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Let ABC be a triangle with G1, G2, G3 the mid-points of BC, AC and AB, respectively. Also let M be the centroid of the triangle. It is given that the circumcircle of ∆MAC touches the side AB of the triangle at point A.

(A)  AG1/b = (P)  2/√3
(B)  Maximum value of Sn ∠CAM + Sn ∠CBM = (q)  √3/2
(C)  a2 + b2/c2=  (r) √2
(D) If (sin∠CAM + sin∠CBM) is maximum then c2/ab = (s) 2

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Correct option  (A) → (q); (B) → (p); (C) → (s); (D) → (r)

(G3A)2 = (G3M) (G3C)


So, (C) → (s).

Now in ΔAG1C, we have

sin ∠CAM = asinC/2(AG1)

and in ΔBCG2

sin ∠CBM = b sinC/2(BG2)

From Eq. (1), we have

a2 + b2 = 2c2 ......(3)

and from Eq. (2) and (3), we have

So, (B)→ (p)

Again, sin∠CAM + sin∠CBM is maximum when C = π/4

Also from Eq. (5), we have 

c2 = 2ab cos C

which implies

c2/ab = 2 cos C = √2

So, (D)→ (r).

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