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We use the standard technique for dealing with problems involving incircles. We let the incircle be tangent to the triangle at points which split the sides into segments of length $x$ and $y$, $x$ and $z$, and $y$ and $z$, such that the sides of the triangle have lengths $x+y$, $x+z$, $y+z$. Now we have right triangles with angles of $A/2$, $B/2$, $C/2$, and with legs $x$ and $r$, $y$ and $r$, and $z$ and $r$. Also note that $s = (2x+2x+2z)/2 = x+y+z$. We rewrite the equation $\left( \cot \dfrac{A}{2} \right)^2 + \left( 2\cot \dfrac{B}{2} \right)^2 + \left( 3\cot \dfrac{C}{2} \right)^2 = \left( \dfrac{6s}{7r} \right)^2$ as $\dfrac{z}{r}^2 + 4\dfrac{y}{r}^2 + 9\dfrac{x}{r}^2 = \left(\dfrac{6(x+y+z)}{7r}\right)^2$ We can multiply out by $r^2$ and simplify to $49(z^2+4y^2+9x^2) = 36(x+y+z)^2$ The most obvious approach is to expand, which we take: $49z^2 + 196y^2 + 441x^2 = 36x^2 + 36y^2 + 36z^2 + 72(xy+xz+yz)$ or $13z^2 + 160y^2 + 405x^2 = 72xy + 72xz + 72yz$ If we try to make squares, we'll need to break up the numbers $13,160,405$ into sums of squares. We can easily see that $13 = 2^2+3^3$ and after some computations that $160=12^2+4^2$ and $405 = 9^2+18^2$. Thus we can factor: $(2z-18x)^2 + (3z-12y)^2 + (9x-4y)^2 = 0$ And we conclude that $z=9x, z=4y, x=4/9y$ Putting $k=y/9$ we get $x=4k, z=36k, y=9k$ and the sides of the triangle must be $13k, 40k, 45k$. Thus the triangle is similar to the triangle with side lenghts $13,40,45$, and we have completed the problem.

Just an interesting side note to solve the $49(x^2+4y^2+9z^2) = 36(x+y+z)^2$ condition. We can apply Cauchy to get $49(x^2+4y^2+9z^2)\left(1+\frac{1}{4}+\frac{1}{9}\right) \ge 49(x+y+z)^2$ $49(x^2+4y^2+9z^2) \ge 36(x+y+z)^2$ So by the equality condition on Cauchy, we know $x = \frac{2y}{\frac{1}{2}} = \frac{3z}{\frac{1}{3}}$ $x = 4y = 9z$. $\frac{x}{36} = \frac{y}{9} = \frac{z}{4} = k$. So $x = 36k, y = 9k, z = 4k$, giving sides of $13k, 40k, 45k$.

Solution: We first prove a lemma. Lemma: $\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\frac{s}{r}$. Proof: Note that $\cot\frac{A}{2}=\frac{s-a}{r}$. We have similar relations for $\cot\frac{B}{2}$ and $\cot\frac{C}{2}$. Hence our sum is simply \[\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\frac{(s-a)+(s-b)+(s-c)}{r}=\frac{3s-2s}{r}=\frac{s}{r}. \] $\Box$ Now we employ Cauchy and our Lemma to get \[\left(\left(\cot\frac{A}{2}\right)^2+\left( 2\cot\frac{B}{2}\right)^2+\left( 3\cot\frac{C}{2}\right)^2\right) \left(1+\frac{1}{4}+\frac{1}{9}\right)\] \[\ge\left(\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}\right)^2=\frac{s^2}{r^2}.\] The left side of the inequality simplifies to \[\frac{49}{36}\left(\left(\cot\frac{A}{2}\right)^2+\left(2\cot\frac{B}{2}\right)^2+\left(3\cot\frac{C}{2}\right)^2\right),\] so our inequality is actually \[\left(\cot\frac{A}{2}\right)^2+\left(2\cot\frac{B}{2}\right)^2+\left( 3\cot\frac{C}{2}\right)^2\ge\frac{(6s)^2}{(7r)^2}. \] Since $\left(\cot\frac{A}{2}\right)^2+\left(2\cot\frac{B}{2}\right)^2+\left(3\cot\frac{C}{2}\right)^2=\frac{(6s)^2}{(7r)^2}$, the inequality is an equality case. Therefore, \[ \frac{\cot\frac{A}{2}}{\cot\frac{B}{2}}=4,\] \[\frac{\cot\frac{B}{2}}{\cot\frac{C}{2}}=\frac{9}{4}, \]and \[\frac{\cot\frac{C}{2}}{\cot\frac{A}{2}}=\frac{1}{9}. \] Since we proved that $\cot\frac{A}{2}=\frac{s-a}{r}$, our system simplifies to \[\frac{-a+b+c}{a-b+c}=4, \]\[\frac{a-b+c}{a+b-c}=\frac{9}{4}\]\[\frac{a+b-c}{-a+b+c}=\frac{1}{9}.\] This is equivalent to the three equations \[5a-5b+3c=0, \] \[5a+13b-13c=0, \]and \[5a+4b-5c=0. \]Subtracting the third from the second, we find that $\frac{9}{8}b=c$. Using this fact, we can reduce the first equation to \[5a-5b+\frac{27}{8}b=5a-\frac{13}{8}\implies \frac{40}{13}a=b. \]Therefore, our triple $(a,b,c)$ is \[(a,b,c)=\left(a,\frac{40}{13}a,\frac{45}{13}a\right). \] Clearly the minimal $a$ such that all sides are integers is $13$, so our triangle $\Delta ABC$ is similar to the triangle with side lengths $13, 40, 45$, which also has the smallest integer side lengths. We are done. $\blacksquare$

Wow. I was laughing at how un-geometric this problem looked, as it was the only geometry problem on the USAMO 2002. Then I actually tried it and realized it was not geometry. Let $x = s-a$, $y = s-b$, $z = s-c$ in the usual fashion, then the equation reads \[ x^2 + 4y^2 + 9z^2 = \left( \frac67(x+y+z) \right)^2. \]However, by Cauchy-Schwarz, we have \[ \left( 1 + \tfrac14 + \tfrac19 \right)\left( x^2 + 4y^2 + 9z^2 \right) \ge \left( x+y+z \right)^2 \]with equality if and only if $1 : \tfrac12 : \tfrac13 = x : 2y : 3z$, id est $x : y : z = 1 : \tfrac14 : \tfrac19 = 36 : 9 : 4$. This is equivalent to $y+z : z+x : x+y = 13 : 40 : 45$. $\blacksquare$ You can tell this is not a geometry problem because eliminate the cotangents right away to get an algebra problem... and then you realize the problem claims that one equation can determine three variables up to scaling, at which point you realize it has to be an inequality (otherwise degrees of freedom don't work). So of course, Cauchy-Schwarz...

Denote $x, y, z$ as the intouch lengths from $A, B, C$ respectively. The given equation is \[ x^2+4y^2+9z^2 = 36(x+y+z)^2 = (6x+6y+6z)^2 \]Additionally, Cauchy yields \[ (x^2+(2y)^2+(3z)^2) \ge (6x+6y+6z)^2 \]So equality holds in Cauchy, thus $x=6k, y=\frac{3k}{2}, z=\frac{2k}{3}$. Therefore the answer is $13:40:45$.

How about Jessen's inequality?

MithsApprentice wrote: Let $ABC$ be a triangle such that \[ \left( \cot \dfrac{A}{2} \right)^2 + \left( 2\cot \dfrac{B}{2} \right)^2 + \left( 3\cot \dfrac{C}{2} \right)^2 = \left( \dfrac{6s}{7r} \right)^2, \]where $s$ and $r$ denote its semiperimeter and its inradius, respectively. Prove that triangle $ABC$ is similar to a triangle $T$ whose side lengths are all positive integers with no common divisors and determine these integers. Let \[ \begin{cases} x=s-a, \\ y=s-b, \\ z=s-c. \end{cases} \]Then we have $$\left(x^2+(2y)^2+(3z)^2\right)\left(1+\frac{1}{4}+\frac{1}{9}\right) \ge (x+y+z)^2$$proving that \[ \left( \cot \dfrac{A}{2} \right)^2 + \left( 2\cot \dfrac{B}{2} \right)^2 + \left( 3\cot \dfrac{C}{2} \right)^2 \ge \left( \dfrac{6s}{7r} \right)^2, \]hence $x:y:z=36:9:4$. Consequently, our desired triangle has side lengths $13, 40, 45$.

We know $\cot(A/2) = \tfrac{s-a}{r}$, and letting $x=s-a,y=s-b,z=s-c$, we get \[ x^2+4y^2+9z^2 = \frac{36}{49}(x+y+z)^2. \]We can rewrite this as \[ (x^2+4y^2+9z^2)(k_1^2+k_2^2+k_3^2) = (k_1x+2k_2y+3k_3z)^2 \]where $k_1=6/7,k_2=3/7,k_3=2/7$, making use of the fact that $k_1^2+k_2^2+k_3^2=1$. However, this is the equality case of Cauchy, hence $\frac{x}{k_1}=\frac{2y}{k_2} = \frac{3z}{k_3}$. This tells us that the ratio of the sides of the triangles are rational numbers, implying the desired conclusion.

Let $x=s-a, y=s-b, z= s-c.$ Then the equation is \[x^2+4y^2+9z^2 = \frac{36}{49}(x+y+z)^2\]But by Cauchy-Schwarz, we know that \[\left( 1 + \frac{1}{4} + \frac{1}{9} \right)\left( x^2 + 4y^2 + 9z^2 \right) \ge (x+y+z)^2 .\]The equality case occurs when $x:y:z = 1 : \tfrac{1}{4} : \tfrac{1}{9}.$ If we scale this up by $36,$ we have $x:y:z = 36:9:4.$ Finally, using the definitions of $x,y,z$ tells us our triangle is one with side lengths $13,40,45.$ $\blacksquare$

Sorry if this is a stupid question but is the proof of similarity in the fact that we used a general form of the side lengths and the relations between $\cos \frac{x}{2}$ to the parts of a triangle to derive the similar triangle T?

baldeagle123 wrote: Sorry if this is a stupid question but is the proof of similarity in the fact that we used a general form of the side lengths and the relations between $\cos \frac{x}{2}$ to the parts of a triangle to derive the similar triangle T? I'm not sure If I understand your question, but what I did was use Cauchy to find triangle similar to $T$. We got that the ratio of the sides are $1:\tfrac14:\tfrac19.$ But: USAMO 2002/2 wrote: Prove that triangle $ABC$ is similar to a triangle $T$ whose side lengths are all positive integers with no common divisors and determine these integers. So we scale up $36$ so that this holds. Since all sides were scaled up by an equal amount, it remains similar.

Observe that \[s=r[\cot A/2+\cot B/2+\cot C/2].\]Let $\cot A/2=x,\cot B/2=y,\cot C/2=z$ so we have \[x^2+4y^2+9z^2=36(x+y+z)^2/49.\]Rearranging, we have \[13x^2+160y^2+405z^2=72[xy+yz+zx].\]Rewrite this equation as follows: \[(2x-18z)^2+(3x-12y)^2+(9z-4y)^2=0.\]Hence we have $y=x/4,z=x/9$. This implies $x:y:z=36:9:4$. Since \[\cot A/2+\cot B/2:\cot B/2+\cot C/2:\cot C/2+\cot A/2=c:a:b,\]we get $a:b:c=13:40:45$, so we are done.

We set $x = s - a$, $y = s - b$, and $z = s - c$ where $s$ is the semiperimeter. Recall that $\tan \left( \frac x 2 \right) = \pm \sqrt{\frac{1 - \cos x}{1 + \cos x}}$. Using these facts, we plug in and the equation reduces to basically \begin{align*} \frac{x}{yz} + \frac{4y}{xz} + \frac{9z}{xy} = \frac{36}{49} \cdot \frac{(x + y + z)^2}{xyz}, \end{align*}which simplifies to \begin{align*} x^2 + 4y^2 + 9z^2 = \frac{36}{49} (x + y + z)^2. \end{align*}However, observe the following application of Cauchy-Schwarz: \begin{align*} (x^2 + 4y^2 + 9z^2)(6^2 + 3^2 + 2^2) \geq 36(x + y + z)^2, \end{align*}so we know that for equality to hold we need $\frac x 6 = \frac{2y}{3} = \frac{3z}{2}$, or $x = 4y = 9z$. It's easy from here to check that the smallest integer solution for $(a, b, c)$ is $(13, 40, 45)$.

This is rather amusing - hence this post. Solved with L567. Let $x,y,z$ be $s-a, s-b,s-c$, respectively, and notice by Sine Rule in the triangles involving vertices, incenter and intouch points we have that $$49(x^2+4y^2+9z^2) = 36(x+y+z)^2$$which by the CS equality case means that $(x,y,z) = (4,9,36)$ and consequently we have that $\triangle ABC$ is similar to the triangle with side-lengths $13,40,45$. $\blacksquare$

Recall that $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ if $A+B+C=\pi$. Let $[ABC]$ be the area of triangle $ABC$. Notice that $$\frac{s}{r}=\frac{[ABC]^2}{sr^3}=\frac{s-a}{r}\cdot \frac{s-b}{r}\cdot \frac{s-c}{r}=\tan\left(90-\frac{A}{2}\right)\tan\left(90-\frac{B}{2}\right)\tan\left(90-\frac{C}{2}\right)=\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2},$$and also that $$\left(\frac{s}{r}\right)^2 = \left(\left( \cot \dfrac{A}{2} \right)^2 + \left( 2\cot \dfrac{B}{2} \right)^2 + \left( 3\cot \dfrac{C}{2} \right)^2 \right)\left(1+\frac{1}{4}+\frac{1}{9}\right) \geq \left(\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}\right)^2.$$Clearly equality is achieved, thus $\cot \frac{A}{2}=36n,\cot \frac{B}{2}=9n,\cot \frac{C}{2}=4n$ for some number $n$. But we also know that $36n+9n+4n=36n\cdot 9n\cdot 4n$, so $n=\frac{7}{36}$. The rest is easy, resulting in triangle $ABC$ being similar to a triangle with side lengths of $13$, $40$, and $45$.

USAMO 2002 P2 wrote: Let $ABC$ be a triangle such that \[ \left( \cot \dfrac{A}{2} \right)^2 + \left( 2\cot \dfrac{B}{2} \right)^2 + \left( 3\cot \dfrac{C}{2} \right)^2 = \left( \dfrac{6s}{7r} \right)^2, \]where $s$ and $r$ denote its semiperimeter and its inradius, respectively. Prove that triangle $ABC$ is similar to a triangle $T$ whose side lengths are all positive integers with no common divisors and determine these integers. Let $x=s-a$, $y=s-b$, and $z=s-c$, we have that $$\cot \frac{A}{2}=\frac{s-a}{r}=\frac{x}{r},\quad \cot \frac{B}{2}=\frac{s-b}{r} =\frac{y}{r}, \quad\cot \frac{C}{2}=\frac{s-c}{r}=\frac{z}{r}\implies$$$$x^2+4y^2+9z^2=\frac{36}{49}(x+y+z)^2$$By Cauchy-Schwarz Inequality we get that $$(x^2+4y^2+9z^2)(1+\frac{1}{2}+\frac{1}{3})\geq (x+y+z)^2\implies x^2+4y^2+9z^2\geq \frac{36}{49}(x+y+z)^2$$Since we know that these two expressions are equal we can consider the equality case: $$x:2y:3z=1:\frac{1}{2}:\frac{1}{3}\implies x:y:z=1:\frac{1}{4}:\frac{1}{9}=36:9:4\implies x+y:y+z:z+x=\boxed{45:11:40}.\blacksquare$$

Write $x=s-a, y=s-b, z=s-c$. Then, the expression becomes \[(\frac{x}{r})^2 + (2\cdot \frac{y}{r})^2 + (3\cdot \frac{z}{r})^2 = (\frac67 \cdot \frac{x+y+z}{r})^2\]But by Cauchy, \[x^2 + (2y)^2 + (3z)^2\cdot (1 + \frac12^2 + \frac13^2) =\lVert \begin{bmatrix} x \\ 2y \\ 3z\end{bmatrix} \rVert^2\cdot \lVert \begin{bmatrix} 1\\ \frac12 \\ \frac13\end{bmatrix} \rVert^2 \geq (x+y+z)^2\]Note that $1+\frac14+\frac19 = \frac{49}{36}$, so $\frac{x}{r})^2 + (2\cdot \frac{y}{r})^2 + (3\cdot \frac{z}{r})^2\geq \frac{36}{49} \frac{x+y+z}{r}^2$. Thus, the equality case of Cauchy must hold true, so $x:2y:3z = 1: \frac12: \frac13$, so $x:y:z = 1:\frac14 : \frac19 = 36:9:4$, and thus $BC:AC:AB = y+z:x+z:x+y = 13:40:45$ and we're done. $\blacksquare$.

Let $D,E,F$ be the contact points of the incircle on $BC,AC,AB.$ Let $AF=AE=u,BF=BD=v,CD=CE=w.$ We have: \begin{align*} \cot\frac{A}{2} &=\frac{u}{r} \\ 2\cot\frac{B}{2} &= \frac{2v}{r} \\ 3\cot\frac{C}{2} &= \frac{3w}{r} \end{align*}The equation becomes $\tfrac{u^2+4v^2+9w^2}{r^2}=\tfrac{36(u+v+w)^2}{49r^2}$ which rearranges to $13u^2+160v^2+405w^2-72wu-72uv-72vw=0.$ In fact, we can complete the square to get \[(2u-18w)^2+(3u-12v)^2+(9w-4v)^2=0\]Therefore, we conclude $u:v:w=36:9:4.$ Our triangle is $BC=13k,AC=40k,AB=45k.$

Using $\cot \frac A2 = \frac{s-a}r$, the equation is equivalent to $$(s-a)^2+4(s-b)^2+9(s-c)^2 = \frac{36s^2}{49}.$$But $$((s-a)^2+4(s-b)^2+9(s-c)^2)(6^2+3^2+2^2) \geq (18s-6(a+b+c))^2 = 36s^2,$$so equality must hold everywhere. Thus the side lengths must be in the ratio $13:40:45$.

Note that $\cot{\tfrac{A}2}=\tfrac{s-a}r$, $\cot{\tfrac{B}2}=\tfrac{s-b}r$, $\cot{\tfrac{C}2}=\tfrac{s-c}r$. Then, \begin{align*} \left(\frac{s-a}{r}\right)^2+\left(2 \cdot \frac{s-b}{r}\right)^2+\left(3 \cdot \frac{s-c}{r}\right)^2&=\left(\frac{6s}{7r}\right)^2\\ \implies (s-a)^2+4(s-b)^2+9(s-c)^2&=\frac{36}{49} s^2. \end{align*}But by Cauchy, \begin{align*} ((s-a)^2+4(s-b)^2+9(s-c)^2)\left(\left(\frac11\right)^2+\left(\frac12\right)^2+\left(\frac13\right)^2\right) &\ge s^2\\ \implies (s-a)^2+4(s-b)^2+9(s-c)^2 &\ge \frac{36}{49} s^2, \end{align*}so we have an equality. Therefore, $(s-a) : 2(s-b) : 3(s-c)=1 : \tfrac12 : \tfrac13$, so $(s-a) : (s-b) : (s-c)=36 : 9 : 4$. Hence, the answer is $(13, 40, 45)$.

$\color{magenta} \boxed {\textbf{SOLUTION P2}}$ $\color{red} \textbf{Geo Marabot Solve 6}$ The equation turns into, $$x^2 + 4y^2 + 9z^2 = \frac{36}{49} \left(x+y+z \right)^2$$By Cauchy-Schwarz Inequality, $$\left( 1 + \tfrac14 + \tfrac19 \right)\left( x^2 + 4y^2 + 9z^2 \right) \ge \left( x+y+z \right)^2 $$Equality holds if and only if $$1 : \tfrac12 : \tfrac13 = x : 2y : 3z \implies x : y : z = 1 : \tfrac14 : \tfrac19 = 36 : 9 : 4 \implies a : b : c= 13 : 40 : 45\blacksquare$$

Denote the side opposite $A$ by $a$, and define $b$ and $c$ similarly. Refer to the diagram, where $I$ is the incentre of triangle $ABC$, and $D$ the foot of the altitude from $I$ onto $AB$. Observe that $$\cot \frac{A}{2} = \frac{AD}{AI} = \frac{s-a}{r}.$$Similarly, $$\cot \frac{B}{2} = \frac{s-b}{r}, \cot \frac{C}{2} = \frac{s-c}{r}.$$Substituting these into the original equation gives us $$\left(\frac{s-a}{r} \right)^2 + \left(2 \cdot \frac{s-b}{r} \right) + \left( 3 \cdot \frac{s-c}{r} \right) = \left( \frac{6s}{7r} \right)^2$$$$\implies (s-a)^2 + 4(s-b)^2 + 9(s-c)^2 = \frac{36}{49} s^2.$$But observe that $$(s-a)^2 + 4(s-b)^2 + 9(s-c)^2 = \frac{(s-a)^2}{1} + \frac{(s-b)^2}{\frac{1}{4}} + \frac{(s-c)^2}{\frac{1}{9}}$$$$\ge \frac{(s-a+s-b+s-c)^2}{1+\frac{1}{4} + \frac{1}{9}}$$$$= \frac{s^2}{\frac{49}{36}}$$$$=\frac{36}{49}s^2,$$so equality must hold in the step with Titu's lemma, i. e. $s-a = 4(s-b) = 9(s-c)$. Simplifying, we get $b+c-a = 4(a+c-b) = 9(a+b-c)$. WLOG let $c = 1$. Then solving the resultant system of equations gives us $a = \frac{13}{45}, b = \frac{8}{9}$. Therefore, $a:b:c = \frac{13}{45}:\frac{40}{45}:1 = 13:40:45$, and so triangle $ABC$ is similar to one with sides $13, 40, 45$. $\square$

Let $I$ be the incenter of $ABC$, and $D, E, F$ be its $A$-, $B$- and $C$-intouch points respectively. Now let \[x = AE = AF, y = BF = BD, z = CD = CE\text{.}\]Therefore we have \begin{align*} \left(\frac{x}{r}\right)^2 + \left(\frac{2y}{r}\right)^2 + \left(\frac{3z}{r}\right)^2 &= \left(\frac{6(x + y + z)}{7r}\right)^2 \\ (x^2 + (2y)^2 + (3z)^2)\left(\frac{1}{1} + \frac{1}{4} + \frac{1}{9}\right) &= (x + y + z)^2 \end{align*}which is the equality case of Cauchy-Schwarz. Therefore \begin{align*} x : 2y : 3z &= 1 : \frac{1}{2} : \frac{1}{3} \\ x : y : z &= 1 : \frac{1}{4} : \frac{1}{9} \\ x : y : z &= 36 : 9 : 4\text{.} \end{align*}Now we have \begin{align*} AB : BC : CA &= (x + y) : (y + z) : (z + x) \\ &= 45 : 13 : 40\text{.} \end{align*}Therefore $ABC$ is similar to a triangle with side lengths $45$, $13$ and $40$.

Using the picture, the given equation simplifies into $a^2+(2b)^2+(3c)^2=\frac{(6a+6b+6c)^2}{49}$. By Cauchy-Schwarz, $(a^2+(2b)^2+(3c)^2)(6^2+3^2+2^2)\ge (6a+6b+6c)^2$ with equality iff $\frac{a}{6}=\frac{2b}{3}=\frac{3c}{2}$ so $a=36k$, $b=9k$, and $c=4k$, so triangle $ABC$ must be similar to a triangle $T$ with side lengths $\boxed{13, 40, 45}$.

Let the incircle of $\triangle ABC$ be tangent to $AB$, $BC$, and $CA$ at $X$, $Y$, and $Z$ such that $a = AZ = AX$, $b = BX = BY$, and $c = CY = CZ$. Then, \begin{align*} \cot \dfrac{A}{2} = \dfrac{a}{r}, \qquad \cot \dfrac{B}{2} = \dfrac{b}{r}, \qquad \cot \dfrac{C}{2} = \dfrac{c}{r}. \end{align*}Substituting to the given equation yields: \begin{align*} \left(\dfrac{a}{r}\right)^2 + \left(\dfrac{2b}{r}\right)^2 + \left(\dfrac{3c}{r}\right)^2 = \left(\dfrac{6\left(a+b+c\right)}{7r}\right)^2 \end{align*}But, by Cauchy Schwarz, note that \begin{align*} \left(1^2 + \left(\dfrac{1}{2}\right)^2 + \left(\dfrac{1}{3}\right)^2\right) \left(\left(\dfrac{a}{r}\right)^2 + \left(\dfrac{2b}{r}\right)^2 + \left(\dfrac{3c}{r}\right)^2\right) \geq \dfrac{(a+b+c)^2}{r^2}, \end{align*}but the equality case of this resulting inequality is equivalent to the given equation. Therefore, $a = 4b = 9c$. Then, $(a,b,c) = (36t, 9t, 4t)$. So, clearly $(a+b,$ $b+c,c+a) = (45t, 40t, 13t)$. Thus, $\triangle ABC$ is similar to a triangle with side lengths $13$, $40$, and $45$ which clearly are all integers whose greatest common divisor is one.

For brevity let $\cot \left(\dfrac{A}{2} \right) = x$ and define similarly for others. Note that by considering the segments that the intouch points partition the sides of $ABC$, we obtain the relation $$s = r(x + y + z) \implies x + y + z = \frac{s}{r}.$$Cauchy-Schwarz yields \[ \bigg[x^2 + (2y)^2 + (3z)^2 \bigg] \left[1 + \frac{1}{2^2} + \frac{1}{3^2} \right] \ge (x + y + z)^2 = \frac{s^2}{r^2} \]\[ \iff x^2 + (2y)^2 + (3z)^2 \ge \left( \frac{6s}{7r} \right)^2. \]The condition of the problem implies equality, so $x = 1$, $y = \dfrac{1}{4}$, $z = \dfrac{1}{9}$. The ratio of the sides of $ABC$ is $x + y : y + z : z + x = \dfrac{5}{4} : \dfrac{13}{36} : \dfrac{10}{9} = 45:13:40$ so our answer is $\{13, 40, 45 \}$.

Note that $\cot\left(\frac A2\right) = \frac{s-a}{r}$ (and similarly for $\cot\left(\frac B2\right), \cot\left(\frac C2\right)$). By Cauchy-Schwarz, we have: \begin{align*}\left(\left( \frac{s-a}{r}\right)^2 + \left(2\cdot \frac{s-b}{r}\right)^2+\left(3\cdot \frac{s-c}{r}\right)^2\right)(6^2+3^2+2^2)&\ge \left( \frac{6s}{r}\right)^2.\\ \implies \left( \frac{s-a}{r}\right)^2 + \left(2\cdot \frac{s-b}{r}\right)^2+\left(3\cdot \frac{s-c}{r}\right)^2&\ge \left( \frac{6s}{7r}\right)^2\\ \implies \left(\cot\frac A2\right) ^2 +\left(2\cot\frac B2\right)^2+\left(3\cot\frac C2\right)^2 &\ge \left( \frac{6s}{7r}\right)^2 \end{align*}Thus, we must be in the equality case of Cauchy, or \begin{align*}\cot\left(\frac A2\right) : 2\cot\left(\frac B2\right) : 3\cot\left(\frac C2\right) &= 6 : 3 : 2\\ \implies\cot\left(\frac A2\right) : \cot\left(\frac B2\right) : \cot\left(\frac C2\right) &= 1 : \frac 14 : \frac19. \end{align*}Therefore, the ratio of the side lengths must be $\frac 54 : \frac {10}9 : \frac {13}{36} = 45 : 40 : 13$. Our answer is $\boxed{13, 40, 45}$. $\blacksquare$

In search of a way to eliminate the coefficients in front of the $\cot^2 \frac A2$ terms, we apply Cauchy-Schwarz in the following manner: \[\left(1 + \frac14 + \frac19\right)\left(\left(\cot \frac A2\right)^2 + \left(2\cot \frac B2\right)^2 + \left(3\cot \frac C2\right)^2\right) \geq \left(\cot \frac A2 + \cot \frac B2 + \cot \frac C2\right)^2.\]This seems to suggest that $\cot \frac A2 + \cot \frac B2 + \cot \frac C2 = \frac sr$, since dividing both sides by $1 + \frac14 + \frac19 = \left(\frac76\right)^2$ gives \[\left(\frac67\left(\cot \frac A2 + \cot \frac B2 + \cot \frac C2\right)\right)^2\]on the RHS, which would imply equality in the above Cauchy application. This is in fact true, since \[r \cot \frac A2 + r \cot \frac B2 + r \cot \frac C2 = (s - a) + (s - b) + (s - c) = s\]by dropping the perpendicular from the incenter to each of the sides. Thus $\cot \frac A2 = 4\cot \frac B2 = 9\cot \frac C2$, and multiplying by $2r$ yields \[b + c - a = 4(c + a - b) = 9(a + b - c).\]Let $k$ denote the common value. Scaling and adding these quantities pairwise, we obtain $72a = 13k, 18b = 10k, 8c = 5k$, from which it's easy to discover that $(13, 40, 45)$ is the requested triple. So we're done. $\square$

cool Let the tangents to the incircle from $A, B, C$ have lengths $a, b, c$ respectively. Then our condition is equivalent to $$\frac{a^2}{r^2}+\frac{4b^2}{r^2}+\frac{9c^2}{r^2} = \frac{36(a+b+c)}{49r^2} \implies (a^2+4b^2+9c^2)(6^2+3^2+4^2)=(6a+6b+6c)^2,$$which is in fact the equality case of Cauchy-Schwartz. Thus, for some constant $k,$ we have that $$\frac{a}{6}=\frac{2b}{3}=\frac{3c}{2}=k \implies (a, b, c) = (6k, \frac32k, \frac23k) \implies (AB, BC, CA) = k\cdot (\frac{15}{2}, \frac{13}{6}, \frac{20}{3}).$$Therefore, the three side lengths are $(45, 13, 40).$

OOK THIS IS KICKED TO. \[\left(\left(\cot\frac A2\right)^2+\left(2\cot\frac B2\right)^2+\left(3\cot\frac C2\right)^2\right)\left(\left(\frac67\right)^2+\left(\frac37\right)^2+\left(\frac27\right)^2\right)\ge\left(\frac{6s}{7r}\right)^2\]BY CAUCHY . EQUALITY IS WHEN $s-a:s-b:s-c=36:9:4$ SO \[\boxed{a:b:c=13:40:45}\]SUSUS.

cj13609517288 wrote: OOK THIS IS KICKED TO. \[\left(\left(\cot\frac A2\right)^2+\left(2\cot\frac B2\right)^2+\left(3\cot\frac C2\right)^2\right)\left(\left(\frac67\right)^2+\left(\frac37\right)^2+\left(\frac27\right)^2\right)\ge\left(\frac{6s}{7r}\right)^2\]BY CAUCHY . EQUALITY IS WHEN $s-a:s-b:s-c=36:9:4$ SO \[\boxed{a:b:c=13:40:45}\]SUSUS. Orz how u so orz