inequality on acute triangles

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N.T.TUAN

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N.T.TUAN

#1 h Apr 26, 2007, 6:25 AM • 6 Y

Y by kittenwarrior, Adventure10, Mango247, and 3 other users

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.

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silouan

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silouan

#2 h Apr 26, 2007, 1:14 PM • 3 Y

Y by Adventure10, mathstudent5, Mango247

Could somebody compute the following two sums as a function of known things?(If yes the problem is over

)
So please compute $\angle{AP_{A}I}+\angle{AQ_{A}I}$ and $\angle{AP_{A}I}-\angle{AQ_{A}I}$

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gemath

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gemath

#3 h Apr 26, 2007, 1:37 PM • 2 Y

Y by Adventure10, Mango247

Hey nice problem

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?

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silouan

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silouan

#4 h Apr 26, 2007, 1:43 PM • 2 Y

Y by Adventure10, Mango247

gemath wrote:

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?

Could you post your way ?Thank you

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gemath

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gemath

#5 h Apr 26, 2007, 1:58 PM • 1 Y

Y by Adventure10

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.

Attachments:

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silouan

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silouan

#6 h Apr 26, 2007, 2:08 PM • 1 Y

Y by Adventure10

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

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e.lopes

349 posts

e.lopes

#7 h Apr 26, 2007, 4:52 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

My proof is basically the same of yours, gemath!

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

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Virgil Nicula

7054 posts

Virgil Nicula

#8 h Apr 26, 2007, 7:20 PM • 2 Y

Y by Adventure10, Mango247

:spider: Sorry Gemath, you commit some small mistakes !

Lemma (Gemath). 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 $\triangle ABC$ .

Proof. 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 $[IQ$ :

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

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

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

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

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gemath

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gemath

#9 h Apr 27, 2007, 2:24 AM • 2 Y

Y by Adventure10, Mango247

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.

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Virgil Nicula

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Virgil Nicula

#10 h Apr 27, 2007, 7:07 AM • 1 Y

Y by Adventure10

gemath wrote:

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.

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.

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gemath

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gemath

#11 h Apr 27, 2007, 9:39 AM • 2 Y

Y by Adventure10, Mango247

Ok that's right I made a small mistake

, 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 ?

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Little Gauss

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Little Gauss

#12 h Jul 16, 2007, 2:42 PM • 1 Y

Y by Adventure10

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

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Virgil Nicula

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Virgil Nicula

#13 h Aug 21, 2007, 4:02 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

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 $\triangle ABC$ .

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.

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

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

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

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

Proof II. Apply the Pythagoras' theorem 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 $\triangle ABC$ .

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

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epitomy01

240 posts

epitomy01

#14 h Dec 1, 2007, 7:59 AM • 2 Y

Y by Adventure10, Mango247

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

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The QuattoMaster 6000

1184 posts

The QuattoMaster 6000

#15 h Aug 24, 2008, 2:46 AM • 5 Y

Y by Adventure10, Mango247, and 3 other users

Solution

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naman12

1358 posts

naman12

#37 h May 8, 2020, 6:31 AM • 1 Y

Y by Aryan-23

Solved with Aryan-23.

$[asy] import cse5; import olympiad; import geometry; pointpen = black; pathpen = black; real p=0.216,q=0.716; pair A=dir(100),B=dir(213),C=dir(342),O=circumcenter(A,B,C),P=p*O+(1-p)*A,Q=q*O+(1-q)*A,I=incenter(A,B,C),M=extension(A,I,O,WP(B--C)),IA=rotate(180,M)*I,BB=extension(IA,rotate(90,IA)*A,A,B),CC=extension(IA,rotate(90,IA)*A,A,C),D=extension(CC,rotate(90,CC)*A,BB,rotate(90,BB)*A); D(D("A",A,A)--D("B",B,B)--D("C",C,C)--cycle); D(circumcircle(A,B,C),blue); D(incircle(A,B,C),red); D(CP(P,A),orange); D(CP(Q,A),deepgreen); D(CP(D,BB),purple); DPA(B--BB^^C--CC); D("P_A",P,S); D("Q_A",Q,S); [/asy]$

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.

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GeronimoStilton

1521 posts

GeronimoStilton

#38 h Jan 13, 2021, 1:12 AM

Y by

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.

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jacoporizzo

615 posts

jacoporizzo

#39 h Apr 9, 2021, 3:01 AM • 3 Y

Y by Mango247, Mango247, Mango247

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 $\blacksquare$

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guptaamitu1

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guptaamitu1

#40 h Oct 18, 2021, 7:22 PM

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Here's a very computational solution with $\sqrt{bc}$ inversion and inversion distance formula (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.

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.

Claim (Key Claim): $\displaystyle P_AQ_A = \frac{R \cdot \sin \alpha \cos^2 \alpha}{\cos \beta \cos \gamma}$

Proof: Let $\Gamma$ be the $A$ -mixtilinear excircle 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,

$\Phi$ 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.
$\displaystyle P_AQ_A = P_A^* Q_A^* \cdot \frac{bc}{AP_A^* \cdot A Q_A^*}$ .

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. $\blacksquare$

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rafaello

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rafaello

#41 h Jan 23, 2022, 8:52 PM

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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 $x=y=z\Longleftrightarrow a=b=c$ .

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kamatadu

465 posts

kamatadu

#42 h May 21, 2023, 7:38 AM • 1 Y

Y by HoripodoKrishno

Thanks to Tafi_ak :love:

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

We already have some exposition on this configuration from ELMO 2016 P6 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 Tafi_ak

, 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

.

This post has been edited 1 time. Last edited by kamatadu, May 21, 2023, 7:40 AM

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HamstPan38825

8857 posts

HamstPan38825

#43 h Aug 16, 2023, 2:43 AM

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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!

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signifance

140 posts

signifance

#44 h Oct 5, 2023, 12:36 AM

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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.

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mihaig

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mihaig

#45 h Oct 5, 2023, 9:22 AM

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Splendid topic

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OronSH

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OronSH

#46 h Nov 3, 2023, 5:48 PM

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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.

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john0512

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john0512

#47 h Nov 12, 2023, 3:17 AM • 1 Y

Y by OronSH

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 $a=b=c.$

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shendrew7

793 posts

shendrew7

#48 h Dec 30, 2023, 12:42 AM • 1 Y

Y by GeoKing

Overlay at $A$ with radius $s-a$ . Then

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 $S_A^*$ .

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. $\blacksquare$

This post has been edited 1 time. Last edited by shendrew7, Dec 30, 2023, 12:49 AM

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IAmTheHazard

5000 posts

IAmTheHazard

#49 h Dec 30, 2023, 3:01 AM • 1 Y

Y by GeoKing

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. $\blacksquare$

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Mathandski

734 posts

Mathandski

#50 h Nov 16, 2024, 5:43 AM

Y by

Lengthy trig bash

Subjective Rating (MOHs) $\text{[math]}$

Note that the equation is homogeneous. Without loss of generality (WLOG), assume $R = 2$ . We prove that,
$P_A Q_A \cdot P_B Q_B \cdot P_C Q_C \leq 1$ with equality if and only if $\triangle ABC$ is equilateral. WLOG, assume that $\angle B > \angle C$ .

Let $r$ be the inradius and define $x = AP_A$ . For $S_A$ to be externally tangent to $\omega$ and internally tangent to $S$ , this occurs if and only if $P_A \in AO$ and $P_A I = x + r$ . We know
$\angle IAO = \frac{\angle A}{2} - \angle CAO = \frac{\angle A}{2} - 90^\circ + \angle B = \frac{\angle B}{2} - \frac{\angle C}{2}.$
Using the law of cosines,
$(x + r)^2 = x^2 + AI^2 - 2x \cdot AI \cdot \cos\left(\frac{B}{2} - \frac{C}{2}\right),$ which simplifies to:
$x = \frac{AI^2 - r^2}{2(r + AI \cos\left(\frac{B}{2} - \frac{C}{2}\right))}.$
Substituting values:
$x = \frac{64 \sin^2\frac{B}{2} \sin^2\frac{C}{2} - 64 \sin^2\frac{A}{2} \sin^2\frac{B}{2} \sin^2\frac{C}{2}}{2(8 \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2} + 8 \sin\frac{B}{2} \sin\frac{C}{2} \cos\left(\frac{B}{2} - \frac{C}{2}\right))}.$
Simplifying further:
$x = \frac{4 \sin^2\frac{B}{2} \sin^2\frac{C}{2} (1 - \sin^2\frac{A}{2})}{\sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2} + \sin\frac{B}{2} \sin\frac{C}{2} \cos\left(\frac{B}{2} - \frac{C}{2}\right)},$ $x = \frac{4 \sin\frac{B}{2} \sin\frac{C}{2} \cos^2\frac{A}{2}}{\sin\frac{A}{2} + \cos\left(\frac{B}{2} - \frac{C}{2}\right)},$ $x = \frac{4 \sin\frac{B}{2} \sin\frac{C}{2} \cos^2\frac{A}{2}}{\sin\frac{A}{2} + \sin\left(\frac{A}{2} + C\right)},$ $x = \frac{4 \sin\frac{B}{2} \sin\frac{C}{2} \cos^2\frac{A}{2}}{2 \sin\left(\frac{A}{2} + \frac{C}{2}\right) \cos\frac{C}{2}},$ $x = 2 \tan\frac{B}{2} \tan\frac{C}{2} \cos^2\frac{A}{2}.$
On the other hand, let $y$ be the length of $AQ_A$ . We know that $Q_A$ lies on $AO$ and $IQ_A = y - r$ . Using similar steps:
$(y - r)^2 = y^2 + AI^2 - 2y \cdot AI \cdot \cos\left(\frac{B}{2} - \frac{C}{2}\right),$ which simplifies to:
$y = \frac{AI^2 - r^2}{2(AI \cos\left(\frac{B}{2} - \frac{C}{2}\right) - r)}.$
After substitution:
$y = 2 \cos^2\frac{A}{2}.$
Therefore:
$P_A Q_A = 2 \cos^2\frac{A}{2} (1 - \tan\frac{B}{2} \tan\frac{C}{2}),$ $P_A Q_A = 2 \cos^2\frac{A}{2} \frac{\tan\frac{B}{2} + \tan\frac{C}{2}}{\tan\left(\frac{B}{2} + \frac{C}{2}\right)},$ $P_A Q_A = \sin A (\tan\frac{B}{2} + \tan\frac{C}{2}),$ $P_A Q_A = \sin A \frac{\sin\frac{B}{2} \cos\frac{C}{2} + \cos\frac{B}{2} \sin\frac{C}{2}}{\cos\frac{B}{2} \cos\frac{C}{2}},$ $P_A Q_A = \sin A \frac{\cos\frac{A}{2}}{\cos\frac{B}{2} \cos\frac{C}{2}}.$
The desired product is:
$\frac{\sin A \sin B \sin C}{\cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2}} = 8 \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2}.$
Since $\sin(x)$ has a second derivative $-\sin(x)$ , it is concave for $x \in (0, \frac{\pi}{2})$ . By Jensen's inequality:
$8 \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2} \leq 8 \sin^3(30^\circ) = 1.$
Equality holds if and only if $A = B = C = 60^\circ$ as desired.

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Ilikeminecraft

330 posts

Ilikeminecraft

#51 h Feb 23, 2025, 2:45 AM

Y by

what. who rmbrs half of these area formulas lmao
Claim: $[ABC] = \frac{abc}{4R}.$
Proof: We use LOS. Note that $a = 2R\sin\angle A$ and its symmetric equations. Multiplying, $abc = 8R^3\sin\angle A\sin\angle B\sin\angle C.$ Furthermore, note that $\frac12ab\sin\angle C = [ABC],$ and its symmetric equations. Thus, $\frac1{64}a^3b^3c^3 = R^3[ABC]^3,$ which finishes.

Let $K$ be the area of $ABC.$
Invert about $A$ with radius $s - a$ and reflect across the $A$ -bisector, where $s$ is the semiperimeter. The incircle doesn’t move. Obviously, $AP_AQ_AO$ map to the altitude of $A$ to $BC.$ Hence, $\omega_A, S_A$ inverts to the line parallel to $BC$ that is tangent to the incircle. Hence, $P_A, Q_A$ invert to the reflections of $A$ across $S_A^*$ and $\omega_A^*.$ Hence, if $r$ is the inradius, then $4r$ is $P_A^*Q_A^*.$ If $h_A$ is the length of the altitude from $A$ to $BC$ , using the inversion distance formula, we get $P_AQ_A = \frac{(s-a)^2}{h_A(h_{A}-2r)}\cdot r = \frac{(s - a)^2r}{\frac{2K}{a}(\frac{2K}{a} - 2\frac{K}{s})} = \frac{(s-a)ra^2s}{4K^2} = \frac{(s-a)a^2}{4K}.$

Now, we take the cyclic product:
$\begin{align*} 8\prod \frac{(s - a)a^2}{4K} & = \frac{(abc)^2(s-a)(s-b)(s-c)}{8K^3} \\ & = \frac{(abc)^2rs(s-a)(s-b)(s-c)}{8K^3sr} \\ & \stackrel{\text{by Heron's and our claim}}= \frac{16K^2R^2r\cdot K^2}{8K^4} \\ & = 2R^2r \end{align*}$ which finishes by Euler's theorem, which states that $R\geq 2r$ with equality when equilateral.

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