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Question Number 111432 by ajfour last updated on 03/Sep/20

Commented by ajfour last updated on 03/Sep/20

$$Findperimeterof△OAB.$$

Commented by ajfour last updated on 04/Sep/20

$$mrWSir,canyoupleaseconfirm$$$$answertothisquestion?$$

Answered by ajfour last updated on 04/Sep/20

Commented by ajfour last updated on 04/Sep/20

$$\cos \alpha =\frac{1}{a+1}$$$$\sin \beta =\frac{a}{a+1}=1-\cos \alpha =\frac{b}{1+2a+b}$$$$\alpha +\beta =60°$$$$\Rightarrow \sin \beta =\frac{a}{a+1}=\frac{\sqrt{3}}{2}\left(\frac{1}{a+1}\right)-\frac{1}{2}\sqrt{1-\frac{1}{{(a+1)}^2}}$$$$\Rightarrow 2a=\sqrt{3}-\sqrt{{(a+1)}^2-1}$$$$\Rightarrow 3+4a^2-4\sqrt{3}a=a^2+2a$$$$\Rightarrow 3a^2-(4\sqrt{3}+2)a+3=0$$$$\Rightarrow a=\frac{2\sqrt{3}+1}{3}-\frac{\sqrt{{(2\sqrt{3}+1)}^2-9}}{3}$$$$\Rightarrow a=\frac{2\sqrt{3}+1-2\sqrt{1+\sqrt{3}}}{3}\approx 0.38611$$$$1+\frac{1}{a}=1+\frac{2a+1}{b}$$$$\Rightarrow b=a(2a+1)\approx 0.68425$$$$.....$$

Commented by mr W last updated on 06/Feb/25

$$\sin \beta =\frac{a}{a+1}=1-\cos \alpha \ne \frac{b}{1+2a+b}$$$$theyarenotcollinear!$$

Answered by 1549442205PVT last updated on 05/Sep/20

$$\left(\mathrm{See}\mathrm{the}\mathrm{figure}\mathrm{below}\right)$$$$\mathrm{HJ}=\mathrm{HQ}=\mathrm{x};\mathrm{EP}=\mathrm{EG}=\mathrm{EF}=\mathrm{y}$$$$\mathrm{Denote}\mathrm{by}\mathrm{x}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circle}$$$$\left(\mathrm{B}\right),\mathrm{by}\mathrm{y}-\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{small}\mathrm{circle}$$$$\mathrm{Apply}{\mathrm{Steward}}^{\prime }\mathrm{s}\mathrm{theorem}\mathrm{we}\mathrm{have}:$$$${(1+\mathrm{x})}^2(1+\mathrm{y})+2^2(\mathrm{x}+\mathrm{y})-{(1+\mathrm{y})}^2(1+\mathrm{x}+2\mathrm{y})$$$$=(1+\mathrm{x}+2\mathrm{y})(1+\mathrm{y})(\mathrm{x}+\mathrm{y})$$$$\iff 1+2\mathrm{x}+{\mathrm{x}}^2+\mathrm{y}+2\mathrm{xy}+{\mathrm{x}}^2\mathrm{y}+4\mathrm{x}+4\mathrm{y}-1$$$$-2\mathrm{y}-{\mathrm{y}}^2-\mathrm{x}-2\mathrm{xy}-{\mathrm{xy}}^2-2\mathrm{y}-{4\mathrm{y}}^2-{2\mathrm{y}}^3$$$$=\mathrm{x}+{\mathrm{x}}^2+4\mathrm{xy}+{\mathrm{x}}^2\mathrm{y}+{3\mathrm{xy}}^2+{3\mathrm{y}}^2+{2\mathrm{y}}^3$$$$\iff {4\mathrm{y}}^3+{8\mathrm{y}}^2+{4\mathrm{xy}}^2+4\mathrm{xy}-4\mathrm{x}=0\left(1\right)$$$$\mathrm{Since}\mathrm{NE}//\mathrm{HQ},\mathrm{by}{\mathrm{Thalet}}^{\prime }\mathrm{theorem}$$$$\mathrm{we}\mathrm{have}:\frac{\mathrm{NE}}{\mathrm{HQ}}=\frac{\mathrm{OE}}{\mathrm{OH}}\iff \frac{\mathrm{y}}{\mathrm{x}}=\frac{1+\mathrm{y}}{1+\mathrm{x}+\mathrm{y}}$$$$\iff \mathrm{x}+\mathrm{xy}=\mathrm{y}+\mathrm{xy}+{\mathrm{y}}^2\iff \mathrm{x}={2\mathrm{y}}^2+\mathrm{y}\left(2\right)$$$$\mathrm{Replace}\left(2\right)\mathrm{into}\left(1\right)\mathrm{we}\mathrm{get}:$$$${4\mathrm{y}}^3+{8\mathrm{y}}^2+{4\mathrm{y}}^2({2\mathrm{y}}^2+\mathrm{y})-4\mathrm{y}({2\mathrm{y}}^2+\mathrm{y})=0$$$$\iff {8\mathrm{y}}^4+{16\mathrm{y}}^3+{4\mathrm{y}}^2-4\mathrm{y}=0$$$$\iff {2\mathrm{y}}^3+{4\mathrm{y}}^2+\mathrm{y}-1=0$$$$\iff (\mathrm{y}+1)({2\mathrm{y}}^2+2\mathrm{y}-1)=0$$$$\iff {2\mathrm{y}}^2+2\mathrm{y}-1=0;{\mathrm{\Delta }}^{\prime }=1+2=3$$$$\Rightarrow \mathrm{y}=\frac{-1+\sqrt{3}}{2}.\mathrm{Replace}\mathrm{into}\left(2\right)\mathrm{we}\mathrm{get}$$$$\mathrm{x}=({2\mathrm{y}}^2+\mathrm{y})=2.\frac{2-\sqrt{3}}{2}+\frac{\sqrt{3}-1}{2}=\frac{3-\sqrt{3}}{2}$$$$\mathrm{Thus},\mathrm{x}=\frac{3-\sqrt{3}}{2}\approx 0.6339$$$$\mathrm{y}=\frac{\sqrt{3}-1}{2}\approx 0.3660.\mathrm{Consequently},\mathrm{the}$$$$\mathrm{perimeter}\mathrm{of}\mathrm{triangle}\mathrm{OAB}\mathrm{equal}\mathrm{to}$$$$4+2\mathbf{x}+2\mathbf{y}=4+3-\sqrt{3}+\sqrt{3}-1=6$$

Commented by 1549442205PVT last updated on 04/Sep/20

Commented by ajfour last updated on 04/Sep/20

$$questionisalrightsir.$$

Commented by 1549442205PVT last updated on 05/Sep/20

$$\mathrm{Thank}\mathrm{you},\mathrm{Sir}.\mathrm{I}\mathrm{mistaked}\mathrm{and}\mathrm{corrected}$$$$\mathrm{E}\mathrm{is}\mathrm{vertex}\mathrm{of}\mathrm{isosceles}\mathrm{triangle}.\mathrm{But}$$$$\mathrm{my}\mathrm{solution}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{use}\mathrm{this}\mathrm{hypothesis}$$

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