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Triangle with sin^2 A + sin^2 B = sin^2 C, sin A, sin B, sin C in AP and Inradius 2 | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
In a triangle ABCABC, sin⁡2A+sin⁡2B=sin⁡2C\sin^{2}A+\sin^{2}B=\sin^{2}C and sin⁡A, sin⁡B, sin⁡C\sin A,\ \sin B,\ \sin C are in A.P. in that order. If the inradius of the triangle is 22, then which of the following is/are correct?
AThe area of triangle ABCABC is 2424
BThe circumradius of triangle ABCABC is 55
CThe distance between the incentre and the circumcentre is 5\sqrt5correct
Dtan⁡A2tan⁡B2=13\tan\dfrac{A}{2}\tan\dfrac{B}{2}=\dfrac13
Solution
Step 1: sin⁡A:sin⁡B:sin⁡C=a:b:c\sin A:\sin B:\sin C=a:b:c, so
sin⁡2A+sin⁡2B=sin⁡2C  ⟹  a2+b2=c2\sin^{2}A+\sin^{2}B=\sin^{2}C \implies a^{2}+b^{2}=c^{2}
 Hence C=90∘C=90^{\circ} and cc is the largest side. Step 2: sin⁡A, sin⁡B, sin⁡C\sin A,\ \sin B,\ \sin C in A.P.   ⟹  2b=a+c\implies 2b=a+c, i.e. c=2b−ac=2b-a.
a2+b2=(2b−a)2  ⟹  3b2=4ab  ⟹  b=4a3,c=5a3a^{2}+b^{2}=(2b-a)^{2} \implies 3b^{2}=4ab \implies b=\frac{4a}{3},\quad c=\frac{5a}{3}
 So a:b:c=3:4:5a:b:c=3:4:5. Let a=3ka=3k, b=4kb=4k, c=5kc=5k, k>0k>0. Step 3:
Δ=12ab=6k2,s=a+b+c2=6k,r=Δs=k\Delta=\frac12ab=6k^{2},\qquad s=\frac{a+b+c}{2}=6k,\qquad r=\frac{\Delta}{s}=k
 r=2  ⟹  k=2r=2 \implies k=2, so a=6a=6, b=8b=8, c=10c=10 and Δ=24\Delta=24. (1) is correct. Step 4: C=90∘C=90^{\circ}, so the hypotenuse ABAB is a diameter of the circumcircle:
R=c2=5R=\frac{c}{2}=5
 (2) is correct. Step 5: Take C(0,0)C(0,0), B(6,0)B(6,0), A(0,8)A(0,8). The circumcentre OO is the mid point of ABAB, i.e. O(3,4)O(3,4). The incircle has radius 22 and touches both axes, so its centre is I(2,2)I(2,2).
OI=(3−2)2+(4−2)2=5OI=\sqrt{(3-2)^{2}+(4-2)^{2}}=\sqrt5
 (3) is correct. Step 6: sin⁡A=35\sin A=\dfrac35, cos⁡A=45\cos A=\dfrac45, sin⁡B=45\sin B=\dfrac45, cos⁡B=35\cos B=\dfrac35.
tan⁡A2=sin⁡A1+cos⁡A=3/59/5=13,tan⁡B2=sin⁡B1+cos⁡B=4/58/5=12\tan\frac{A}{2}=\frac{\sin A}{1+\cos A}=\frac{3/5}{9/5}=\frac13,\qquad \tan\frac{B}{2}=\frac{\sin B}{1+\cos B}=\frac{4/5}{8/5}=\frac12
tan⁡A2tan⁡B2=16≠13\tan\frac{A}{2}\tan\frac{B}{2}=\frac16\ne\frac13
 (4) is not correct. Answer: (1), (2), (3).
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