inequalities geometry circumcircle function trigonometry inradius incenter

Let $ABC$ be an acute triangle with $\omega,S$, and $R$ being its incircle, circumcircle, and circumradius, respectively. Circle $\omega_{A}$ is tangent internally to $S$ at $A$ and tangent externally to $\omega$. Circle $S_{A}$ is tangent internally to $S$ at $A$ and tangent internally to $\omega$. Let $P_{A}$ and $Q_{A}$ denote the centers of $\omega_{A}$ and $S_{A}$, respectively. Define points $P_{B}, Q_{B}, P_{C}, Q_{C}$ analogously. Prove that
\[8P_{A}Q_{A}\cdot P_{B}Q_{B}\cdot P_{C}Q_{C}\leq R^{3}\; , \]
with equality if and only if triangle $ABC$ is equilateral.

Could somebody compute the following two sums as a function of known things?(If yes the problem is over :D )
So please compute $\angle{AP_{A}I}+\angle{AQ_{A}I}$ and $\angle{AP_{A}I}-\angle{AQ_{A}I}$

Hey nice problem  :D
We can show $P_{A}Q_{A}+P_{B}Q_{B}+P_{C}Q_{C}\le\frac{3}{2}R$ Indeed
we can show $P_{A}Q_{A}=\frac{a^{2}(b+c-a)}{2S}$ thus
$\sum P_{A}Q_{A}\le\frac{3}{2}R\Leftrightarrow \sum\frac{a^{2}(b+c-a)}{2S}\le \frac{3}{2}R\Leftrightarrow a^{3}+b^{3}+c^{3}+3abc\ge \sum (b+c)a$
Trivial by Schur!
But I think the hard problem here is show $P_{A}Q_{A}=\frac{a^{2}(b+c-a)}{2S}$ I have a solution by calculation can anybody show a nice proof?

[quote="gemath"] But I think the hard problem here is show $P_{A}Q_{A}=\frac{a^{2}(b+c-a)}{2S}$ I have a solution by calculation can anybody show a nice proof?[/quote]

Could you post your way ?Thank you

Ok here is my solution for that indentity for simple I change notation $P_{A},Q_{A}$ into $P,Q$
We easily show $IP+OP=R+r,IQ+OQ=R-r,PQ=OP-OQ$ here we only use triangle $AIO$ noate that $P\in AO,OA=R$ and $IP+OP=R+r$ that's all for we calculate now let $\frac{OP}{OA}=x=\frac{OP}{R}\Rightarrow \frac{PA}{R}=1-x$ now by Stewart's theorem in triangle $IOA$ we have $\frac{1}{x}IO^{2}+\frac{1}{1-x}IA^{2}=(\frac{1}{x}+\frac{1}{1-x})IP^{2}+R^{2}$ and using $IP=R+r-OP=(1-x)R+r,OI^{2}=R^{2}-2Rr$ we easily show $OP=Rx=R\frac{4Rr+r^{2}}{IA^{2}+4Rr}$ similarly for $Q\in OA$ let $\frac{OQ}{OA}=y$ but note that $IQ=(1-y)R-r$ we have $OQ=Ry=R\frac{r^{2}}{IA^{2}}$ now using $PQ=OP-OQ=R\frac{IA^{2}-r^{2}}{IA^{2}+4Rr}$ now note that $IA^{2}=\frac{(p-a)^{2}}{\cos^{2}\frac{A}{2}}=\frac{sbc}{s-a}$ with $s$ is semiperimeter and $4Rr=\frac{abc}{p}$ we easily seen my indentity.

Many thanks .Your way is straightforward .
Now could you find something for the angles I wrote above ?

My proof is basically the same of yours, gemath!

Very Nice problem. Now, i'm trying to find one solution using inversion!

:spider: [size=150][color=darkblue][b]Sorry Gemath[/b], you commit some small mistakes ![/color] [/size]

[color=darkred][u][b]Lemma (Gemath).[/b][/u] Let $ABC$ be an acute triangle with the incircle $\omega =C(I,r)$ and the circumcircle $\rho =C(O,R)$.
The circles $C_{1}=C(P,r_{1})$ and $C_{2}=(Q,r_{2})$ are tangent internally to the circle $\rho$ in the same point $A$
The circle $w$ is tangent externally to the circle $C_{1}$ and is tangent internally to the circle $C_{2}$.
Prove that $\boxed{PQ=\frac{a^{2}(p-a)}{4S}}$, where $2p=a+b+c$ and $S\equiv [ABC]$- the area of $\triangle ABC$.[/color]

[color=darkblue][b]Proof.[/b] Prove easily that $\{\begin{array}{ccccc}IP=r+r_{1}& , & IQ=r_{2}-r\\\\ PO=R-r_{1}& , & PA=r_{1}\\\\ QO=R-r_{2}& , & QA=r_{2}\\\\ OA=R & , & \boxed{PQ=r_{2}-r_{1}}\end{array}$. The relations $\{\begin{array}{ccc}IO^{2}=R(R-2r) & , & IA^{2}=\frac{bc(p-a)}{p}\\\\ IA^{2}-r^{2}=(p-a)^{2}& , & IA^{2}+4Rr=bc\\\\ p(p-a)+(p-b)(p-c)=bc\end{array}$ are well-known.

Denote $IO=m$, $IA=n$ and apply the Stewart's theorem in $\triangle OIA$ for the cevian-rays $[IP$ and $[IQ$ :

$\{\begin{array}{c}m^{2}r_{1}^{2}+n^{2}(R-r_{1})=R(r+r_{1})^{2}+Rr_{1}(R-r_{1})\\\\ m^{2}r_{2}+n^{2}(R-r_{2})=R(r_{2}-r)^{2}+Rr_{2}(R-r_{2})\end{array}\|$ $\implies$ $r_{1}(R^{2}+2Rr+n^{2}-m^{2})=R(n^{2}-r^{2})=r_{2}(R^{2}-2Rr+n^{2}-m^{2})$ $\implies$

$r_{1}(R^{2}+2Rr+bc-4Rr-R^{2}+2Rr)=$ $R(p-a)^{2}=$ $r_{2}(R^{2}-2Rr+bc-4Rr-R^{2}+2Rr)$ $\implies$ $r_{1}bc=R(p-a)^{2}=r_{2}(bc-4Rr)$ $\implies$

$PQ=r_{2}-r_{1}=$ $\frac{4R^{2}r(p-a)^{2}}{bc(bc-4Rr)}=$ $\frac{4R^{2}pr(p-a)^{2}}{b^{2}c^{2}(p-a)}=$ $\frac{4R^{2}pra^{2}(p-a)^{2}}{16R^{2}p^{2}r^{2}(p-a)}$ $\implies$ $\boxed{\ PQ=\frac{a^{2}(p-a)}{4S}\ }$.

[b]Remark.[/b] $8\cdot \prod PQ=\frac{8(abc)^{2}(p-a)(p-b)(p-c)}{(4S)^{3}}=\frac{8(4RS)^{2}Sr}{(4S)^{3}}=$ $R^{2}\cdot 2r$ $\implies$ $\{\begin{array}{c}\boxed{\ r\le \sqrt [3]{\prod PQ}\le \frac{R}{2}\ }\\\\ \boxed{\ r\le\frac{1}{3}\cdot \sum PQ\le \frac{R^{2}}{4r}\ }\end{array}$[/color]

I think my result is true $P_{A}Q_{A}=\frac{a^{2}(b+c-a)}{2S}=\frac{a^{2}(s-a)}{4S}=\frac{a^{2}}{4r_{a}}$ but maybe I made some mistake when I wrote this problem here, thanks for your remark Mr Virgil. An I think the inequality $\sum P_{A}Q_{A}\le \frac{3}{2}R$ is true, it is Schur? and note that we can get $\sum \frac{P_{A}Q_{A}}{r_{a}}\ge \frac{9}{16}$ ($r_{a},r_{b},r_{c}$ are exradius of $ABC$) from Iran inequality.

[quote="gemath"][color=purple]I think my result is true $\boxed{P_{A}Q_{A}=\frac{a^{2}(b+c-a)}{2S}=\frac{a^{2}(s-a)}{4S}}=\frac{a^{2}}{4r_{a}}$ but maybe I made some mistake when I wrote this problem here, thanks for your remark Mr Virgil. An I think the inequality $\sum P_{A}Q_{A}\le \frac{3}{2}R$ is true, it is Schur? and note that we can get $\sum \frac{P_{A}Q_{A}}{r_{a}}\ge \frac{9}{16}$ ($r_{a},r_{b},r_{c}$ are exradius of $ABC$) from Iran inequality.[/color][/quote]

Gemath, if $2s=a+b+c$, then the relation $\frac{a^{2}(b+c-a)}{2S}=\frac{a^{2}(s-a)}{4S}$ is falsely. Correctly, $P_{A}Q_{A}=\frac{a^{2}(s-a)}{4S}$. I used no Schur in the my proof.

Ok that's right I made a small mistake  :blush: , 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 ?

This problem was proposed by Sung-Yoon Kim (MIT).

[color=darkred]Let $ ABC$ be an acute triangle with the incircle $ \omega = C(I,r)$ and the circumcircle $ \rho = C(O,R)$.
The circles $ C_{1}= C(P,r_{1})$ and $ C_{2}= (Q,r_{2})$ are tangent internally to the circle $ \rho$ in the same point $ A$
The circle $ w$ is tangent externally to the circle $ C_{1}$ and is tangent internally to the circle $ C_{2}$.
Prove that $ \boxed{PQ =\frac{a^{2}(p-a)}{4S}}$, where $ 2p = a+b+c$ and $ S\equiv [ABC]$- the area of $ \triangle ABC$.[/color]

[color=darkblue]Prove easily that $ \{\begin{array}{ccccc}IP = r+r_{1}& , & IQ = r_{2}-r\\ \\ PO = R-r_{1}& , & PA = r_{1}\\ \\ QO = R-r_{2}& , & QA = r_{2}\\ \\ OA = R & , &\boxed{PQ = r_{2}-r_{1}}\end{array}$. The relations $ \{\begin{array}{ccc}IO^{2}= R(R-2r) & , & IA^{2}=\frac{bc(p-a)}{p}=\frac{4Rr(p-a)}{a}\\ \\ IA^{2}-r^{2}= (p-a)^{2}& , & IA^{2}+4Rr = bc\\ \\ p(p-a)+(p-b)(p-c) = bc.\end{array}$ are well-known.

[b]Proof I.[/b] Denote $ IO = m$, $ IA = n$. Apply the [b]Stewart's theorem[/b] in $ \triangle OIA$ for the cevian-rays $ [IP$ and $ [IQ$ :

$ \{\begin{array}{c}m^{2}r_{1}^{2}+n^{2}(R-r_{1}) = R(r+r_{1})^{2}+Rr_{1}(R-r_{1})\\ \\ m^{2}r_{2}+n^{2}(R-r_{2}) = R(r_{2}-r)^{2}+Rr_{2}(R-r_{2})\end{array}\|$ $ \implies$ $ r_{1}(R^{2}+2Rr+n^{2}-m^{2}) = R(n^{2}-r^{2}) = r_{2}(R^{2}-2Rr+n^{2}-m^{2})$ $ \implies$

$ r_{1}(R^{2}+2Rr+bc-4Rr-R^{2}+2Rr) =$ $ R(p-a)^{2}=$ $ r_{2}(R^{2}-2Rr+bc-4Rr-R^{2}+2Rr)$ $ \implies$ $ r_{1}bc = R(p-a)^{2}= r_{2}(bc-4Rr)$ $ \implies$

$ PQ = r_{2}-r_{1}=$ $ \frac{4R^{2}r(p-a)^{2}}{bc(bc-4Rr)}=$ $ \frac{4R^{2}pr(p-a)^{2}}{b^{2}c^{2}(p-a)}=$ $ \frac{4R^{2}pra^{2}(p-a)^{2}}{16R^{2}p^{2}r^{2}(p-a)}$ $ \implies$ $ \boxed{\ PQ =\frac{a^{2}(p-a)}{4S}\ }$.

[b]Proof II.[/b] Apply the [b]Pythagoras' theorem[/b] in the triangles :
$ \triangle\ IOP\blacktriangleright$ $ (r+r_{1})^{2}= (R-r_{1})^{2}+(R^{2}-2Rr)-2(R-r_{1})\cdot$ $ OI\cdot\frac{(R^{2}-2Rr)+R^{2}-\frac{4Rr(p-a)}{a}}{2R\cdot IO}\implies$ $ \boxed{r_{1}=\frac{R(p-a)^{2}}{bc}}\ .$
$ \triangle\ IOQ\blacktriangleright$ $ (r_{2}-r)^{2}= (R-r_{2})^{2}+(R^{2}-2Rr)-2(R-r_{2})\cdot$ $ IO\cdot\frac{(R^{2}-2Rr)+R^{2}-\frac{4Rr(p-a)}{a}}{2R\cdot IO}\implies$ $ \boxed{r_{2}=\frac{Rp(p-a)}{bc}}\ .$ Therefore,

$ \frac{r_{1}}{p-a}=\frac{r_{2}}{p}=\frac{R(p-a)}{bc}=\frac{PQ}{a}$, i.e. $ \boxed{PQ =\frac{aR(p-a)}{bc}}=\frac{a^{2}(p-a)}{4S}$. Observe that $ \frac{r_{1}}{r}=\frac{r_{2}}{r_{a}}$, where $ r_{a}$ is the $ A$- exinradius of $ \triangle ABC$.

[b]Remark.[/b] Prove easily that two more interesting relations : $ \boxed{\sum PQ = R+r\ }$ and $ r_{1}= R-\frac{a(4R+r)}{4p}$ , $ \boxed{r_{2}=\frac{ r_{b}+r_{c}}{4}}$.[/color]

Another solution:
if $ \angle BAC \equal{} 2A, \angle ABC \equal{} 2B \angle ACB \equal{} 2C$, then it's clear that $ \angle IAO \equal{} B\minus{}C$. It's also clear, that if $ O$ is the circumcentre of $ ABC$, that $ A, P_{A}, Q_{A}, O$ are collinear, and $ IP_{A} \equal{} AP_{A}\plus{}r$. But using the Cosine Rule on $ IAP_{A}$, it's clear that $ (IP_{A})^2 \equal{} IA^2 \plus{} AP_{A}^2 \minus{} 2(IA)(AP_{A}) cos (B\minus{}C) \equal{} AP_{A}^2 \plus{} r^2 \plus{} 2(AP_{A})r$; now using the fact that $ IA^2 \minus{} r^2 \equal{} (\frac {b\plus{}c\minus{}a}{2})^2$, we have
$ AP_{A} \equal{} \frac {(b\plus{}c\minus{}a)^2} {8(IA cos(B\minus{}C)\plus{}r)}$.
Similarly, $ AQ_{A} \equal{} \frac {(b\plus{}c\minus{}a)^2} {8(IA cos(B\minus{}C) \minus{} r)}$
Using the fact that $ IA \equal{} \frac {r}{sin a}$, gives us that
$ P_{A}Q_{A}$ = $ AP_{A} \minus{} AQ_{A} \equal{} (\frac {(b\plus{}c\minus{}a)^2 sin A}{8r})( \frac{1}{cos(B\minus{}C) \minus{} sinA} \minus{} \frac{1}{cos(B\minus{}C)\plus{}sin A}) \equal{} \frac {(b\plus{}c\minus{}a)^2 (sin A)^2}{16 r sin B sin C cos B cos C}$, using the fact that $ sin A \equal{} cos (B\plus{}C)$.
Thus,
$ 8 (P_{A} Q_{A})(P_{B} Q_{B})(P_{C} Q_{C}) \equal{}  \frac {(b\plus{}c\minus{}a)^2 (a\plus{}b\minus{}c)^2 (a\plus{}c\minus{}b)^2} {2^9 r^3 (cos A cos B cos C)^2} \equal{} \frac {r^3} {8 (sin A sin B sin C)^2}$ (using the fact that $ b\plus{}c\minus{}a \equal{} \frac {2r} {tan A}$.
Since $ (sin A)^2 \equal{} \frac {1 \minus{} cos 2A} {2}$,
It suffices to prove now that:
$ \frac {1}{(1\minus{} cos 2A)(1\minus{}cos 2B)(1\minus{} cos 2C)} \leq (\frac {R}{r})^3$
Using the Cosine Rule, ( $ cos 2A \equal{} \frac {b^2 \plus{} c^2 \minus{} a^2} {2bc}$) we have,
$ 1\minus{} cos 2A \equal{} \frac {(a\minus{}b\plus{}c)(a\minus{}c\plus{}b)} {2bc}$,
and using the fact that $ \frac {R}{r} \equal{} \frac {2abc}{(a\minus{}b\plus{}c)(a\minus{}c\plus{}b)(b\minus{}a\plus{}c)}$,
our inequality is reduced to
$ 8 ( \frac {abc}{(a\minus{}b\plus{}c)(a\minus{}c\plus{}b)(b\minus{}c\plus{}a)} )^2 \leq 8 ( \frac {abc}{(a\minus{}b\plus{}c)(a\minus{}c\plus{}b)(b\minus{}c\plus{}a)} )^3$,
which is clearly true since $ abc \geq (a\minus{}b\plus{}c)(a\minus{}c\plus{}b)(b\plus{}c\minus{}a)$

[hide="Solution"]
Let $ s$, $ a$, $ b$, $ c$, $ r$, $ R$, $ [ABC]$, and $ h_a$ have their usual meanings.  Let $ r_a$ be the $ A$-excircle radius of $ \triangle ABC$.  Invert around $ A$ with radius $ s\minus{}a$ that maps a point $ X$ to $ X'$.  As $ s\minus{}a$ is the length of the tangent from $ A$ to the incircle of $ \triangle ABC$, we see that the incircle of $ \triangle ABC$ remains as it was before.  Now, the $ \omega_A'$ turns into a line tangent to the incircle and parallel to $ B'C'$.  Let $ \omega_A'$ meet $ AB$ and $ AC$ at $ X$ and $ Y$ respectively.  $ S_A'$ is also a line parallel to $ C'B'$ and tangent to the incircle, but it is below $ A$ since the radius of $ S_A$ is less than that of $ \omega_A$.  Let $ S_A'$ meet $ AB$ and $ AC$ at $ P$ and $ Q$ respectively.  Now, as $ P_A$ was the intersection between the perpendicular bisectors of $ AX'$ and $ AY'$, we see that $ P_A'$ is the foot of the altitude from $ A$ to $ XY$, but dilated by a factor of $ 2$ around $ A$.  Simiarly, $ Q_A'$ is the foot of the altitude from $ A$ to $ PQ$, but dilated by a factor of $ 2$ around $ A$.  Let the feet of the altitudes from $ A$ to $ PQ$ and $ XY$ be $ R$ and $ Z$ respectively.  Now, as $ XY\parallel B'C'\parallel PQ$, we have that $ \triangle ABC\sim \triangle AC'B'\sim \triangle AQP\sim \triangle AYX$.  Furthermore, $ \triangle APQ$ and $ \triangle ABC$ have the same inradius (in fact, the same incircle), so $ \triangle ABC\cong \triangle AQP$.  Thus, $ AR\equal{}h_a$.  Furthermore, the $ A$-excircle of $ \triangle AXY$ is the incircle of $ \triangle ABC$.  Thus, the scale factor between $ \triangle AXY$ and $ \triangle ABC$ is $ \frac{r}{r_a}\equal{}\frac{s\minus{}a}{s}$.  It follows that $ AZ\equal{}\frac{h_a(s\minus{}a)}{s}$.  This means that $ AP_A'\equal{}\frac{2h_a(s\minus{}a)}{s}$ and $ AQ_A'\equal{}2h_a$.  Thus, $ AP_A\equal{}(s\minus{}a)^2\cdot \frac{s}{2h_a(s\minus{}a)}$ and $ AQ_A\equal{}(s\minus{}a)^2\cdot \frac{1}{2h_a}$.  This means that \[ P_AQ_A\equal{}AP_A\minus{}AQ_a\equal{}\frac{(s\minus{}a)^2}{2h_a}\cdot (\frac{s}{s\minus{}a}\minus{}1)\equal{}\frac{a(s\minus{}a)}{2h_a}\equal{}\frac{a^2(s\minus{}a)}{[ABC]}\] Similarly, $ P_BQ_B\equal{}\frac{b^2(s\minus{}b)}{[ABC]}$ and $ P_CQ_C\equal{}\frac{c^2(s\minus{}c)}{[ABC]}$.  Then \[ P_CQ_C\cdot P_BQ_B\cdot P_AQ_A\le 8R^3\iff\] \[ \frac{a^2b^2c^2(s\minus{}a)(s\minus{}b)(s\minus{}c)}{[ABC]^3}\le \frac{a^3b^3c^3}{8[ABC]^3}\iff\] \[ 8(s\minus{}a)(s\minus{}b)(s\minus{}c)\le abc\iff 8s(s\minus{}a)(s\minus{}b)(s\minus{}c)\le sabc\iff\] \[ 8[ABC]^2\le abcs\iff \frac{2[ABC]}{s}\le \frac{abc}{4[ABC]}\iff 2r\le R\] which is true. [/hide]

Let $\angle AOI = \angle P_AOI = \angle Q_AOI = \alpha$.  In the $\triangle OP_A I$, $IP_A= R - OP_A + r$. So by cosine rule $OP_A ^2 + OI^2 - 2 \cdot OI \cdot OP_A \cdot \cos \alpha = (R - OP_A + r)^2$.  We know that $OI^2 = R^2 - 2Rr$.

So we get   [color=#008040][b](1)[/b][/color]  $OP_A = \frac{4Rr + r^2 }{2R + 2r - 2 \cdot OI \cdot \cos \alpha } $

In $\triangle OQ_A I$, we have $Q_A I = (R - OQ_A - r)^2$. So by cosine rule $OQ_A ^2 + OI^2 - 2 \cdot OI \cdot OQ_A \cdot \cos \alpha = (R - OQ_A - r)^2$.

So we get  [color=#008040][b](2)[/b][/color] $OQ_A = \frac{r^2}{ 2R - 2r - 2 \cdot OI \cdot \cos \alpha } $

But from $ \triangle AIO$, we have $R^2 + OI^2 -  2\cdot OI \cdot R \cdot \cos \alpha = AI^2 $

Which gives $- 2 \cdot IO \cdot R \cdot \cos \alpha = AI^2 + 2Rr - 2R^2$. Plugging this in equation   [color=#008040][b](1)[/b][/color] and   [color=#008040][b](2)[/b][/color] yields $OP_A = \frac{Rr^2 + 4R^2 r}{ 4Rr + AI^2}$ and $OQ_A = \frac{Rr^2}{AI^2 } $.

So  [color=#008040][b](3)[/b][/color]  $P_A Q_A = OP_A - OQ_A = P_A Q_A = 4R^2r \left ( \frac{AI^2 - r^2}{AI^2 (4Rr + AI^2)} \right ) $.

Now we use the following relations  $AI^2 - r^2 = (s-a)^2$. $AI^2 = \frac{(s-a)bc}{s}$ and $4Rr = \frac{abc}{s}  $.

So $AI^2 + 4Rr = bc$. Plugging these relations in [color=#008040][b](3)[/b][/color] we get $P_A Q_A =  \frac{a\cdot R\cdot (s-a)}{bc}$

So  the desired inequality becomes $(b+c-a)(c+a-b)(a+b-c) \le abc $, which is true. Equality holds iff $a=b=c$, i.e when the triangle is equilateral.

Do you remember a famous problem which was also proposed by USA ?

Suppose we have a triangle $ ABC $ and a circle tangent to $ AB ~,~ AC $ respectively at $ E ~,~ F $ and the circumcircle internally ( externally ) at $ T $  . Then the incentre ( excentre opposite $A$ ) of $ \Delta ABC $ is the mid-pt of segment $ EF $ .

It can be proved by joining $ TE $ and $ TF $ and applying Pascal's theorem on $ A B' C T B C' $ , where $ B' ~,~ C' $ are the mid-pts of arc $ AC ~,~ AB $ , which are also the points lying on $ TF ~,~ TE $ respectively .

Let $ R_1 ~,~ R_2 $ be the radius of $ \omega_A ~,~ S_A $ respectively , $ I_A $ the excentre opposite $A$ , $M$ the mid-pt of the chord of $\omega$ which is the polar of $A$  . Now we know that $ \frac{P_A Q_A}{R } = \frac{ R_2 - R_1 }{R} = AM \left( \frac{1}{AI} - \frac{1}{AI_A} \right) $

$ P_A Q_A = 2R~\frac{ (a/2)(s-a) }{bc}  = \frac{aR(s-a)}{bc} $

Therefore , the inequality is equivalent to :

$ 8(s-a)(s-b)(s-c) = (a+b-c)(a-b+c)(-a+b+c) \leq abc $ which is known as Pados' inequality .

It turns out we can compute $P_AQ_A$ explicitly. Let us invert around $A$ with radius $s-a$ (hence fixing the incircle) and then compose this with a reflection around the angle bisector of $\angle BAC$. We denote the image of the composed map via \[ \bullet \mapsto \bullet^\ast \mapsto \bullet^+. \] We overlay this inversion with the original diagram.
Let $P_AQ_A$ meet $\omega_A$ again at $P$ and $S_A$ again at $Q$. Now observe that $\omega_A^\ast$ is a line parallel to $S^\ast$; that is, it is perpendicular to $\overline{PQ}$. Moreover, it is tangent to $\omega^\ast = \omega$.
Now upon the reflection, we find that $\omega^+ = \omega^\ast = \omega$, but line $\overline{PQ}$ gets mapped to the altitude from $A$ to $\overline{BC}$, since $\overline{PQ}$ originally contained the circumcenter $O$ (isogonal to the orthocenter). But this means that $\omega_A^\ast$ is none other than the $\overline{BC}$! Hence $P^+$ is actually the foot of the altitude from $A$ onto $\overline{BC}$.
By similar work, we find that $Q^+$ is the point on $\overline{AP^+}$ such that $P^+Q^+ = 2r$.
[asy]
size(8.5cm); pair A = dir(140); pair B = dir(210); pair C = dir (330); filldraw(A--B--C--cycle, invisible, grey); dot("$A$", A, A); dot("$B$", B, B); dot("$C$", C, C); draw(incircle(A,B,C), black+1); pair D = foot(A,B,C); dot("$P^+$", D, dir(-45), brown); draw(A--D, brown); pair I = incenter(A,B,C); pair O = circumcenter(A,B,C);
real a = abs(B-C); real b = abs(C-A); real c = abs(A-B); real s = 0.5*(a+b+c); real ha = abs(A-D); real k = ((s-a)**3) / (s*ha*abs(A-I)**2);
pair PA = k*b*c/2*dir(O-A)+A; dot("$P_A$", PA, dir(255), blue); draw(CP(PA,A), lightblue); pair P = Drawing("P", 2*PA-A, dir(180));
pair QA = PA + dir(PA-A)*( (s-a)*a**2 / (a*b*c) ); dot("$Q_A$", QA, dir(-80)*1.4, blue); draw(CP(QA,A), lightblue); pair Q = Drawing("Q", 2*QA-A, dir(-70));
draw(unitcircle, grey); draw(A--Q); real r = abs(I-foot(I,B,C)); Drawing("I", I, dir(90)); pair X = I + dir(O-A)*r; pair P1 = foot(X,P,Q); draw(Line(X,P1,2.5), red, Arrows(TeXHead)); dot("$P^\ast$", P1, dir(5), red); draw(CP(A, foot(I,A,B)), dashed+deepgreen);
pair Q2 = D+dir(A-D)*2*r; dot("$Q^+$", Q2, dir(145), brown); draw(Q2--(I+r*dir(90)), brown);
[/asy]
Now we can compute all the lengths directly.  We have that \[ AP_A = \frac{1}{2} AP = \frac{(s-a)^2}{2AP^+} = \frac{1}{2} (s-a)^2 \cdot \frac{1}{h_a} \] and \[ AQ_A = \frac{1}{2} AQ = \frac{(s-a)^2}{2AQ^+} = \frac{1}{2} (s-a)^2 \cdot \frac{1}{h_a-2r} \] where $h_a = \frac{2K}{a}$ is the length of the $A$-altitude, with $K$ the area of $ABC$ as usual. Now it follows that \[ P_AQ_A = \frac{1}{2} (s-a)^2 \left( \frac{2r}{h_a(h_a-2r)} \right). \] This can be simplified, as \[ h_a - 2r = \frac{2K}{a} - \frac{2K}{s} = 2K \cdot \frac{s-a}{as}. \] Hence \[ P_AQ_A = \frac{a^2rs(s-a)}{4K^2} = \frac{a^2(s-a)}{4K}. \] Hence, the problem is just asking us to show that \[ a^2b^2c^2(s-a)(s-b)(s-c) \le 8 (RK)^3. \] Using $abc=4RK$ and $(s-a)(s-b)(s-c) = \frac 1s K^2 = rK$, we find that this becomes \[ 2(s-a)(s-b)(s-c) \le RK \iff 2r \le R \] which follows immediately from $IO^2=R(R-2r)$. Alternatively, one may rewrite this as Schur's Inequality in the form \[ abc \ge (-a+b+c)(a-b+c)(a+b-c). \]

[hide]
Let $r, r_1, r_2$ denote the radii of the incircle of $\triangle ABC,$ $\omega_A, S_A$ respectively. Note that $IP_A= r+r_1, OP_A= R - r_1,$ $IQ_A= r_2-r,$ and $OQ_A= R-r_2.$ Also, let $s$ be the semiperimeter of $\triangle ABC.$ By cosine rule on $\triangle OP_AI,$
\begin{align*}
IP_A^2 & = OP_A^2 + OI^2 - 2 \cdot OP_A \cdot OI \cos \angle AOI \\
\implies (r+r_1)^2 &= (R-r_1)^2 + R^2-2Rr - 2(R-r_1)\sqrt{R^2-2Rr} \cos \angle AOI \\
\implies r^2 + r_1^2 + 2rr_1 &= R^2 + r_1^2 - 2Rr_1 + R^2 - 2Rr - 2(R-r_1)\sqrt{R^2-2Rr} \cos \angle AOI \\
\implies 2rr_1 & = 2R^2 +  2Rr_1 - 2Rr - 2(R-r_1) \sqrt{R^2-2Rr} \cos \angle AOI \\
\implies  2rr_1& = 2R^2 + 2Rr_1 - 2(R-r_1) \sqrt{R^2-2Rr} \dfrac{R^2 + R^2-2Rr + \dfrac{bc}{s(s-a)}}{2R\sqrt{R^2-2Rr}} \\
\implies r_1 & =  R - \dfrac{R(4R+r)}{4Rr + \dfrac{bc}{s(s-a)}}. \end{align*}

Similarly we can prove that $r_2= R - \dfrac{Rsr^2(s-a)}{bc}.$ After some simplifications, it follows that
\[P_AQ_A= r_2-r_1=  \dfrac{Ra(s-a)}{bc}.\]
We obtain analogous expressions for $P_BQ_B$ and $P_CQ_C.$ We need to show
\begin{align*}\dfrac{8R^3 abc(s-a)(s-b)(s-c)}{(ab)(bc)(ca)} & \leq R^3 \\
\iff 8 (s-a)(s-b)(s-c) & \leq abc \\
\iff (a+b-c)(b+c-a)(c+a-b) & \leq abc, \end{align*}
which is true. Equality holds iff $a=b=c.$ $\blacksquare$ [/hide]

Let $r,r_a,r_i,r_e$ be the radius of the incircle, A-excircle,  A-internal and A-external mixtilinear circles respectively.

Note that $\omega$ is the ex-mixtilinear wrt $S_A, AB, AC$ and in-mixtilinear wrt $\omega_A, AB, AC$.

Dilate $S_A$ to $S$ and  $\omega_A$ to $S$ to obtain:
$\frac{AP_A}{R}=\frac{r}{r_e}$ , $\frac{AQ_A}{R}=\frac{r}{r_i}$
Subtract the last 2 to get $\frac{P_AQ_A}{R}=\frac{r}{r_i}-\frac{r}{r_e}$

It can be easily shown (from well-known lemmas about mixtilinear circles) that $\frac{r}{r_i}=cos^2\frac{A}{2}$ ,
$\frac{r_a}{r_e}=cos^2\frac{A}{2}$ and $\frac{r}{r_a}=\frac{s-a}{s}$

Combining the last 4 we get $\frac{P_AQ_A}{R}=\frac{a}{s}cos^2\frac{A}{2}$

So the inequality becomes $\frac{8abc}{s^3}cos^2\frac{A}{2}cos^2\frac{B}{2}cos^2\frac{C}{2} \leq 1$

This is however obvious by multipliing
$s^3 \geq \frac{27}{8}abc$ (AM-GM) and
$cos^2\frac{A}{2}cos^2\frac{B}{2}cos^2\frac{C}{2} \leq (\frac{\sqrt3}{2})^6$ (Jensen)

[hide="Solution"]Consider the homotethy $\mathcal{H}$ centered at $A$ which sends the circle $S_A$ to $S$. This will send the incircle to a circle tangent internally to the circle $\mathcal H (S_A)=S$ and tangent to the lines$\mathcal H (AB)=AB$ and $\mathcal H (AC)=AC$, so the incircle is mapped to the interior $A$ mixtilinear circle. Therefore, $\dfrac{AQ_A}{R}=\dfrac{r}{r_1}$, where $r_1$ is the radius of the aforementioned circle.  Analogously, $\dfrac{AP_A}{R}=\dfrac{r}{r_2}$, where $r_2$ is the radius of the other $A$ mixtlinear circle.
Now, let $D,E,F$ be the tangency points of the interior mixtlinear circle with $(ABC),AC,AB$. As $AF$ is tangent to the circle, $\widehat{FDE}=\widehat{AFE}=90^\circ-\dfrac{\hat{A}}{2}$. Applying Casey's theorem for $(DEF),A,B,C$, we get that $AF=AE=\dfrac{2bc}{a+b+c}=\dfrac{bc}{p}$. Applying the sine theorem in $\triangle{AFE}$ and $\triangle{DFE}$, we get that
$$r_1=\dfrac{EF}{2\sin{\left ( 90^\circ- \dfrac{\hat{A}}{2} \right )}}=\dfrac{bc}{p}\tan{\dfrac{\hat{A}}{2}}$$
Abusing of notation, forget about the old $D,E,F$ and redefine them to be the tangency points of the exterior $A$ mixtlinear circle with $(ABC),AC,AB$. We have then $$\widehat{FDE}=180^\circ-\widehat{DFE}-\widehat{DEF}=180^\circ-\widehat{AFE}=90^\circ+\dfrac{\hat{A}}{2}$$
It is easy to see that the exterior $A$ mixtlinear circle is actually the image of the incircle when composing a $\sqrt{bc}$ inversion with a rotation wrt to the bisector of $A$, hence $(p-a)AF=bc$. Applying the sine theorem in $\triangle{AFE}$ and $\triangle{DFE}$, we get that
$$r_2=\dfrac{EF}{2\sin{\left ( 90^\circ+\dfrac{\hat{A}}{2} \right )}}=\dfrac{bc}{p-a}\tan{\dfrac{\hat{A}}{2}}$$

We infer that $P_AQ_A=Rr \left | \dfrac{1}{r_1}-\dfrac{1}{r_2} \right |=\dfrac{aRr}{bc\tan{\dfrac{\hat{A}}{2}}}$  . So we have now to prove that

$$\dfrac{8R^3r^3}{abc \tan{\dfrac{\hat{A}}{2}} \tan{\dfrac{\hat{B}}{2}} \tan{\dfrac{\hat{C}}{2}}} \le R^3\Leftrightarrow 8r^3\le abc \tan{\dfrac{\hat{A}}{2}} \tan{\dfrac{\hat{B}}{2}} \tan{\dfrac{\hat{C}}{2}}$$

But $\tan{\dfrac{\hat{A}}{2}}=\dfrac{r}{p-a}$, so we are left to prove that
$$8(p-a)(p-b)(p-c)\le abc\Leftrightarrow 8xyz\le (x+y)(y+z)(z+x)$$ which is true by AM-GM.[/hide]

[color=darkred][b][u]PP[/u].[/b] Let an acute $\triangle ABC$  with the incircle $ \omega = C(I,r)$ and the circumcircle $ \rho = C(O,R)$.
The circles $ c_{1}= C\left(P,r_{1}\right)$ and $ c_{2}= \mathbb C\left(Q,r_{2}\right)$ are tangent internally to $ \rho$ in the same point $ A$. The circle $ w$ is tangent externally to the circle $ c_{1}$ and is tangent internally to the circle $ c_{2}$. Prove that $PQ =\frac{a^{2}(p-a)}{4S}$, where $p$ is the semiperimeter  and $ S=[ABC]$ is the area for$ \triangle ABC$.[/color]

[color=darkblue][b][u]Remark[/u].[/b] Prove easily that $\left \{\begin{array}{ccccc}
IP = r+r_{1} & ; & IQ = r_{2}-r\\\\
PO = R-r_{1} & ; & PA = r_{1}\\\\
QO = R-r_{2} & ; & QA = r_{2}\\\\
OA = R & ; & PQ = r_{2}-r_{1}\end{array}\right\|$. The relations $\left \{\begin{array}{ccc}
IO^{2}= R(R-2r) & ; & IA^{2}=\frac{bc(p-a)}{p}=\frac{4Rr(p-a)}{a}\\\\
IA^{2}-r^{2}= (p-a)^{2} & ; & IA^{2}+4Rr = bc\\\\
p(p-a)+(p-b)(p-c) = bc.& ; & p(p-a)(p-b)(p-c) = S^2\end{array}\right\|$ are well-known.

[b][u]Proof I[/u].[/b] Denote $\left\{\begin{array}{ccc}
IO & = & m\\\
 IA & = & n\end{array}\right\|$. Apply the [u][b]Stewart[/b]'s theorem[/u] in $ \triangle OIA$ for the cevian-rays $ [IP$ and $ [IQ\ :\ \left\{\begin{array}{ccc}
m^{2}r_{1}^{2}+n^{2}(R-r_{1}) & = & R(r+r_{1})^{2}+Rr_{1}(R-r_{1})\\\\
m^{2}r_{2}+n^{2}(R-r_{2}) & = &  R(r_{2}-r)^{2}+Rr_{2}(R-r_{2})\end{array}\right\|$ $ \implies$ $ r_{1}(R^{2}+2Rr+n^{2}-m^{2}) = R(n^{2}-r^{2}) = r_{2}(R^{2}-2Rr+n^{2}-m^{2})$ $ \implies$

$ r_{1}(R^{2}+2Rr+bc-4Rr-R^{2}+2Rr) =$ $ R(p-a)^{2}=$ $ r_{2}(R^{2}-2Rr+bc-4Rr-R^{2}+2Rr)$ $ \implies$ $ r_{1}bc = R(p-a)^{2}= r_{2}(bc-4Rr)$ $ \implies$

$ PQ = r_{2}-r_{1}=$ $ \frac{4R^{2}r(p-a)^{2}}{bc(bc-4Rr)}=$ $ \frac{4R^{2}pr(p-a)^{2}}{b^{2}c^{2}(p-a)}=$ $ \frac{4R^{2}pra^{2}(p-a)^{2}}{16R^{2}p^{2}r^{2}(p-a)}$ $ \implies$ $ \boxed{\ PQ =\frac{a^{2}(p-a)}{4S}\ }$.

[b][u]Proof II[/u].[/b] Apply the [u][b]Pythagoras[/b]' theorem[/u] in the triangles :
$ \triangle\ IOP\blacktriangleright$ $ (r+r_{1})^{2}= (R-r_{1})^{2}+(R^{2}-2Rr)-2(R-r_{1})\cdot$ $ OI\cdot\frac{(R^{2}-2Rr)+R^{2}-\frac{4Rr(p-a)}{a}}{2R\cdot IO}\implies$ $ \boxed{r_{1}=\frac{R(p-a)^{2}}{bc}}\ .$

$ \triangle\ IOQ\blacktriangleright$ $ (r_{2}-r)^{2}= (R-r_{2})^{2}+(R^{2}-2Rr)-2(R-r_{2})\cdot$ $ IO\cdot\frac{(R^{2}-2Rr)+R^{2}-\frac{4Rr(p-a)}{a}}{2R\cdot IO}\implies$ $ \boxed{r_{2}=\frac{Rp(p-a)}{bc}}\ .$ Therefore,

$ \frac{r_{1}}{p-a}=\frac{r_{2}}{p}=\frac{R(p-a)}{bc}=\frac{PQ}{a}$, i.e. $ \boxed{PQ =\frac{aR(p-a)}{bc}}=\frac{a^{2}(p-a)}{4S}$. Observe that $ \frac{r_{1}}{r}=\frac{r_{2}}{r_{a}}$, where $ r_{a}$ is the $ A$- exinradius of $ \triangle ABC$.

[b]Remark.[/b] Prove easily that two more interesting relations : $ \boxed{\sum PQ = R+r\ }$ and $ r_{1}= R-\frac{a(4R+r)}{4p}$ , $ \boxed{r_{2}=\frac{ r_{b}+r_{c}}{4}}$.[/color]

Let $AP_A=x$ and $AQ_A=y$. Then by Stewart's on $AIO$ with $P_A$ on $O$ we get $x(R-x)\cdot R + (r+x)^2\cdot R = (R^2-2Rr)x+AI^2(R-x)$. Solving, $x=\frac{R(AI^2-r^2)}{4rR+AI^2}$. Note that $AI^2-r^2$ is the power of $A$ wrt the incircle, so $AI^2-r^2=(s-a)^2$. Similarly, note $AI^2=\frac{bc}{s}\cdot (s-a)$. Thus $x=\frac{R(s-a)^2}{4bc(s-a)/s+4rR}$. Similarly, by Stewarts on $AIO$ but with $Q_A$ instead, we get $y(R-y)\cdot R + (r-y)^2\cdot R = (R^2-2Rr)y+AI^2(R-y)$, or $y=\frac{R(s-a)^2}{4bc(s-a)/s}$. Now $y-x= R(s-a)^2 \frac{4rR}{\frac{bc}{s}(s-a)\left[\frac{bc}{s}(s-a)+4rR\right]}=\frac{4rR^2s^2(s-a)}{bc(bc(s-a)+4rRs)}$. Note $4rRs=4R[ABC]=abc$ so $y-x=\frac{4rR^2s(s-a)}{b^2c^2}$.

The inequality we want to prove becomes $512r^3R^6s^3(s-a)(s-b)(s-c)\le R^3a^4b^4c^4$. Recall that Schur's gives $8(s-a)(s-b)(s-c)\le abc$, with equality for $a=b=c$. On the other hand, $(4rRs)^3=(abc)^3$. Multiplying, we get $512r^3R^3s^3(s-a)(s-b)(s-c)\le a^4b^4c^4$ as desired. $\square$

Consider an inversion about $A$ with radius $\sqrt{AB\cdot AC}$ followed by a reflection across the angle bisector of $\angle BAC$. This takes the incircle to the $A$-mixtilinear-excircle and the circumcircle of $ABC$ to $BC$. Since $\omega_A$ and $S_A$ are circles that are tangent to the circumcircle of $ABC$, their image is parallel to $BC$. Furthermore, they are tangent to the $A$-mixtilinear-excircle. Finally, $AP_A$ and $AQ_A$ map to the altitudes from $A$ to the tangents to the mixtilinear excircle that are parallel to $BC$. Denote the distance from $A$ to these tangents by $d_1$ and $d_2$. We have that $2P_AQ_A=\frac{bc}{d_1}-\frac{bc}{d_2}=\frac{bc(d_2-d_1)}{d_1d_2}=\frac{2bcR_A}{d_1d_2}$.

Let $h_a$ be the length of the $A$-altitude, $r$ be the inradius of $ABC$, and $r_a$ be the $A$-exradius. By similar triangles, $d_1=\frac{h_aR_A}{r_a}$ and $d_2=\frac{h_aR_A}{r}$. Hence, $2P_AQ_A=\frac{2bcrr_a}{h_a^2R_A}$. Multiplying similar expressions for $2P_BQ_B$ and $2P_CQ_C$, it suffices to show that $\frac{8a^2b^2c^2r^3r_ar_br_c}{h_a^2h_b^2h_c^2R_AR_BR_C}\leq R^3$ or $8a^2b^2c^2r^3r_ar_br_c\leq R^3h_a^2h_b^2h_c^2R_AR_BR_C$.

Because the $A$-excenter lies on the polar of $A$ with respect to $\Omega_A$, we can get $R_A=\frac{r_a}{\cos^2\frac{A}{2}}$ with similar triangles. Hence, it remains to show that $R^3h_a^2h_b^2h_c^2\geq8r^3a^2b^2c^2\cos^2\frac{A}{2}\cos^2\frac{B}{2}\cos^2\frac{C}{2}$. Multiplying both sides by $a^2b^2c^2$, the inequality becomes $8R^3[ABC]^6\geq r^3a^4b^4c^4\cos^2\frac{A}{2}\cos^2\frac{B}{2}\cos^2\frac{C}{2}$. Since $[ABC]=\frac{abc}{4R}$, we can further reduce this to $[ABC]^3\geq8r^3abc\cos^2\frac{A}{2}\cos^2\frac{B}{2}\cos^2\frac{C}{2}$. Since $[ABC]=rs$, this is equivalent to $(a+b+c)^3\geq 64abc\cos^2\frac{A}{2}\cos^2\frac{B}{2}\cos^2\frac{C}{2}$. By AM-GM, $(a+b+c)^3\geq27abc$ so it suffices to show that $\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\leq\frac{3\sqrt{3}}{8}$. Note that $f(x)=\cos x$ is concave on $[0,\frac{\pi}{2}]$. By Jensen's inequality, $\cos\frac{A}{2}+\cos\frac{B}{2}+\cos\frac{C}{2}\leq\frac{3\sqrt{3}}{2}$. Hence, we are done by AM-GM. Equality holds iff $a=b=c$ on the first AM-GM, so equality holds iff $ABC$ is equilateral.

Here's a much cleaner solution:

The central claim is that $\frac{2P_AQ_A}R=\frac{a(b+c-a)}{bc}$.

Let $M$ be the midpoint of arc $BC$ not containing $A$, and let $AD$ meet $\omega_A$ and $S_A$ at $X_A$ and $Y_A$. Let $I$ and $DEF$ be the incenter and intouch triangle of $ABC$, and let $D'$ be the reflection of $D$ across $I$. Let $AI$ meet $D'E,D'F,DE,DF,EF$ at $S,T,U,V,W$. By similar circles, we have that $\frac{2P_AQ_A}R=\frac{2X_AY_A}{AM}$. We claim that $X_A$ and $Y_A$ are the midpoints of $ST$ and $UV$. This follows from the configuration of [url=https://artofproblemsolving.com/community/c6h1262194p6556907]2016 ELMO P6[/url]. Hence, we have $2X_AY_A=2AY_A-2AX_A=AU+AV-AS-AT=US+VT$. We also have that $US=EW\left(\tan\frac B2+\cot\frac B2\right)=\frac{2EW}{\sin B}=\frac{(b+c-a)\sin\frac A2}{\sin B}$ and similarly $VT=\frac{(b+c-a)\sin\frac A2}{\sin C}$. Thus, \begin{align*}\frac{2X_AY_A}{AM}&=\frac{(b+c-a)\sin\frac A2(\sin B+\sin C)}{2R\cos\left(\frac{B-C}2\right)\sin B\sin C}\\&=\frac{2(b+c-a)\sin\frac A2\sin\left(\frac{B+C}2\right)\cos\left(\frac{B-C}2\right)}{2R\cos\left(\frac{B-C}2\right)\sin B\sin C}\\&=\frac{(b+c-a)\sin A}{2R\sin B\sin C}\\&=\frac{a(b+c-a)}{bc}\end{align*} as desired.

To finish, note that the inequality is now equivalent to $(b+c-a)(c+a-b)(a+b-c)\le abc$, which is just Schur's inequality. Equality holds for positive $a,b,c$ iff $a=b=c$, as desired.

In fact, the stronger inequality $P_AQ_A+P_BQ_B+P_CQ_C=R+r\le\frac{3R}2$ holds and can be proven similarly.

Please tell me there's a better way to construct $\omega_A$ other than
[quote=wu2481632 is way too extra]
Let M_A be the midpoint of minor arc BC
Let D be the A-intouch point
similarly define M_B, E
By Canada 2007/5 we know that M_AD and M_BE concur at the insimilicenter of the incircle and circumcircle
Now take O as the circumcenter, I as the incenter
it is well-known that (O, similitude center; I, similitude center) forms a harmonic bundle)
so we can find the external similitude center by constructing the circle with diameter OI and constructing the polar of the insimilicenter
next, we know that A is an external similitude center of omega_A and the circumcircle
by Monge's theorem the three exsimilicenters are collinear
and one lies on the incircle
so we intersect the line joining A and the exsimilicenter of incircle and circumcircle with the incircle
which determines our third similitude center
we then draw the tangent from this point to the incirlce
and the tangent from A to the circumcircle
and draw perpendiculars to both tangents at the points of tangency
which helps us find the center
of omega_A
at which point the rest is trivial[/quote]

In my defense, this process takes about 2 minutes on geogebra and you explicitly said this was allowed.

[asy]

import cse5; import olympiad;
pathpen = black; pointpen = black;
size(13cm);

pen n_purple = rgb(0.7,0.4,1),
n_blue = rgb(0,0.6,1),
n_green = rgb(0,0.4,0),
n_orange = rgb(1,0.4,0.1),
n_red = rgb(1,0.2,0.4);

pair O = origin, A= dir(115), B=dir(210), C=dir(330), I=incenter(A,B,C), D=foot(I,B,C), E=foot(I,C,A), F=foot(I,A,B), XA=I+abs(I-D)*(A-O), YA=2*I-XA, A1=-A, PA1=foot(XA,A,O), QA1=foot(YA,A,O), K=foot(I,A,O), M=dir(-90);

DPA(A--B--C--cycle);
D(circumcircle(A,B,C), n_blue);
D(incircle(A,B,C), n_red);
D(A--A1, n_orange);
D(CP(A,E), blue);
D((XA+(PA1-XA)*6)--(XA-(PA1-XA)*4));
D((YA+(QA1-YA)*6)--(YA-(QA1-YA)*4));
D(I--K, n_purple);
D(A--M, n_orange);
D(circumcircle(I,M,A1), n_green);

D("A",A,A);
D("B",B,B);
D("C",C,C);
D("O",O,dir(0));
D("I",I,dir(160));
D("D",D,dir(90));
D("E",E,2*dir(90));
D("F",F,2*dir(210));
D("Y_A",XA,XA);
D("X_A",YA,dir(I-YA));
D("A'",A1,A1);
D("Q_A''",PA1,2*PA1);
D("P_A''",QA1,2*QA1);
D("K",K,dir(60));
D("M",M,M);
[/asy]

Let $P_A'$ be the image of $P_A$ under a homothety of ratio $2$ at $A$ (which lies on $\omega_A$), and let $P_A''$ be the image of $P_A'$ under an inversion at $A$ with radius $s-a$ (which fixes the incircle). Let $Q_A''$ be defined similarly.

The image of $\omega_A$ under this inversion is the further line tangent to the incircle perpendicular to $AO$, the image of $S_A$ is the closer line (to $A$). Then $P_A'',Q_A''$ are the intersections of these lines with $AO$. Let $K$ be the foot of $I$ onto $AO$. Then $AQ_A''=AK-r,AP_A''=AK+r$. Then $2P_AQ_A=P_A'Q_A'=AQ_A'-AP_A'=(s-a)^2(\frac{1}{AQ_A''}-\frac{1}{AP_A''})=\frac{2r(s-a)^2}{AK^2-r^2}$.

Extend $AI$ to meet $S$ again at $M$. Let $A'$ be the antipode of $A$ on $S$. We have $IKA'M$ is cyclic, so $AK = \frac{AI\cdot AM}{AA'}$. By Ptolemy's theorem we have that $AM\cdot a = BM\cdot b+CM\cdot c = (b+c)\cdot MI$, so $AK=\frac{AI\cdot MI}{2R}\cdot\frac{b+c}{a}=\frac{b+c}{a}\cdot\frac{R^2-OI^2}{2R}=\frac{b+c}{a}\cdot r$. So $AK^2-r^2=r^2\Bigg(\bigg(\frac{b+c}{a}\bigg)^2-1\Bigg)=r^2\Bigg(\frac{(b+c-a)(a+b+c)}{a^2}\Bigg)=r^2\frac{2(s-a)\cdot 2s}{a^2}$.

Thus $2P_AQ_A=\frac{2r(s-a)^2}{r^2\frac{2(s-a)\cdot 2s}{a^2}}=\frac{a^2(s-a)}{2rs}$. So
\[8P_AQ_A\cdot P_BQ_B\cdot P_CQ_C=\frac{(abc)^2(s-a)(s-b)(s-c)}{8K^3}=\frac{(abc)^2s(s-a)(s-b)(s-c)}{8K^3s}=\frac{(abc)^2}{8Ks}=\frac{abcR}{2s}=\frac{2abcrR}{4K}\]
\[=2rR^2=(R^2-OI^2)R=R^3-R\cdot OI^2\le R^3.\]$\square$

A dif. Sol (hopefully). I avoid details of the computations to visualise the beauty of the prob. with geo tastes  !!!

Take an inversion with radius $(s-a)$ and reflection W. r. t. angle bisector of $BAC$
Say $AP_AQ_A$ intersects with $\omega_{A}$
and $S_A$ at $P$ and $Q$ respectively. Under the map $P$ goes to the feet of $A$ altitude and $Q$ goes to a point on $A$ altitude s. t. the distance between the  2 new points is $2r$ where $r$ is inradius. Now all the distance are nicely comparable since we can find the distances in the inverted diagram.
The inequality is transformed into $R-2r$ is nonzero. So equally for an equilateral triangle  !

Solved with [url=https://artofproblemsolving.com/community/user/435766]Aryan-23[/url].

[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]

Consider a $\sqrt{bc}$ inversion. It is well-known that the incircle goes to the $A$-mixtillinear excircle. Now, as $\omega_A$ and $S_A$ pass through $A$, they get inverted to parallel lines tangent to the said excircle. Now, we have that the centers $P_A'$ and $Q_A'$ are the reflections of $A$ in $S_A'$ and $\omega_A'$. Thus, we get
\[P_AQ_A=AQ_A-AP_A=\dfrac{bc}{AQ_A'}-\dfrac{bc}{AP_A'}\]
However, if $d_1$ and $d_2$ are the distances from $A$ to $S_A'$ and $\omega_A'$, respectively, then we get that
\[P_AQ_A=\dfrac 12\left(\dfrac{bc}{d_1}-\dfrac{bc}{d_2}\right)=\dfrac 12\left(\dfrac{bc(d_1-d_2)}{d_1d_2}\right)\]
Now, we get that as $\omega_A'\parallel S_A$, we must have that $d_1-d_2=2R_a$, where $R_a$ is the radius of the said $A$-mixtillinear excircle. We note the well-known property that the tangents from $A$ to the mixtillinear excircle are on the perpendicular line from $I_A$ (the $A$-excircle) to $AI$, so thus $I_A$ is the inverse of $A$ with respect to the $A$-excircle. In particular, we have that
\[R_a=\dfrac{r_a}{\cos^2\frac 12 A}\]
By similar triangles, we must have that as $BC$ and $B'C'$ (where $B'=\omega_A'\cap AB$ and similarly with $C$) are antiparallel, so we must have that
\[d_1=\dfrac{R_a\cdot h_a}{r_a}\]
by comparing the excircles. Similarly, we get
\[d_2=\dfrac{R_a\cdot h_a}{r}\]
by comparing incircles. Finally, we also note
\[h_a=\dfrac{bc}{2R}\]
so thus we get
\[P_AQ_A=\dfrac{bcR_arr_a}{R_a^2h_a^2}=\dfrac{4Rr\cos^2\frac 12 A}{bc}\]
However, the well-known formula $\cos^2\frac 12A=\frac{s(s-a)}{bc}$, we get
\[P_AQ_A=\dfrac{4Rrs(s-a)}{b^2c^2}=\dfrac{4R\Delta(s-a)}{b^2c^2}\]
Multiplying the symmetric versions, we get
\[8P_AQ_A\cdot P_BQ_B\cdot P_CQ_C=\dfrac{512R^3\Delta^3(s-a)(s-b)(s-c)}{a^4b^4c^4}\]
Using the well-known formula
\[\Delta=\dfrac{abc}{4R}\]
we have
\[8P_AQ_A\cdot P_BQ_B\cdot P_CQ_C=\dfrac{8R^3(s-a)(s-b)(s-c)}{abc}\]
However, Schur's Inequality guarantees that
\[8(s-a)(s-b)(s-c)\leq abc\]
so thus we have
\[8P_AQ_A\cdot P_BQ_B\cdot P_CQ_C\leq R^3\]
completing the proof.

We use the machinery of inversion to determine $P_AQ_A$ etc. Let the intouch triangle be $DEF$, define $a=BC,b=CA,c=AB,s=(a+b+c)/2$. Let the image of $P_A,Q_A$ under transformation of inversion about the circle with radius $AE$ and center $A$ and then reflection over the angle bisector of $\angle BAC$, which we call $\tau$, be $P_A',Q_A'$. Clearly $\tau$ fixes $\omega$. Let the intersection of $AI$ and $\omega$ closer to $A$ be $Q$ and the other be $P$. Then $P_A'$ is the reflection of $A$ over the line $BC$ and $Q_A'$ is the reflection of $A$ over the line parallel to $BC$ through the antipode of $D$ wrt $\omega$, which we subsequently call $\ell$. Let $r$ be the inradius. Then the distance from $A$ to $BC$ is $\frac{2sr}{a}$ so
\[2P_AQ_A=\frac{(s-a)^2}{2(s-a)r/a}-\frac{(s-a)^2}{2sr/a} = \frac{a(s-a)}{2r}\cdot (1-(s-a)/s)=\frac{a^2(s-a)}{2rs}.\]
Similar results hold for $2P_BQ_B,2P_CQ_C$. It is well-known that $R=\frac{abc}{4[ABC]}$ where $[ABC]$ is the area of $\triangle ABC$: for one proof, use the Law of Sines and the identity $[ABC]=\frac{1}{2}ab\sin C$. Thus it suffices to establish
\[\frac{a^2b^2c^2(s-a)(s-b)(s-c)}{(2[ABC])^3}\le \frac{a^3b^3c^3}{(4[ABC])^3}.\]
Let $s-a=x,s-b=y,s-c=z$ for positive $x,y,z$, then the result amounts to showing $8xyz\le (x+y)(y+z)(z+x)$ which is immediate by AM-GM on $x+y,y+z,z+x$. Moreover, equality holds exactly when $x=y=z\implies a=b=c$, or when $\triangle ABC$ is equilateral, as desired.

Denote $I,O,x,y,r,s$ as the incenter and circumcenter of $ABC$, radiuses of $\omega_A, S_A,\omega,$ and the semiperimeter of $ABC$, respectively. We first find $P_AQ_A$, from which $P_BQ_B$ and $P_CQ_C$ follow symmetrically. In terms of $x,y,$ $P_AQ_A$ is equivalent to $x-y$. Now by Stewarts Theorem on $AIO$ and cevian $IP_A,$ $$AI^2 (R-x) + OI^2  x = x(R-x)R+(x+r)^2 R$$ $$\Rightarrow AI^2(R-x)+R(R-2r)x = x(R-x)R+(x+r)^2R$$ $$\Rightarrow AI^2R-x(AI^2-R^2+2Rr) = Rx^2+2Rrx+Rr^2+R^2x-Rx^2$$ $$\Rightarrow x(4Rr+AI^2)=R(AI^2-r^2)=R(s-a)^2$$ $$\Rightarrow x = \frac{R(s-a)^2}{4Rr+AI^2}$$ To solve for $y$ now, we again use Stewarts, this time on $AIQ_A$ and cevian $IP_A:$ $$AI^2(y-x)+(y-r)^2x=(x+r)^2y+x(y-x)y$$ $$\Rightarrow AI^2(y-x)-2rxy+r^2x=2rxy+r^2y$$ $$\Rightarrow (s-a)^2(y-x)+r^2(y-x)-2rxy+r^2x=2rxy+r^2y$$ $$\Rightarrow (s-a)^2(y-x)=4rxy$$ $$\Rightarrow y = \frac{x(s-a)^2}{(s-a)^2-4rx}$$ So, $$P_AQ_A = y-x = \frac{x(s-a)^2}{(s-a)^2-4rx} - x = \frac{4rx^2}{(s-a)^2-4rx} = \frac{\frac{4r(R^2(s-a)^4)}{4Rr+AI^2}}{(s-a)^2 - \frac{4Rr(s-a)^2}{4Rr+AI^2}}$$ $$= \frac{4R^2r(s-a)^2}{(1-\frac{4Rr}{4Rr+AI^2})(4Rr+AI^2)^2} = \frac{4R^2r(s-a)^2}{AI^2(4Rr+AI^2)}$$ Hence, it suffices to show $$8\prod_{\text{cyc}} \frac{4R^2r(s-a)^2}{AI^2(4Rr+AI^2)} \leq R^3$$ $$\Rightarrow 512R^3r^3(s-a)^2(s-b)^2(s-c)^2 \leq \prod_{\text{cyc}}(AI^2(4Rr+AI^2))$$ Using $K=rs$, the $LHS$ simplifies to $512R^3r^7s^2$. Also, $$ \sqrt{\frac{(s-b)(s-c}{bc}} = \sin{(\frac{a}{2})} = \frac{r}{AI} \Rightarrow AI^2 = \frac{r^2bc}{(s-b)(s-c)}$$ $$\Rightarrow AI^2+4Rr = \frac{r^2bc}{(s-b)(s-c)} + 4Rr = r(\frac{rbc(s-a)}{(s-a)(s-b)(s-c)} + \frac{abc}{rs})$$ $$ = r(\frac{rbc(s-a)+rabc}{r^2s}) = (\frac{bcs}{s}) = bc$$ So, the inequality reduces to $$512R^3r^7s^2 \leq \prod_{\text{cyc}} (\frac{r^2bc}{(s-b)(s-c)} bc) =  \frac{r^6 a^4b^4c^4}{(s-a)^2(s-b)^2(s-c)^2}$$ $$\Rightarrow 512R^3r^7s^2 \leq \frac{r^6 (4Rrs)^4}{r^4s^2}$$ $$\Rightarrow 512R^3r^{11}s^4 \leq r^6 (256R^4r^4s^4)$$ $$\Rightarrow 2r\leq R$$ which is just Euler's Inequality, so we are done $\blacksquare$

[i]Here's a very computational solution with $\sqrt{bc}$ inversion and inversion distance formula [/i] (different from post #37).
The solution may appear long but the advantage is that there is no tricky step and if we do no silly mistake, then we are just done.
[rule]
Let $r$ be the inradius of $\triangle ABC$ and $2 \alpha,2 \beta, 2 \gamma$ denote the measure of angles $\angle CAB,\angle ABC,BCA$, respectively.

[color=#f00][b]Claim[/b] (Key Claim)[b]:[/b][/color] $\displaystyle P_AQ_A = \frac{R \cdot \sin \alpha \cos^2 \alpha}{\cos \beta \cos \gamma}$

[b][color=#00f]Proof:[/color][/b] Let $\Gamma$ be the $A$-mixtilinear [b]excircle[/b] with center $T$ and radius $R_A$. Let $X,Y$ be the projection of $T$ onto lines $AB,AC$, respectively. It is well known that the midpoint of segment $XY$ is $I_A$, the $A$-excenter of $\triangle ABC$. Thus,
$$\sin 2 \alpha \cdot AT_A = XY = 2AI_A \cdot \tan \alpha ~ \implies ~ \boxed{AT_A = \frac{AI_A}{\cos^2 \alpha}}$$
Let $\Phi$ be the $\sqrt{bc}$ inversion (followed by reflection in angle bisector, of course). Denote by $Z^*$ the image of a point $Z$ under $\Phi$. Then,
[list]
[*] $\Phi$ swaps $\omega \longleftrightarrow \Gamma$ and $S \longleftrightarrow \overline{BC}$.
[*] $\Phi$ maps $\omega_A,S_A$ to some lines $\ell_1,\ell_2$ (respectively) both parallel to $\overline{BC}$ and tangent to $\Gamma$ such that $\ell_2$ is closer to $A$ than $\ell_1$.
[*] $P_A^*,Q_A*$ are reflection of $A$ in lines $\ell_1,\ell_2$, respectively.
[*] $\displaystyle P_AQ_A = P_A^* Q_A^* \cdot \frac{bc}{AP_A^* \cdot A Q_A^*}$.
[/list]
Let $\ell_3$ be the line passing through $T$ parallel to $\overline{BC}$ and $\theta = \beta - \gamma$. Now $d(\ell_2,\ell_3) = d(\ell_3,\ell_1) = R_A$ and $d(A,\ell_3) = AT_A \cdot \cos \theta$. Also, $R_A = AT_A \cdot \sin \alpha$. Thus,
\begin{align*}
P_A^* Q_A^* = 4R_A ~,~ AP_A^* = 2(AT_A \cdot \cos \theta + R_A) ~,~ AQ_A^* = 2(AT_A \cdot \cos \theta - R_A) \\
\implies P_AQ_A = P_A^* Q_A^* \cdot \frac{bc}{AP_A^* \cdot A Q_A^*} = \frac{\sin \alpha}{\cos^2 \theta - \sin^2 \alpha} \cdot \frac{bc}{AT_A} = \frac{r \cdot \cos^2 \alpha}{\cos^2 \theta - \sin^2 \alpha}
\end{align*}
But we also have
$$\cos^2 \theta - \sin^2 \alpha = \Big(\sin(90^\circ + \beta - \gamma) - \sin \alpha \Big)\Big( \sin (90^\circ + \beta - \gamma) + \alpha \Big)$$
Now using Product-Sum Identities (in ch-5 of EGMO) we obtain
\begin{align*}
\sin(90^\circ + \beta - \gamma) - \sin \alpha = 2 \cdot \sin \left( \frac{(90^\circ + \beta - \gamma) - \alpha}{2} \right) \cdot \cos \left(  \frac{(90^\circ + \beta - \gamma) + \alpha}{2} \right) = 2 \sin \beta \sin \gamma \\
 \sin (90^\circ + \beta - \gamma) + \sin \alpha  =  2 \cdot \sin \left( \frac{(90^\circ + \beta - \gamma) + \alpha}{2} \right) \cdot \cos \left( \frac{(90^\circ + \beta - \gamma) - \alpha}{2} \right) = 2 \cos \gamma \cos \beta
\end{align*}
Hence
$$\cos^2 \theta - \sin^2 \alpha = 4 \cos \beta \cos \gamma \sin \beta \sin \gamma  $$
So we finally have that
$$P_AQ_A = \frac{r \cdot \cos^2 \alpha}{\cos^2 \theta - \sin^2 \alpha} = \frac{r \cdot \cos^2 \alpha}{4  \cos \beta \cos \gamma \sin \beta \sin \gamma }$$
Lastly, it is well know that
$$r = 4R \cdot \sin \alpha \sin \beta \sin \gamma$$
Hence,
$$P_AQ_A  = \frac{r \cdot \cos^2 \alpha}{4  \cos \beta \cos \gamma \sin \beta \sin \gamma } = \frac{R \cdot \sin \alpha \cos^2 \alpha}{\cos \beta \cos \gamma}$$
This proves our claim. $\square$

So using our claim, we obtain that the given inequality is equivalent to the following
$$\sin \alpha \sin \beta \sin \gamma \le \frac{1}{8} \qquad{(1)}$$
The most simple way to finish from here is take $\log$ on both sides and using Jensen's inequality. But we also present a nice synthetic type proof: \\
We know that
$$\sin \alpha \sin \beta \sin \gamma = \frac{r}{4R}$$
So $(1)$ is equivalent to
$$2r \le R$$
Which is well known to be true (and we also know that equality holds if and only if circumcenter and incenter of $\triangle ABC$ coincide, which can happen iff and only if $\triangle ABC$ is equilateral).

This completes the proof of the problem. $\blacksquare$

Let $H$ be foot from $A$ to $\overline{BC}$. Let $K,L$ be feet from $I$ to $\overline{AO},\overline{AH}$, respectively. Let $R,r,r_1,r_2$ be the circumradius, inradius, radius of $\omega_A$ and radius of $S_A$, respectively.
[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]By Law of Cosines on $\triangle OIP_A$, \begin{align*}
IP_A^2&=OI^2+OP_A^2-2OP_A\cdot OI\cdot \cos{\angle AOI}\implies \\
(r_1+r)^2&=R(R-2r)+(R-r_1)^2-2(R-r_1)(R-AK)\implies \\
r_1&=\frac{2R\cdot AK-r^2-2Rr}{2(AK+r)}.
\end{align*}
Similarly, by Law of Cosines on $\triangle OIQ_A$, \begin{align*}
r_2=\frac{2R\cdot AK-r^2-2Rr}{2(AK-r)}.
\end{align*}
Hence, \begin{align*}
r_2-r_1=\frac{2R\cdot AK-r^2-2Rr}{2(AK-r)}-\frac{2R\cdot AK-r^2-2Rr}{2(AK+r)}=\frac{(2R\cdot AK-r^2-2Rr)r}{AK^2-r^2}.
\end{align*}
Moreover, since $AK=AL=AH-r=\frac{bc}{2R}-r$, we have \begin{align*}
P_AQ_A=r_2-r_1=\frac{(bc-r^2-4Rr)\cdot 4R^2r}{b^2c^2-4bcRr}.
\end{align*}
Similarly, we find $P_BQ_B$ and $P_CQ_C$, hence we would like to show that \begin{align*}
a^2b^2c^2(bc-4Rr)(ab-4Rr)(ac-4Rr)\geq \\
R^3r^3\cdot 2^9\cdot (bc-r^2-4Rr)(ab-r^2-4Rr)(ac-r^2-4Rr).
\end{align*}
Now, using $Rr=\frac{abc}{2(a+b+c)}$ and $r^2=\frac{(s-a)(s-b)(s-c)}{s}$, we would like to show the following,
\begin{align*}
abc(s-a)(s-b)(s-c)\geq\\ 2^3 \prod_{cyc} \left(bc-\frac{(s-a)(s-b)(s-c)-abc}{s}\right).
\end{align*}
Now, for simplicity, we do a sub $x=s-a$, $y=s-b$ and $z=s-c$ and we would like to show that
\begin{align*}
(x+y+z)^3(x+y)(y+z)(z+x)xyz\geq\\ 2^3\prod_{cyc}\left((x+y+z)(x+y)(y+z)-xyz-(x+y)(y+z)(z+x)\right),
\end{align*}which is actually equivalent to \begin{align*}
(x+y)(y+z)(z+x)\geq 2^3 xyz,
\end{align*}which is true by AM-GM. In fact, the equality holds iff $x=y=z\Longleftrightarrow a=b=c$.

Thanks to [b]Tafi_ak[/b]  :love:  for sharing with me this identity which essentially solves the inequality part which I wasn't being able to do,\[\sin\left(\dfrac{A}{2}\right)\sin\left(\dfrac{B}{2}\right)\sin\left(\dfrac{C}{2}\right)=\dfrac{r}{4R}.\]

[img]https://i.imgur.com/Qj8ZNzz.png[/img]

We already have some exposition on this configuration from [url=https://artofproblemsolving.com/community/c6h1262194p27770686]ELMO 2016 P6[/url] part (b). Now we perform a $\sqrt{bc}$ Inversion to get that the image of $\odot(S_A)$ which is a line is tangent to the mixtilinear incircle of $\triangle ABC$ apart from being tangent to the image of $\odot(I)$.

This helps us to characterize the image of $\odot(I)$ properly, that is, we can now say that the image of $\odot(I)$ is the excircle of the triangle $\triangle AB'C'$ that has it's incircle as the $A-$mixtilinear circle of $\triangle ABC$. Also note that as all the circles are tangent to $\odot(ABC)$ at $A$, we can say that $\overline{A-P_A-Q_A-O}$ are collinear. Now note that the centers of the circles $\odot(\omega_A)$ and $\odot(S_A)$ are just the midpoints of the intersection of $AO$ with $\odot(\omega_A)$ and $\odot(S_A)$ respectively. So their images under this $\sqrt{bc}$ Inversion are going to be the reflections of $A$ over the intersections of the $A-$altitude with the images of $\odot(\omega_A)$ and $\odot(S_A)$ respectively.

Let $\ell_1$ and $\ell_2$ be the images of $\odot(S_A)$ and $\odot(\omega_A)$ respectively. Now take the foot from the incenter and do a bunch of sine law span in those triangles to get the radius of the mixtilinear incircle as $=\dfrac{r}{\cos^2\left(\dfrac{A}{2}\right)}$.

Now let the intersection of $\ell_1$ and $\ell_2$ with the $A-$altitude be $X$ and $Y$ respectively. Also, let $H_A$ denote the foot of the $A-$altitude. Now taking a homothety at $A$ that sends $\triangle ABC\mapsto\triangle AB'C'$, we get that $\dfrac{AH_A}{AX}=\dfrac{r}{\dfrac{r}{\cos^2\left(\dfrac{A}{2}\right)}}=\cos^2\left(\dfrac{A}{2}\right)$ as their incircles also get mapped. This gives that $AX=\dfrac{AH_A}{\cos^2\left(\dfrac{A}{2}\right)}$.

Now we do some more trig spam to get that the radius of the excircle of $\triangle ABC$ is $=s\tan\left(\dfrac{A}{2}\right)$. Now again considering our previous homothety, we get that $\dfrac{s\tan\left(\dfrac{A}{2}\right)}{KL}=\cos^2\left(\dfrac{A}{2}\right)\implies KL=\dfrac{s\sin\left(\dfrac{A}{2}\right)}{\cos^3\left(\dfrac{A}{2}\right)}$.

This homothety again finally gives us that $\dfrac{AX}{AY}=\dfrac{\dfrac{r}{\cos^2\left(\dfrac{A}{2}\right)}}{KL}=\dfrac{\dfrac{r}{\cos^2\left(\dfrac{A}{2}\right)}}{\dfrac{s\sin\left(\dfrac{A}{2}\right)}{\cos^3\left(\dfrac{A}{2}\right)}}=\dfrac{r}{s\tan\left(\dfrac{A}{2}\right)}\implies AY=\dfrac{AX\cdot s\tan\left(\dfrac{A}{2}\right)}{r}=\dfrac{AH_A\cdot s\sin\left(\dfrac{A}{2}\right)}{\cos^3\left(\dfrac{A}{2}\right)\cdot r}$.

Now we finally use our Inversion-Distance Formula to get $P_AQ_A=\dfrac{1}{2}\cdot\dfrac{bc}{AX\cdot AY}\cdot XY=\dfrac{AB\cdot AC}{AX\cdot AY}\cdot KL=\dfrac{AB\cdot AC\cdot \cos^2\left(\dfrac{A}{2}\right)\cdot r}{AH_A^2}=\dfrac{r\cdot BC^2}{AB\cdot AC\cdot 4\sin^2\left(\dfrac{A}{2}\right)}$. Now literally multiply all the cycles of our current equality to get that what we want to prove is equivalent to proving $R^3\cdot 8\sin^2\left(\dfrac{A}{2}\right)\sin^2\left(\dfrac{B}{2}\right)\sin^2\left(\dfrac{C}{2}\right)\ge r^3$.

Now thanks to [b]Tafi_ak[/b] :D , I got this identity from him :love: ,\[\sin\left(\dfrac{A}{2}\right)\sin\left(\dfrac{B}{2}\right)\sin\left(\dfrac{C}{2}\right)=\dfrac{r}{4R},\]and also we have our Euler's Inequality that says $R\ge 2r$ where the inequality holds iff $O\equiv I$ which is when we have an equilateral triangle. Now we just follow through to finish off the rest of the problem.

\begin{align*}
    R^3\cdot 8\sin^2\left(\dfrac{A}{2}\right)\sin^2\left(\dfrac{B}{2}\right)\sin^2\left(\dfrac{C}{2}\right)&=R^3\cdot 8\left(\dfrac{r}{4R}\right)^2\\
    &=\dfrac{R\cdot r^2}{2}\\
    &\ge r^3
,\end{align*}
which is what we wanted and we are done :D .

Consider an inversion with radius $s-a$ about $A$ and swap about the $\angle A$-bisector. Define $P$ to be the $A$-antipode in $\omega_A$ and $Q$ similarly, and let $K, L$ be the images of $P, Q$ (typing stars is hard). The conditions mean that $K$ is the foot of the $A$-altitude, and $L$ is the intersection of the $A$-altitude with the tangent to $\omega$ parallel to $\overline{BC}$.

Thus, as $AP \cdot AK = (s-a)^2$, we have $AP = \frac{(s-a)^2}{h_A}$ and $AQ = \frac{(s-a)^2}{h_A-2r}$. The rest is computation: this implies $$P_AQ_A = \frac{(s-a)^2 r}{h_A(h_A-2r)} = \frac{(s-a)^2 \cdot \frac Ks}{\frac{2K}a \cdot \left(\frac{2K}a-\frac{2K}s\right)} = \frac{a^2(s-a)}{4K}.$$ So then
\begin{align*}
\prod P_AQ_A  &\leq R^3 \\
\iff \frac{a^2b^2c^2 \prod (s-a)}{64K^3} &\leq R^3 \\
\iff a^2b^2c^2 \prod (s-a) &\leq (4KR)^3 \\
\iff \prod (s-a) &\leq abc.
\end{align*}
This is just Schur, thus done!

One of the worst problems ever..
Invert with radius s-a followed by a reflection over AI (with $a,h_a$ 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 $\omega_a,S_a$, and s,r the semiperimeter and inradius. S goes to a line parallel to BC (since B',C' lie on AC,AB and satisfy $AC'B'=ABC$), while the incircle is orthogonal and hence preserved so $\omega_a$ that's tangent to the preimage must be a parallel line to the after image and also tangent to $\omega$, so it's BC; in particular, E inverts to D since the inverse lies on BC and satisfies $CAE=CAO=BAD$. Similarly, $S_a$ inverts to the other line l parallel to BC tangent to $\omega$, so F inverts to $AD\cap l$. The problem is bash from here: $AE = \frac{(s-a)^2}{h_a}$ and $AF = \frac{(s-a)^2}{h_a-2r}$ implies \[P_aQ_a=\frac{AF}2-\frac{AE}2= \frac{(s-a)^2 r}{h_A(h_A-2r)}=\frac{(s-a)^2\frac{[ABC]}s}{\frac{2[ABC]}a\left(\frac{2[ABC]}a-\frac{2[ABC]}s\right)}= \frac{a^2(s-a)}{4[ABC]}\]\[\implies8\prod P_AQ_A\le R^3\iff8a^2b^2c^2\prod(s-a)\le(4[ABC]R)^3\iff8\prod(s-a)\le abc\]\(\stackrel{a=x+y,b=y+z,c=z+x}{\iff}8xyz\le(x+y)(y+z)(z+x),\) which is evident by, say, AM-GM over all terms after expanding.
    $\textbf{Remark.}$ 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.

Consider the inversion at $A$ orthogonal to the incircle followed by a reflection over the angle bisector. Then $P_A,Q_A$ are sent to the reflections of $A$ over $BC$ and the other line parallel to $BC$ tangent to the incircle. Then, by Inversion Distance formula we get that $P_AQ_A=\frac{r(s-a)^2}{h_a(h_a-2r)},$ where $r$ is the inradius, $s$ is the semiperimeter and $h_a$ is the length of the altitude from $A$ to $BC.$ Then, letting $K$ be the area of $ABC$ we get that $h_a=\frac{2K}{a}.$ We proceed similarly for $B,C.$ We then get the following by using Heron's and the inradius and circumradius formula:

\begin{align*}
8P_AQ_A \cdot P_BQ_B \cdot P_CQ_C \le R^3 &\iff \frac{8r^3(s-a)^2(s-b)^2(s-c)^2}{\frac{8K^3}{abc}(\frac{2K}{a}-2r)(\frac{2K}{b}-2r)(\frac{2K}{c}-2r)} \le R^3 \\
&\iff \frac{r^3 \frac{K^4}{s^2}}{\frac{K^2}{4R} \cdot 8K^3 \cdot (\frac{1}{a}-\frac{1}{s})(\frac{1}{b}-\frac{1}{s})(\frac{1}{c}-\frac{1}{s})} \le R^3 \\
&\iff \frac{r^3 R}{2Ks^2 \cdot \frac{(s-a)(s-b)(s-c)}{abcs^3}} \le R^3 \\
&\iff \frac{r^3}{2K \cdot \frac{K^2}{4KRs^2}} \le R^2 \\
&\iff \frac{2r^3s^2}{K^2} \le R \\
&\iff 2r \le R,
\end{align*}
which is well known, e.g. by constructing the nine-point circle.

Let $P_{2A}$ and $Q_{2A}$ denote the reflections of $A$ over $P_A$ and $Q_A$ respectively, which both lie on their respective circles as well as line $AO.$ The problem then reduces to showing that $$\prod_{cyc}P_{2A}Q_{2A}\leq R^3.$$

The main claim is as follows:\\

\begin{claim}
$$P_{2A}Q_{2A}=\frac{a^2(-a+b+c)}{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}.$$
\end{claim}

Perform $\sqrt{bc}$ inversion, and denote image points with $'$. The image of the incircle is the circle tangent to $AB',AC',(AB'C')$. Thus, $Q_{2A}'$ is the intersection of the top parallel tangent to $BC$ and $\omega'$ and the line $AO'$, which is the altitude, and $P_{2A}'$ is the same intersection but with the bottom parallel tangent. Let $T_B'$ and $T_C'$ be the tangents of $\omega'$ to $AB'$ and $AC'$ Note that $$AT_B'=AT_C'=\frac{bc}{s-a}=\frac{2bc}{b+c-a}.$$

If we take a homothety sending $\omega'$ to the excircle, then $Q_{2A}'$ gets sent to the foot from $A$ to $B'C'$, which has length say $L$. Thus, $$AQ_{2A}'=L\cdot\frac{\frac{bc}{s-a}}{s}=\frac{4bc}{(a+b+c)(-a+b+c)}\cdot L.$$ Thus, $$AQ_{2A}=\frac{(a+b+c)(-a+b+c)}{4L}$$ since the radius of inversion is $\sqrt{bc}$.\\

Now, if we take a homothety sending $\omega$ to the excircle, $P_{2A}'$ gets sent to the foot from $A$ to $B'C'$. Thus, $$AP_{2A}'=L\cdot\frac{\frac{bc}{s-a}}{s-a}=\frac{4Lbc}{(b+c-a)^2}.$$ Thus, $$AP_{2A}=\frac{(b+c-a)^2}{4L}.$$ This means that $$P_{2A}Q_{2A}=AQ_{2A}-AP_{2A}$$ $$=\frac{2a(b+c-a)}{4L}=\frac{a(-a+b+c)}{2L}.$$ However, of course we have $$L=\frac{2[ABC]}{a}=\frac{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{2a}.$$ Therefore, plugging back in we have $$P_{2A}Q_{2A}=\frac{a^2(-a+b+c)}{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}.$$ Thus, we wish to show that $$\prod_{cyc}P_{2A}Q_{2A}\leq R^3$$ $$\frac{a^2b^2c^2(-a+b+c)(a-b+c)(a+b-c)}{((a+b+c)(a+b-c)(a-b+c)(-a+b+c))^{3/2}}\leq R^3$$

$$\frac{a^2b^2c^2}{(a+b+c)\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}\leq R^3$$

$$\frac{a^2b^2c^2}{(a+b+c)4A}\leq R^3.$$ Substituting $R=\frac{abc}{4A}$ we have $$\frac{a^2b^2c^2}{(a+b+c)4A}\leq \frac{a^3b^3c^3}{64A^3}$$ $$\frac{1}{a+b+c}\leq \frac{abc}{16A^2}$$ $$16A^2\leq (a+b+c)(abc)$$ $$(a+b+c)(a+b-c)(a-b+c)(-a+b+c)\leq (a+b+c)abc$$ $$(a+b-c)(a-b+c)(-a+b+c)\leq abc.$$ The left hand side expands out to $$(\sum_{sym}a^2b)-(a^3+b^3+c^3)-2abc,$$ so it suffices to show that $$a^3+b^3+c^3+3abc\geq \sum_{sym}a^2b,$$ which is just Schur's inequality. Since $a,b,c$ are nonzero, equality occurs if and only if $a=b=c.$

Overlay at $A$ with radius $s-a$. Then
[list]
[*] The incircle remains fixed, while the other three circle get mapped to parallel lines.
[*] $AP_AQ_AO$ is mapped to the altitude from $A$ to $BC$.
[*] $\omega_A$ is mapped to a line tangent to the incircle and perpendicular to the altitude, which is $BC$.
[*] $S_A$ is simiarily mapped to a line parallel to $BC$ tangent to the incircle that is not $BC$.
[*] $P_A$ maps to the reflection of $A$ over $\omega_A^*$, and $Q_A$ maps to the reflection of $A$ over $S_A^*$.
[/list]

Consequently, the inversion distance formula says
\[P_AQ_A = P_A^*Q_A^* \cdot \frac{(s-a)^2}{AP_A^* \cdot AQ_A^*} = \frac{(s-a)^2 \cdot r}{h_a(h_a-2r)} = \frac{a^2 \cdot (s-a)}{4rs}.\]

Thus the LHS can be rewritten as
\[8 \cdot \prod_{\text{cyc}} \frac{a^2 \cdot (s-a)}{4[ABC]} = \frac{(abc)^2 \cdot s(s-a)(s-b)(s-c)}{8 s [ABC]^3} = \frac{(4Rrs) \cdot (4R \cdot [ABC]) \cdot [ABC]^2}{8 s [ABC]^3} = 2R^2r,\]

which is less than or equal to $R^3$ if and only if $R \ge 2r$, which is true and has equality when $\triangle ABC$ is equilateral. $\blacksquare$

what a bizarre problem

Invert at $A$ fixing $\omega$ and then reflect over the $\angle A$-bisector. This has radius $s-a$ and sends $S_A$ and $\omega_A$ to $\overline{BC}$ and the other tangent to $\omega$ parallel to $\overline{BC}$. Moreover, $2P_AQ_A$ equals the distance from $A$ to the $A$-antipode of $S_A$, minus the distance from $A$ to the $A$-antipode of $\omega_A$, and these points get sent to the feet of the altitudes from $A$ to the lines that the circles are mapped to. Hence
$$2P_AQ_A=(s-a)^2\left(\frac{1}{d(A,\overline{BC})-2r}-\frac{1}{d(A,\overline{BC})}\right)=(s-a)^2\frac{2r}{h_A(h_A-2r)}.$$
It turns out that this actually equals something nice, but we will not realize this and instead turn our brains off. Letting $K=[ABC]$, this implies
\begin{align*}
8(P_AQ_A)(P_BQ_B)(P_CQ_C)&=\frac{8r^3(s-a)^2(s-b)^2(s-c)^2}{h_Ah_Bh_C(h_A-2r)(h_B-2r)(h_C-2r)}\\
&=\frac{8r^3K^4}{s^2h_Ah_Bh_C(h_A-2r)(h_B-2r)(h_C-2r)}\\
&=\frac{8r^3K^4}{s^2\frac{8K^3}{abc}(\frac{2K}{a}-2r)(\frac{2K}{b}-2r)(\frac{2K}{c}-2r)}\\
&=\frac{r^3Ka^2b^2c^2}{s^2(2K-2ar)(2K-2br)(2K-2cr)}\\
&=\frac{Ka^2b^2c^2}{s^2(2s-2a)(2s-2b)(2s-2c)}\\
&=\frac{Ka^2b^2c^2}{8sK^2}\\
&=\frac{a^2b^2c^2}{16K^2}\cdot \frac{2K}{s}\\
&=R^2(2r).
\end{align*}
By Euler's Theorem, $R \geq 2r$ with equality iff $\triangle ABC$ is equilateral, finishing the problem. $\blacksquare$


[quote="thanhnam2902"][color=darkblue]Let $ ABC$ is triangle. Find max of $ \boxed{A \equal{} \frac {\sqrt [3]{sinA} \plus{} \sqrt [3]{sinB} \plus{} \sqrt [3]{sinC}} {\sqrt [3]{cos\frac {A}{2}} \plus{} \sqrt [3]{cos\frac {B}{2}} \plus{} \sqrt [3]{cos\frac {C}{2}}}}$[/color][/quote]
By Chebyshev :
$ \sqrt [3]{sinA} \plus{} \sqrt [3]{sinB} \plus{} \sqrt [3]{sinC} \equal{} \sqrt[3]{2}(\sqrt[3]{sin(A/2)}\sqrt[3]{cos(A/2)}\plus{}\sqrt[3]{sin(B/2)}\sqrt[3]{cos(B/2)}\plus{}\sqrt[3]{sin(C/2)}\sqrt[3]{cos(C/2)}) \le \frac{\sqrt[3]{2}}{3}(\sqrt[3]{sin(A/2)}\plus{}\sqrt[3]{sin(B/2)}\plus{}\sqrt[3]{sin(C/2)})(\sqrt[3]{cos(A/2)}\plus{}\sqrt[3]{cos(B/2)}\plus{}\sqrt[3]{cos(C/2)})\equal{}> A \le \frac{\sqrt[3]{2}}{3}(\sqrt[3]{sin(A/2)}\plus{}\sqrt[3]{sin(B/2)}\plus{}\sqrt[3]{sin(C/2)})$;
$ f(x)\equal{} \sqrt[3]{sin(x/2)}$ - convex then   $ 0 \le x \le \pi \equal{}>A \le \sqrt[3]{2}\sqrt[3]{sin(\pi/6)}\equal{}1$;

[i]  Easy problem[/i] :lol:

[i]
   With $ a , b$ are nonnegative real numbers , we have :

   $ (a \plus{} b)^{3} \leq 4(a^3 \plus{} b^3)$

   Deduce $ ( \sqrt [3]{sinA} \plus{} \sqrt [3]{sinB} )^{3} \leq 4( sinA \plus{} sinB) \equal{} 8sin \frac {A \plus{} B}{2} cos \frac {A \minus{} B}{2} \leq 8 cos \frac {C}{2}$

    => $ \sqrt [3]{sinA} \plus{} \sqrt [3]{sinB} \leq 2 \sqrt [3]{ cos \frac {C}{2} }$

    =>      $ \sqrt [3]{sinB} \plus{} \sqrt [3]{sinC} \leq 2 \sqrt [3]{ cos \frac {A}{2} }$

     =>   $ \sqrt [3]{sinC} \plus{} \sqrt [3]{sinA} \leq 2 \sqrt [3]{ cos \frac {B}{2} }$

   => $ \sum \sqrt [3]{ sinA} \leq \sum \sqrt [3]{ cos \frac {A}{2}}$ :wink: [/i]

    $ max_{A}^{}  \equal{} 1$

[i]  Hùynh Võ Phương An[/i]

Thank you very much. It's nice solution.


Let $ \mathbb{N}_0$ denote the set of nonnegative integers. Find all  functions $ f$ from $ \mathbb{N}_0$ to itself such that
\[ f(f(m) \plus{} n) \equal{} f(f(m)) \plus{} f(n)\qquad \text{for all} \; m, n \in \mathbb{N}_0.
\]
[hide="Remark"]This problem is similar to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=60429]an IMO problem[/url]. I haven't solved this though.[/hide]

nobody gives it a try? :(

btw, is there any functional equation which is really unsolvable? :roll:


Hello, we have a ball of mass m

[code]
^ y
|
|
|_____> x
                    ||
                    ||
                    ||
                    ||
                    X O (center of mass and rotating pivot)
                    ||
                    ||
                    ||
                    ||                 <---o (ball)
                                          v
[/code]

The bar has mass M and length L and the ball has the speed drawn in picture.

I want to compute the angular momentum of the system with respect to O right now,Many solved exercise say it is $ \frac{L}{2}mv\hat{\mathbf k}$ but how is it possible if we don't know the distance between the ball and the bar? Do we assume the ball is already "near" the bar extremum right before the hit?

[hide="hint"]
angular momentum $ \vec L \equal{} m\vec r*\vec v \equal{} mvr\sin\theta\vec k \equal{} mv (r\sin\theta)\vec k...$ [/hide]


geometry trapezoid inequalities trigonometry quadratics AMC AIME

Let $ ABCD$ be an isosceles trapezoid with $ \overline{AD}\parallel{}\overline{BC}$ whose angle at the longer base $ \overline{AD}$ is $ \dfrac{\pi}{3}$. The diagonals have length $ 10\sqrt {21}$, and point $ E$ is at distances $ 10\sqrt {7}$ and $ 30\sqrt {7}$ from vertices $ A$ and $ D$, respectively. Let $ F$ be the foot of the altitude from $ C$ to $ \overline{AD}$. The distance $ EF$ can be expressed in the form $ m\sqrt {n}$, where $ m$ and $ n$ are positive integers and $ n$ is not divisible by the square of any prime. Find $ m \plus{} n$.

[hide="Cheap"] Assume that AB=BC=CD (this obviously isn't always the case, but the problem statement implies that the answer will always be the same anyway). We see that AD=2AB, since we can break up our trapezoid into 3 equilateral triangles.

Consider $ \Delta ABC$, and drop perpendicular $ BB'$ down to $ AC$. From 30-60-90 triangles, $ BC\equal{}AB\equal{}CD\equal{}10\sqrt{7}$, and $ AD\equal{}20\sqrt{7}$.

Now, we have $ EA\equal{}10\sqrt{7}$, $ ED\equal{}30\sqrt{7}$, $ AD\equal{}20\sqrt{7}$. By the Triangle Inequality, $ E,A,D$ are collinear.

Now, we easily find by 30-60-90 triangles again that $ DF\equal{}5\sqrt{7}$. Thus, $ EF\equal{}ED\minus{}FD\equal{}25\sqrt{7}$, giving an answer of $ 25\plus{}7\equal{}\boxed{032}$.[/hide]

[hide]
[asy]size(400);
defaultpen(1);
pair A=(0,0), D=(4,0), B= A+2 expi(1/3*pi), C= D+2expi(2/3*pi), E=(-4/3,0), F=(3,0);
draw(F--C--B--A);
draw(E--A--D--C);
draw(A--C,dashed);
draw(circle(A,abs(C-A)),dotted);
label("$A$",A,S);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,SE);
label("$E$",E,W);
label("$F$",F,S);
clip(currentpicture,(-1.5,-1)--(5,-1)--(5,3)--(-1.5,3)--cycle);[/asy]

Applying the triangle inequality to $ ADE$, we see that
\[ AD \ge 20\sqrt{7} . \]
On the other hand, if $ AD$ is strictly greater than $ 20\sqrt{7}$, then the circle with radius $ 10\sqrt{21}$ and center $ A$ does not touch $ DC$, which implies that $ AC> 10\sqrt{21}$, a contradiction.  Hence
\[ AD \equal{} 20\sqrt{7}. \]
It follows that $ A,D,E$ are collinear, and also that $ ADC$ and $ ACF$ are 30-60-90 triangles.  Hence $ AF \equal{} 15\sqrt{7}$, and
\[ EF\equal{} EA \plus{} AF \equal{} 10\sqrt{7} \plus{} 15\sqrt{7} \equal{} 25\sqrt{7} . \]
Hence the answer to this problem is 25+7, or 32.  $ \blacksquare$[/hide]

Extend rays $ AB$ and $ DC$ to meet at $ G$. It follows that $ ADG$ is equilateral. Since $ AC$ is a cevian of this triangle, $ AC\ge h$, so \[ 10\sqrt{21}\ge AD\frac{\sqrt{3}}{2}\implies AD\le 20\sqrt{7}\] By the triangle inequality, \[ DE\le AE\plus{}AD\implies AD\ge 20\sqrt{7}\] Equality occurs when $ D$, $ A$, $ E$ form a line and $ AC$ is the altitude to $ DG$. It is easy to see that $ 2DC\equal{}DG\equal{}DA$ so $ DC\equal{}10\sqrt{7}$. Then $ DF\equal{}DC\sin 60^\circ\equal{}5\sqrt{3}$. It follows that \[ EF\equal{}DE\minus{}DF\equal{}30\sqrt{7}\minus{}5\sqrt{7}\equal{}25\sqrt{7}\] It follows that the answer is $ 25\plus{}7\equal{}\boxed{032}$.

isosceles trapezoid means cyclic quad. Lucky i've seen these a lot :D

Why is assuming that E is on AD incorrect? there is nothing in the problem that says against it. I solved for it that way, using some algebra stuff, and than pythagorean theorem. but when i tried using quadratic formula, I kept getting a negative discriminant.

I also assumed directly that E was on AD, and I also got $ 25\sqrt{7}$ for EF, 25+7=32.

[hide="Whats wrong with this solution"]Since BC is a free variable, let BC be a degenerate segment, giving an equilateral triangle. The foot of the perpendicular from C onto AD is just the midpoint of AD, and now use Stewart's to get $ 5 \sqrt{119} \implies 124$[/hide]

[quote="mihail911"]i didnt. :(
dropping the altitude and callling where it meets AD at F. call FD x. by our nifty 30-60-90 things, we know the altitude is x\sqrt{3}. because i assumed E was on AD, i got that AD=40\sqrt{7}. thus AF=40\sqrt{7}-x. because the diagonal is 10\sqrt{21}, i used the pythagorean theorem to try and solve for x. however, when i used quadratic formula, i got a negative discriminant which implies imaginary x??!?!? by my method, FD should equal 5\sqrt{7}. if i use this value, i get AF=37\sqrt{7} which wouldn't satisfy the pythagorean statement i set up.  :maybe:[/quote]

It's because I assumed E was on the extension of line AD through A. Therefore, AD=20sqrt(7), and ACD and ABD are 2 30-60-90 triangles, with side lengths 10sqrt(7), 10sqrt(21), and 20sqrt(7).

[quote="The Zuton Force"][hide="Whats wrong with this solution"]Since BC is a free variable, let BC be a degenerate segment, giving an equilateral triangle. The foot of the perpendicular from C onto AD is just the midpoint of AD, and now use Stewart's to get $ 5 \sqrt {119} \implies 124$[/hide][/quote]

thats what I did but i checked after and the triangle wouldnt exist...

[quote="33286"]At first I thought that AB=BC, but I looked at the problem again and realized that it didn't say that, so I ended up not getting it...I might have gotten it right had I not done so.

Lessons from this year's AIME:

1. Never check your work: I decided to do 14 and 15 instead, and got them right.
2. Don't reread the problem.[/quote]

Wow...those sound pretty counterintuitive  :P

[quote="The Zuton Force"][hide="Whats wrong with this solution"]Since BC is a free variable, let BC be a degenerate segment, giving an equilateral triangle. The foot of the perpendicular from C onto AD is just the midpoint of AD, and now use Stewart's to get $ 5 \sqrt {119} \implies 124$[/hide][/quote]
What's wrong is that BC is not a free variable.  Read my solution or Altheman's, and you will see that BC is fixed.

[quote="CatalystOfNostalgia"][hide="Cheap"] Assume that AB=BC=CD (this obviously isn't always the case, but the problem statement implies that the answer will always be the same anyway). We see that AD=2AB, since we can break up our trapezoid into 3 equilateral triangles.

Consider $ \Delta ABC$, and drop perpendicular $ BB'$ down to $ AC$. From 30-60-90 triangles, $ BC \equal{} AB \equal{} CD \equal{} 10\sqrt {7}$, and $ AD \equal{} 20\sqrt {7}$.

Now, we have $ EA \equal{} 10\sqrt {7}$, $ ED \equal{} 30\sqrt {7}$, $ AD \equal{} 20\sqrt {7}$. By the Triangle Inequality, $ E,A,D$ are collinear.

Now, we easily find by 30-60-90 triangles again that $ DF \equal{} 5\sqrt {7}$. Thus, $ EF \equal{} ED \minus{} FD \equal{} 25\sqrt {7}$, giving an answer of $ 25 \plus{} 7 \equal{} \boxed{032}$.[/hide][/quote]

So actually the answer is not always the same. Your guess was very lucky because $ AB\equal{}BC\equal{}CD$ . [i]must[/i] be true.

It can be solved very trivially using pythagorean theorem by just drawing the hieght i think.

[quote="FibonacciFan"][quote="Boy Soprano II"]then the circle with radius $ 10\sqrt {21}$ and center $ A$ does not touch $ DC$,[/quote]

Why is this true?[/quote]
Let $ C'$ be the foot of the perpendicular from $ A$ to the line $ DC$.  Then $ AC'$ is the shortest possible distance from $ A$ to a point on line $ DC$.  Also, $ ADC'$ is a 30-60-90 triangle, so if $ AD > 20 \sqrt{7}$, then $ AC' \equal{} AD \cdot \frac{\sqrt{3}}{2} > 20\sqrt{7} \cdot \sqrt{3}/2 \equal{} 10\sqrt{21}$.

Why does the triangle inequality show that A,D,F are collinear. I see how it makes sense, and thats what I did to solve it but my solution wasnt rigorous :huh:

[quote="mathemonster"]Why does the triangle inequality show that A,D,F are collinear.[/quote]
It doesn't.  It's in the problem statement; F is the altitude from C to AD.

Sorry to be reviving such an old thread, but I have a question about the statement

 if $ AD$ is strictly greater than $ 20\sqrt{7}$, then the circle with radius $ 10\sqrt{21}$ and center $ A$ does not touch $ DC$, which implies that $ AC> 10\sqrt{21}$, a contradiction.

How does the inequality $ AD>20\sqrt{7}$ lead to drawing the circle, and how do the dimensions of this circle imply that $ AC>10\sqrt{21}$?  Could somebody please elaborate?  Thanks.

The triangle inequality applied to $ \triangle ADE$ implies that $ AD\ge 20\sqrt{7}$.
Suppose that $ AD>20\sqrt{7}$. Look at $ \triangle ABD$:[asy]defaultpen(fontsize(8));
int i;
pair A=(0,0), F=16*expi(pi/3), D=(20,0);
draw(A--D);
draw(A--F, blue+1);
for(i=2;i<16;i+=2){
draw(D--i*expi(pi/3),linetype("8 8"));
label("A",A,(-1,-1));label("D",D,(1,-1));
}[/asy]as $ B$ varies along the blue line, which is at angle $ \pi/3$ with respect to $ AD$, the least possible distance $ BD$ is when $ B$ is the foot of the perpendicular from $ D$ onto the blue line. At this configuration, $ BD \equal{} \frac{\sqrt{3}}{2}AD>10\sqrt{21}$. But we have a contradiction since $ BD \equal{} 10\sqrt{21}$.

The above is equivalent to obtaining a contradiction by noting that:
1) $ B$ is on the circle with radius $ 10\sqrt{21}$ with center $ D$
2) This circle does not touch the blue line when $ AD>20\sqrt{7}$.

For solving this problem quickly, you can fix the length of BC to fit the 30-60-90 ratios right? (Since it doesn't mention anything about BC)


quadratics
inequalities
trigonometry
absolute value

We know that $\sin^2{x}+a\cos{x}+a^2 \ge 1+\cos{x}$, $x \in R$.
Find the range of $a$ if $a<0$.

[quote="honglei"]It is too easy.[/quote]
If so, why don't you show us your work!? ;)  ;)

[quote="mna851"]It is pretty easy you can just bring everything to one side of the equation and factor.[/quote]
Yes, we can just use a trig identity to transform $\sin{x}$ to $\cos{x}$. Then do some calculations...
So please post your solution and answers if you can solve it.

So any one got the answer?? This problem has not been solved yet...

[quote="nr1337"][hide]
Note that sin^2 x = 1-cos^2 x, bring everything to one side, find that cos x = (a-1 +- sqrt(-(1+a)(3a-1)))/2 and then note that -1<= a <= 1/3.  a < 0 (problem statement), so -1 <= a <0.

It's not hard to see that  (a-1 + sqrt(-(1+a)(3a-1)))/2 never has a value that has an absolute value greater than one.[/hide][/quote]
huh?
I wonder why I keep getting $a \leq -2$...


function
inequalities
inequalities proposed

$a,b,c,d >0$ satisfying $a+b+c+d=1$.Prove that:
               $(ab+bc+cd+da)(\frac{a}{b^{2}+b}+\frac{b}{c^{2}+c}+\frac{c}{d^{2}+d}+\frac{d}{a^{2}+a}) \geq \frac{4}{5}$
Conjecture:
$a_{1},a_{2},...,a_{n}>0$ satisfying $a_{1}+a_{2}+...+a_{n}=1$ ($n \in N$ and $n \geq 3$).Prove that:
       $(a_{1}a_{2}+a_{2}a_{3}+...+a_{n}a_{1})(\sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}^{2}+a_{i+1}}) \geq \frac{n}{n+1}$
I have proved for the case $n=3,4$ :)   :) but for $n \geq 5$,i don't have the solution :(  :( .I hope some one can answer my conjecture

Quite nice problem... :wink:
Let the function $f(x)=\frac{1}{x^{2}+x}$  . By Jensen we have
$af(b)+bf(c)+cf(d)+df(a)\geq f(ab+bc+cd+da)=\frac{1}{(ab+bc+cd+da)^{2}+ab+bc+cd+da}$

So $LHS\geq \frac{1}{ab+bc+cd+da+1}$
So it remains to prove that $\frac{1}{ab+bc+cd+da+1}\geq \frac{4}{5}$

which is trivial .
I think this method proves also the general one

[quote="silouan"]So $LHS\geq \frac{1}{ab+bc+cd+da+1}$ [/quote]

That's not true... $a=b=c=0, d=1 \to 0 \geq 1$

[quote="nhat"]
Conjecture:
$a_{1},a_{2},...,a_{n}>0$ satisfying $a_{1}+a_{2}+...+a_{n}=1$ ($n \in N$ and $n \geq 3$).Prove that:
       $(a_{1}a_{2}+a_{2}a_{3}+...+a_{n}a_{1})(\sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}^{2}+a_{i+1}}) \geq \frac{n}{n+1}$
I have proved for the case $n=3,4$ :)   :) but for $n \geq 5$,i don't have the solution :(  :( .I hope some one can answer my conjecture[/quote]

I think that the conjecture can solved with cauchy!!!! :roll:
  Also i think that silouan can solved this problems!

[quote="Fitim"]
  Also i think that silouan can solved this problems![/quote]

I don't understand you . What do you mean ??
Also I think my solution is valid.....

[quote="silouan"]Quite nice problem... :wink:
Let the function $f(x)=\frac{1}{x^{2}+x}$  . By Jensen we have
$af(b)+bf(c)+cf(d)+df(a)\geq f(ab+bc+cd+da)=\frac{1}{(ab+bc+cd+da)^{2}+ab+bc+cd+da}$

So $LHS\geq \frac{1}{ab+bc+cd+da+1}$
So it remains to prove that $\frac{1}{ab+bc+cd+da+1}\geq \frac{4}{5}$

[/quote]

It is a valid solution. The first part is weight Jense. For the second part, I just want to add a little detail to his solution. That is
$\frac{1}{ab+bc+ca+da+1}\geq \frac45$
equivalent to $(a-b+c-d)^{2}\geq 0$


geometry
circumcircle
perpendicular bisector
geometry proposed

Given triangle ABC with the shortest side BC and let P be a point of AB such that PCB=BAC and Q be a point on AC such that QBC=BAC. Prove that the line through the centers of the circumcircles of triangle ABC and APQ is perpendicular to BC

$\angle BAC=\angle PCB=\angle QBC$ implies that circles $\odot(AQB)$ and $\odot(APC)$ are tangent to $ BC$ through $B,C.$ Hence, $ BC^2 \equal{} BP \cdot BA$ and $ BC^2 \equal{} BQ \cdot CA$ $\Longrightarrow$ $ B,C$ have equal power with respect to $\odot (APQ)$ $\Longrightarrow$ $ B,C$ are equidistant from the center of $\odot (APQ).$ Centers of $\odot (ABC)$ and $\odot (APQ)$ lie on the perpendicular bisector of $BC.$


linear algebra
matrix
algebra unsolved
algebra

Let A is an nXn invertible matrix. Prove that the sum of the entries of the first row of A is the reciprocal of the sum of the entris of the first row of the matrix which is the inverse of A.

That is not true in general. For a 2 x 2 matrix this is equivalent to

$ a_{11} \equal{} 0$ or $ a_{11} \plus{} a_{12} \equal{} a_{21} \plus{} a_{22}$

For a larger matrix a sufficient condition would be that the sums of the entries in each row of A are the same.


Prove that for all triangle ABC we have:

a^3 * b^4 * c^5 >= 25* \sqrt 5/27 * (2*S)^6,
where a,b,c are the sides and S denote the area of this traingle.
I have already solved this problem, but I am looking for a nice solution.

Hi Bao,

The constant  \sqrt 5/27 is sqrt(5/27) or sqrt(5)/27?

Namdung

Hi, it's sqrt(5)/27.
I'm looking for a nice solution, since mine is quite long and not nice

:o does such a dirty question really have a nice solution?

Why dirty? I think is very nice prb! The symmetric ones sometimes are boring. I will try and I'm sure that there is nice solution with weigted AM-GM.

well i am nowhere near a solution but this is what i found
given inequality equivalent to proving
(sinA)^3*(sinB) 2 *(sinC)  \leq   27/ \sqrt (3125)
then i used weighed am-gm and then generalized jensen inequality for the lhs and came up with this
(sinA)^3*(sinB) 2 *(sinC)  \leq [sin{60+(A-C)/6}]^6
BUT this seems weak,duzn't it
anyone else had any luck?

Here is my sln. It is not so nice, but I hope that my result will help somebody else (for example Harazi) find the nice one.

Let us find maximum of  f(x, y, z) = (sinx)^3.(siny)^2.sinz with x, y, z from (0, Pi), x+y+z = Pi.

Consider  F(x,y,z,k) = f(x,y,z) + k(x+y+z-Pi)
System dF/dx = 0, dF/dy = 0, dF/dz = 0, dF/dk = 0 is equivalent to
 3(sinx)^2.cosx.(siny)^2.sinz = -k
 (sinx)^3.2sinycosy.sinz = - k
 (sinx)^3.(siny)^2.cosz = -k
 x+y+z-Pi=0
Divide two first eqn, we get: (3/2)tgy = tgx
Divide second eqn to the third eqn, we get tgy = 2tgz => tgx = 3tgz

From the last eqn, we get tgx + tgy + tgz = tgx.tgy.tgz => 6tgz = 6(tgz)^3 => tgz = 1, tgx = 3, tgy = 2

A trivial check show that (tgx=3, tgy=2, tgz=1) is the point of maximum.
sinz = 1/sqrt(2), siny=2/sqrt(5), sinz = 3/sqrt(10) => max = 27/(25*sqrt(5)).

The numbers are quite good, so I hope we can find more elementary proof.

Namdung


search
inequalities unsolved
inequalities

[b]1/[/b]  Let a,b,c be real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$. Prove that:
a) $ |a^3 \plus{} b^3 \plus{} c^3 \minus{} abc|\leq 2\sqrt {2}$  (1)
b) $ a \plus{} b \plus{} c\leq 2 \plus{} abc$                         (2)

[b]2/[/b] Let a,b,c be non-nagetive real numbers such that a+b+c=2. Prove that $ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \plus{} abc\leq 2$
[b]3/[/b] Let a,b,c be real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 9$. Prove that: $ 2(a \plus{} b \plus{} c) \minus{} abc\leq 10$
[b]4/[/b] Let a,b,c be non-nagetive real numbers. Prove that:
a) $ a\sqrt {a^2 \plus{} 2bc} \plus{} b\sqrt {b^2 \plus{} 2ca} \plus{} c\sqrt {c^2 \plus{} 2ab}\geq\sqrt {3}(ab \plus{} bc \plus{} ca)$
b) $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\geq\frac {a \plus{} b}{c^2 \plus{} ab} \plus{} \frac {b \plus{} c}{a^2 \plus{} bc} \plus{} \frac {c \plus{} a}{b^2 \plus{} ca}$

1(a) squaring both sides
  (b) [url]http://www.mathlinks.ro/viewtopic.php?p=338233#p338233[/url]
3. [url]http://www.mathlinks.ro/viewtopic.php?search_id=1623653548&t=27862[/url]


euh... what have you proven now?  :maybe:

[quote=sahadian]$\sqrt[3]{2-\sqrt{a}}=A$
$\sqrt[3]{2+\sqrt{a}}=B$
if $A+B$ is integer so $(A+B)^3$is integer with a little Calculation We conclude that $AB$ is  integer so we have $\sqrt[3]{4-a}$ is integer  so for $a\leq 4$ we should check 3 and 4
if a=3, $2<A+B<3 $
if a=4 ,$ A+B=\sqrt[3]{4}$
if a>4  an $A+B$ is integer $A+B=1$ so with a little Calculation We conclude that$AB=-1$so $\sqrt[3]{4-a}=-1$ so $a=5$
so a=5 is only answer[/quote]

it just gives AB is rational, doesn't it?????

[quote=shinichiman][quote="Peter"]For what positive numbers $a$ is \[\sqrt[3]{2+\sqrt{a}}+\sqrt[3]{2-\sqrt{a}}\] an integer?[/quote]
[b]Property.[/b] If $a+b+c=0$ then $a^3+b^3+c^3=3abc$.
[b]Proof.[/b] We have $a^3+b^3+c^3-3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.

[b]Solution.[/b] Let $\sqrt[3]{2+\sqrt{a}}+\sqrt[3]{2-\sqrt{a}}=b$.
We have $b + \left( - \sqrt[3]{2+ \sqrt a} \right) + \left( - \sqrt[3]{2- \sqrt a} \right) =0$
Therefore $b^3 - \left( 2+ \sqrt a \right) - \left( 2- \sqrt a \right) = 3b \sqrt[3]{ (2+ \sqrt a)(2 - \sqrt a)}$.
Or $b^3-4= 3b \sqrt[3]{4-a} \Leftrightarrow b^3-3xb-4=0 \qquad (1)$ (we let $\sqrt[3]{4-a}=x$).
Thus, we need to find $a$ such that $\sqrt[3]{4-a}$ and $b$ is an integer.
If $b$ is an integer then from $(1)$ we obtain $b|4$.
+) $b=1$ then $x=-1$ implies $a=5$.
+) $b=-1$ then $x= \dfrac{-5}{3}$ isn't integer.
+) $b=2$ ......
So the only solution is $a=5$.[/quote]

it just gives you that $ \sqrt[3]{4-a}$ is rational!


calculus
integration
algebra proposed
algebra

Let {x_n} :
 $ x_1=1 ; x_n=(1+\frac{3}{n})x_n+2-\frac{3}{n} $ , $ n \geq 1 , n \in N $
 Prove that : $ x_n $ is integer for all n

i assume you actually meant $ x_{n+1}=(1+\frac{3}{n})x_n+2-\frac{3}{n} $...

Rewrite the given equation as $nx_{n+1} = (n+3)x_n + 2n-3.$

Let $y_n = x_n + n - 1$ then $ny_{n+1} = (n+3)y_n$, which is $y_{n+1} = \frac {n+3} n y_n.$
Since $y_1 = 1$ it easily follows that $y_n = \frac {n(n+1)(n+2)} 6.$
Therefore $x_n = \frac {n(n+1)(n+2)} 6 - n + 1$ is an integer.

Pierre.

as i conjectured even before getting the answer... the recursion very likely lends itself to a combinatorial interpretation (involving expected values, probabilities, etc.), from which the result would follow... though i haven't thought of one yet...


calculus
CMU
college
MIT
geometry
Carnegie Mellon
complex analysis

I'm looking for some advice on the quality of math programs at these two schools. How do they stack up in terms of courses, faculty, student math entusiasm, etc.?

I know very little about either of those schools, so this comment is not directed so much at you as it is at all similar questions of this form: seeking the opinions of others is a good idea, but I hope you're also planning on engaging in some direct empirical tests -- going to campus, sitting in on classes, talking to students and professors, etc.

Best of luck!

I assume you're asking about undergrad programs.

I went to Carnegie Mellon for undergrad.  They have a very strong [url=http://www.math.cmu.edu/ug/ungrd/special.html]honors undergrad math sequence[/url]: a year of Apostol's [i]Calculus[/i] followed by an integrated year-long course covering undergrad analysis and algebra.  I got into MIT's Ph.D. program after doing my undergrad at CMU, for whatever that's worth.

On the down side, the math department faculty at CMU has a reputation for poor instructional quality.  In some instances, that reputation is deserved; in other cases, it's mainly engineering students complaining about the difficulty of their math classes and taking it out on the professors.  (Note that this is not an issue with the Math Studies program referenced above -- generally the better teaching professors are involved with it.)

If you get to the point where you're looking at taking graduate courses and/or doing research, you should know that the department has a very applied leaning -- there is not much "pure" math research going on there.

I don't know much about Washington University.

I strongly agree with JBL -- the best way to decide is to visit both campuses and decide for yourself which one you like better.  It really doesn't matter too much which is "better" on paper, you have to decide which one is better for you, and this may be for reasons that have nothing to do with the math department.

So what did you do when you got to grad level work DPatrick?

After I finished Math Studies, I took the standard first-year grad-school stuff: algebra, real and complex analysis, topology.  (If you want to really know, CMU's list of grad courses is [url=http://www.math.cmu.edu/grad/courses/index.html]here[/url]: I took 610/611, 620/621, 623, 630, and 651, basically two courses during each of my final 3 semesters.)  My interest was in algebra, and CMU doesn't really have much beyond the basic first-year algebra class.  At MIT, I ended up going into algebraic geometry, and you won't find a shred of that at CMU as far as I can tell.  If you want to do nonlinear analysis, however, CMU is one of the best places to be.

As a undergrad, this doesn't have much effect unless you expect to be taking classes beyond the first-year grad level and/or doing research.  And you can always walk down the street and take classes at the University of Pittsburgh (which has a decent math department of its own).

Why did you pick Carnegie Mellon over other schools?(I hear you are a rickoid...so MIT?)

My dad went to Carnegie Mellon so it's one of the schools that I am seriously considering, along with schools like MIT so any comparison would help. Thanks.

[quote="agolsme"]Why did you pick Carnegie Mellon over other schools?(I hear you are a rickoid...so MIT?)[/quote]First of all, I am a Rickoid, but RSI was not at MIT when I attended (it was at George Mason University).

One non-trivial reason why I attended CMU was that they gave me a scholarship.  I certainly could have afforded to attend MIT, but it would have required a lot of loans.  I graduated from CMU debt-free. :)

But the main reason is that I simply liked it better.  Basically, the people that I met at CMU actually seemed to care whether I went there or not.  The people at MIT could not have cared less.  (Note: this was nearly 20 years ago.  MIT Admissions has changed a lot since then.)  CMU seemed, on balance, a much friendlier place than MIT, which seemed (to me at the time) a bit cold and impersonal.  It's also smaller and more well-rounded -- my roommate for 2.5 years was a music major -- whereas MIT is large and everybody is a techie.  Plus I liked the CMU campus and the students I met, and Pittsburgh was a bit closer to home than Boston, and had more of a small-town feel (I grew up in a small town).

My tough choice actually was between CMU and Caltech, which I really liked -- the people there were friendly and I liked the campus and the curriculum -- but Caltech is a bit, um, weird, and it was a long way from home.

The moral of the story (if there is one) is that in my opinion you shouldn't decide where to go to school solely based on which school is "better" or which has a better math department or a better reputation.  Visit the campuses, talk to students and professors and administrators, and decide which school you like the best.  Don't underestimate the intangibles like location, campus size, etc.

When you say Caltech is weird, what do you mean? :huh:

I'm interested now!

I'm not sure if this is too late, but I thought I'd add my two cents.

I'm currently attending Carnegie Mellon. I absolutely love it, to say the least. I chose CMU over Cornell, Berkley, Case Western (full ride) and UPitt (Full Ride).

In terms of my major, I've been bouncing around quite a bit. Currently I'm an ECE (Electrical & Computer Engineering) major, but I intend to pursue math and computer science as much as I can. DPatrick hit the nail on the head when he mentioned that the math professors get a bad wrap because of the engineers complaining. Engineering students have a lot on their plate and when they have to concentrate a bit more on discrete issues, they complain.

Personally, every prof I've had so far went beyond my expectations. My teacher for concepts of math (first discrete math course) really inspired me to continue with math. I feel as if CMU's curriculum is a well-kept-secret.

Let me know if you have any specific questions.

For what it's worth, CMU is one of the very best universities in North America for applied math (the Center for Nonlinear Analysis there does very, very good work).


function
trigonometry
geometry
rectangle
integration
algebra
partial fractions

Evaluate
$\sum_{k=-\infty}^{k=\infty}\frac{1}{(z+k)^{2}+a^{2}}$.

This question comes at the end of a section about "partial fraction expansions" of entire functions (mittag-leffler and all that). However, I can't figure out how to use these things.

Nevermind: I figured it out immediately after posting, as typical. Disregard.

when you want to evaluate $\sum_{\mathbb{Z}}f(z)$, where $f$ has some good properties, it is often a good idea to look at $F(z)=f(z) \frac{ \cos(\pi z)}{ \sin(\pi z)}$ and to integrate on a rectangle with vertices $(N+\frac{1}{2})(\pm 1 \pm i)$

There's playing around in the neighborhood of the Poisson summation formula.

For the time being, let $x$ be real and let $f(x)=\frac1{\pi}\frac{a}{a^{2}+x^{2}}.$ Then it's one of the better-known formulas of Fourier analysis that

$\widehat{f}(\xi)=\int_{-\infty}^{\infty}f(x)e^{-2\pi i\xi x}\,dx=e^{-2\pi a|x|}.$

Now let $g(x)=\sum_{k=-\infty}^{\infty}f(x+k)=\frac{a}{\pi}\sum_{k=-\infty}^{\infty}\frac{1}{(x+k)^{2}+a^{2}}.$

The resulting $g$ is periodic of period $1$ and can be represented as a Fourier series. Here's where the Poisson summation formula comes in: we can write

$g(x)=\sum_{n=-\infty}^{\infty}\widehat{g}(n)e^{2\pi inx}$ where $\widehat{g}(n)=\widehat{f}(n)=e^{-2\pi a|n|}.$

So now,

$\sum_{k=-\infty}^{\infty}\frac{1}{(x+k)^{2}+a^{2}}=\frac{\pi}{a}g(x)=\frac{\pi}{a}\sum_{n=-\infty}^{\infty}e^{-2\pi a|n|}e^{2\pi i nx}$

$=\frac{\pi}{a}\cdot \frac{1-e^{-4\pi a}}{1-2e^{-2\pi a}\cos x+e^{-4\pi a}}.$

My original claim is that this is the correct forumla for real $x,$ but we can extend this by analytic continuation, as long as we avoid the poles at $k\pm ai$

Thanks!

My solution was to do partial fractions then subtract and add a convergence term $\frac{1}{k}$ so that you can split the sum into two other sums, and their values are 'well-known' (they can be found by logarithmically differentiating $\frac{\sin \pi z}{\pi z}= \prod_{k=1}^\infty\left(1-\frac{z^{2}}{k^{2}}\right)$)


Prove that:$\frac{a^{6}}{b^{2}c}+\frac{b^{6}}{c^{2}a}+\frac{c^{6}}{a^{2}b}\geq 3$.
where $a;b;c$ are positive real numbers and satisfying $ab+bc+ca=3$

? It is a stupid question to put in "Open section". Clearly, the left hand side is greater than $a^{3}+b^{3}+c^{3}\ge ab+bc+ca=3$.

Sorry didn't see this problem.

The first way that came to me was with extended Cauchy:
\[\left(\sum a^{2}c\right)\cdot\left(\sum \frac{a}{b}\right)\cdot\left(\sum \frac{b}{c}\right)\geq \left(\sum \sqrt[3]{a^{2}c\cdot \frac{a}{b}\cdot\frac{b}{c}}\right)^{3}=\left(\sum a\right)^{3}\]
is true by extended Cauchy (it's either called Holder's or Minkowski's). Thus,
\[\left(\sum \frac{a}{b}\right)^{3}abc\geq \left(\sum a\right)^{3}\iff \sum \frac{a}{b}\ge \sum a \]

[quote="chien than"]I have a problem,we can use AM-GM:
Prove that:$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a+b+c$
where $a;b;c>0$ and $abc=1$
(This problem is very easy!)[/quote]

this was posted before.


1) Propose a synthesis of 4,4-dimethylcyclopentenone.

2) Outline a synthesis of 4-butenylmagnesium bromide.

3) Starting with the two compounds above, propose a reasonable synthesis of the following ring system:

[quote="carcul"]1) Propose a synthesis of 4,4-dimethylcyclopentenone.
[/quote]
Here's one method:
[hide="click to reveal the hidden content"]
Probably intramolecular aldol condensation of 2,2-dimethyl 4-oxo pentanal.
[/hide]

Very good, that's one possibility. Keep going.


Thank you for the abstract package and hints. I read it more carefully and I could even make a dot appear after "Abstract", to be exactly as in the model.

Now, something completely new for me. I have to insert 6 pictures or graphics. I made them with Maple 11 and exported them as EPS files. They represent the graphics of some functions. [b]I used the following packages:
[/b]
\usepackage[dvips]{color}
\usepackage{psfig,epsfig}
\usepackage{eucal}

[b]And I used the commands:[/b] [quote] \begin{center}\begin{tabular}{cc}
\epsfxsize=5cm \epsfysize=5cm
 \epsffile{Fig.1..eps} &
\epsfxsize=5cm \epsfysize=5cm
\epsffile{Fig.2..eps}\\
\epsfxsize=5cm \epsfysize=5cm
 \epsffile{Fig.3..eps} &
\epsfxsize=5cm \epsfysize=5cm
\epsffile{Fig.4..eps}\\
\end{tabular}
\end{center}
\begin{center}\begin{tabular}{cc}
\epsfxsize=5cm \epsfysize=5cm
 \epsffile{Fig.5..eps} &
\epsfxsize=5cm \epsfysize=5cm
\epsffile{Fig.6..eps}\\
\end{tabular}
\end{center}  [/quote]

When I converted (I use Texnic center now) from Latex to PDF, the pictures didn't appear, only blank spaces instead of these.
When I converted from Latex to DVI, the message was: "The document cannot be rendered properly because the current render method does not support PostScript specials. Temporarily switch to the Dvips render method? yes or no"

I chose yes, and the pictures appeared, 2 by two, in pairs, one pair below the other as expected, but something was different in the graphic's lines as I saw in Maple's windows.

Do I needsome PostScriptum related soft, or what is the problem?

Jean

Conversion using pdflatex ie LaTeX=>PDF cannot read eps files. Convert the files to pdf first (or jpg, gif, png) then you can include them as graphics without involving postscript, using \usepackage{graphicx} and \includegraphics


geometry
modular arithmetic
rectangle
combinatorics unsolved
combinatorics

A retangle can be divided into n equal squares and in n + 98 equal squares. The area of the retangle is n. Find the all the possible values for the sides of the retangle.

I don't understand why we need area $n$...
The problem reduces to the following one:
find all $n$ s.t. $\frac{n+98}{n}=\frac{a^2}{b^2}$, $a,b\in\mathbb{N}$.
It is clear $a>b$. Suppose WLOG $(a,b)=1)$.
If $(n,n+98)=1$, then $n+98=a^2$ and $n=b^2$ -- impossible, since $a^2-b^2=2\pmod{4}$.
If $(n,n+98)\neq 1$, then $(n,n+98)=2,7,14,49,98$.
1) $(n,n+98)=2$ $\Longrightarrow$ $a^2-b^2=49$, i.e. $a=25$, $b=24$ and $n=2\cdot 24^2$.
2) $(n,n+98)=7$ $\Longrightarrow$ $a^2-b^2=14$ -- impossible.
3) $(n,n+98)=14$ $\Longrightarrow$ $a^2-b^2=7$, i.e. $a=4$, $b=3$ and $n=14\cdot 3^2$.
4) $(n,n+98)=49$ $\Longrightarrow$ $a^2-b^2=2$ -- impossible.
5) $(n,n+98)=98$ $\Longrightarrow$ $a^2-b^2=1$ -- impossible.

[quote="Myth"]
I don't understand why we need area $n$...
[/quote]

 You have found $n$ but you still not find the sides of the rectangle yet!! Isn't it? :D

Ok! I forgot to do that final step :blush:
And I am too lazy to do it now :blush:
Only reason why I wrote this solution is that problem have been lying here too much time.


geometry
incenter
circumcircle
inequalities
inequalities unsolved

[quote="Big Bolzano"]Prove that :
$AI^2 + BI^2 + CI^2 \geqslant 3(abc)^{\frac{2}{3}} - \frac{6abc}{a+b+c}$
$I$ - incenter[/quote]

We can find that
$AI^2=\frac{cb(c+b-a)}{a+b+c}$
$BI^2=\frac{ac(a+c-b)}{a+b+c}$
$CI^2=\frac{ab(a+b-c)}{a+b+c}$
(We can prove it using the circumcircle of the triangle. We can note there some similarities)

So we can rewrite this inequality in this way:
$\frac{cb(c+b-a)}{a+b+c}+\frac{ac(a+c-b)}{a+b+c}+\frac{ab(a+b-c)}{a+b+c}\ge 3(abc)^{\frac{2}{3}} - \frac{6abc}{a+b+c} \Leftrightarrow$
$\frac{cb(c+b+a)}{a+b+c}+\frac{ac(a+c+b)}{a+b+c}+\frac{ab(a+b+c)}{a+b+c}\ge 3(abc)^{\frac{2}{3}} \Leftrightarrow$
$cb+ac+ab \ge 3(abc)^{\frac{2}{3}}$

But it is the inequality of arithmetic mean and geometric mean.

V:)