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this problem is very nice, my solution is as follows. let $x,y,z$ be $AP,BQ,CR$ respectively and suppose wlog $x\geq y\geq z$. now we have that $a=y+z, b=x+z, c=y+z$, so we have $x=s-a, y=s-b, z=s-c$. from the area formulas we have that $\displaystyle r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}=\sqrt\frac{xyz}{x+y+z}$ or $6\sqrt{\frac{x+y+z}{xyz}}=\frac{6}{r}$ returning to the problem we have that $\displaystyle\frac{5}{x}+\frac{5}{y}+\frac{2}{z}=6\sqrt{\frac{x+y+z}{xyz}}$, squaring both sides and multiplying by $(xyz)^2$ gives $4x^2y^2+25y^2z^2+25x^2z^2-16xy^2z-16x^2yz+14xyz^2=0$. we may see the equation above as a quadratic in $z$. so it turns into $z^2(25x^2+25y^2+14xy)-z(16xy(x+y))+4x^2y^2=0$, using the fact that the minimum of a cuadratic is $\displaystyle\frac{-\Delta}{4a}$, where $\Delta$ is the discriminant of the equation we have that $\displaystyle 0=z^2(25x^2+25y^2+14xy)-z(16xy(x+y))+4x^2y^2\geq\frac{-(256x^2y^2(x+y)^2-4(25x^2+25y^2+14xy)(4x^2y^2))}{4(25x^2+25y^2+14xy)}=\frac{36(x^2y-xy^2)^2}{25x^2+25y^2+14xy}\geq0$ $\Rightarrow 36(x^2y-xy^2)^2=0 \Rightarrow x=y$. we've proved that all of them are isosceles. now, substituting $x=y$ in the equation above gives $64x^2z^2-32x^3z+4x^4=0 \Rightarrow 4x^2(x-4z)^2=0 \Rightarrow x=y=4z$ $\Rightarrow a=y+z=4k+k=5k, b=x+z=4k+k=5k, c=x+y=4k+4k=8k$ $\Rightarrow (a,b,c)=(5k,5k,8k)$ and finally we're done.

Possibly faster finish from $ 4x^{2}y^{2}+25y^{2}z^{2}+25x^{2}-16xy^{2}z-16x^{2}yz+14xyz^{2}=0$. We will show that the left hand side is $ \ge0$. Let $ a=xz,b=xy,c=yz$. It turns into $ 4b^{2}+25c^{2}+25a^{2}-16bc-16ab+14ac\ge0$. Consider smoothing $ a$ and $ c$. Say that $ b$ and $ a+c$ are fixed, and if $ a\neq c$, consider replacing $ a$ and $ c$ with $ \dfrac{a+c}{2},\dfrac{a+c}{2}$. We don't care about the $ 4b^{2}-16bc-16ab=4b^{2}-16b(a+c)$ because this is fixed. Note that $ 25c^{2}+25a^{2}+14ac=25(a+c)^{2}-64ac$, so when we move $ a$ and $ c$ together so they are equal, $ ac$ goes up, and the whole sum goes down (strictly). Thus we only need to consider $ a=c$, in which the inequality turns into $ (b-4c)^{2}\ge0$. Thus, the LHS is at least 0, with equality if and only if $ a=c,b=4c$. The result follows.

This is equivalent yet less ugly, even though it uses Lagrange multipliers. There's probably a simple noncalculus way to do the inequality; I'm just not good enough at inequalities. Let $ A$, $ B$, and $ C$ be the angles of the triangle; let $ a = \tan\frac {A}{2} = \frac {r}{AP}$, $ b = \tan\frac {B}{2} = \frac {r}{BQ}$, and $ c = \tan\frac {C}{2} = \frac {r}{CR}$. If we WLOG let $ A\le B\le C$ and therefore $ a\le b\le c$, $ AP\ge BQ\ge CR$. Our condition for $ S$ is now $ 5a + 5b + 2c = 6$. $ \frac {A}{2} + \frac {B}{2} + \frac {C}{2} = \frac {\pi}{2}$, so $ \Re((1 + ai)(1 + bi)(1 + ci)) = 0$; this is equivalent to $ ab + bc + ca = 1$. We wish to minimize $ f(a,b,c) = 5a + 5b + 2c$ under the constraint $ ab + bc + ca = 1$. This gives $ F(a,b,c,\lambda) = 5a + 5b + 2c - \lambda(ab + bc + ca - 1)$. \[ \frac {\partial F}{\partial a} = 5 - \lambda(b + c) = 0 \\ \frac {\partial F}{\partial b} = 5 - \lambda(a + c) = 0 \\ \frac {\partial F}{\partial c} = 2 - \lambda(a + b) = 0 \]So $ a = b$ at the minimum by the first two and by the third $ c = 4a$. Substituting into the constraint, we have $ a^2 + 4a^2 + 4a^2 = 1$, or $ a = \frac13$. This leaves $ b = \frac13$ and $ c = \frac43$ with a minimum of $ 6$ for $ 5a + 5b + 2c$. Since $ \frac {A}{2}$ ranges from $ 0$ to $ \frac {\pi}{2}$, its tangent uniquely determines it. Because $ a$ is fixed across all triangles in $ S$, as there is only one $ a$ at which the minimum occurs, $ A$ is fixed as well, as are $ B$ and $ C$. All the triangles are thus similar. Since $ A = B$, they are all isosceles too.

Upon trying is problem it can also finished with $ 4b^{2}+25c^{2}+25a^{2}-16bc-16ab+14ac=7(a+b-\frac{c}{2})^2+18(a-\frac{b}{4})^2+18(c-\frac{b}{4})^2\ge0 $ after CatalystOfNostalgia's substitution.

Isn't there a solution with a more geometric idea ?

I don't think there's a solution with a more geometrical idea, because of the super-contrived condition making it clear that one must find an equality case in an inequality. Similar to 2003 USAMO 2.

We will prove the inequality \[ \frac{2}{AP} + \frac{5}{BQ} + \frac{5}{CR} \ge \frac 6r \]with equality when $AP : BQ : CR = 1 : 4 : 4$. This implies the problem statement. Letting $x= AP$, $y = BQ$, $z = CR$, the inequality becomes \[ \frac2x + \frac5y + \frac5z \ge 6\sqrt{\frac{x+y+z}{xyz}}. \]Squaring both sides and collecting terms gives \[ \frac{4}{x^2} + \frac{25}{y^2} + \frac{25}{z^2} + \frac{14}{yz} \ge \frac{16}{xy} + \frac{16}{xz}. \]If we replace $x=1/a$, $y=4/b$, $z=4/c$, then it remains to prove the inequality \[ 64a^2 + 25(b+c)^2 \ge 64a(b+c) + 36bc \]where equality holds when $a=b=c$. This follows by two applications of AM-GM: \begin{align*} 16 \left( 4a^2 + (b+c)^2 \right) &\ge 64a(b+c) \\ 9(b+c)^2 &\ge 36bc. \end{align*}

mssmath wrote: Upon trying is problem it can also finished with $ 4b^{2}+25c^{2}+25a^{2}-16bc-16ab+14ac=7(a+b-\frac{c}{2})^2+18(a-\frac{b}{4})^2+18(c-\frac{b}{4})^2\ge0 $ after CatalystOfNostalgia's substitution. Sorry for the revive, but I don't believe this works. However, $4a^2 + 25b^2 + 25c^2 -16ab-16ac + 14bc = 4(a-2b-2c)^2 + 9(b-c)^2 \geq 0$ does.

I've been trying to cauchy this for a long time due to the trig problem in USAMO 2002, but apparently quadratic formula trivializes it: Making the substitutions that campos and v_Enhance did leads to the equation: $$4a^2 + 25b^2 + 25c^2 -16ab - 16ac +14bc = 0$$ Applying the quadratic formula with respect to a gives: $$ a = \frac{16b + 16c + \sqrt{(16b + 16c)^2 - 16(25b^2 + 25c^2 + 14 bc)}}{8} = \frac{16b + 16c + \sqrt{-(3b-3c)^2}}{8} = 2b + 2c $$ Subtracting $$4(a-2b-2c)^2$$from the original equation gives: $$4a^2 + 25b^2 + 25c^2 -16ab - 16ac +14bc - 4(a-2b-2c)^2 = (3b-3c)^2 = 0$$ So $b = c \rightarrow y = z$. From this, substituting y = z into $ \frac2x + \frac5y + \frac5z \ge 6\sqrt{\frac{x+y+z}{xyz}}. $ Gives 4x = y = z, which gives equality at $AP : BQ : CR = 1 : 4 : 4 \rightarrow AB : BC : AC = 5 : 8 : 5$, as desired.

I am seeing solutions and getting frustrated.....I came till the part... $\displaystyle\frac{5}{x}+\frac{5}{y}+\frac{2}{z}=6\sqrt{\frac{x+y+z}{xyz}}$ n then, I started believing that MAA wouldn't do something so bad to me as to give me full-length quadratic.... Why am I even alive?

A proof using some (healthy) amount of linear algebra. Suppose without loss of generality that ${\rm min}(|AP|,|BQ|,|CR|)=|AP|$. Let the side lengths of this triangle be $m,n,k$ with $u=\frac{m+n+k}{2}$ being the semiperimeter. Then, $|AP|=u-m$, $|BQ|=u-n$, and $|CR|=u-k$ hold. Using now Heron's formula, we obtain $$ u\cdot r=\sqrt{u(u-m)(u-n)(u-k)}. $$Let now $x=u-m$, $y=u-n$, and $z=u-k$. Observe that $u=x+y+z$. With this, we thus obtain $$ r=\sqrt{\frac{xyz}{x+y+z}}. $$Upon taking squares, we now conclude $$ \frac{4}{x^2}+\frac{25}{y^2}+\frac{25}{z^2}+\frac{14}{xz}=\frac{16}{xy}+\frac{16}{yz}. $$Let now $a=x^{-1}$, $b=y^{-1}$, and $c=z^{-1}$. We then obtain that $$ 4a^2+25b^2+25c^2-16ab-16ac+14bc=0. $$We now claim $$ 4a^2+25b^2+25c^2-16ab-16ac+14bc\geqslant 0, $$with equality iff $a=4b=4c$. This will then yield the desired claim. For this we appeal to a matrix-vector product notation. Note that $$ 4a^2+25b^2+25c^2-16ab-16ac+14bc = \begin{pmatrix}a&b&c\end{pmatrix}\begin{pmatrix}4 & -8 & -8 \\ -8 & 25 & 7 \\ -8 & 7 & 25\end{pmatrix}\begin{pmatrix}a\\ b \\ c\end{pmatrix}\triangleq \xi^T M\xi. $$Now, a calculation yields the eigenvalues of $M$ to be $0,18$, and $36$. Since they are all non-negative, $M$ is positive semidefinite; hence $z^TMz\geqslant 0$ for every $z\in\mathbb{R}^3$. We now study the equality case. For this, we rely on eigendecomposition of the matrix. A further inspection reveals that the eigenvectors $v_2,v_3$ corresponding respectively to the eigenvalues $18$ and $36$ are $$ v_2 = \begin{pmatrix}0\\-1 \\ 1\end{pmatrix}\quad\text{and}\quad v_3 = \begin{pmatrix}-1\\2 \\ 2\end{pmatrix}. $$Hence, $M=18v_2v_2^T + 36v_3v_3^T$, and thus $$ \xi^T M\xi = 18(-b+c)^2+36(-a+2b+2c)^2. $$In particular, for equality to hold, it must hold that $b=c$; and $-a+2b+2c=0$, that is, $a=4b=4c$, as claimed earlier.

Is this Titu? Let $BC=a,CA=b,AB=c$ and $s=(a+b+c)/2$. WLOG $a=\max(a,b,c)$. Then the given is \[\frac{6}{r}=\frac{2}{s-a}+\frac{5}{s-b}+\frac{5}{s-c}.\]Let $x=s-a,y=s-b,z=s-c$. By Heron's Formula, we have \[r=\frac{[ABC]}{s}=\sqrt{\frac{xyz}{x+y+z}}.\]Thus we can write \[\frac{6\sqrt{x+y+z}}{\sqrt{xyz}}=2/x+5/y+5/z.\]By scaling we can WLOG $x=1$. Then $6\sqrt{yz+(y+z)yz}=2yz+5y+5z$. Squaring yields \[36yz+36(y+z)yz=4y^2z^2+20(y+z)yz+50yz+25y^2+25z^2\implies 0=4y^2z^2-16(y+z)yz+14yz+25y^2+25z^2=\]\[(2yz-4y-4z)^2+(3y-3z)^2.\]Thus $y=z,2y^2=8y\implies y=z=4$. Thus $a=8,b=c=5$. Thus all possible triangles $ABC$ are similar to the isosceles $5-5-8$ triangle.

It seems that every problem on an olympiad which asks for solutions to a single real-number equation is an inequality. Here is a solution that does not rely on finding a sum of squares and expanding a large quadratic. I will first prove the inequality: $$\frac{2}{AP}+\frac{5}{BQ}+\frac{5}{CR} \geq \frac{6}{r}.$$We can see that if $s$ is the semiperimeter and $a,b,c$ are the side lengths as usual, we have $AP=s-c,BQ=s-b,CR=s-a$. Also, we have the well-known formula $A=rs \implies \tfrac{1}{r}=\tfrac{s}{A}$, where $A$ is the area. Further, by Heron's we have $A=\sqrt{s(s-a)(s-b)(s-c)}$. Substsituting $p=s-c,q=s-b,r=s-a$ (where $p,q,r \in \mathbb{R^+}$) and noting that $p+q+r=s$, the inequality becomes: $$\frac{2}{p}+\frac{5}{q}+\frac{5}{r} \geq 6\sqrt{\frac{p+q+r}{pqr}}.$$Now, making the substitution $\tfrac{1}{p}=x,\tfrac{1}{q}=y,\tfrac{1}{r}=z$, the inequality becomes: $$2x+5(y+z)\geq 6\sqrt{x(y+z)+yz},$$where $x,y,z \in \mathbb{R^+}$. To prove this inequality, we first perform a bit of smoothing. If we keep $y+z$ constant and vary $y,z$ individually, we can see that the LHS remains constant and the RHS attains its maximum value when $y=z$. As such we can prove the stronger inequality (obtained by putting $z=y$): $$2x+10y\geq 6\sqrt{2xy+y^2} \iff x+5y \geq 3\sqrt{2xy+y^2}.$$Squaring and simplifying, this is equivalent to: $$x^2-8xy+16y^2=(x-4y)^2 \geq 0,$$which is true. Further, equality holds if and only if $x=4y=4z$, so $\tfrac{1}{x}=\tfrac{1}{4y}=\tfrac{1}{4z} \implies 4p=q=r$. Let $p=k$, so $q=r=4k$. Then we have $s=9k$, from which we can recover $c=8k,a=b=5k$. Observe that this inequality and its equality case implies that the equation holds if and only if $\min\{AP,BQ,CR\}=AP$, and $a:b:c=5:5:8$. Since $a:b:c=5:5:8$ implies that $\min\{AP,BQ,CR\}=AP$ always, we obtain that the equation holds when $AB:BC:CA=5:5:8$ and permutations, so all triangles in $S$ are isosceles and similar to each other. $\blacksquare$

ew Let $x = s - a$, $y = s - b$, $z = s - c$, and we say that WLOG $x$ is the smallest. Then our equality reduces to \begin{align*} \frac 2 x + \frac 5 y + \frac 5 z = 6 \frac{\sqrt{x + y + z}}{\sqrt{xyz}}. \end{align*}Multiplying, we get \begin{align*} 2 \sqrt{ \frac{yz}{x}} + 5 \sqrt{\frac{xz}{y}} + 5 \sqrt{\frac{xy}{z}} = 6 \sqrt{x + y + z}. \end{align*}Noting that both sides are homogeneous, we set $x = 1$, which yields \begin{align*} 6 \sqrt{1 + y + z} &= 2 \sqrt{yz} + 5 \sqrt{\frac y z} + 5 \sqrt{\frac z y} \\ &= \frac{2yz + 5y + 5z}{\sqrt{yz}} \implies \\ (2yz + 5y + 5z)^2 &= 36yz(1 + y + z). \end{align*} However, observe that \begin{align*} (2yz + 5y + 5z)^2 - 36yz(1 + y + z) &= \\ 4y^2z^2 + 25y^2 + 25z^2 + 14yz - 16yz(y + z) &= \\ (2yz - 4y - 4z)^2 + (3y - 3z)^2 &\geq 0. \end{align*}This forces $y = z$ as well as $yz = 2y + 2z$, so our solution is $(x, y, z) = (1, 4, 4)$. If we solve for $a, b, c$ and then dehomogenize, we get $(8k, 5k, 5k)$ as our solution set.

WLOG let $AP\leq BQ\leq CR$. Then the condition (after some simplifying) is equivalent to $$2\tan \frac{A}{2}+5\tan \frac{B}{2}+5\tan \frac{C}{2}=6,$$and since $$\tan \frac{A}{2}=\frac{1}{\tan \left(\frac{B}{2}+\frac{C}{2}\right)}=\frac{1-\tan \tfrac{B}{2}\tan \tfrac{C}{2}}{\tan \tfrac{B}{2}+\tan \tfrac{C}{2}},$$the condition is now equivalent to $$\frac{2-2\tan \tfrac{B}{2}\tan \tfrac{C}{2}}{\tan \tfrac{B}{2}+\tan \tfrac{C}{2}}+5\tan \frac{B}{2}+5\tan \frac{C}{2}=6.$$ Set $x=3\tan \frac{B}{2}-1$ and $y=3\tan \frac{C}{2}-1$. We may verify that this condition is equivalent to $$5x^2+8xy+5y^2=0.$$Now suppose WLOG that $y\neq 0$; then we may divide this by $y^2$ to get $$5\left(\frac{x}{y}\right)^2+8\left(\frac{x}{y}\right)+5=0$$which has no solutions. Thus $x=0,y=0$, which easily works, so that $\tan \frac{B}{2},\tan \frac{C}{2}=\frac{1}{3}$. From here the rest is easy; the isosceles triangle in question has side lengths in the ratio $AB:BC:CA=5:8:5$ (and permutations!).

This is my solution (pretty similar to ChrisWren's one ) Lemma: We know that for a given triangle $\Delta ABC$ if $P,Q,R$ are the tangency points of the incircle with the sides $AB,BC,CA$ respectively,then $AP=s-a,BQ=s-b,CR=s-c$ where $a,b,c$ are side lengths of $AB,BC,CA$ respectively and $s$ is the semi-perimeter of the triangle . and also $\frac{r}{s-a}=tan(\frac{A}{2}),$ $\frac{r}{s-b}=tan(\frac{B}{2})$,$\frac{r}{s-c}=tan(\frac{C}{2})$ Claim:If a triangle $\Delta XYZ \in \mathbb{S}$ , then it is an isosceles triangle with sides $(5k,5k,8k) $ (in some order ) PROOF:Consider a triangle $\Delta ABC\in \mathbb{S}$ w.l.o.g, let $a\geq b\geq c$ which directly implies that $s-a \leq s-b \leq s-c $ Now as $\Delta ABC\in \mathbb{S}$, \[\frac{5}{s-a}+\frac{5}{s-b}+\frac{5}{s-c}-\frac{3}{s-a}=\frac{6}{r}\]\[\implies \frac{2r}{s-a}+\frac{5r}{s-b}+\frac{5r}{s-c}=6 \]\[\implies 2 tan \frac{A}{2}+5 tan \frac{B}{2}+5 tan \frac{C}{2}=6..(*)\]\[\implies 2 \frac{1}{cot(\frac{\pi}{2}-\frac{B+C}{2})}+5 tan \frac{B}{2}+5 tan \frac{C}{2}=6\]\[\implies \frac{2}{tan({\frac{B}{2}+\frac{C}{2})}}+5 (tan\frac{B}{2}+ tan\frac{C}{2})=6\]\[\implies \frac{2(1-tan\frac{B}{2}tan\frac{C}{2})}{tan\frac{B}{2}+ tan\frac{C}{2}}+5 (tan\frac{B}{2}+ tan\frac{C}{2})=6\]\[\implies \frac{2(1-xy)}{(x+y)}+5(x+y)=6\]\[\implies 2-2xy +5(x+y)^2=6(x+y)\](where $x= tan \frac{B}{2}$ and $y= tan \frac{C}{2}$) \[\implies 5x^2+5y^2+8xy+2-6x-6y=0\]\[\implies 5x^2+x(8y-6)+(5y^2-6y+2)=0 .....(i)\]\[\implies x=\frac{6-8y \pm \sqrt{36+64y^2-96y-100y^2+120y-40}}{10}\]\[\implies x=\frac{6-8y \pm \sqrt{-(6y-2)^2}}{10}\] Now note that as $(6y-2)^2\geq 0 $ and $x\in \mathbb{R}$ Discriminant of $(i)$ must be $0$ (otherwise $x\notin \mathbb{R}$) hence \[6y-2=0\implies y= \frac{1}{3} \implies x= \frac{6- \frac{8}{3}}{10}=\frac{1}{3}\implies \boxed{tan\frac{B}{2}=\tan\frac {C}{2}=\frac{1}{3}}....(ii)\]As $a\geq b\geq c$ ,so,$A\geq B \geq C$ which implies that both $B,C$ is less than $\frac{\pi}{2}$ hence \[\boxed {B=C}\]and the triangle $\Delta ABC$ is isosceles triangle. Putting the value of $tan \frac{B}{2} $ and $tan{\frac{C}{2}}$ in $(*)$ we get $tan \frac{A}{2}=\frac{4}{3}$ This implies that \[\frac{r}{s-a}:\frac{r}{s-b}:\frac{r}{s-c}:=4:1:1\]\[\frac{1}{s-a}:\frac{1}{s-b}:\frac{1}{s-c}:=4:1:1\]\[\implies (s-a):(s-b):(s-c)=1:4:4\]\[\implies s=9k \ \text{for some} \ k\in \mathbb{R} \implies \boxed{ a=8k,b=c=5k}\] Hence our claim is proved to be true.

Firstly, note that we have $AP = s-a = x, BQ = s-b = y, CR = s-c = z$ for values $x,y,z$. WLOG take $a \leq b \leq c$, so our equality rewrites to $$5\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) - \frac{3}{z} = \frac{5}{x} + \frac{5}{y} + \frac{2}{z} = \frac{6}{r}.$$But $r = \frac{\sqrt{sxyz}}{s}$, so we have $$\frac{5}{x} + \frac{5}{y} + \frac{2}{z} = \frac{6s}{\sqrt{sxyz}} = \frac{6\sqrt{s}}{\sqrt{xyz}}.$$Multiplying by $xyz$ gives $$2xy+5xz+5yz = 6\sqrt{sxyz}.$$But $s = x+y+z$, so squaring gives $$(2xy+5xz+5yz)^2 = 36xyz(x+y+z).$$Using the minimum-of-a-quadratic formula it can be shown that $(2xy+5xz+5yz)^2 - 36xyz(x+y+z)$ can only be $0$ when $x = y$. From there we can easily obtain $x = y = 4z$, and upon obtaining $a, b, c$ the similarity follows.

Assume without loss of generality $a=\max(a,b,c)$. We have $\frac2{s-a}+\frac5{s-b}+\frac5{s-c}=\frac6r$. Since $r=\sqrt{\frac{(s-a)(s-b)(s-c)}s}$, we have $$2(s-b)(s-c)+5(s-a)(s-c)+5(s-a)(s-b)=6\sqrt{s(s-a)(s-b)(s-c)}.$$Let $x=s-a$, $y=s-b$, and $z=s-c$. Then, $2yz+5xy+5xz=6\sqrt{xyz(x+y+z)}$. Squaring gives $$4y^2z^2+25x^2y^2+25x^2z^2+20xy^2z+20xyz^2+50x^2yz=36xyz(x+y+z),$$or $$4y^2z^2+25x^2y^2+25x^2z^2-16xy^2z-16xyz^2+14x^2yz=0.$$This is equivalent to $x^2(25y^2+14yz+25z^2)-16xyz(y+z)+4y^2z^2=0$. The discriminant is $256y^2z^2(y^2+2yz+z^2)-16y^2z^2(25y^2+14yz+25z^2)$. This is nonnegative when $16y^2+32yz+16z^2\geq25y^2+14yz+25z^2$, or $y=z$. Therefore, $64x^2y^2-32xy^3+4y^4=0$, which implies $y=4x$. This means $s=x+y+z=9x$, so $a=8x$, $b=5x$, and $c=5x$. Therefore, the triangle must be isosceles, and it is similar to the triangle with sides $5$, $5$, and $8$.

Let $u,v,w$ be $AP,BQ,CR$, respectively. WLOG assume $\min(u,v,w)=u.$ Since \[\sqrt{\frac{uvw}{u+v+w}}=A=rs=r(u+v+w),\]we have $r=\sqrt{\frac{xyz}{x+y+z}}.$ The equation in the problem becomes: \[\frac{2}{u}+\frac{5}{v}+\frac{5}{w}=6\sqrt{\frac{u+v+w}{uvw}}.\]When the equation is squared and multiplied by $(uvw)^2$, we get \[25u^2v^2+25w^2u^2+4w^2v^2+14u^2vw-16uvw^2-16uv^2w=0.\]Now we have $9(uv-wu)^2+4(2uv+2uw-vw)^2=0$ which implies that $uv=wu$, so $v=w$. Another thing it implies is $2uv+2uw=vw\implies 4uv=v^2$ so $v=4u.$ Thus, the lengths $u,v,w$ are in the ratio $4:4:1$ in some order, which makes the sides of the triangle have ratio $8:5:5$ in some order, and we're done.

Let $(AP, BQ, CR) = (x, y, z)$ respectively, and assume $x = \min(x, y, z)$. Then we wish to show that, \begin{align*} 5 \sum_{cyc} \frac1x = \frac 6r + \frac 3x \end{align*}Note that from Herons we have, \begin{align*} r &= \frac{[ABC]}{x+y+z}\\ &= \sqrt{\frac{xyz}{x+y+z}} \end{align*}Then we wish to find all solutions to, \begin{align*} 5 \left(\frac{xy+yz+xz}{xyz} \right) &= 6\sqrt{\frac{x+y+z}{xyz}} + \frac 3x\\ 5(xy + yz + xz) &= 6 \sqrt{xyz(x+y+z)} + 3yz\\ 5xy + 5xz + 3yz &= 6 \sqrt{xyz(x+y+z)}\\ 25x^2y^2 + 25x^2z^2 + 9y^2z^2 + 50x^2yz + 20xy^2z + 20xyz^2 &= 36x^2yz + 36xy^2z + 36xyz^2\\ x^2(25y^2 + 25z^2 + 14yz) - 16x(y^2z + yz^2) + 4y^2z^2 &= 0 \end{align*}Now from the discriminant we find we must have, \begin{align*} 16y^2 + 32yz + 16z^2 \geq 25y^2 + 14yz + 25z^2\\ 18yz \geq 9y^2 + 9z^2\\ 0 \geq (y-z)^2 \end{align*}whence we must have equality at $y = z$. Substituting this back we find $y= 4x$, from which we conclude $ABC$ has sides of the form $$a:b:c = 8x:5x:5x$$