393. Some problems with circles.
by Virgil Nicula, Apr 3, 2014, 5:26 PM
PP1 (Ruben Dario). Let 
be a circle with diameter ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
and 
which are interior tangent to 
in 
, 
respectively and which
are exterior tangent one to other in
so that 
, 
doesn't separate 
and 
is tangent in 
to . Find ![$[TB]=f(R,r)$](//latex.artofproblemsolving.com/e/9/f/e9f76a368c532f1d8750af4f0dcf5d3ff8be62df.png)
.
Proof. Let the diameter![$[AD]$](//latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
of 
and 
. Thus, 
and apply the Pytagoras' theorem to 
-rightangled 
.
Remark. Denote
. Thus, 
and prove easily that 
.
Denote
. Thus, 
with the discriminant
. Therefore, 
and
. If 
and 
so that 
is the common exterior tangent of and ,
then
and the ratio 
.
Particular case. For
obtain that 
and 
.
PP2 (Edson Curahua Ortega). Let an isosceles trapezoid
with 
and incircle 
. Denote the midpoint 
of .
For
let 
so that 
is tangent to . Denote 
and 
. Prove that the division 
is harmonically.
Proof 1 (metric). Denote
so that 
and 
. For 
denote 
.
Apply the generalized Pythagoras' theorem in
. Therefore, 
, i.e. the division
is harmonically. Observe that
is the bisector of 
.
Remark. From
and 
obtain that the area ![$[BRQ]=[BQS]-[BRS]=$](//latex.artofproblemsolving.com/c/0/7/c07278ccfdc4edecb0e78ae2b2dd5c9de0145be8.png)
![$\frac 12\cdot BS\cdot\left[\delta_{BS}(Q)-\delta_{BS}(R)\right]=$](//latex.artofproblemsolving.com/1/9/d/19df0096772d779b9b86b0677a12af0881615b72.png)
![$\boxed{\ [BRQ]=\frac {b(x+a)\sqrt{ab}}{x+b}\ }$](//latex.artofproblemsolving.com/d/0/6/d0617aa987a428785c3f2d656eda05ba07ff9e79.png)
.
In the particular case
![$[BRQ]=$](//latex.artofproblemsolving.com/1/8/2/182f4d8a2db717dfe05835fa3fc4d49e6cbf528e.png)
![$\frac {(a+b)\sqrt {ab}}{2}=\frac {[ABCD]}4$](//latex.artofproblemsolving.com/8/e/1/8e199db0a0387c5abcc3509caf755f3e834097f6.png)
.
Proof 2 (proiective).
PP3 (Julio Orihuela). Let
be an -right-angled triangle with the incircle 
and denote 
. Consider the circle
with the center 
and the radius ![$[EA]$](//latex.artofproblemsolving.com/8/0/0/80090b8f6097098c5b0ba3180165d904cabb51d3.png)
. Let 
be two points so that 
is tangent to . Prove that 
.
Proof.
and 
PP4 Let be a triangle and let 
be a point for which 
. The second tangent from the point
to the incircle of
touches it at the point . Denote the intersection 
. Prove that 
.
Proof. Let the incircle of and 
. Suppose w.l.o.g. 
and let 
Thus, 
. Therefore,
. Apply the Brianchon's theorem to 
i.e. there is 
.
Apply Menelaus' theorem to 
in 
.
Apply the Menelaus' theorem to the transversal 
in 
. Observe that 
.
Apply the Menelaus' theorem to the transversal 
in 
![$\boxed{\frac{XM}{m\left[m(p-c)-(p-b)\right]}=\frac{XC}{(1+m)(p-b)}=\frac{a}{(1+m)\left[m^{2}(p-c)+(p-b)\right]}}\ \ (5)$](//latex.artofproblemsolving.com/2/6/c/26c8fa754ca54285b1d337765a9cbfbe8e18969e.png)
.
Therefore,
.
Application. In a scalene the rays 
are the angle bisectors 
. Points 
on the incircle of are so that 
are tangent to the incircle and 
. Let 
be the midpoints of sides
respectively. Prove that the lines 
intersect on the incircle of the triangle .
PP5 (Balkan MO 1995). The circles
and 
, 
, intersect at 
and 
such that
. The line 
meets 
at 
and 
, and 
at 
and 
(in the order , , , ). The line
meets at 
and 
at 
. The line 
meets at 
and 
at 
. Prove that 
.
Proof. This problem is a nice application of the harmonical division. See the lemma from here.
The division 
is harmonically ;
The angles 
, 
, 
, 
, 
hava same measure, 
.
(see Camaronsou's) 
.
PP6. Let with the incenter 
and 
. Prove that 
and 
.
Proof 1.
is cyclic. Let circumcircle of and 
. Suppose w.l.o.g. 
, 
.
Apply the powers of
and 
w.r.t. 
.
Proof 2. is cyclically 
=(a+b)(a+c)\left[(b+c)^2-a^2\right]\iff$](//latex.artofproblemsolving.com/3/0/8/308fbbe6d377103e800e574f79ead00a83700644.png)
![$(b+c)\left[a(b+c)+\left(b^2+c^2\right)\right]\iff$](//latex.artofproblemsolving.com/5/3/8/538975a2889d02028df3c7846bfdfb3dd61c78c8.png)
![$a(a+b)(a+c)=(b+c)\left[b(a+b)+c(a+c)\right]\iff$](//latex.artofproblemsolving.com/c/2/8/c28145d8215b870cd9981d2e9ba620116ec2bc77.png)
.
Proof 3 (Leo Giugiuc). Denote
. Apply the theorem of Cosines to the triangles 
and 
. In conclusion, 
![$(y+z-x)\left[a^2(y-z)-x\left(b^2-c^2\right)\right]=0$](//latex.artofproblemsolving.com/b/a/5/ba55773ad2d7a2ba9f8be23f4d199f9af78d7d58.png)
. Observe that 
![$\frac {a(b-c)}{(a+b)(a+c)}\cdot\left[a(a+b+c)-(a+b)(a+c)\right]=$](//latex.artofproblemsolving.com/2/d/5/2d51ea8631d2461a4a6db23e8e1817257ff91754.png)
.
Hence
, i.e. 
.
Proof 4.
. Let 
. Theorem of Cosines to 
![$bc(2a+b+c)(b+c)=(a+b)(a+c)\left[(b+c)^2-a^2\right]$](//latex.artofproblemsolving.com/a/6/4/a64f6ea46f1d59f7acc55c289a319e4ce9e4caf5.png)
![$a(a+b)(a+c)=a(b+c)^2+b^3+c^3+bc(b+c)=(b+c)\left[b(a+b)+c(a+c)\right]\implies$](//latex.artofproblemsolving.com/4/c/9/4c95f8a18885f91842a98e3639aa78b427c9139c.png)
.
Proof 5. Denote
. Is well-known that 
and 
. Apply the Stewart's relation to the cevians 
in the triangles 
respectively
![$b\left[ac-\frac {ab^2c}{(a+c)^2}\right]+c\cdot\left(\frac {ab}{a+c}\right)^2=$](//latex.artofproblemsolving.com/b/c/1/bc1d4b0fecdab69fe30241f7a22488f6d940d740.png)
![$c\left[ ab-\frac {abc^2}{(a+b)^2} \right]+b\cdot\left(\frac {ac}{a+b}\right)^2\implies$](//latex.artofproblemsolving.com/9/6/6/9661dfca5cf681f35737d35b45136a3ade2b0250.png)
![$a^3+a^2(b+c)=a\left[(b+c)^2-bc\right]+(b+c)\left(b^2+c^2\right)\implies$](//latex.artofproblemsolving.com/e/6/9/e69c0756122cb565f2b8e2194afcfd72381cdb57.png)
![$a\left[a^2+a(b+c)+bc\right]=a(b+c)^2+(b+c)\left(b^2+c^2\right)\implies$](//latex.artofproblemsolving.com/4/6/8/4687bf05352a68399a9772e32152450183d3a07f.png)
![$a(a+b)(a+c)=(b+c)\left[a(b+c)+\left(b^2+c^2\right)\right]\implies$](//latex.artofproblemsolving.com/f/2/b/f2bbb1018b7c8cedf92d12ce3fb055fe2f43ef06.png)
![$a(a+b)(a+c)=(b+c)\left[b(a+b)+c(a+c)\right]\implies$](//latex.artofproblemsolving.com/f/e/b/feb42d63843672574b77086f28c8cce8f37bc588.png)
. Observe that 
and 
.
Extension 1. Let with incenter and 
so that 
. Prove 
is cyclic 
.
Proof . Denote the circumcircle of and . Suppose w.l.o.g. and . Apply the powers of the points and w.r.t.
the circle
.
Extension 2. Let with 
and 
. Prove is cyclic 
.
Proof . Denote
, where 
. Apply the Ceva's theorem to 
.
Let be the circumcircle of and . Suppose w.l.o.g. and . Apply the powers of the points and w.r.t. the circle 
.
PP7. Let be an equilateral triangle with the incircle . Denote the tangent points 
and 
. Let 
so that
doesn't separate the points and . Denote 
. Prove that 
and 
.
Proof.
a.s.o.
Remark. If
, then 
and 
. In the particular case 
and 
obtain that 
. Let 
and 
.
Prove easily that
. Indeed, and 
.
An easy extension. Let be an -isosceles triangle with 
and with the incircle . Denote the tangent points and .
Let so that doesn't separate the points and . Denote . Prove that and 
.
Proof. . Observe that 
.
Therefore,
.
PP8. Let with the circumcircle 
and let 
be a point on the small arc 
.
Denote
.
Prove that
the line 
is the -symmedian in 
.
Lemma. Let![$\triangle ABC\ ,\ M\in [BC]$](//latex.artofproblemsolving.com/4/d/d/4ddb96ea5e649e3697d9b7acba4200b23daf9b12.png)
and denote the distance 
from 
to 
. Then 
Proof of the proposed problem. Denote the midpoint of the side ![$ [AB]$](//latex.artofproblemsolving.com/d/7/a/d7a8027c238eec9cf67de0f7ec6cb1df4df49a61.png)
, the intersections 
and 
.Observe that 
and the
quadrilaterals
, 
are cyclically, i.e. 
. Therefore, 
, 
and
(lines 
, 
are antiparallels to 
, in 
). Apply the upper lemma in the triangles 
and 
to the cevians 
, 
respectively :
. In conclusion,
. Observe that in this case 
,
i.e. in this case the point
is the foot of the -symmedian in the triangle 
. We can apply this property to the problem from here
PP9. Let a-isosceles with the circumcircle . For 
between and (and on
the opposite side of the line to ) denote 
so that 
. Show that 
.
An equivalent enunciation. Let with the circumcircle . Let the midpoint of the side ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and the diameter ![$[NS]$](//latex.artofproblemsolving.com/8/0/d/80de18ed66832d94ce5323f567fc5b59dc992316.png)
of so that
line
separates 
and 
. Denote 
so that 
. Prove that 
and 
.
Proof 1. Apply Ptolemy's theorem to
. Prove easily 
, i.e. 
. Thus, 
. Since 
is cyclically, obtain that 
, i.e.in the -right triangle 
we have .
Proof 2. Denote
. Well-known relations 
.
PP10. Let an acute with incircle 
and circumcircle 
. The circles 
and 
are tangent internally to 
in
the same . 
is tangent externally to 
and is tangent internally to 
. Prove that 
, where 
and ![$ S\equiv [ABC]$](//latex.artofproblemsolving.com/8/d/f/8df17304ca8c5a866f0e34f9089fe302737d3f65.png)
- the area of .
and 
. Thus, 
.
Proof 1. Let
, 
. Apply Stewart's theorem in 
for the cevian-rays 
and 
.
Proof 2. Apply the Pythagoras' theorem in the triangles:
Thus,
, i.e. 
. Observe that 
, where 
is the - exinradius of .
Remark. Prove easily that two more interesting relations :
and 
, 
.
PP11. Let two secant circles
, where 
. The tangent of in intersect in and the tangent of in intersect
in . A ray by inside of 
intersect again , and the circumcircle of 
in , 
and 
respectively. Prove that 
.
Proof. Let
. Thus, 
. With Steiner's theorem in 
and 
obtain : 
. Thus,
, what is truly. See here an equivalent enunciation of this nice and difficult problem.
PP12. Let be the incircle of -right . The circle touches at 
,
and 
. Denote 
. Prove that 
.
Proof. Denote the midpoint of ![$[PD]$](//latex.artofproblemsolving.com/3/b/3/3b3ab695b20d0a266cb5c4fad99fa342facb80ef.png)
, i.e. 
and 
, i.e. 
. Define 
and
, i.e. 
, 
Prove easily that 
. Observe that 
. But 
is cyclically, i.e. 
.
From the evident relation
and the Stewart's relation applied to the cevians 
obtain that
![$m[(x+y)^2+(x+z)^2]+n(y^2+z^2)-2mx^2=$](//latex.artofproblemsolving.com/d/7/f/d7fe6a9ac88da3f0d4f946d40e611a5bd6b2ac52.png)
Therefore, from the relations
and 
results 
. But 
. In conclusion, 
, what is truly.
PP13. Let such that 
. Let ![$D\in [AC]$](//latex.artofproblemsolving.com/c/5/a/c5acd77baad8d15d98170c2ac60f572eeeed5507.png)
such that 
. The incircle of touches its
sides in![$K\in [AB]$](//latex.artofproblemsolving.com/f/7/c/f7c55ea4f0642b6ee4395cbe8c000c6fec53e3ef.png)
and ![$L\in [AC]$](//latex.artofproblemsolving.com/c/4/5/c45e50a5a8063a6607d6afd834b15cea829eb243.png)
. Let 
be the incircle of 
. Denote 
. Prove that 
.
Proof.. The incircle
touches in 
, 
and 
. Observe that 
.
Thus,
. Since 
obtain that 
. Denote 
. The
property
is well-known (see the lower remark). Thus, 
is cyclically. Analogously 
, i.e. 
. Observe that 
. Apply the Menelaus' theorem to transversal 
in 
: 
. So the quad. 
is cyclically 
or 
be a circle with diameter ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
and 
which are interior tangent to 
in 
, 
respectively and whichare exterior tangent one to other in
so that 
, 
doesn't separate 
and 
is tangent in 
to . Find ![$[TB]=f(R,r)$](http://latex.artofproblemsolving.com/e/9/f/e9f76a368c532f1d8750af4f0dcf5d3ff8be62df.png)
.Proof. Let the diameter
![$[AD]$](http://latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
of 
and 
. Thus, 
and apply the Pytagoras' theorem to 
-rightangled 
.Remark. Denote
. Thus, 
and prove easily that 
.Denote
. Thus, 
with the discriminant
. Therefore, 
and
. If 
and 
so that 
is the common exterior tangent of and ,then
and the ratio 
.Particular case. For
obtain that 
and 
.PP2 (Edson Curahua Ortega). Let an isosceles trapezoid
with 
and incircle 
. Denote the midpoint 
of .For
let 
so that 
is tangent to . Denote 
and 
. Prove that the division 
is harmonically.Proof 1 (metric). Denote
so that 
and 
. For 
denote 
.Apply the generalized Pythagoras' theorem in
. Therefore, 
, i.e. the division is harmonically. Observe that
is the bisector of 
.Remark. From
and 
obtain that the area ![$[BRQ]=[BQS]-[BRS]=$](http://latex.artofproblemsolving.com/c/0/7/c07278ccfdc4edecb0e78ae2b2dd5c9de0145be8.png)
![$\frac 12\cdot BS\cdot\left[\delta_{BS}(Q)-\delta_{BS}(R)\right]=$](http://latex.artofproblemsolving.com/1/9/d/19df0096772d779b9b86b0677a12af0881615b72.png)
![$\boxed{\ [BRQ]=\frac {b(x+a)\sqrt{ab}}{x+b}\ }$](http://latex.artofproblemsolving.com/d/0/6/d0617aa987a428785c3f2d656eda05ba07ff9e79.png)
.In the particular case
![$[BRQ]=$](http://latex.artofproblemsolving.com/1/8/2/182f4d8a2db717dfe05835fa3fc4d49e6cbf528e.png)
![$\frac {(a+b)\sqrt {ab}}{2}=\frac {[ABCD]}4$](http://latex.artofproblemsolving.com/8/e/1/8e199db0a0387c5abcc3509caf755f3e834097f6.png)
.Proof 2 (proiective).
PP3 (Julio Orihuela). Let
be an -right-angled triangle with the incircle 
and denote 
. Consider the circle
with the center 
and the radius ![$[EA]$](http://latex.artofproblemsolving.com/8/0/0/80090b8f6097098c5b0ba3180165d904cabb51d3.png)
. Let 
be two points so that 
is tangent to . Prove that 
. Proof.
and 
PP4 Let
be a triangle and let 
be a point for which 
. The second tangent from the point to the incircle of
touches it at the point . Denote the intersection 
. Prove that 
.Proof. Let the incircle
of and 
. Suppose w.l.o.g. 
and let 
Thus, 
. Therefore,
. Apply the Brianchon's theorem to 
i.e. there is 
.
Apply Menelaus' theorem to 
in 
.
Apply the Menelaus' theorem to the transversal 
in 
. Observe that 
.
Apply the Menelaus' theorem to the transversal 
in 
![$\boxed{\frac{XM}{m\left[m(p-c)-(p-b)\right]}=\frac{XC}{(1+m)(p-b)}=\frac{a}{(1+m)\left[m^{2}(p-c)+(p-b)\right]}}\ \ (5)$](http://latex.artofproblemsolving.com/2/6/c/26c8fa754ca54285b1d337765a9cbfbe8e18969e.png)
.Therefore,
.Application. In a scalene
the rays 
are the angle bisectors 
. Points 
on the incircle of
are so that 
are tangent to the incircle and 
. Let 
be the midpoints of sides
respectively. Prove that the lines 
intersect on the incircle of the triangle .PP5 (Balkan MO 1995). The circles
and 
, 
, intersect at 
and 
such that
. The line 
meets 
at 
and 
, and 
at 
and 
(in the order , , , ). The line
meets at 
and 
at 
. The line 
meets at 
and 
at 
. Prove that 
.Proof. This problem is a nice application of the harmonical division. See the lemma from here.
The division 
is harmonically ; The angles 
, 
, 
, 
, 
hava same measure, 
.
(see Camaronsou's) 
.PP6. Let
with the incenter 
and 
. Prove that 
and 
.Proof 1.
is cyclic. Let circumcircle of and 
. Suppose w.l.o.g. 
, 
.Apply the powers of
and 
w.r.t. 
.Proof 2.
is cyclically 
=(a+b)(a+c)\left[(b+c)^2-a^2\right]\iff$](http://latex.artofproblemsolving.com/3/0/8/308fbbe6d377103e800e574f79ead00a83700644.png)
![$(b+c)\left[a(b+c)+\left(b^2+c^2\right)\right]\iff$](http://latex.artofproblemsolving.com/5/3/8/538975a2889d02028df3c7846bfdfb3dd61c78c8.png)
![$a(a+b)(a+c)=(b+c)\left[b(a+b)+c(a+c)\right]\iff$](http://latex.artofproblemsolving.com/c/2/8/c28145d8215b870cd9981d2e9ba620116ec2bc77.png)
.Proof 3 (Leo Giugiuc). Denote
. Apply the theorem of Cosines to the triangles 
and 
. In conclusion, 
![$(y+z-x)\left[a^2(y-z)-x\left(b^2-c^2\right)\right]=0$](http://latex.artofproblemsolving.com/b/a/5/ba55773ad2d7a2ba9f8be23f4d199f9af78d7d58.png)
. Observe that 
![$\frac {a(b-c)}{(a+b)(a+c)}\cdot\left[a(a+b+c)-(a+b)(a+c)\right]=$](http://latex.artofproblemsolving.com/2/d/5/2d51ea8631d2461a4a6db23e8e1817257ff91754.png)
.Hence
, i.e. 
.Proof 4.
. Let 
. Theorem of Cosines to 
![$bc(2a+b+c)(b+c)=(a+b)(a+c)\left[(b+c)^2-a^2\right]$](http://latex.artofproblemsolving.com/a/6/4/a64f6ea46f1d59f7acc55c289a319e4ce9e4caf5.png)
![$a(a+b)(a+c)=a(b+c)^2+b^3+c^3+bc(b+c)=(b+c)\left[b(a+b)+c(a+c)\right]\implies$](http://latex.artofproblemsolving.com/4/c/9/4c95f8a18885f91842a98e3639aa78b427c9139c.png)
.Proof 5. Denote
. Is well-known that 
and 
. Apply the Stewart's relation to the cevians 
in the triangles 
respectively
![$b\left[ac-\frac {ab^2c}{(a+c)^2}\right]+c\cdot\left(\frac {ab}{a+c}\right)^2=$](http://latex.artofproblemsolving.com/b/c/1/bc1d4b0fecdab69fe30241f7a22488f6d940d740.png)
![$c\left[ ab-\frac {abc^2}{(a+b)^2} \right]+b\cdot\left(\frac {ac}{a+b}\right)^2\implies$](http://latex.artofproblemsolving.com/9/6/6/9661dfca5cf681f35737d35b45136a3ade2b0250.png)
![$a^3+a^2(b+c)=a\left[(b+c)^2-bc\right]+(b+c)\left(b^2+c^2\right)\implies$](http://latex.artofproblemsolving.com/e/6/9/e69c0756122cb565f2b8e2194afcfd72381cdb57.png)
![$a\left[a^2+a(b+c)+bc\right]=a(b+c)^2+(b+c)\left(b^2+c^2\right)\implies$](http://latex.artofproblemsolving.com/4/6/8/4687bf05352a68399a9772e32152450183d3a07f.png)
![$a(a+b)(a+c)=(b+c)\left[a(b+c)+\left(b^2+c^2\right)\right]\implies$](http://latex.artofproblemsolving.com/f/2/b/f2bbb1018b7c8cedf92d12ce3fb055fe2f43ef06.png)
![$a(a+b)(a+c)=(b+c)\left[b(a+b)+c(a+c)\right]\implies$](http://latex.artofproblemsolving.com/f/e/b/feb42d63843672574b77086f28c8cce8f37bc588.png)
. Observe that 
and 
.Extension 1. Let
with incenter and 
so that 
. Prove 
is cyclic 
.Proof . Denote the circumcircle
of and . Suppose w.l.o.g. and . Apply the powers of the points and w.r.t.the circle
.Extension 2. Let
with 
and 
. Prove is cyclic 
.Proof . Denote
, where 
. Apply the Ceva's theorem to 
.Let
be the circumcircle of and . Suppose w.l.o.g. and . Apply the powers of the points and w.r.t. the circle 
.PP7. Let
be an equilateral triangle with the incircle . Denote the tangent points 
and 
. Let 
so that
doesn't separate the points and . Denote 
. Prove that 
and 
.Proof.
a.s.o.Remark. If
, then 
and 
. In the particular case 
and 
obtain that 
. Let 
and 
.Prove easily that
. Indeed, and 
.An easy extension. Let
be an -isosceles triangle with 
and with the incircle . Denote the tangent points and .Let
so that doesn't separate the points and . Denote . Prove that and 
.Proof.
. Observe that 
.Therefore,
.PP8. Let
with the circumcircle 
and let 
be a point on the small arc 
.Denote
.Prove that
the line 
is the -symmedian in 
.Lemma. Let
![$\triangle ABC\ ,\ M\in [BC]$](http://latex.artofproblemsolving.com/4/d/d/4ddb96ea5e649e3697d9b7acba4200b23daf9b12.png)
and denote the distance 
from 
to 
. Then 
Proof of the proposed problem. Denote the midpoint
of the side ![$ [AB]$](http://latex.artofproblemsolving.com/d/7/a/d7a8027c238eec9cf67de0f7ec6cb1df4df49a61.png)
, the intersections 
and 
.Observe that 
and thequadrilaterals
, 
are cyclically, i.e. 
. Therefore, 
, 
and
(lines 
, 
are antiparallels to 
, in 
). Apply the upper lemma in the triangles 
and 
to the cevians 
, 
respectively :
. In conclusion,
. Observe that in this case 
,i.e. in this case the point
is the foot of the -symmedian in the triangle 
. We can apply this property to the problem from herePP9. Let a
-isosceles with the circumcircle . For 
between and (and onthe opposite side of the line
to ) denote 
so that 
. Show that 
.An equivalent enunciation. Let
with the circumcircle . Let the midpoint of the side ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and the diameter ![$[NS]$](http://latex.artofproblemsolving.com/8/0/d/80de18ed66832d94ce5323f567fc5b59dc992316.png)
of so thatline
separates 
and 
. Denote 
so that 
. Prove that 
and 
.Proof 1. Apply Ptolemy's theorem to
. Prove easily 
, i.e. 
. Thus, 
. Since 
is cyclically, obtain that 
, i.e.in the -right triangle 
we have .Proof 2. Denote
. Well-known relations 
.PP10. Let an acute
with incircle 
and circumcircle 
. The circles 
and 
are tangent internally to 
inthe same
. 
is tangent externally to 
and is tangent internally to 
. Prove that 
, where 
and ![$ S\equiv [ABC]$](http://latex.artofproblemsolving.com/8/d/f/8df17304ca8c5a866f0e34f9089fe302737d3f65.png)
- the area of .
and 
. Thus, 
.Proof 1. Let
, 
. Apply Stewart's theorem in 
for the cevian-rays 
and 
.Proof 2. Apply the Pythagoras' theorem in the triangles:
Thus,
, i.e. 
. Observe that 
, where 
is the - exinradius of .Remark. Prove easily that two more interesting relations :
and 
, 
.PP11. Let two secant circles
, where 
. The tangent of in intersect in and the tangent of in intersect in . A ray by inside of 
intersect again , and the circumcircle of 
in , 
and 
respectively. Prove that 
.Proof. Let
. Thus, 
. With Steiner's theorem in 
and 
obtain : 
. Thus, 
, what is truly. See here an equivalent enunciation of this nice and difficult problem.PP12. Let
be the incircle of -right . The circle touches at 
,
and 
. Denote 
. Prove that 
.Proof. Denote the midpoint
of ![$[PD]$](http://latex.artofproblemsolving.com/3/b/3/3b3ab695b20d0a266cb5c4fad99fa342facb80ef.png)
, i.e. 
and 
, i.e. 
. Define 
and
, i.e. 
, 
Prove easily that 
. Observe that 
. But 
is cyclically, i.e. 
.From the evident relation
and the Stewart's relation applied to the cevians 
obtain that
![$m[(x+y)^2+(x+z)^2]+n(y^2+z^2)-2mx^2=$](http://latex.artofproblemsolving.com/d/7/f/d7fe6a9ac88da3f0d4f946d40e611a5bd6b2ac52.png)
Therefore, from the relations
and 
results 
. But 
. In conclusion, 
, what is truly.PP13. Let
such that 
. Let ![$D\in [AC]$](http://latex.artofproblemsolving.com/c/5/a/c5acd77baad8d15d98170c2ac60f572eeeed5507.png)
such that 
. The incircle of touches itssides in
![$K\in [AB]$](http://latex.artofproblemsolving.com/f/7/c/f7c55ea4f0642b6ee4395cbe8c000c6fec53e3ef.png)
and ![$L\in [AC]$](http://latex.artofproblemsolving.com/c/4/5/c45e50a5a8063a6607d6afd834b15cea829eb243.png)
. Let 
be the incircle of 
. Denote 
. Prove that 
.Proof.. The incircle
touches in 
, 
and 
. Observe that 
.Thus,
. Since 
obtain that 
. Denote 
. Theproperty
is well-known (see the lower remark). Thus, 
is cyclically. Analogously 
, i.e. 
. Observe that 
. Apply the Menelaus' theorem to transversal 
in 
: 
. So the quad. 
is cyclically 
or 
This post has been edited 171 times. Last edited by Virgil Nicula, Jan 25, 2016, 8:32 AM
August 2018
January 2018