Maximum squarelength triangle vector

Geometry Level pending

Point A A lies on circle x 2 + y 2 = 5 2 x^2+y^2=5^2 , point B B lies on circle x 2 + y 2 = 4 2 x^2+y^2=4^2 and C C lies on circle x 2 + y 2 = 3 2 x^2+y^2=3^2 . Find the maximum of A B 2 + A C 2 + B C 2 AB^2+AC^2+BC^2 .


The answer is 150.

Yuriy Kazakov by Yuriy Kazakov , 61, Russia

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1 solution

Let O O be the origin of coordinates, ∠ A O B = α , ∠ B O C = β , ∠ C O A = 2 π − ( α + β ) \angle {AOB}=α, \angle {BOC}=β, \angle {COA}=2π-(α+β) . Then

l = ∣ A B ‾ ∣ 2 + ∣ B C ‾ ∣ 2 + ∣ C A ‾ ∣ 2 = 100 − 40 cos ⁡ α − 24 cos ⁡ β + 30 cos ⁡ ( α + β ) l=|\overline {AB}|^2+|\overline {BC}|^2+|\overline {CA}|^2=100-40\cos α-24\cos β+30\cos (α+β) .

This will be maximum when

∂ l ∂ α = 0 , ∂ l ∂ β = 0 ⟹ 40 sin ⁡ α = 30 sin ⁡ ( α + β ) , 24 sin ⁡ β = 30 sin ⁡ ( α + β ) ⟹ sin ⁡ β = 5 3 sin ⁡ α , cos ⁡ α = − 4 5 , cos ⁡ β = 0 , cos ⁡ ( α + β ) = 3 5 \dfrac{\partial l}{\partial α}=0, \dfrac{\partial l}{\partial β}=0\implies 40\sin α=30\sin (α+β), 24\sin β=30\sin (α+β)\implies \sin β=\dfrac{5}{3}\sin α, \cos α=-\dfrac{4}{5}, \cos β=0, \cos (α+β)=\dfrac{3}{5} and so

l m a x = 100 + 40 × 4 5 + 30 × 3 5 = 150 l_{max}=100+40\times \dfrac{4}{5}+30\times \dfrac{3}{5}=\boxed {150} .

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