find-the-greatest-possible-square-insribed-in-a-triangle-with-sides-a-b-c-

Question Number 39458 by MJS last updated on 06/Jul/18

find the greatest possible square insribed in

a triangle with sides a b c

Commented by MJS last updated on 06/Jul/18

the side s of the square not necessarily on

one of the sides of the triangle

Commented by ajfour last updated on 06/Jul/18

I n △ P C R

\frac{s \sqrt{2}}{\sin γ} = \frac{b - y}{\sin (θ + \frac{π}{4})} \dots . (i)

I n △ S R A

\frac{s}{\sin α} = \frac{y}{\sin (β - θ)} \dots . . (i i)

e l i m i n a t i n g y a m o n g s t (i), (i i)

\frac{s \sqrt{2} \sin (θ + \frac{π}{4})}{\sin γ} = b - \frac{s \sin (β - θ)}{\sin α}

\Rightarrow \boldsymbol s = \frac{\boldsymbol b}{\frac{\sqrt{2} \sin (\boldsymbol θ + \frac{\boldsymbol π}{4})}{\sin \boldsymbol γ} + \frac{\sin (\boldsymbol β - \boldsymbol θ)}{\sin \boldsymbol α}}

= \frac{b}{\frac{\sin θ + \cos θ}{\sin γ} + \frac{\sin β \cos θ - \cos β \sin θ}{\sin α}}

= \frac{b \sin γ \sin α}{\sin θ (\sin α - \cos β \sin γ) + \cos θ (\sin α + \sin β \sin γ)}

\boldsymbol s = \frac{b \sin γ \sin α}{A \sin (θ + ϕ)}

\boldsymbol w h e r e

\tan \boldsymbol ϕ = \frac{\sin α + \sin β \sin γ}{\sin α - \cos β \sin γ}

\boldsymbol A = \sqrt{{(\sin α - \cos β \sin γ)}^{2} + {(\sin α + \sin β \sin γ)}^{2}}

f o r s t o b e m a x i m u m

\sin (θ + ϕ) h a s t o b e m i n i m u m .

F o r m a n y o f t h e c a s e s (t h o u g h

n o t a l l),

\sin (θ + ϕ) i s m i n i m u m i f θ = 0

T h e n

\boldsymbol s_{m a x} = \frac{\boldsymbol b \sin \boldsymbol γ \sin \boldsymbol α}{A \sin \boldsymbol ϕ}

= \frac{b \sin γ \sin α}{\sin α + \sin β \sin γ}

a n d s i n c e

\frac{\sin α}{a} = \frac{\sin β}{b} = \frac{\sin γ}{c} = \frac{1}{2 R}

\boldsymbol s_{m a x} = \frac{a b c}{2 a R + b c} = \frac{a}{\frac{2 a R}{b c} + 1} .

T h i s f o r m u l a w i l l h a v e t o b e

a m e n d e d i f \sin (θ + ϕ) < \sin ϕ

f o r n o n z e r o θ .

Commented by ajfour last updated on 06/Jul/18

Commented by MJS last updated on 06/Jul/18

thanks . I already thought there would be

several different cases \dots

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