be real numbers such that 
.
is the interval ![$\left[-\frac{2}{3}, \frac{2}{3}\right]$](http://latex.artofproblemsolving.com/4/f/2/4f22df2b391cf34b9b4f86634575e3e0ad23a386.png)
.
,
,
be positive integers such that 
.
grid as follows:
and a constant 
,
and 
are also friendly polynomials. Prove that 
.
for which the sequence of real numbers with 
for all 
is periodic from the beginning.
Y by kittenwarrior, Adventure10, Mango247, and 3 other users
Let 
be an acute triangle with 
, and 
being its incircle, circumcircle, and circumradius, respectively. Circle 
is tangent internally to 
at 
and tangent externally to 
. Circle 
is tangent internally to at and tangent internally to . Let 
and 
denote the centers of and , respectively. Define points 
analogously. Prove that
![\[8P_{A}Q_{A}\cdot P_{B}Q_{B}\cdot P_{C}Q_{C}\leq R^{3}\; , \]](//latex.artofproblemsolving.com/2/d/a/2da1369c7def7247df61099331d0f72573f8b3fe.png)
with equality if and only if triangle is equilateral.
be an acute triangle with 
, and 
being its incircle, circumcircle, and circumradius, respectively. Circle 
is tangent internally to 
at 
and tangent externally to 
. Circle 
is tangent internally to at and tangent internally to . Let 
and 
denote the centers of and , respectively. Define points 
analogously. Prove that![\[8P_{A}Q_{A}\cdot P_{B}Q_{B}\cdot P_{C}Q_{C}\leq R^{3}\; , \]](http://latex.artofproblemsolving.com/2/d/a/2da1369c7def7247df61099331d0f72573f8b3fe.png)
with equality if and only if triangle
is equilateral.
)
and 
Indeed
thus
into 
here we only use triangle 
noate that 
and 
that's all for we calculate now let 
now by Stewart's theorem in triangle 
we have 
and using 
we easily show 
similarly for 
let 
but note that 
we have 
now using 
now note that 
with 
is semiperimeter and 
we easily seen my indentity.
and the circumcircle 
.
and 
are tangent internally to the circle 
in the same point 
is tangent externally to the circle 
and is tangent internally to the circle 
.
,
and ![$S\equiv [ABC]$](http://latex.artofproblemsolving.com/3/e/2/3e28399980b83ac5f9599575b56840d9a050dda6.png)
-
.
.
are well-known.
,
and apply the Stewart's theorem in 
for the cevian-rays 
and 
:
.
![$\{\begin{array}{c}\boxed{\ r\le \sqrt [3]{\prod PQ}\le \frac{R}{2}\ }\\\\ \boxed{\ r\le\frac{1}{3}\cdot \sum PQ\le \frac{R^{2}}{4r}\ }\end{array}$](http://latex.artofproblemsolving.com/d/2/0/d20f5342da50aa400d3856ae407e348e0bb28fb7.png)
but maybe I made some mistake when I wrote this problem here, thanks for your remark Mr Virgil. An I think the inequality 
is true, it is Schur? and note that we can get 
but maybe I made some mistake when I wrote this problem here, thanks for your remark Mr Virgil. An I think the inequality 
,
is falsely. Correctly, 
.
, but I think solution of stronger by Schur is true? And moreover here we don't need acute triangle my solution is true for all triangle ?
be an acute triangle with the incircle 
and the circumcircle 
.
and 
are tangent internally to the circle 
in the same point 
is tangent externally to the circle 
and is tangent internally to the circle 
.
,
and ![$ S\equiv [ABC]$](http://latex.artofproblemsolving.com/8/d/f/8df17304ca8c5a866f0e34f9089fe302737d3f65.png)
-
.
.
are well-known.
,
.
for the cevian-rays 
and 
:
.
Therefore,
,
.
,
is the 
and 
, 
.
,
.
is the circumcentre of 
are collinear, and 
.
,
;
,
.
,
= 
,
.
(using the fact that 
.
,
)
,
,
,
,
,
,
,
,
,![$ [ABC]$](http://latex.artofproblemsolving.com/8/0/4/804b9de1e7a0a77c9e7b272da41c8e1cdff3c514.png)
,
have their usual meanings. Let 
be the 
that maps a point 
to 
.
turns into a line tangent to the incircle and parallel to 
.
and 
at 
respectively. 
is also a line parallel to 
and tangent to the incircle, but it is below 
is less than that of 
.
and 
respectively. Now, as 
was the intersection between the perpendicular bisectors of 
and 
,
is the foot of the altitude from 
,
around 
is the foot of the altitude from 
,
respectively. Now, as 
,
.
and 
.
.
is the incircle of 
.
.
and 
.
and 
.![\[ P_AQ_A=AP_A-AQ_a=\frac{(s-a)^2}{2h_a}\cdot (\frac{s}{s-a}-1)=\frac{a(s-a)}{2h_a}=\frac{a^2(s-a)}{[ABC]}\]](http://latex.artofproblemsolving.com/d/d/a/ddac88ec99c84334c7ef353077c3109156de866a.png)
Similarly, ![$ P_BQ_B=\frac{b^2(s-b)}{[ABC]}$](http://latex.artofproblemsolving.com/3/5/1/35173fc4a4f3f37cda72ff69221fbbc38132b456.png)
and ![$ P_CQ_C=\frac{c^2(s-c)}{[ABC]}$](http://latex.artofproblemsolving.com/f/c/b/fcb6c4a58c425a87d6150600346aa9a252ad84dd.png)
.![\[ P_CQ_C\cdot P_BQ_B\cdot P_AQ_A\le 8R^3\iff\]](http://latex.artofproblemsolving.com/5/7/1/5717afa13dc2e801130a811e059625eeaed7e568.png)
![\[ \frac{a^2b^2c^2(s-a)(s-b)(s-c)}{[ABC]^3}\le \frac{a^3b^3c^3}{8[ABC]^3}\iff\]](http://latex.artofproblemsolving.com/a/5/8/a5836bb6113380d5e50ec875d76b0cc400ca29f6.png)
![\[ 8(s-a)(s-b)(s-c)\le abc\iff 8s(s-a)(s-b)(s-c)\le sabc\iff\]](http://latex.artofproblemsolving.com/c/a/7/ca786cecbf67081d656334e4b5de8acc1b412590.png)
![\[ 8[ABC]^2\le abcs\iff \frac{2[ABC]}{s}\le \frac{abc}{4[ABC]}\iff 2r\le R\]](http://latex.artofproblemsolving.com/d/8/6/d86ac877e8c6768437242cc8e89e48277e457280.png)
which is true.
![[asy]
import cse5;
import olympiad;
import geometry;
pointpen = black;
pathpen = black;
real p=0.216,q=0.716;
pair A=dir(100),B=dir(213),C=dir(342),O=circumcenter(A,B,C),P=p*O+(1-p)*A,Q=q*O+(1-q)*A,I=incenter(A,B,C),M=extension(A,I,O,WP(B--C)),IA=rotate(180,M)*I,BB=extension(IA,rotate(90,IA)*A,A,B),CC=extension(IA,rotate(90,IA)*A,A,C),D=extension(CC,rotate(90,CC)*A,BB,rotate(90,BB)*A);
D(D("A",A,A)--D("B",B,B)--D("C",C,C)--cycle);
D(circumcircle(A,B,C),blue);
D(incircle(A,B,C),red);
D(CP(P,A),orange);
D(CP(Q,A),deepgreen);
D(CP(D,BB),purple);
DPA(B--BB^^C--CC);
D("P_A",P,S);
D("Q_A",Q,S);
[/asy]](http://latex.artofproblemsolving.com/c/8/b/c8b45de63362766105f0563f199c5039136c8401.png)
inversion. It is well-known that the incircle goes to the 
and 
pass through 
and 
are the reflections of 
and 
.![\[P_AQ_A=AQ_A-AP_A=\dfrac{bc}{AQ_A'}-\dfrac{bc}{AP_A'}\]](http://latex.artofproblemsolving.com/9/5/8/9586d6c5d54d070dbd0991b7234f46e790adefd6.png)
However, if 
and 
are the distances from ![\[P_AQ_A=\dfrac 12\left(\dfrac{bc}{d_1}-\dfrac{bc}{d_2}\right)=\dfrac 12\left(\dfrac{bc(d_1-d_2)}{d_1d_2}\right)\]](http://latex.artofproblemsolving.com/4/3/0/430c11d6a8af211c8bc765f9887f06c57aecabc9.png)
Now, we get that as 
,
,
is the radius of the said 
(the 
,![\[R_a=\dfrac{r_a}{\cos^2\frac 12 A}\]](http://latex.artofproblemsolving.com/5/f/b/5fb54da7bf06b057d450095662a6688db1c664c1.png)
By similar triangles, we must have that as 
and 
(where 
and similarly with 
)![\[d_1=\dfrac{R_a\cdot h_a}{r_a}\]](http://latex.artofproblemsolving.com/d/0/e/d0efd4d200cbeca953bb21440fa06e4c11126b17.png)
by comparing the excircles. Similarly, we get![\[d_2=\dfrac{R_a\cdot h_a}{r}\]](http://latex.artofproblemsolving.com/c/0/0/c0000d0f153a6bb5696b0030a5251a8ca52023df.png)
by comparing incircles. Finally, we also note![\[h_a=\dfrac{bc}{2R}\]](http://latex.artofproblemsolving.com/3/f/7/3f73fc329f52f8b2dbbb5b94e726846b581bac99.png)
so thus we get![\[P_AQ_A=\dfrac{bcR_arr_a}{R_a^2h_a^2}=\dfrac{4Rr\cos^2\frac 12 A}{bc}\]](http://latex.artofproblemsolving.com/3/6/9/36915b3298f4e2460c4409fd0d3946b717191852.png)
However, the well-known formula 
,![\[P_AQ_A=\dfrac{4Rrs(s-a)}{b^2c^2}=\dfrac{4R\Delta(s-a)}{b^2c^2}\]](http://latex.artofproblemsolving.com/d/c/6/dc68726bc20b8acdb40f94f7d5eed2615c8d8236.png)
Multiplying the symmetric versions, we get![\[8P_AQ_A\cdot P_BQ_B\cdot P_CQ_C=\dfrac{512R^3\Delta^3(s-a)(s-b)(s-c)}{a^4b^4c^4}\]](http://latex.artofproblemsolving.com/5/8/5/58538da6d7807a255333d54c3645fe848d26e149.png)
Using the well-known formula![\[\Delta=\dfrac{abc}{4R}\]](http://latex.artofproblemsolving.com/5/e/7/5e7287f50cb4f4dab2f5d46e7ea19d573a3571f8.png)
we have![\[8P_AQ_A\cdot P_BQ_B\cdot P_CQ_C=\dfrac{8R^3(s-a)(s-b)(s-c)}{abc}\]](http://latex.artofproblemsolving.com/d/5/5/d552f8110e0b8c30cf63846e104f2a16c504ac97.png)
However, Schur's Inequality guarantees that![\[8(s-a)(s-b)(s-c)\leq abc\]](http://latex.artofproblemsolving.com/2/7/8/278973dd0716379d89e0df95f10568e81cb429a5.png)
so thus we have![\[8P_AQ_A\cdot P_BQ_B\cdot P_CQ_C\leq R^3\]](http://latex.artofproblemsolving.com/5/7/2/5722645f814a4214f7f541c7aac254855d2d1671.png)
completing the proof.
etc. Let the intouch triangle be 
,
.
under transformation of inversion about the circle with radius 
and center 
,
,
.
and the other be 
.
wrt 
.
be the inradius. Then the distance from 
so![\[2P_AQ_A=\frac{(s-a)^2}{2(s-a)r/a}-\frac{(s-a)^2}{2sr/a} = \frac{a(s-a)}{2r}\cdot (1-(s-a)/s)=\frac{a^2(s-a)}{2rs}.\]](http://latex.artofproblemsolving.com/a/2/3/a2377af8027ebc4414debba61d0b1289a6f706d0.png)
Similar results hold for 
.![$R=\frac{abc}{4[ABC]}$](http://latex.artofproblemsolving.com/c/1/3/c137e069ac07a28e5b9d3bcfebd073414f2f228d.png)
where ![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
is the area of ![$[ABC]=\frac{1}{2}ab\sin C$](http://latex.artofproblemsolving.com/7/8/4/7841844a1efeb147d5572a6c2655286c81cf43bd.png)
.![\[\frac{a^2b^2c^2(s-a)(s-b)(s-c)}{(2[ABC])^3}\le \frac{a^3b^3c^3}{(4[ABC])^3}.\]](http://latex.artofproblemsolving.com/b/1/0/b109e8e22b3e4b1753957f8ee0b48931d6eb2b2d.png)
Let 
for positive 
,
which is immediate by AM-GM on 
.
,
as the incenter and circumcenter of 
and the semiperimeter of 
and 
follow symmetrically. In terms of 
.
To solve for 
now, we again use Stewarts, this time on 
and cevian 
So, 
Hence, it suffices to show 
Using 
,
simplifies to 
.
So, the inequality reduces to 
which is just Euler's Inequality, so we are done 
denote the measure of angles 
,
be the 
and radius 
.
be the projection of 
,
is 
Let 
be the 
the image of a point 
under 
and 
.
to some lines 
(respectively) both parallel to 
and tangent to 
is closer to 
.
are reflection of 
.
be the line passing through 
.
and 
.
.
But we also have
Now using Product-Sum Identities (in ch-5 of EGMO) we obtain
Hence
So we finally have that
Lastly, it is well know that
Hence,
This proves our claim. 
The most simple way to finish from here is take 
on both sides and using Jensen's inequality. But we also present a nice synthetic type proof:
So 
is equivalent to
Which is well known to be true (and we also know that equality holds if and only if circumcenter and incenter of 
be foot from 
be feet from 
to 
,
be the circumradius, inradius, radius of ![[asy]import olympiad;import geometry;
size(10cm);defaultpen(fontsize(10pt));
pair A,B,C,I,O,D,E,H,K,L;
A=dir(120);B=dir(205);C=dir(335);O=circumcenter(A,B,C);I=incenter(A,B,C);
real R=abs(A-O);real r=abs(I-foot(I,B,C));real k=(2*(abs(A-foot(A,B,C))-r)*R-r^2-2*R*r)/(2(abs(A-foot(A,B,C))));
real l=(2*(abs(A-foot(A,B,C))-r)*R-r^2-2*R*r)/(2*(abs(A-foot(A,B,C))-2r));
D=(A*(R-k)+O*k)/R;E=(A*(R-l)+O*l)/R;H=foot(A,B,C);K=foot(I,O,A);L=foot(I,A,H);
draw(A--B--C--cycle);draw(circumcircle(A,B,C));draw(incircle(A,B,C));
draw(circle(D,abs(D-A)));draw(circle(E,abs(E-A)));
draw(E--intersectionpoints(line(E,I),incircle(A,B,C))[1]);draw(L--I--foot(I,B,C));draw(A--O);draw(D--I);draw(A--H);draw(I--K);
draw(rightanglemark(I,K,O,1));draw(rightanglemark(I,L,A,1));draw(rightanglemark(C,H,A,1));draw(rightanglemark(C,foot(I,B,C),I,1));
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$I$",I,dir(I));
dot("$O$",O,dir(O));
dot("$P_A$",D,dir(D));
dot("$Q_A$",E,dir(E));
dot("$H$",H,dir(H));
dot("$K$",K,dir(K));
dot("$L$",L,dir(L));
[/asy]](http://latex.artofproblemsolving.com/0/5/a/05a54c9dcc0bec94e7238c5eab88bd945e977b01.png)
By Law of Cosines on 
,
Similarly, by Law of Cosines on 
,
Hence, 
Moreover, since 
,
Similarly, we find 
Now, using 
and 
,
Now, for simplicity, we do a sub 
,
and 
and we would like to show that
which is actually equivalent to 
which is true by AM-GM. In fact, the equality holds iff 
.
for sharing with me this identity which essentially solves the inequality part which I wasn't being able to do,![\[\sin\left(\dfrac{A}{2}\right)\sin\left(\dfrac{B}{2}\right)\sin\left(\dfrac{C}{2}\right)=\dfrac{r}{4R}.\]](http://latex.artofproblemsolving.com/1/a/f/1af79948d59a2c6e55e68a440cdc8914257db8db.png)
which is a line is tangent to the mixtilinear incircle of 
.
that has it's incircle as the 
m
at 
are collinear. Now note that the centers of the circles 
and 
with 
.
and 
respectively. Also, let 
denote the foot of the 
,
as their incircles also get mapped. This gives that 
.
.
.
.
.
.![\[\sin\left(\dfrac{A}{2}\right)\sin\left(\dfrac{B}{2}\right)\sin\left(\dfrac{C}{2}\right)=\dfrac{r}{4R},\]](http://latex.artofproblemsolving.com/6/5/2/652ce42584b7cf9dd73f81f12f9450cac74356ee.png)
and also we have our Euler's Inequality that says 
where the inequality holds iff 
which is when we have an equilateral triangle. Now we just follow through to finish off the rest of the problem.
which is what we wanted and we are done 
about 
-
be the images of 
(typing stars is hard). The conditions mean that 
is the foot of the 
is the intersection of the 
,
and 
.
So then
This is just Schur, thus done!
and cyclic variants the length and altitude length usual definition). Let D be the foot of altitude from A, E,F the antipodes of A wrt 
,
)
that's tangent to the preimage must be a parallel line to the after image and also tangent to 
.
inverts to the other line l parallel to BC tangent to 
.
and 
implies ![\[P_aQ_a=\frac{AF}2-\frac{AE}2= \frac{(s-a)^2 r}{h_A(h_A-2r)}=\frac{(s-a)^2\frac{[ABC]}s}{\frac{2[ABC]}a\left(\frac{2[ABC]}a-\frac{2[ABC]}s\right)}= \frac{a^2(s-a)}{4[ABC]}\]](http://latex.artofproblemsolving.com/2/9/3/2932e111dab945bf3ff66c8d4794fa37d25b19aa.png)
![\[\implies8\prod P_AQ_A\le R^3\iff8a^2b^2c^2\prod(s-a)\le(4[ABC]R)^3\iff8\prod(s-a)\le abc\]](http://latex.artofproblemsolving.com/d/7/e/d7e2b26afcd74bc92c849bed0de790a2dae0f2f3.png)
which is evident by, say, AM-GM over all terms after expanding.
Defining the new points is motivated by the fact that 1. segment connecting A-antipode is diameter, altitude are well known lengths, and because this is inversion unit (but also the circles' tangency suggests that), we look for radii, upon s-a looks fine to work with.
where 
is the length of the altitude from 
Then, letting 
We proceed similarly for 
We then get the following by using Heron's and the inradius and circumradius formula:
which is well known, e.g. by constructing the nine-point circle.
and 
denote the reflections of 
and 
respectively, which both lie on their respective circles as well as line 
The problem then reduces to showing that 
\end{claim}
.
.
is the intersection of the top parallel tangent to 
and the line 
,
is the same intersection but with the bottom parallel tangent. Let 
and 
be the tangents of 
and 
Note that 
Thus, 
since the radius of inversion is 
Thus, 
This means that 
However, of course we have ![$$L=\frac{2[ABC]}{a}=\frac{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{2a}.$$](http://latex.artofproblemsolving.com/d/2/9/d292c105446a73a9de0bdf6ee01e0dca9aaae90a.png)
Therefore, plugging back in we have 
Substituting 
we have 
The left hand side expands out to 
so it suffices to show that 
which is just Schur's inequality. Since 
are nonzero, equality occurs if and only if 
is mapped to the altitude from 
,
.![\[P_AQ_A = P_A^*Q_A^* \cdot \frac{(s-a)^2}{AP_A^* \cdot AQ_A^*} = \frac{(s-a)^2 \cdot r}{h_a(h_a-2r)} = \frac{a^2 \cdot (s-a)}{4rs}.\]](http://latex.artofproblemsolving.com/e/b/0/eb033ed3a904c36ef751a988a493f28f807bebb4.png)
![\[8 \cdot \prod_{\text{cyc}} \frac{a^2 \cdot (s-a)}{4[ABC]} = \frac{(abc)^2 \cdot s(s-a)(s-b)(s-c)}{8 s [ABC]^3} = \frac{(4Rrs) \cdot (4R \cdot [ABC]) \cdot [ABC]^2}{8 s [ABC]^3} = 2R^2r,\]](http://latex.artofproblemsolving.com/a/0/4/a04eeb18ac6a5b65ce5d55eddc83b87683058a57.png)
if and only if 
,
equals the distance from 
It turns out that this actually equals something nice, but we will not realize this and instead turn our brains off. Letting ![$K=[ABC]$](http://latex.artofproblemsolving.com/e/8/b/e8b7f4b298b7acdef16cf10905ad88615ca235ce.png)
,
By Euler's Theorem, 
with equality iff 
.![\[
P_A Q_A \cdot P_B Q_B \cdot P_C Q_C \leq 1
\]](http://latex.artofproblemsolving.com/f/9/f/f9f8e9e62a3198e22e203389ff1b394e7c47148f.png)
with equality if and only if 
.
.
and 
.![\[
\angle IAO = \frac{\angle A}{2} - \angle CAO = \frac{\angle A}{2} - 90^\circ + \angle B = \frac{\angle B}{2} - \frac{\angle C}{2}.
\]](http://latex.artofproblemsolving.com/9/6/a/96a7386a1d55b697141362f60e88c437f850f3e2.png)
![\[
(x + r)^2 = x^2 + AI^2 - 2x \cdot AI \cdot \cos\left(\frac{B}{2} - \frac{C}{2}\right),
\]](http://latex.artofproblemsolving.com/f/0/b/f0b2dc50deab502a43d0386ae695ea7b58d2b38d.png)
which simplifies to:![\[
x = \frac{AI^2 - r^2}{2(r + AI \cos\left(\frac{B}{2} - \frac{C}{2}\right))}.
\]](http://latex.artofproblemsolving.com/b/c/f/bcfaab6af615aa21c1e95b5960f17ccab7ea0c1f.png)
![\[
x = \frac{64 \sin^2\frac{B}{2} \sin^2\frac{C}{2} - 64 \sin^2\frac{A}{2} \sin^2\frac{B}{2} \sin^2\frac{C}{2}}{2(8 \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2} + 8 \sin\frac{B}{2} \sin\frac{C}{2} \cos\left(\frac{B}{2} - \frac{C}{2}\right))}.
\]](http://latex.artofproblemsolving.com/2/7/5/275dc6082029778bc473c401d81786bdf0e49e6e.png)
![\[
x = \frac{4 \sin^2\frac{B}{2} \sin^2\frac{C}{2} (1 - \sin^2\frac{A}{2})}{\sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2} + \sin\frac{B}{2} \sin\frac{C}{2} \cos\left(\frac{B}{2} - \frac{C}{2}\right)},
\]](http://latex.artofproblemsolving.com/7/5/2/7527d9163f3c89d012a23c73b388ffd4cf72d926.png)
![\[
x = \frac{4 \sin\frac{B}{2} \sin\frac{C}{2} \cos^2\frac{A}{2}}{\sin\frac{A}{2} + \cos\left(\frac{B}{2} - \frac{C}{2}\right)},
\]](http://latex.artofproblemsolving.com/e/0/a/e0ac5eece3d2875698c8e75f599ddc7d7e419a05.png)
![\[
x = \frac{4 \sin\frac{B}{2} \sin\frac{C}{2} \cos^2\frac{A}{2}}{\sin\frac{A}{2} + \sin\left(\frac{A}{2} + C\right)},
\]](http://latex.artofproblemsolving.com/0/d/4/0d40ce440a0c95b605f3eee92f683526eb9ec5bc.png)
![\[
x = \frac{4 \sin\frac{B}{2} \sin\frac{C}{2} \cos^2\frac{A}{2}}{2 \sin\left(\frac{A}{2} + \frac{C}{2}\right) \cos\frac{C}{2}},
\]](http://latex.artofproblemsolving.com/e/8/3/e83da1ad877241e89b42d4d642a565ffa0e93dcd.png)
![\[
x = 2 \tan\frac{B}{2} \tan\frac{C}{2} \cos^2\frac{A}{2}.
\]](http://latex.artofproblemsolving.com/4/b/e/4be3c5bd90faa3def670ff0780b1764a57744d09.png)
.
.![\[
(y - r)^2 = y^2 + AI^2 - 2y \cdot AI \cdot \cos\left(\frac{B}{2} - \frac{C}{2}\right),
\]](http://latex.artofproblemsolving.com/7/8/5/785293a017b2c1b075c19141edc08113856b0052.png)
which simplifies to:![\[
y = \frac{AI^2 - r^2}{2(AI \cos\left(\frac{B}{2} - \frac{C}{2}\right) - r)}.
\]](http://latex.artofproblemsolving.com/7/3/e/73e55bc6eb8a03a8db7a7683b83d008d521a6b9a.png)
![\[
y = 2 \cos^2\frac{A}{2}.
\]](http://latex.artofproblemsolving.com/e/4/0/e4052f9c8979fc974b1a00cf1782175b35cfd00a.png)
![\[
P_A Q_A = 2 \cos^2\frac{A}{2} (1 - \tan\frac{B}{2} \tan\frac{C}{2}),
\]](http://latex.artofproblemsolving.com/b/7/b/b7b2d87fa93683578b1b6a3a9ea256fca66ec217.png)
![\[
P_A Q_A = 2 \cos^2\frac{A}{2} \frac{\tan\frac{B}{2} + \tan\frac{C}{2}}{\tan\left(\frac{B}{2} + \frac{C}{2}\right)},
\]](http://latex.artofproblemsolving.com/a/1/c/a1c1b8a0b424146aaa834879309e4abedbcf4ade.png)
![\[
P_A Q_A = \sin A (\tan\frac{B}{2} + \tan\frac{C}{2}),
\]](http://latex.artofproblemsolving.com/0/7/4/074e104402d16f64e328d0de4a2e695e81513c10.png)
![\[
P_A Q_A = \sin A \frac{\sin\frac{B}{2} \cos\frac{C}{2} + \cos\frac{B}{2} \sin\frac{C}{2}}{\cos\frac{B}{2} \cos\frac{C}{2}},
\]](http://latex.artofproblemsolving.com/d/4/1/d41d58998b0cb3e4705d09a60145f896bcb8021a.png)
![\[
P_A Q_A = \sin A \frac{\cos\frac{A}{2}}{\cos\frac{B}{2} \cos\frac{C}{2}}.
\]](http://latex.artofproblemsolving.com/9/4/7/947bc4c2683d1a0ea560ebda468f931c84063b3b.png)
![\[
\frac{\sin A \sin B \sin C}{\cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2}} = 8 \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2}.
\]](http://latex.artofproblemsolving.com/2/9/e/29edd305234b5cd67a82888b469450eb93b49f60.png)
has a second derivative 
,
.![\[
8 \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2} \leq 8 \sin^3(30^\circ) = 1.
\]](http://latex.artofproblemsolving.com/6/c/7/6c7bca5078c0134fbb09f37c64758bc283c853f1.png)
as desired.![$[ABC] = \frac{abc}{4R}.$](http://latex.artofproblemsolving.com/b/8/3/b83dc05cd5d3f19efa96aeff31943410f5df0316.png)
and its symmetric equations. Multiplying, 
Furthermore, note that ![$\frac12ab\sin\angle C = [ABC],$](http://latex.artofproblemsolving.com/1/9/3/1934b14b6a16a5427de7bc669e56f7edd80ca75d.png)
and its symmetric equations. Thus, ![$\frac1{64}a^3b^3c^3 = R^3[ABC]^3,$](http://latex.artofproblemsolving.com/e/c/6/ec668ef89816b3b2f5b6ad87efcb9b9220196d1f.png)
which finishes.
and reflect across the 
inverts to the line parallel to 
invert to the reflections of 
Hence, if 
is 
If 
is the length of the altitude from 
which finishes by Euler's theorem, which states that 
with equality when equilateral.
