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Question Number 185101 by emmanuelson123 last updated on 17/Jan/23

Answered by Frix last updated on 17/Jan/23

$$\mathrm{In}\mathrm{triangle}a,b,c:$$$$D=\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$$$$$$$$a^2+b^2-2ab\cos \gamma =c^2\Rightarrow$$$$\cos \gamma =\frac{a^2+b^2-c^2}{2ab}\Rightarrow$$$$\sin \gamma =\frac{D}{2ab}$$$$\sin \frac{x}{2}=\sqrt{\frac{1-\sqrt{1-{\sin }^{2}x}}{2}}\Rightarrow$$$$\left[\mathrm{Similar}\mathrm{for}\mathrm{the}\mathrm{other}\mathrm{angles}\right]$$$$\sin \frac{\alpha }{2}=\frac{\sqrt{a^2-{(b-c)}^2}}{2\sqrt{bc}}$$$$\sin \frac{\beta }{2}=\frac{\sqrt{b^2-{(a-c)}^2}}{2\sqrt{ac}}$$$$\sin \frac{\gamma }{2}=\frac{\sqrt{c^2-{(a-b)}^2}}{2\sqrt{ab}}$$$$\mathrm{We}\mathrm{know}\mathrm{that}$$$$r=\frac{D}{2(a+b+c)}\mathrm{and}R=\frac{abc}{D}$$$$\Rightarrow$$$$\frac{D}{2(a+b+c)}=4\frac{abc}{D}\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\sin \frac{\gamma }{2}$$$$\iff$$$$\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\sin \frac{\gamma }{2}=\frac{D^2}{8abc(a+b+c)}$$$$\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\sin \frac{\gamma }{2}=\frac{(b+c-a)(a+c-b)(a+b-c)}{8abc}$$$$\mathrm{Lhs}:$$$$\frac{\sqrt{a^2-{(b-c)}^2}}{2\sqrt{bc}}\times \frac{\sqrt{b^2-{(a-c)}^2}}{2\sqrt{ac}}\times \frac{\sqrt{c^2-{(a-b)}^2}}{2\sqrt{ab}}=$$$$=\frac{\sqrt{{(b+c-a)}^2{(a+c-b)}^2{(a+b-c)}^2}}{8\sqrt{a^2{b}^2{c}^2}}=$$$$=\frac{(b+c-a)(a+c-b)(a+b-c)}{8abc}=\mathrm{Rhs}$$$$\mathrm{q}.\mathrm{e}.\mathrm{d}.$$

Answered by mnjuly1970 last updated on 17/Jan/23

$$2p=2R(sin\left(A\right)+sin\left(B\right)+sin\left(C\right))$$$$=2R(2sin\left(\frac{A+B}{2}\right)cos\left(\frac{A-B}{2}\right)+2sin\left(\frac{A+B}{2}\right)cos\left(\frac{A+B}{2}\right))$$$$=4Rcos\left(\frac{C}{2}\right)(2cos\left(\frac{A}{2}\right).cos\left(\frac{B}{2}\right))$$$$\Rightarrow p:=4Rcos\left(\frac{A}{2}\right)cos\left(\frac{B}{2}\right)cos\left(\frac{C}{2}\right)$$$$r:=\frac{S}{p}=\frac{\frac{1}{2}bcsin\left(A\right)}{4Rcos\left(\frac{A}{2}\right)cos\left(\frac{B}{2}\right)cos\left(\frac{C}{2}\right)}$$$$\underset{c=2Rsin\left(C\right)}{\stackrel{b=2Rsin\left(B\right)}{=}}4Rsin\left(\frac{A}{2}\right)sin\left(\frac{B}{2}\right)sin\left(\frac{C}{2}\right)$$$$\therefore r⩽\frac{R}{2}\Rightarrow sin\left(\frac{A}{2}\right)sin\left(\frac{B}{2}\right)sin\left(\frac{C}{2}\right)⩽\frac{1}{8}$$

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