Let ABC be a triangle with G1,G2,G3 as the mid-points of BC(a),AC(b) and AB(c) respectively. Also, let M be the centroid of the triangle. It is given that the circumcircle of ΔMAC touches the side AB of the triangle at point A, then
A
AG1b=√32
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B
max{sin∠CAM+sin∠CBM}=2√3
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C
a2+b2c2=2
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D
c2ab=√2 if sin∠CAM+sin∠CBM is maximum
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Solution
The correct option is Dc2ab=√2 if sin∠CAM+sin∠CBM is maximum (G3A)2=(G3M)(G3C) ∴(c2)2=13(G3C)2 ∴c24=13(2b2+2a2−c24)⇒a2+b2=2c2⋯(1)
Now, in ΔAG1C sin∠CAM=asinC2(AG1)
and in ΔBCG2 sin∠CBM=bsinC2(BG2)
Also, AG1=√2b2+2c2−a24=√32b [Using Eq.(1)]
and BG2=√2a2+2c2−b24=√32a
Now, sin∠CAM+sin∠CBM=2Δ√3(a2+b2a2b2) =1√3(a2+b2ab)sinC⋯(2)
Also, cosC=a2+b2−c22ab⇒a2+b2=4abcosC
So, from Eq.(2) sin∠CAM+sin∠CBM=2√3sin2C≤2√3 (where C =π4)