Question

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.

Column-I	Column-II
(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

Answer 1

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

(G₃A)₂ = (G₃M) (G₃C)

Therefore,

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).