. Obtain that 
.
Particular case.
the length of the altitude is 
.
PP3. Let of , and 
and 
be the incentres of 
, 
respectively. Suppose that 
, 
, , are concyclic. Prove that 
Proof. Let
and we put 
. Thus, 
. Since 
is a cyclical quadrilateral,
giving 
. 
bisects 
and 
bisects 
. Hence 
. In let the inradius, the altitude from 
, the area be 
, 
, 
respectively. In let the
corresponding quantities be
, 
, 
. In 
, 
let the altitude from 
be 
, 
respectively. Then 
. Hence 
. From get 
. From 
obtain 
. Now 
. So 
and similarly
. From 
it follows 
.
PP4. Let
with the circumcircle and the bisector 
, where for which denote 
.
Circle with the diameter![$[DE]$](//latex.artofproblemsolving.com/4/f/5/4f55b2be1d3d9963afec61b4973bfecc6141b1ff.png)
cuts again at the point 
. Prove that 
is the -symmedian of .
Lemma. Let with the circumcircle . Denote the following points : the midpoint 
of ![$ [BC]$](//latex.artofproblemsolving.com/3/5/5/3550468aa97af843ef34b8868728963dec043efe.png)
; the point 
for which 
; the point 
for which
; the 
-symmedian 
, where 
; the point 
for which 
. Then 
, i.e. the quadrilateral 
is cyclically.
Proof. Denote
- the tangent in 
to 
. Denote 
. Observe that 
and 
. From the first well-known property, 
and
is the -symmedian in 
. From the second well-known property, the division 
is harmonically. Since 
results
. Show easily that 
. Thus, 
, i.e. 
. From the third well-known property 
, obtain
, i.e. 
. Since 
get 
and in consequence, 
.
Since and 
obtain 
, i.e. is cyclicall quadrilateral. In conclusion, .
Remark. Here are another interesting relations :
. The proposed problem is an immediate consequence of the above lemma.
PP5 (Iran TST 2007). Let
be the incircle of the triangle . Let and 
be two points such that 
and 
is tangent to the circle .
, 
touch the circle at , respectively. Denote the middlepoint of ![$[PQ]$](//latex.artofproblemsolving.com/2/1/c/21ca08816cf8b23ddf756ce9ae098ad327f2443d.png)
and 
. Prove that the line 
is tangent to the circle (Ali Khezeli)
Proof. Denote
, 
and the midpoints 
, of the segments ![$[AD]$](//latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
, ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
respectively. Observe that 
, 
. From the well-known properties
obtain that 
, i.e. 
and the line is tangent to the circle because the lines , 
are symmetrically w.r.t the line 
.
Suppose w.l.o.g. 
. Denote the intersection 
. Prove easily that 
, 
,
. Apply the Menelaus' theorem to the transversal 
and the triangle 
, i.e. 
.
The line 
is the polar 
of the point w.r.t. the circle and the point 
belongs to the polar 
of the point 
w.r.t.
the circle and 
the polar of is 
. Otherwise prove easily that 
.
PP6. Let an acute with the circumcircle . Let 
and the projections , of 
on , respectively. Prove that the midpoint of is the orthocenter of 
.
Proof.
, 
are isogonals w.r.t. 
, i.e. 
and 
is cyclic, i.e. 
. Therefore, 
. Since 
and 
are cyclic
, i.e. 
is the -altitude of . Hence is the orthocenter of (ARMO - 2013).
with semiperimeter 
and incircle 
For 
let the incircles 
of 
respectively. Prove that :
Prove easily that the relation 
where ![$\delta=\frac{r}{a}\cdot \left[ p-\sqrt{p(p-a)}\right]$](http://latex.artofproblemsolving.com/d/6/6/d661005ea113b4346819d5da662ad51e5ffea5a7.png)
, with equality iff 
The circles 
are tangent to the line 
in the same point. In this case denote the middlepoints 
of the side ![$[BC]\ ,$](http://latex.artofproblemsolving.com/6/5/6/6563ca77d31eb676ef646fa9c37ea181b812d864.png)
the segment
![$[AM]$](http://latex.artofproblemsolving.com/1/f/9/1f9b22599237fb6240a50b5f75e8f6ced1292374.png)
respectively ; the point 
for which 
Prove easily that 
Proof. Suppose w.l.o.g. that 
Denote : the tangent points 
of the circles with the line respectively ; the tangent points 
of the circles with the line
respectively ; 
Prove easily that : 
and 
Thus,
![$\left[(b-c)+(x-y)\right]^{2}+4r_{1}r_{2}=[a-(x+y)]^{2}$](http://latex.artofproblemsolving.com/3/1/4/314ea7977a02f33789e74435e3cccdfdf935e0b7.png)
Proof 1 (with derivatives). Let 
or 
Prove easily that 
Let 
The relation
becomes 
and the problem 
i.e. 
Remark
Let 
Prove easily that 
Thus, for any 
, where
Coming back to the initial problem get 
i.e. ![$\boxed{\ \delta =\frac{r}{a}\cdot \left[ p-\sqrt{p(p-a)}\right]\ }\ .$](http://latex.artofproblemsolving.com/2/3/5/2354e1652fbc38a7fb56d2bd2c225d7966fb9f37.png)
Proof 2 (without derivatives). Using the same notations from the previous proof obtain :
Thus, 
![$k(x+y)^{2}\ge 4[(x+y)-1]$](http://latex.artofproblemsolving.com/7/b/5/7b5a9829e4a3a419dd6237ed71987b08205629db.png)
and 
because 
We have equality iff 
a.s.o.Remark 1. In the case
![$r_{1}=r_{2}=\delta =\frac{r}{a}\cdot \left[p-\sqrt{p(p-a)}\right]$](http://latex.artofproblemsolving.com/3/d/9/3d92b9c1c67a89e0855d7d36fa98c0cefd9cd277.png)
add and another relations : 
From the Stewart's theorem applying for obtain 
Remark 2. In same case
![$r_1=r_2=\delta=\frac{r}{a}\cdot \left[ p-\sqrt{p(p-a)}\right]$](http://latex.artofproblemsolving.com/0/a/9/0a9d7bd303eecab9d93a6dd6db908f12c9752977.png)
. Let 
. Thus, 
![$\frac {\frac{r}{a}\cdot \left[ p-\sqrt{p(p-a)}\right]}{\frac {(r-\delta)(p-c)}r}=$](http://latex.artofproblemsolving.com/a/7/c/a7cbd7319cd87cf6b9746d52dde6cfecf231e345.png)
![$\frac {r^2\left[p-\sqrt{p(p-a)}\right]}{(p-c)\left[ar-pr+r\sqrt{p(p-a)}\right]}=$](http://latex.artofproblemsolving.com/4/1/8/418eef6fdcb1973c874c4e46f99bdee0ea3c6934.png)
![$\frac {r\left[p-\sqrt{p(p-a)}\right]}{(p-c)\left[\sqrt{p(p-a)}-(p-a)\right]}=$](http://latex.artofproblemsolving.com/e/2/1/e21798fc72d171c12ca0d839330eae018bda233a.png)
. On other hand, 
. Proof 1. Using the above notations, the circles are tangent in the same point to the line 
Proof 2.
the triangle 
is a degenerated circumscribed quadrilateral 
(from a well-known property) the incircles of the triangles
are tangent in the same point of the "diagonal" 
![$[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)-$](http://latex.artofproblemsolving.com/9/5/0/950e8ac7d65c3ca95fd760e14e857ebe7e98f1f6.png)
![$[(r-r_{1})(p-b)+(r-r_{2})(p-c)]^{2}=4r^{2}r_{1}r_{2}$](http://latex.artofproblemsolving.com/0/4/1/0416777ab4653a1b6cfa392e185c2b6e2be2b9e5.png)
Otherwise.
![$(p-b)r_{1}-(p-c)r_{2}=r[(p-b)-(p-c)]$](http://latex.artofproblemsolving.com/9/3/3/933825ca2f65cc6e3f3254b1e15390ccdd2173c6.png)
,i.e.
From the relations 
and 
obtain 
Denote the intersections 
Prove easily that 
, 
, 
, 
Apply the Menelaus's theorem to the transversals
.From the relations
and 
obtain 
, i.e. 
and 
Apply the Menelaus' theorem to the transversal 
in 
, i.e. 
Remark 
Denote the tangent points 
, 
of the incircle 
and the intersection 
Thus, 
i.e. 
Thus, 
i.e. 
The points 
, 
are conjugatew.r.t.
Thus, 
and 
the ray 
is the bisector of the angle 
I hope for no mistakes in the my proof ! A similar identity. For the triangle
denote the length 
of the 
- altitude and the - exincircle 
For a point denote the -excircles 
,
of the triangles 
, 
respectively. Prove that exists the identity 
, which is equivalently with 
PP2. Let an equilateral
and for 
denote the incircles 
. Find the length of the side .Proof. Suppose w.l.o.g.
. Denote 
. Prove easily that 
and 
. Observe that 
. Obtain that 
.Particular case.
the length of the altitude is 
.PP3. Let
of , and 
and 
be the incentres of 
, 
respectively. Suppose that 
, 
, , are concyclic. Prove that 
Proof. Let
and we put 
. Thus, 
. Since 
is a cyclical quadrilateral,
giving 
. 
bisects 
and 
bisects 
. Hence 
. In let the inradius, the altitude from 
, the area be 
, 
, 
respectively. In let thecorresponding quantities be
, 
, 
. In 
, 
let the altitude from 
be 
, 
respectively. Then 
. Hence 
. From get 
. From 
obtain 
. Now 
. So 
and similarly
. From 
it follows 
. PP4. Let
with the circumcircle and the bisector 
, where for which denote 
.Circle with the diameter
![$[DE]$](http://latex.artofproblemsolving.com/4/f/5/4f55b2be1d3d9963afec61b4973bfecc6141b1ff.png)
cuts again at the point 
. Prove that 
is the -symmedian of .Lemma. Let
with the circumcircle . Denote the following points : the midpoint 
of ![$ [BC]$](http://latex.artofproblemsolving.com/3/5/5/3550468aa97af843ef34b8868728963dec043efe.png)
; the point 
for which 
; the point 
for which
; the 
-symmedian 
, where 
; the point 
for which 
. Then 
, i.e. the quadrilateral 
is cyclically.Proof. Denote
- the tangent in 
to 
. Denote 
. Observe that 
and 
. From the first well-known property, 
and
is the -symmedian in 
. From the second well-known property, the division 
is harmonically. Since 
results
. Show easily that 
. Thus, 
, i.e. 
. From the third well-known property 
, obtain
, i.e. 
. Since 
get 
and in consequence, 
.Since
and 
obtain 
, i.e. is cyclicall quadrilateral. In conclusion, .Remark. Here are another interesting relations :
. The proposed problem is an immediate consequence of the above lemma.PP5 (Iran TST 2007). Let
be the incircle of the triangle . Let and 
be two points such that 
and 
is tangent to the circle .
, 
touch the circle at , respectively. Denote the middlepoint of ![$[PQ]$](http://latex.artofproblemsolving.com/2/1/c/21ca08816cf8b23ddf756ce9ae098ad327f2443d.png)
and 
. Prove that the line 
is tangent to the circle (Ali Khezeli)Proof. Denote
, 
and the midpoints 
, of the segments ![$[AD]$](http://latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
, ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
respectively. Observe that 
, 
. From the well-known properties
obtain that 
, i.e. 
and the line is tangent to the circle because the lines , 
are symmetrically w.r.t the line 
.
Suppose w.l.o.g. 
. Denote the intersection 
. Prove easily that 
, 
,
. Apply the Menelaus' theorem to the transversal 
and the triangle 
, i.e. 
.
The line 
is the polar 
of the point w.r.t. the circle and the point 
belongs to the polar 
of the point 
w.r.t.the circle
and 
the polar of is 
. Otherwise prove easily that 
.PP6. Let an acute
with the circumcircle . Let 
and the projections , of 
on , respectively. Prove that the midpoint of is the orthocenter of 
.Proof.
, 
are isogonals w.r.t. 
, i.e. 
and 
is cyclic, i.e. 
. Therefore, 
. Since 
and 
are cyclic 
, i.e. 
is the -altitude of . Hence is the orthocenter of (ARMO - 2013).This post has been edited 42 times. Last edited by Virgil Nicula, Nov 13, 2015, 8:48 AM
August 2018
January 2018