Find the sides of this triangle.

If $d$ is the difference in the arithmetic progression of the sides, then $b = c + d$ and $a = c + 2d$ , and if $x$ is the smallest angle, then the biggest angle is $2x$ and the third angle is $180° - 3x$ .

Then by the law of sines, $\frac{\sin x}{c} = \frac{\sin (180° - 3x)}{c + d} = \frac{\sin 2x}{c + 2d}$ .

Using trigonometric identities, $\sin 2x = 2 \sin x \cos x = 2 \sin x \sqrt{1 - \sin^2 x}$ , and $\sin (180° - 3x) = \sin 3x = 3\sin x - 4\sin^3 x = \sin x(3 - 4 \sin^2 x)$ . These can be substituted in the above law of sines equations for $\frac{\sin x}{c} = \frac{\sin x(3 - 4 \sin^2 x)}{c + d} = \frac{2 \sin x \sqrt{1 - \sin^2 x}}{c + 2d}$ , which solves to $\sin^2 x = \frac{7}{16}$ and $d = \frac{c}{4}$ .

Since $a$ , $b$ , and $c$ are integers in an arithmetic progression, the difference $d$ must also be an integer, and since $d = \frac{c}{4}$ , $c$ must be divisible by $4$ . Since $a$ , $b$ , and $c$ are co-prime, $c = 4$ , which means $d = \frac{c}{4} = \frac{4}{4} = 1$ , $b = c + d = 4 + 1 = 5$ and $a = c + 2d = 4 + 2 \cdot 1 = 6$ .

Therefore, $a + b + c = 6 + 5 + 4 = \boxed{15}$ .

For $\triangle ABC$ , suppose D is the intersection point of BC and the angle bisector of A. As the lengths of AC, AB and BC are in an arithmetic progression, assume $AC= a-d , AB= a , BC=a+d$

$\because \triangle ACB\sim \triangle DAB\quad (AAA)\\ \therefore \frac { DA }{ AC } =\frac { AB }{ CB } =\frac { DB }{ AB } \quad (corr. sides, \sim \triangle s)$ Then we have, $\frac { AD }{ a } =\frac { a-d }{ a+d } =\frac { DB }{ a-d }$ Also, we know that $CD=AD \quad (base\angle s,isos.\triangle )\\$

Finally, consider the length of BC, we have, $a(\frac { a-d }{ a+d } )+\frac { { (a-d) }^{ 2 } }{ a+d } = a+d\\ a(a-d)={ (a+d) }^{ 2 }-{ (a-d) }^{ 2 }\\ a=5d\quad (\because a\neq 0)\\ \\$ Therefore, $AC:AB:BC=4:5:6$

draw $AD$ that bisect $\angle BAC$ , $\Delta ABC \sim \Delta DBA$

use sides $1,k,k^2-1$ from $\Delta DBA$ to form an arithmetic progression ( $a+c=2b$ and 2 other permutations), solve for $k$

$k=-1, \frac{3}{2}, 0, 2, \frac{-1 \pm \sqrt{13}}{2}$

$k=\frac{3}{2}$ is the only reasonable solution, which work out to be $4:5:6$

edit: only one permutation possible, read replies.

Let $C=\theta, A=2\theta, B=180-3\theta$ .

$\sin ^{2} (2\theta)-\sin^{2}(\theta)=\sin(180-3\theta)\sin(\theta)$

$a^{2}-c^{2}=bc$

As $a+c=2b$ we get $\frac{a}{c}=\frac{3}{2}$ .

Thus, $a:b:c=6:5:4$

$6+5+4=\boxed{15}$

Using Law of Sines $\implies \dfrac{\sin(2\beta)}{\sin(\beta)} = \dfrac{a}{a + 2r} \implies \cos(\beta) = \dfrac{a}{2(a + 2r)}$

Using Law of cosines with included angle $\beta \implies a^2 + 4ar + 4r^2 = a^2 +2ar + r^2 + a^2 - \dfrac{a^2(a + r)}{a + 2r} \implies$ $r(6r^2 + 7ar + a^2) = 0 \:\ r \neq 0 \implies r = -a, -\dfrac{a}{6} \:\ r \neq -a \implies r = -\dfrac{a}{6} \implies (a, b, c) = (a, \dfrac{5a}{6}, \dfrac{2a}{3})$

$\implies a:b:c = 6:5:4 \implies a + b + c = \boxed{15}.$

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the proportion of the sides are: a : b : c = sin(A) : sin(B) : sin(C)

= 3√7/8 : 5√7/16 : √7/4 = 6 : 5 : 4

Answer: The sides of the triangle are in proportion as 6 : 5 : 4

By the sine law, $\frac{a}{\sin 2 \alpha} = \frac{c}{\sin \alpha}$ , from which it follows that $\cos \alpha = \frac {a}{2c}$ (use the identity for $\sin 2 \alpha$ ).

Because $a,b,c$ form an arithmetic progression, $b=\frac{a+c}{2}$ , so now by the law of cosines:

$c^2=(\frac{a+c}{2})^2 + a^2 -2a(\frac{a+c}{2})(\frac{a}{2c})$ .

Simplify to get

$3c^3=3a^2c+2ac^2-2a^3$ .

If we now divide by $c^3$ and let $\frac{a}{c} =q$ , we obtain the equation:

$2q^3-3q^2-2q+3=0$

For which the solutions are $1.5,1,-1$ . The only reasonable option is $q=1.5$ , and so

$a:b:c = 1:1.25:1.5 = 4:5:6$

b= ( a+c ) /2 ,therefore a=2b-c ...✳︎ Let raise both side of this equation to the square, a^2=4b^2-4bc+c^2 ...① Then, if we define D for the intersection of BC and the bisector of angle A, we get DB = ac/( a+c ) follows from angle bisector theorem. Moreover, △ABC~△DBA, so c : a = ac/( b+c ) : c Therefore, c^2 = a^2c/( b+c ) c = a^2/( b+c ) ( divide both side by c) So, a^2 = bc+c^2 ...②

The equation 4b^2-4bc+c^2=bc+c^2 follows from ①,② So, 4b^2=5bc That is c = (4/5)b ...③ Let substitute ③ for ✳︎, a= 2b - (4/5)b = (6/5)b ...④ From ③,④ we get a : b: c = (4/5) : 1 : (6/5) = 4 : 5 : 6 So, a+b+c =4+5+6=15

a/sinA=b/sinB=c/sinC= 2R ——-(1). As the sides are in AP, 2b = (a + c)

Using (1), sin(B) = (sin(A) + sin(C))/2, we get:

sin(B) = sin{(A + C)/2}*cos{(A - C)/2} ----- (2)

By the given condition, taking angle C as the least, angle A as the greatest, ∠A=2∠C. So, A + C = 3C and A - C = C and ∠B=180−(∠A+∠C)=180−3C; so sin(B)=sin(180−3C)=sin(3C)

Substituting this in (2), sin(3C) = sin(3C/2)cos(C/2)

i.e. 2sin(3C/2)cos(3C/2) - sin(3C/2)cos(C/2) = 0

i.e. sin(3C/2){2cos(3C/2) - cos(C/2)} = 0

So either sin(3C/2) = 0 or 2cos(3C/2) - cos(C/2) = 0 If sin(3C/2) = 0, then C = 0, which is not possible; hence this is rejected.

So 2cos(3C/2) - cos(C/2) = 0 or {cos(3C/2)}/cos(C/2) = 1/2 {4cos³(C/2) - 3cos(C/2)}/cos(C/2) = 1/2

4cos²(C/2) - 3 = 1/2 cos²(C/2) = 7/8.. Hence cos(C/2) = √(7/8) only;

so sin(C/2) = √{1 - cos²(C/2)} = √(1/8)

so cos(C) = 2cos²(C/2) - 1 = 3/4

Substitute in the above, sin(C) = 2sin(C/2)cos(C/2) = 2{√(1/8)}{√(7/8)} = (√7)/4

sin(A) = sin(2C) = 2sin(C)cos(C) = 2{ (√7)/4)(3/4) = 3√7/8

sin(B) = sin(3C) = {3sin(C) - 4sin³(C)} = 3√7/4 - 4(√7/4)³ = 5√7/16

By sine rule, the proportions are a:b:c :: sin(A) : sin(B): sin(C) = 6 : 5 : 4.

Let a=b-d, b & c=b+d be the 3 sides, and A, B & C be the 3 angles. C =2 A. We know that sinC/c =sinA/a. This leads to sin(2A)/(b+d) = sinA/(b-d) . This gives 2sinAcosA= (b+d)/(b-d) sinA. This gives cosA = (b+d)/(2(b-d)). Also a^2 = b^2+c^2 -2.b.c.cosA. Substituting for a, c and cos (A) and simplifying we get b = 5d. Hence the sides are 4d, 5d and 6d. Hence the a+b+c =4+5+6=15