Relation between sides and angles, Solutions of triangles Questions in English

(A) In $\triangle ABC$,by the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$.
Given $\angle C = 60^{\circ}$,so $\cos 60^{\circ} = \frac{1}{2} = \frac{a^2+b^2-c^2}{2ab}$.
This implies $ab = a^2+b^2-c^2$,or $c^2 = a^2+b^2-ab$.
Now,consider the expression $E = \frac{a}{b+c} + \frac{b}{c+a} = \frac{a(c+a) + b(b+c)}{(b+c)(c+a)} = \frac{ac+a^2+b^2+bc}{bc+ab+c^2+ac}$.
Substituting $a^2+b^2 = c^2+ab$ into the numerator:
$E = \frac{ac + (c^2+ab) + bc}{bc+ab+c^2+ac} = \frac{ac+c^2+ab+bc}{ac+ab+c^2+bc} = 1$.

(C) Let $\frac{b+c}{9}=\frac{c+a}{10}=\frac{a+b}{11}=k$.
Then $b+c=9k$,$c+a=10k$,and $a+b=11k$.
Adding these gives $2(a+b+c)=30k$,so $a+b+c=15k$.
Thus,$a=(a+b+c)-(b+c)=15k-9k=6k$,$b=(a+b+c)-(c+a)=15k-10k=5k$,and $c=(a+b+c)-(a+b)=15k-11k=4k$.
Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{25k^2+16k^2-36k^2}{2(5k)(4k)} = \frac{5k^2}{40k^2} = \frac{1}{8}$.
$\cos B = \frac{a^2+c^2-b^2}{2ac} = \frac{36k^2+16k^2-25k^2}{2(6k)(4k)} = \frac{27k^2}{48k^2} = \frac{9}{16}$.
$\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{36k^2+25k^2-16k^2}{2(6k)(5k)} = \frac{45k^2}{60k^2} = \frac{3}{4}$.
Therefore,$\frac{\cos A+\cos B}{\cos C} = \frac{\frac{1}{8}+\frac{9}{16}}{\frac{3}{4}} = \frac{\frac{2+9}{16}}{\frac{3}{4}} = \frac{11}{16} \times \frac{4}{3} = \frac{11}{12}$.
Hence,option $C$ is correct.

(B) In $\triangle ABC$,$AD$ is the median to $BC$,so $BD = DC = a/2$. Given $AD \perp AC$,in right-angled $\triangle ADC$,by Pythagoras theorem,$AD^2 + b^2 = (a/2)^2$,so $AD^2 = a^2/4 - b^2$.
By Apollonius theorem on $\triangle ABC$,$c^2 + b^2 = 2(AD^2 + (a/2)^2)$.
Substituting $AD^2$,we get $c^2 + b^2 = 2(a^2/4 - b^2 + a^2/4) = a^2 - 2b^2$,which implies $a^2 = 3b^2 + c^2$ (Eq. $iii$).
Now,consider the expression $3ca \cos A \cos C + 2a^2$.
Using cosine rule,$\cos A = (b^2 + c^2 - a^2)/(2bc)$ and $\cos C = (a^2 + b^2 - c^2)/(2ab)$.
Substituting these,the expression becomes $3ca \cdot [(b^2 + c^2 - a^2)/(2bc)] \cdot [(a^2 + b^2 - c^2)/(2ab)] + 2a^2$.
$= (3/4b^2) \cdot (b^2 + c^2 - a^2)(a^2 + b^2 - c^2) + 2a^2$.
Since $a^2 - c^2 = 3b^2$,we have $b^2 + c^2 - a^2 = -2b^2$ and $a^2 + b^2 - c^2 = 4b^2$.
$= (3/4b^2) \cdot (-2b^2)(4b^2) + 2a^2 = -6b^2 + 2a^2 = 2(a^2 - 3b^2)$.
From Eq. $iii$,$a^2 - 3b^2 = c^2$.
Thus,the expression equals $2c^2$.

(C) Given the ratio of sides as $a: b: c = 3: 5: 7$,let $a = 3x, b = 5x, c = 7x$.
Using the cosine rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.
Substituting the values:
$\cos A = \frac{(5x)^2 + (7x)^2 - (3x)^2}{2(5x)(7x)} = \frac{25x^2 + 49x^2 - 9x^2}{70x^2} = \frac{65x^2}{70x^2} = \frac{13}{14}$.
$\cos B = \frac{(3x)^2 + (7x)^2 - (5x)^2}{2(3x)(7x)} = \frac{9x^2 + 49x^2 - 25x^2}{42x^2} = \frac{33x^2}{42x^2} = \frac{11}{14}$.
Therefore,$\cos A + \cos B = \frac{13}{14} + \frac{11}{14} = \frac{24}{14} = \frac{12}{7}$.

(D) Let the sides of the triangle be $a, b, c$ where $a = 80$ and the angle $B = 60^{\circ}$.
Given $b + c = 90$.
Using the Law of Cosines: $b^2 = a^2 + c^2 - 2ac \cos(B)$.
Substituting the values: $b^2 = 80^2 + c^2 - 2(80)(c) \cos(60^{\circ})$.
Since $\cos(60^{\circ}) = 0.5$,we have $b^2 = 6400 + c^2 - 80c$.
Substitute $b = 90 - c$ into the equation: $(90 - c)^2 = 6400 + c^2 - 80c$.
$8100 - 180c + c^2 = 6400 + c^2 - 80c$.
$1700 = 100c$.
$c = 17$.
Then $b = 90 - 17 = 73$.
The sides are $80, 73, 17$. The shortest side is $17$.

(B) Given $b \cos \theta = c - a$,we have $\cos \theta = \frac{c - a}{b}$.
Since $\theta$ is an acute angle,$\sin \theta = \sqrt{1 - \cos^2 \theta} = \frac{\sqrt{b^2 - (c - a)^2}}{b}$.
Thus,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{b^2 - (c - a)^2}}{c - a}$.
Therefore,$(c - a) \tan \theta = \sqrt{b^2 - (c - a)^2} = \sqrt{b^2 - c^2 - a^2 + 2ac}$.
Using the cosine rule $b^2 = a^2 + c^2 - 2ac \cos B$,we substitute $b^2 - c^2 - a^2 = -2ac \cos B$.
So,$(c - a) \tan \theta = \sqrt{2ac - 2ac \cos B} = \sqrt{2ac(1 - \cos B)}$.
Using the identity $1 - \cos B = 2 \sin^2 \frac{B}{2}$,we get:
$(c - a) \tan \theta = \sqrt{2ac \cdot 2 \sin^2 \frac{B}{2}} = \sqrt{4ac \sin^2 \frac{B}{2}} = 2 \sqrt{ca} \sin \frac{B}{2}$.

(A) Given the sides $a: b: c = 4: 5: 6$,let $a = 4k$,$b = 5k$,and $c = 6k$ for some constant $k > 0$.
Using the Law of Cosines:
$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{(5k)^2 + (6k)^2 - (4k)^2}{2(5k)(6k)} = \frac{25 + 36 - 16}{60} = \frac{45}{60} = \frac{3}{4}$.
$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{(4k)^2 + (6k)^2 - (5k)^2}{2(4k)(6k)} = \frac{16 + 36 - 25}{48} = \frac{27}{48} = \frac{9}{16}$.
$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{(4k)^2 + (5k)^2 - (6k)^2}{2(4k)(5k)} = \frac{16 + 25 - 36}{40} = \frac{5}{40} = \frac{1}{8}$.
Now,find the ratio $\cos A : \cos B : \cos C = \frac{3}{4} : \frac{9}{16} : \frac{1}{8}$.
Multiply by the least common multiple of the denominators,which is $16$:
$\cos A : \cos B : \cos C = (\frac{3}{4} \times 16) : (\frac{9}{16} \times 16) : (\frac{1}{8} \times 16) = 12 : 9 : 2$.

(B) We have,\\ $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$ \\ $= (a^2 + b^2 - 2ab) \cos^2 \frac{C}{2} + (a^2 + b^2 + 2ab) \sin^2 \frac{C}{2}$ \\ $= (a^2 + b^2)(\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2}) - 2ab(\cos^2 \frac{C}{2} - \sin^2 \frac{C}{2})$ \\ $= (a^2 + b^2)(1) - 2ab \cos C$ \\ $= a^2 + b^2 - 2ab \cos C$ \\ Since by the Law of Cosines,$c^2 = a^2 + b^2 - 2ab \cos C$,\\ Therefore,the expression is equal to $c^2$.

(C) Given: Area $\Delta = 4\sqrt{5}$, $b = 6$, and $\tan \frac{B}{2} = \frac{\sqrt{5}}{4}$.
Using the formula $\tan \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$, where $s = \frac{a+b+c}{2}$.
We know $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$, so $(s-a)(s-c) = \frac{\Delta^2}{s(s-b)}$.
Substituting this into the tangent formula: $\tan^2 \frac{B}{2} = \frac{\Delta^2}{s^2(s-b)^2}$.
Given $\tan \frac{B}{2} = \frac{\sqrt{5}}{4}$, so $\tan^2 \frac{B}{2} = \frac{5}{16}$.
$\frac{5}{16} = \frac{(4\sqrt{5})^2}{s^2(s-6)^2} = \frac{80}{s^2(s-6)^2}$.
$s^2(s-6)^2 = \frac{80 \times 16}{5} = 256$.
$s(s-6) = 16$ or $s(s-6) = -16$ (impossible).
$s^2 - 6s - 16 = 0 \implies (s-8)(s+2) = 0$. Since $s > 0$, $s = 8$.
$s = \frac{a+b+c}{2} = 8 \implies a+c+6 = 16 \implies a+c = 10$.
Also, $\Delta = \frac{1}{2}ac \sin B = 4\sqrt{5}$.
Using $\tan \frac{B}{2} = \frac{\sqrt{5}}{4}$, $\cos B = \frac{1 - \tan^2(B/2)}{1 + \tan^2(B/2)} = \frac{1 - 5/16}{1 + 5/16} = \frac{11/16}{21/16} = \frac{11}{21}$.
$\sin B = \sqrt{1 - (11/21)^2} = \frac{\sqrt{441-121}}{21} = \frac{\sqrt{320}}{21} = \frac{8\sqrt{5}}{21}$.
$\frac{1}{2}ac \left(\frac{8\sqrt{5}}{21}\right) = 4\sqrt{5} \implies ac = 21$.
We have $a+c = 10$ and $ac = 21$. The roots of $x^2 - 10x + 21 = 0$ are $x = 3, 7$.
The sides are $3, 6, 7$. The smallest side is $3$.

(B) We know the formulas for the exradii of a triangle: $r = \frac{\Delta}{s-a}$,$r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Substituting these into the expression $\sqrt{\frac{r r_2}{r_3 r_1}}$:
$\sqrt{\frac{\left(\frac{\Delta}{s}\right) \left(\frac{\Delta}{s-b}\right)}{\left(\frac{\Delta}{s-c}\right) \left(\frac{\Delta}{s-a}\right)}} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$.
Using the half-angle formula $\tan^2(B/2) = \frac{(s-a)(s-c)}{s(s-b)}$,we get $\sqrt{\frac{(s-a)(s-c)}{s(s-b)}} = \tan(B/2)$.

(B) We know that $r_1 = \frac{\Delta}{s-a}$ and $r_3 = \frac{\Delta}{s-c}$.
So,$r_1 + r_3 = \Delta \left( \frac{1}{s-a} + \frac{1}{s-c} \right) = \Delta \left( \frac{s-c+s-a}{(s-a)(s-c)} \right) = \Delta \left( \frac{2s-a-c}{(s-a)(s-c)} \right)$.
Since $2s = a+b+c$,we have $2s-a-c = b$.
Thus,$r_1 + r_3 = \frac{\Delta b}{(s-a)(s-c)}$.
Also,$1 + \cos B = 1 + \frac{a^2+c^2-b^2}{2ac} = \frac{2ac+a^2+c^2-b^2}{2ac} = \frac{(a+c)^2-b^2}{2ac} = \frac{(a+c-b)(a+c+b)}{2ac} = \frac{(2s-2b)(2s)}{2ac} = \frac{2(s-b)s}{ac}$.
Substituting these into the expression:
$\frac{2(r_1+r_3)}{ac(1+\cos B)} = \frac{2 \cdot \frac{\Delta b}{(s-a)(s-c)}}{ac \cdot \frac{2s(s-b)}{ac}} = \frac{2 \Delta b}{2s(s-a)(s-b)(s-c)} = \frac{\Delta b}{s(s-a)(s-b)(s-c)}$.
Since $\Delta^2 = s(s-a)(s-b)(s-c)$,the expression becomes $\frac{\Delta b}{\Delta^2} = \frac{b}{\Delta}$.

(B) Given $r_1=4, r_2=8, r_3=24$.
We know that $\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{4} + \frac{1}{8} + \frac{1}{24} = \frac{6+3+1}{24} = \frac{10}{24} = \frac{5}{12}$.
Thus,$r = \frac{12}{5}$.
We know that $\Delta^2 = r r_1 r_2 r_3 = \frac{12}{5} \times 4 \times 8 \times 24 = \frac{9216}{5}$.
So,$\Delta = \sqrt{\frac{9216}{5}} = \frac{96}{\sqrt{5}}$.
Using $r = \frac{\Delta}{s}$,we get $s = \frac{\Delta}{r} = \frac{96}{\sqrt{5}} \times \frac{5}{12} = 8\sqrt{5}$.
Since $r_1 = \frac{\Delta}{s-a}$,we have $4 = \frac{96/\sqrt{5}}{8\sqrt{5}-a}$.
$4(8\sqrt{5}-a) = \frac{96}{\sqrt{5}} \Rightarrow 32\sqrt{5} - 4a = \frac{96}{\sqrt{5}}$.
$4a = 32\sqrt{5} - \frac{96}{\sqrt{5}} = \frac{160-96}{\sqrt{5}} = \frac{64}{\sqrt{5}}$.
$a = \frac{16}{\sqrt{5}}$.

(A) We know that $r = \frac{\Delta}{s}$,$r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Consider the expression $rr_1 + r_2 r_3 = \frac{\Delta^2}{s(s-a)} + \frac{\Delta^2}{(s-b)(s-c)}$.
Since $\Delta^2 = s(s-a)(s-b)(s-c)$,we have $\frac{\Delta^2}{s(s-a)} = (s-b)(s-c)$ and $\frac{\Delta^2}{(s-b)(s-c)} = s(s-a)$.
Thus,$rr_1 + r_2 r_3 = (s-b)(s-c) + s(s-a)$.
Expanding this,we get $s^2 - s(b+c) + bc + s^2 - sa = 2s^2 - s(a+b+c) + bc$.
Since $2s = a+b+c$,we have $2s^2 - s(2s) + bc = 2s^2 - 2s^2 + bc = bc$.
Therefore,$rr_1 + r_2 r_3 = bc$,which implies $bc - r_2 r_3 = rr_1$.

(B) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$ and $\Delta^2 = s(s-a)(s-b)(s-c)$.
Substituting these into the expression:
$\frac{r_2(r_1+r_3)}{\sqrt{r_1 r_2+r_2 r_3+r_3 r_1}} = \frac{\frac{\Delta}{s-b}(\frac{\Delta}{s-a} + \frac{\Delta}{s-c})}{\sqrt{\frac{\Delta^2}{(s-a)(s-b)} + \frac{\Delta^2}{(s-b)(s-c)} + \frac{\Delta^2}{(s-c)(s-a)}}}$
$= \frac{\frac{\Delta^2}{s-b} \cdot \frac{s-c+s-a}{(s-a)(s-c)}}{\Delta \sqrt{\frac{s-c+s-a+s-b}{(s-a)(s-b)(s-c)}}}$
$= \frac{\Delta \cdot b}{(s-a)(s-b)(s-c)} \cdot \sqrt{\frac{(s-a)(s-b)(s-c)}{3s-(a+b+c)}}$
Since $a+b+c = 2s$,the denominator inside the square root is $3s-2s = s$.
$= \frac{\Delta \cdot b}{(s-a)(s-b)(s-c)} \cdot \frac{\sqrt{(s-a)(s-b)(s-c)}}{\sqrt{s}}$
$= \frac{\Delta \cdot b}{\sqrt{s(s-a)(s-b)(s-c)}} = \frac{\Delta \cdot b}{\Delta} = b$.

(B) We know that $r_2 = \frac{\Delta}{s-b}$ and $r_3 = \frac{\Delta}{s-c}$.
Also,$\sin^2 \frac{A}{2} = \frac{(s-b)(s-c)}{bc}$.
Thus,$(r_2+r_3) \operatorname{cosec}^2 \frac{A}{2} = \left(\frac{\Delta}{s-b} + \frac{\Delta}{s-c}\right) \times \frac{bc}{(s-b)(s-c)}$.
$= \Delta \left(\frac{s-c+s-b}{(s-b)(s-c)}\right) \times \frac{bc}{(s-b)(s-c)} = \Delta \left(\frac{a}{(s-b)(s-c)}\right) \times \frac{bc}{(s-b)(s-c)}$.
$= \frac{\Delta abc}{(s-b)^2(s-c)^2} = \frac{4R \Delta^2}{(s-b)^2(s-c)^2}$.
$= 4R \left(\frac{\Delta}{(s-b)(s-c)}\right)^2 = 4R \left(\cot \frac{A}{2}\right)^2 = 4R \cot^2 \frac{A}{2}$.

(A) Given $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 3 : 7 : 9$.
Using the formula $\cot \frac{A}{2} = \frac{s(s-a)}{\Delta}$,we have:
$\frac{s(s-a)}{\Delta} : \frac{s(s-b)}{\Delta} : \frac{s(s-c)}{\Delta} = 3 : 7 : 9$.
Multiplying by $\frac{\Delta}{s}$,we get $(s-a) : (s-b) : (s-c) = 3 : 7 : 9$.
Let $s-a = 3k$,$s-b = 7k$,and $s-c = 9k$.
Adding these,we get $3s - (a+b+c) = 19k$. Since $a+b+c = 2s$,we have $3s - 2s = s = 19k$.
Now,$a = s - 3k = 19k - 3k = 16k$.
$b = s - 7k = 19k - 7k = 12k$.
$c = s - 9k = 19k - 9k = 10k$.
Thus,$a : b : c = 16k : 12k : 10k = 16 : 12 : 10 = 8 : 6 : 5$.

(C) Given,in $\triangle ABC$,$r_1 = 2r_2 = 3r_3$.
Using the formula $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,we have:
$\frac{\Delta}{s-a} = 2 \frac{\Delta}{s-b} = 3 \frac{\Delta}{s-c} = k$ (let).
Then,$s-a = \frac{1}{k}$,$s-b = \frac{2}{k}$,and $s-c = \frac{3}{k}$.
Adding these equations:
$(s-a) + (s-c) = \frac{1}{k} + \frac{3}{k} = \frac{4}{k}$.
Since $s-b = \frac{2}{k}$,we have $\frac{4}{k} = 2(s-b)$.
Thus,$2s - a - c = 2s - 2b$,which simplifies to $a+c = 2b$.

(A) We know that in $\triangle ABC$,$\tan \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$,$\tan \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$,and $\tan \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$.
Substituting these values into the expression:
$\left(\tan \frac{A}{2} + \tan \frac{B}{2}\right) \tan \frac{C}{2} = \tan \frac{A}{2} \tan \frac{C}{2} + \tan \frac{B}{2} \tan \frac{C}{2}$
$= \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \cdot \sqrt{\frac{(s-a)(s-b)}{s(s-c)}} + \sqrt{\frac{(s-a)(s-c)}{s(s-b)}} \cdot \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$
$= \frac{s-b}{s} + \frac{s-a}{s}$
$= \frac{2s - a - b}{s}$
Since $2s = a + b + c$,we have $2s - a - b = c$ and $s = \frac{a+b+c}{2}$.
$= \frac{c}{\left(\frac{a+b+c}{2}\right)} = \frac{2c}{a+b+c}$.

(A) Given: $(a-b)(s-c)=(b-c)(s-a)$
Since $a = (s-b) + (s-c)$,$b = (s-a) + (s-c)$,and $c = (s-a) + (s-b)$,we have $(a-b) = (s-b) - (s-a)$ and $(b-c) = (s-c) - (s-b)$.
Substituting these into the equation:
$((s-b)-(s-a))(s-c) = ((s-c)-(s-b))(s-a)$
$(s-b)(s-c) - (s-a)(s-c) = (s-c)(s-a) - (s-b)(s-a)$
Rearranging terms:
$2(s-a)(s-c) = (s-b)(s-c) + (s-b)(s-a)$
Dividing both sides by $(s-a)(s-b)(s-c)$:
$\frac{2}{s-b} = \frac{1}{s-a} + \frac{1}{s-c}$
Multiplying by $\Delta$ (area of the triangle):
$\frac{2\Delta}{s-b} = \frac{\Delta}{s-a} + \frac{\Delta}{s-c}$
Since $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,we get:
$2r_2 = r_1 + r_3$
Thus,$r_1, r_2, r_3$ are in Arithmetic Progression.

(A) Given $\tan \frac{A}{2} = \frac{1}{\sqrt{2}}$.
Using the formula $\cos A = \frac{1-\tan^2 \frac{A}{2}}{1+\tan^2 \frac{A}{2}}$:
$\cos A = \frac{1 - (\frac{1}{\sqrt{2}})^2}{1 + (\frac{1}{\sqrt{2}})^2} = \frac{1 - 1/2}{1 + 1/2} = \frac{1/2}{3/2} = \frac{1}{3}$.
Using the Law of Cosines: $\cos A = \frac{b^2+c^2-a^2}{2bc}$.
$\frac{1}{3} = \frac{5^2 + 6^2 - a^2}{2(5)(6)}$.
$\frac{1}{3} = \frac{25 + 36 - a^2}{60}$.
$20 = 61 - a^2$.
$a^2 = 41$.
$a = \sqrt{41}$.

(C) We know that the area $R = \frac{1}{2}ab \sin C$,so $2R = ab \sin C$,which implies $4R^2 = a^2b^2 \sin^2 C$.
Substituting this into the first term: $\sqrt{a^2b^2 - 4R^2} = \sqrt{a^2b^2 - a^2b^2 \sin^2 C} = \sqrt{a^2b^2(1 - \sin^2 C)} = \sqrt{a^2b^2 \cos^2 C} = ab \cos C$.
Similarly,$\sqrt{b^2c^2 - 4R^2} = bc \cos A$ and $\sqrt{c^2a^2 - 4R^2} = ca \cos B$.
Using the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$,so $ab \cos C = \frac{a^2+b^2-c^2}{2}$.
Thus,the expression becomes: $\frac{a^2+b^2-c^2}{2} + \frac{b^2+c^2-a^2}{2} + \frac{c^2+a^2-b^2}{2} = \frac{a^2+b^2+c^2}{2}$.

(C) Given $2 \Delta^2 = \frac{a^2 b^2 c^2}{a^2+b^2+c^2}$.
Using the relation $\Delta = \frac{abc}{4R}$,we have $a^2 b^2 c^2 = (4R \Delta)^2 = 16 R^2 \Delta^2$.
Substituting this into the given equation: $2 \Delta^2 = \frac{16 R^2 \Delta^2}{a^2+b^2+c^2}$.
Dividing by $2 \Delta^2$,we get $a^2+b^2+c^2 = 8 R^2$.
Using $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$,we get $4R^2(\sin^2 A + \sin^2 B + \sin^2 C) = 8R^2$.
Thus,$\sin^2 A + \sin^2 B + \sin^2 C = 2$.
Using the identity $\sin^2 A + \sin^2 B + \sin^2 C = 2 + 2 \cos A \cos B \cos C$,we have $2 + 2 \cos A \cos B \cos C = 2$.
This implies $2 \cos A \cos B \cos C = 0$,so $\cos A = 0$ or $\cos B = 0$ or $\cos C = 0$.
Therefore,one of the angles must be $90^{\circ}$,which means the triangle is right-angled.

(A) Given expression: $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$
$= (a^2 - 2ab + b^2) \cos^2 \frac{C}{2} + (a^2 + 2ab + b^2) \sin^2 \frac{C}{2}$
$= (a^2 + b^2)(\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2}) + 2ab(\sin^2 \frac{C}{2} - \cos^2 \frac{C}{2})$
Since $\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2} = 1$ and $\cos C = \cos^2 \frac{C}{2} - \sin^2 \frac{C}{2}$,we have:
$= (a^2 + b^2)(1) - 2ab(\cos C)$
Using the Law of Cosines,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$:
$= a^2 + b^2 - 2ab \left( \frac{a^2 + b^2 - c^2}{2ab} \right)$
$= a^2 + b^2 - (a^2 + b^2 - c^2)$
$= c^2$

(B) Let $\triangle ABC$ be a triangle such that $\angle A: \angle B: \angle C = 1: 2: 3$.
Let the ratio constant be $x$,then $\angle A = x, \angle B = 2x, \angle C = 3x$.
Using the angle sum property of a triangle:
$\angle A + \angle B + \angle C = 180^{\circ}$
$x + 2x + 3x = 180^{\circ}$ $\Rightarrow 6x = 180^{\circ}$ $\Rightarrow x = 30^{\circ}$.
Thus,$\angle A = 30^{\circ}, \angle B = 60^{\circ}, \angle C = 90^{\circ}$.
Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$:
$\frac{a}{\sin 30^{\circ}} = \frac{b}{\sin 60^{\circ}} = \frac{c}{\sin 90^{\circ}}$
$\frac{a}{1/2} = \frac{b}{\sqrt{3}/2} = \frac{c}{1}$
Multiplying by $1/2$,we get $a: b: c = 1: \sqrt{3}: 2$.

(C) Given,$s(s-a) = (s-b)(s-c)$.
We know that $\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$ and $\cos \frac{A}{2} = \sqrt{\frac{s(s-a)}{bc}}$.
Squaring both,we get $\sin^2 \frac{A}{2} = \frac{(s-b)(s-c)}{bc}$ and $\cos^2 \frac{A}{2} = \frac{s(s-a)}{bc}$.
Since $s(s-a) = (s-b)(s-c)$,it follows that $\sin^2 \frac{A}{2} = \cos^2 \frac{A}{2}$.
Dividing by $\cos^2 \frac{A}{2}$,we get $\tan^2 \frac{A}{2} = 1$.
Since $\frac{A}{2}$ is an angle of a triangle,$\frac{A}{2} = \frac{\pi}{4}$,which implies $A = \frac{\pi}{2}$.

(A) Given that in a $\triangle ABC$,$\tan(A/2)$,$\tan(B/2)$,and $\tan(C/2)$ are in Arithmetic Progression $(AP)$.
Therefore,$2 \tan(B/2) = \tan(A/2) + \tan(C/2)$.
Using the identity $\tan(A/2) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$,this can be solved,but a more direct trigonometric approach is:
$\tan(B/2) - \tan(A/2) = \tan(C/2) - \tan(B/2)$
$\frac{\sin(B/2)}{\cos(B/2)} - \frac{\sin(A/2)}{\cos(A/2)} = \frac{\sin(C/2)}{\cos(C/2)} - \frac{\sin(B/2)}{\cos(B/2)}$
$\frac{\sin(B/2-A/2)}{\cos(B/2)\cos(A/2)} = \frac{\sin(C/2-B/2)}{\cos(C/2)\cos(B/2)}$
$\frac{\sin((B-A)/2)}{\cos(A/2)} = \frac{\sin((C-B)/2)}{\cos(C/2)}$
Using $A/2 = 90^{\circ} - (B+C)/2$ and $C/2 = 90^{\circ} - (A+B)/2$,we get:
$\frac{\sin((B-A)/2)}{\sin((B+C)/2)} = \frac{\sin((C-B)/2)}{\sin((A+B)/2)}$
$\sin((A+B)/2)\sin((B-A)/2) = \sin((B+C)/2)\sin((C-B)/2)$
Using $2\sin x \sin y = \cos(x-y) - \cos(x+y)$:
$\cos A - \cos B = \cos B - \cos C$
$\cos A + \cos C = 2\cos B$
Thus,$\cos A$,$\cos B$,and $\cos C$ are in Arithmetic Progression.

(A) In $\triangle ABC$,the sum of angles is $180^{\circ}$. Thus,$\angle B = 180^{\circ} - (75^{\circ} + 60^{\circ}) = 45^{\circ}$.
Since $\triangle BAD$ and $\triangle BCD$ share the same altitude from vertex $B$ to the base $AC$,the ratio of their areas is equal to the ratio of their bases: $\frac{\text{Area}(\triangle BAD)}{\text{Area}(\triangle BCD)} = \frac{AD}{CD} = \sqrt{3}$.
Let $\angle ABD = \alpha$,then $\angle DBC = 45^{\circ} - \alpha$.
Using the sine rule in $\triangle BAD$ and $\triangle BCD$:
$\frac{AD}{\sin \alpha} = \frac{BD}{\sin 75^{\circ}}$ and $\frac{CD}{\sin(45^{\circ} - \alpha)} = \frac{BD}{\sin 60^{\circ}}$.
Dividing these equations gives $\frac{AD}{CD} = \frac{\sin \alpha}{\sin(45^{\circ} - \alpha)} \cdot \frac{\sin 60^{\circ}}{\sin 75^{\circ}} = \sqrt{3}$.
$\frac{\sin \alpha}{\sin(45^{\circ} - \alpha)} = \sqrt{3} \cdot \frac{\sin 75^{\circ}}{\sin 60^{\circ}} = \sqrt{3} \cdot \frac{(\sqrt{6} + \sqrt{2})/4}{\sqrt{3}/2} = \frac{\sqrt{6} + \sqrt{2}}{2} = \frac{\sqrt{2}(\sqrt{3} + 1)}{2} = \frac{\sqrt{3} + 1}{\sqrt{2}}$.
Solving $\frac{\sin \alpha}{\sin 45^{\circ} \cos \alpha - \cos 45^{\circ} \sin \alpha} = \frac{\sqrt{3} + 1}{\sqrt{2}}$ leads to $\cot \alpha = \sqrt{3}$,hence $\alpha = 30^{\circ}$.

(B) The area of $\triangle ABC$ is given by $\Delta = \frac{1}{2} a \alpha = \frac{1}{2} b \beta = \frac{1}{2} c \gamma$.
Thus,$\alpha = \frac{2\Delta}{a}$,$\beta = \frac{2\Delta}{b}$,and $\gamma = \frac{2\Delta}{c}$.
Substituting these into the expression:
$\alpha^{-2} + \beta^{-2} + \gamma^{-2} = \frac{a^2}{4\Delta^2} + \frac{b^2}{4\Delta^2} + \frac{c^2}{4\Delta^2} = \frac{a^2+b^2+c^2}{4\Delta^2}$.
Using the identity $\cot A = \frac{b^2+c^2-a^2}{4\Delta}$,we have $b^2+c^2-a^2 = 4\Delta \cot A$.
Summing these for $A, B, C$:
$(b^2+c^2-a^2) + (c^2+a^2-b^2) + (a^2+b^2-c^2) = a^2+b^2+c^2 = 4\Delta(\cot A + \cot B + \cot C)$.
Therefore,$\frac{a^2+b^2+c^2}{4\Delta^2} = \frac{4\Delta(\cot A + \cot B + \cot C)}{4\Delta^2} = \frac{1}{\Delta}(\cot A + \cot B + \cot C)$.

(B) We know that $r_2 = \frac{\Delta}{s-b}$ and $r_3 = \frac{\Delta}{s-c}$.
Also,$\cot \left(\frac{B+C}{2}\right) = \tan \left(\frac{A}{2}\right) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$.
Substituting these values:
$(r_2 + r_3) \cot \left(\frac{B+C}{2}\right) = \left( \frac{\Delta}{s-b} + \frac{\Delta}{s-c} \right) \tan \left(\frac{A}{2}\right)$
$= \Delta \left( \frac{s-c+s-b}{(s-b)(s-c)} \right) \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$
$= \Delta \left( \frac{a}{(s-b)(s-c)} \right) \frac{\sqrt{(s-b)(s-c)}}{\sqrt{s(s-a)}}$
$= \frac{\Delta \cdot a}{\sqrt{s(s-a)(s-b)(s-c)}}$
Since $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$,the expression simplifies to:
$= \frac{\Delta \cdot a}{\Delta} = a$.
Thus,option $B$ is correct.

(D) Given,$\angle A=75^{\circ}, \angle B=45^{\circ}$ and $a=2(\sqrt{3}+1)$.
In $\triangle AOC$,$\tan 60^{\circ} = \frac{x}{y}$ $\Rightarrow \sqrt{3} = \frac{x}{y}$ $\Rightarrow x = \sqrt{3}y$.
Now,$x+y = 2(\sqrt{3}+1)$.
Substituting $x = \sqrt{3}y$,we get $\sqrt{3}y + y = 2(\sqrt{3}+1)$ $\Rightarrow y(\sqrt{3}+1) = 2(\sqrt{3}+1)$ $\Rightarrow y = 2$.
Then,$x = 2\sqrt{3}$.
Now,the area of $\triangle ABC = \text{Area of } \triangle AOB + \text{Area of } \triangle AOC$.
Area $= \frac{1}{2} \times x \times x + \frac{1}{2} \times x \times y = \frac{1}{2}x(x+y)$.
Area $= \frac{1}{2} \times (2\sqrt{3}) \times (2\sqrt{3} + 2) = \sqrt{3} \times 2(\sqrt{3}+1) = 2(3 + \sqrt{3}) = 6 + 2\sqrt{3} \text{ sq units}$.

(B) Given,$3a = b + c$ ... $(i)$
Let $s$ be the semi-perimeter of $\triangle ABC$,so $s = \frac{a + b + c}{2}$.
Substituting the value from $(i)$,we get $s = \frac{a + 3a}{2} = \frac{4a}{2} = 2a$.
We know that $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.
Therefore,$\cot \frac{B}{2} \cot \frac{C}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)} \cdot \frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-a)^2}} = \frac{s}{s-a}$.
Substituting $s = 2a$,we get $\frac{2a}{2a - a} = \frac{2a}{a} = 2$.

(D) Given,the area of triangle $\Delta = b^2-(c-a)^2$.
Using the identity $x^2-y^2 = (x-y)(x+y)$,we have $\Delta = (b-c+a)(b+c-a)$.
Since $2s = a+b+c$,we have $b-c+a = 2s-2c$ and $b+c-a = 2s-2a$.
Thus,$\Delta = (2s-2c)(2s-2a) = 4(s-a)(s-c)$.
By Heron's formula,$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.
Equating the two expressions: $\sqrt{s(s-a)(s-b)(s-c)} = 4(s-a)(s-c)$.
Dividing both sides by $\sqrt{(s-a)(s-c)}$,we get $\sqrt{s(s-b)} = 4\sqrt{(s-a)(s-c)}$.
Therefore,$\tan(\frac{B}{2}) = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}} = \frac{1}{4}$.
Using the double angle formula $\tan B = \frac{2\tan(B/2)}{1-\tan^2(B/2)}$,we get $\tan B = \frac{2(1/4)}{1-(1/4)^2} = \frac{1/2}{1-1/16} = \frac{1/2}{15/16} = \frac{8}{15}$.

(D) Given in a $\triangle ABC$:
$r_1 = \frac{\Delta}{s-a} = 36, r_2 = \frac{\Delta}{s-b} = 18, r_3 = \frac{\Delta}{s-c} = 12$
$\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{36} + \frac{1}{18} + \frac{1}{12} = \frac{1+2+3}{36} = \frac{6}{36} = \frac{1}{6}$
Since $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{s}{\Delta}$,we have $\frac{s}{\Delta} = \frac{1}{6} \Rightarrow \Delta = 6s$
$r_1 = \frac{6s}{s-a} = 36$ $\Rightarrow 6s = 36s - 36a$ $\Rightarrow 36a = 30s$ $\Rightarrow a = \frac{5s}{6}$
$r_2 = \frac{6s}{s-b} = 18$ $\Rightarrow 6s = 18s - 18b$ $\Rightarrow 18b = 12s$ $\Rightarrow b = \frac{2s}{3}$
$r_3 = \frac{6s}{s-c} = 12$ $\Rightarrow 6s = 12s - 12c$ $\Rightarrow 12c = 6s$ $\Rightarrow c = \frac{s}{2}$
Using Heron's formula $\Delta^2 = s(s-a)(s-b)(s-c) = (6s)^2 = 36s^2$
$s(s - \frac{5s}{6})(s - \frac{2s}{3})(s - \frac{s}{2}) = 36s^2$
$s(\frac{s}{6})(\frac{s}{3})(\frac{s}{2}) = 36s^2$
$\frac{s^4}{36} = 36s^2$ $\Rightarrow s^2 = 36^2$ $\Rightarrow s = 36$
$a = \frac{5 \times 36}{6} = 30$
$b = \frac{2 \times 36}{3} = 24$
$a+b = 30+24 = 54$

(B) We know that $\cot \frac{A}{2} = \frac{s(s-a)}{\Delta}$,where $s = \frac{a+b+c}{2}$ and $\Delta$ is the area of the triangle.
Thus,$\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \frac{s(s-a) + s(s-b) + s(s-c)}{\Delta} = \frac{s(3s - (a+b+c))}{\Delta} = \frac{s(3s - 2s)}{\Delta} = \frac{s^2}{\Delta} = \frac{(a+b+c)^2}{4\Delta}$.
Also,$\cot A + \cot B + \cot C = \frac{a^2+b^2+c^2}{4\Delta}$.
Therefore,the ratio is $\frac{(a+b+c)^2}{a^2+b^2+c^2}$.
Substituting $a=3, b=4, c=6$:
Ratio $= \frac{(3+4+6)^2}{3^2+4^2+6^2} = \frac{13^2}{9+16+36} = \frac{169}{61}$.

(A) Given the equation: $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$
Using the half-angle formulas $\cos^2 \frac{C}{2} = \frac{s(s-c)}{ab}$ and $\cos^2 \frac{A}{2} = \frac{s(s-a)}{bc}$:
$a \left( \frac{s(s-c)}{ab} \right) + c \left( \frac{s(s-a)}{bc} \right) = \frac{3b}{2}$
$\frac{s(s-c)}{b} + \frac{s(s-a)}{b} = \frac{3b}{2}$
$\frac{s}{b} (s - c + s - a) = \frac{3b}{2}$
Since $2s = a + b + c$,we have $2s - a - c = b$:
$\frac{s}{b} (b) = \frac{3b}{2} \Rightarrow s = \frac{3b}{2}$
$\frac{a + b + c}{2} = \frac{3b}{2} \Rightarrow a + b + c = 3b$
$a + c = 2b$

(D) Let $\frac{s-a}{4} = \frac{s-b}{5} = \frac{s-c}{6} = k$.
Then $s-a = 4k$,$s-b = 5k$,and $s-c = 6k$.
Adding these gives $3s - (a+b+c) = 15k$. Since $a+b+c = 2s$,we have $3s - 2s = 15k$,so $s = 15k$.
Then $a = s - 4k = 11k$,$b = s - 5k = 10k$,and $c = s - 6k = 9k$.
We know $\sin^2\left(\frac{A}{2}\right) = \frac{(s-b)(s-c)}{bc}$.
Thus,$\sum \sin^2\left(\frac{A}{2}\right) = \frac{(s-b)(s-c)}{bc} + \frac{(s-c)(s-a)}{ca} + \frac{(s-a)(s-b)}{ab}$.
Substituting the values:
$= \frac{(5k)(6k)}{(10k)(9k)} + \frac{(6k)(4k)}{(9k)(11k)} + \frac{(4k)(5k)}{(11k)(10k)} = \frac{30}{90} + \frac{24}{99} + \frac{20}{110} = \frac{1}{3} + \frac{8}{33} + \frac{2}{11}$.
$= \frac{11 + 8 + 6}{33} = \frac{25}{33}$.

(A) Given that,$\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}=k$.
$s-a=11k$,$s-b=12k$,$s-c=13k$.
Adding these,we get $3s-(a+b+c) = 36k$.
Since $a+b+c=2s$,we have $3s-2s=36k$,so $s=36k$.
Using the formula $\tan^2\left(\frac{A}{2}\right) = \frac{(s-b)(s-c)}{s(s-a)}$ and $\tan^2\left(\frac{C}{2}\right) = \frac{(s-a)(s-b)}{s(s-c)}$,
$\tan^2\left(\frac{A}{2}\right)+\tan^2\left(\frac{C}{2}\right) = \frac{(12k)(13k)}{(36k)(11k)} + \frac{(11k)(12k)}{(36k)(13k)}$.
$= \frac{12 \times 13}{36 \times 11} + \frac{11 \times 12}{36 \times 13} = \frac{1}{3} \left( \frac{13}{11} + \frac{11}{13} \right)$.
$= \frac{1}{3} \left( \frac{169+121}{143} \right) = \frac{1}{3} \times \frac{290}{143} = \frac{290}{429}$.

(A) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,where $\Delta$ is the area of the triangle and $s$ is the semi-perimeter.
Given $r_1: r_2: r_3 = 1: 2: 3$,let $r_1 = x, r_2 = 2x, r_3 = 3x$.
Then $s-a = \frac{\Delta}{x}$,$s-b = \frac{\Delta}{2x}$,and $s-c = \frac{\Delta}{3x}$.
Adding these three equations:
$(s-a) + (s-b) + (s-c) = \frac{\Delta}{x} + \frac{\Delta}{2x} + \frac{\Delta}{3x}$
$3s - (a+b+c) = \Delta \left( \frac{6+3+2}{6x} \right)$
Since $a+b+c = 2s$,we have $3s - 2s = \frac{11\Delta}{6x}$,so $s = \frac{11\Delta}{6x}$.
Now,$a = s - (s-a) = \frac{11\Delta}{6x} - \frac{\Delta}{x} = \frac{11\Delta - 6\Delta}{6x} = \frac{5\Delta}{6x}$.
$b = s - (s-b) = \frac{11\Delta}{6x} - \frac{\Delta}{2x} = \frac{11\Delta - 3\Delta}{6x} = \frac{8\Delta}{6x}$.
$c = s - (s-c) = \frac{11\Delta}{6x} - \frac{\Delta}{3x} = \frac{11\Delta - 2\Delta}{6x} = \frac{9\Delta}{6x}$.
Thus,$a: b: c = \frac{5\Delta}{6x} : \frac{8\Delta}{6x} : \frac{9\Delta}{6x} = 5: 8: 9$.

(B) Using the Law of Tangents: $\frac{a-b}{a+b} = \frac{\tan(\frac{A-B}{2})}{\tan(\frac{A+B}{2})}$.
Given $a = 2b$,we have $\frac{2b-b}{2b+b} = \frac{b}{3b} = \frac{1}{3}$.
Also,$|A-B| = \frac{\pi}{3}$,so $\frac{A-B}{2} = \frac{\pi}{6}$.
Thus,$\frac{1}{3} = \frac{\tan(\pi/6)}{\tan((A+B)/2)} = \frac{1/\sqrt{3}}{\tan((A+B)/2)}$.
This implies $\tan(\frac{A+B}{2}) = \sqrt{3}$.
Therefore,$\frac{A+B}{2} = \frac{\pi}{3}$,which means $A+B = \frac{2\pi}{3}$.
Since $A+B+C = \pi$,we have $C = \pi - (A+B) = \pi - \frac{2\pi}{3} = \frac{\pi}{3}$.

(A) Using Napier's Analogy,we have $\tan \left(\frac{A-B}{2}\right) = \frac{a-b}{a+b} \cot \left(\frac{C}{2}\right)$.
Given $\tan \left(\frac{A-B}{2}\right) = \frac{1}{3} \tan \left(\frac{A+B}{2}\right)$.
Since $A+B+C = 180^{\circ}$,we have $\frac{A+B}{2} = 90^{\circ} - \frac{C}{2}$,so $\tan \left(\frac{A+B}{2}\right) = \cot \left(\frac{C}{2}\right)$.
Substituting this into the given equation: $\frac{a-b}{a+b} \cot \left(\frac{C}{2}\right) = \frac{1}{3} \cot \left(\frac{C}{2}\right)$.
Assuming $\cot \left(\frac{C}{2}\right) \neq 0$,we get $\frac{a-b}{a+b} = \frac{1}{3}$.
Cross-multiplying gives $3(a-b) = a+b$,which simplifies to $3a - 3b = a + b$.
Rearranging terms gives $2a = 4b$,or $\frac{a}{b} = \frac{4}{2} = 2$.
Thus,$a : b = 2 : 1$.

(B) Given $r_1 = 2r_2 = 3r_3 = \lambda$ (say).
We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
So,$\frac{\Delta}{s-a} = \lambda$,$\frac{\Delta}{s-b} = \frac{\lambda}{2}$,and $\frac{\Delta}{s-c} = \frac{\lambda}{3}$.
This gives $s-a = \frac{\Delta}{\lambda}$,$s-b = \frac{2\Delta}{\lambda}$,and $s-c = \frac{3\Delta}{\lambda}$.
Adding these,$(s-a) + (s-b) + (s-c) = \frac{\Delta}{\lambda} (1 + 2 + 3) = \frac{6\Delta}{\lambda}$.
$3s - (a+b+c) = \frac{6\Delta}{\lambda}$.
Since $a+b+c = 2s$,we have $3s - 2s = s = \frac{6\Delta}{\lambda}$.
Thus,$s-a = \frac{s}{6}$,$s-b = \frac{2s}{6} = \frac{s}{3}$,and $s-c = \frac{3s}{6} = \frac{s}{2}$.
Solving for sides: $a = s - \frac{s}{6} = \frac{5s}{6}$,$b = s - \frac{s}{3} = \frac{2s}{3}$,$c = s - \frac{s}{2} = \frac{s}{2}$.
Perimeter $P = a+b+c = \frac{5s}{6} + \frac{4s}{6} + \frac{3s}{6} = \frac{12s}{6} = 2s$.
Also,$b = \frac{2s}{3} \implies 3b = 2s = P$. Therefore,the perimeter is $3b$.

(A) Let the two sides be $a$ and $b$. We are given $a + b = x$ and $ab = y$.
Given the condition $x^2 - c^2 = y$,we substitute $x = a + b$:
$(a + b)^2 - c^2 = y$
$a^2 + b^2 + 2ab - c^2 = y$
Since $ab = y$,we have $a^2 + b^2 + 2y - c^2 = y$,which simplifies to $a^2 + b^2 - c^2 = -y$.
By the Law of Cosines,$c^2 = a^2 + b^2 - 2ab \cos C$.
Comparing $a^2 + b^2 - c^2 = 2ab \cos C$ with $a^2 + b^2 - c^2 = -y$ and $ab = y$,we get $2y \cos C = -y$,so $\cos C = -\frac{1}{2}$.
Thus,$C = 120^\circ$.
The circumradius $R$ is given by $R = \frac{c}{2 \sin C}$.
$R = \frac{c}{2 \sin 120^\circ} = \frac{c}{2 (\sqrt{3}/2)} = \frac{c}{\sqrt{3}}$.

(C) We are given $r_1 = 2r_2 = 3r_3 = k$ (let).
Then $r_1 = k$,$r_2 = k/2$,and $r_3 = k/3$.
The exradii are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Thus,$s-a = \frac{\Delta}{k}$,$s-b = \frac{2\Delta}{k}$,and $s-c = \frac{3\Delta}{k}$.
Adding these,$(s-a) + (s-b) + (s-c) = 3s - (a+b+c) = 3s - 2s = s$.
So,$s = \frac{\Delta}{k} (1 + 2 + 3) = \frac{6\Delta}{k}$.
Now,$a = s - (s-a) = \frac{6\Delta}{k} - \frac{\Delta}{k} = \frac{5\Delta}{k}$.
And $b = s - (s-b) = \frac{6\Delta}{k} - \frac{2\Delta}{k} = \frac{4\Delta}{k}$.
Therefore,$a : b = \frac{5\Delta}{k} : \frac{4\Delta}{k} = 5 : 4$.

(A) We know that in a triangle,the area $\Delta = rs = \frac{abc}{4R}$.
Also,the expression $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$ can be written as $\frac{a+b+c}{abc} = \frac{2s}{abc}$.
From the area formula,$abc = 4R\Delta = 4R(rs) = 4Rrs$.
Substituting this into the expression,we get $\frac{2s}{4Rrs} = \frac{1}{2Rr}$.
Given $r=3$ and $R=5$,the value is $\frac{1}{2 \times 5 \times 3} = \frac{1}{30}$.

(C) Let the circumcentre be $O$ and the incentre be $I$. The coordinates of $O$ and $I$ relative to the side $BC$ can be analyzed. The distance of $O$ from $BC$ is $R \cos A$ and the distance of $I$ from $BC$ is $r$. Since $OI$ is parallel to $BC$,their distances from $BC$ must be equal,so $R \cos A = r$.
Using the identity $r = 4R \sin(A/2) \sin(B/2) \sin(C/2)$,we have $\cos A = 4 \sin(A/2) \sin(B/2) \sin(C/2)$.
Using $\cos A = 1 - 2 \sin^2(A/2)$,we get $1 - 2 \sin^2(A/2) = 4 \sin(A/2) \sin(B/2) \sin(C/2)$.
Also,using the property $\cos B + \cos C = 2 \cos((B+C)/2) \cos((B-C)/2) = 2 \sin(A/2) \cos((B-C)/2)$.
Given $OI \parallel BC$,it is a known property that $\cos B + \cos C = 1$.

(D) We know that the exradii of a triangle are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Given $r_1 : r_2 = 3 : 4$,we have $\frac{s-b}{s-a} = \frac{3}{4}$,which implies $4s - 4b = 3s - 3a$,or $s = 4b - 3a$.
Given $r_2 : r_3 = 2 : 3$,we have $\frac{s-c}{s-b} = \frac{2}{3}$,which implies $3s - 3c = 2s - 2b$,or $s = 3c - 2b$.
Equating the two expressions for $s$: $4b - 3a = 3c - 2b \implies 6b = 3a + 3c \implies 2b = a + c$.
This shows that $a, b, c$ are in arithmetic progression.
Using the property $r_1 : r_2 : r_3 = \frac{1}{s-a} : \frac{1}{s-b} : \frac{1}{s-c}$,we have $r_1 : r_2 : r_3 = 3 : 4 : 6$ (since $r_1:r_2=3:4$ and $r_2:r_3=2:3=4:6$).
Thus,$s-a = \frac{k}{3}$,$s-b = \frac{k}{4}$,$s-c = \frac{k}{6}$ for some constant $k$.
Summing these: $3s - (a+b+c) = k(\frac{1}{3} + \frac{1}{4} + \frac{1}{6}) = k(\frac{4+3+2}{12}) = \frac{9k}{12} = \frac{3k}{4}$.
Since $a+b+c = 2s$,we have $3s - 2s = s = \frac{3k}{4}$.
Then $a = s - \frac{k}{3} = \frac{3k}{4} - \frac{k}{3} = \frac{5k}{12}$,$b = s - \frac{k}{4} = \frac{3k}{4} - \frac{k}{4} = \frac{2k}{4} = \frac{6k}{12}$,$c = s - \frac{k}{6} = \frac{3k}{4} - \frac{k}{6} = \frac{7k}{12}$.
Therefore,$a : b : c = 5 : 6 : 7$.

(D) Given that $(r_2-r_1)(r_3-r_1)=2 r_2 r_3$.
Dividing by $r_2 r_3$,we get $(1-\frac{r_1}{r_2})(1-\frac{r_1}{r_3})=2$.
Using $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$,we have $(1-\frac{s-b}{s-a})(1-\frac{s-c}{s-a})=2$.
This simplifies to $(\frac{s-a-s+b}{s-a})(\frac{s-a-s+c}{s-a})=2$,which is $(b-a)(c-a)=2(s-a)^2$.
Since $2(s-a) = b+c-a$,we have $(b-a)(c-a) = \frac{1}{2}(b+c-a)^2$.
Expanding this leads to $b^2+c^2-a^2=0$,so $a^2=b^2+c^2$,meaning $\angle A=90^{\circ}$.
We know $2(r+R) = 2r+2R = (b+c-a) + a = b+c$.
Since $b=2R \sin B$ and $c=2R \sin C$,$b+c = 2R(\sin B + \sin C) = 2R(2 \sin \frac{B+C}{2} \cos \frac{B-C}{2})$.
Since $A=90^{\circ}$,$B+C=90^{\circ}$,so $\sin \frac{B+C}{2} = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$.
Thus,$2(r+R) = 4R(\frac{1}{\sqrt{2}}) \cos \frac{B-C}{2} = 2 \sqrt{2} R \cos \frac{B-C}{2}$.

(C) In $\triangle ABC$,$A+B+C=\pi$.
We know that $r_1 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$ and $r_2 = 4R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}$.
Adding these,$r_1+r_2 = 4R \cos \frac{C}{2} [\sin \frac{A}{2} \cos \frac{B}{2} + \cos \frac{A}{2} \sin \frac{B}{2}]$.
Using the identity $\sin(x+y) = \sin x \cos y + \cos x \sin y$,we get $r_1+r_2 = 4R \cos \frac{C}{2} \sin(\frac{A+B}{2})$.
Since $A+B = \pi - C$,$\sin(\frac{A+B}{2}) = \sin(\frac{\pi}{2} - \frac{C}{2}) = \cos \frac{C}{2}$.
Thus,$r_1+r_2 = 4R \cos^2 \frac{C}{2}$.
Now,$(r_1+r_2) \operatorname{cosec}^2 \frac{C}{2} = \frac{4R \cos^2 \frac{C}{2}}{\sin^2 \frac{C}{2}} = 4R \cot^2 \frac{C}{2}$.