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Question Number 125572 by ajfour last updated on 12/Dec/20

Commented by ajfour last updated on 12/Dec/20

$$IfequilateraltriangleABChas$$$$sidelengths=3units,find{x}_A.$$$$Parabolaeq.isy=x^2.$$

Commented by mr W last updated on 12/Dec/20

$$nicequestionagain!$$$$icannotfindanexactsolution.$$

Answered by mr W last updated on 12/Dec/20

$$A(a,0)witha⩾0$$$$D(h,k)withD=midpointBC$$$$k^2+{(h-a)}^2=\frac{3s^2}{4}...\left(i\right)$$$$eqn.ofBC:$$$$y=k-\frac{h-a}{k}(x-h)=\frac{a-h}{k}x+k-\frac{h(a-h)}{k}$$$$intersection$$$$x^2=\frac{a-h}{k}x+k-\frac{h(a-h)}{k}$$$$x^2-\frac{a-h}{k}x-k+\frac{h(a-h)}{k}=0$$$$x_1+x_2=\frac{a-h}{k}=2h...\left(ii\right)$$$$x_1{x}_2=-k+\frac{h(a-h)}{k}$$$${(y_2-y_1)}^2+{(x_2-x_1)}^2=s^2$$$${(x_2^2-x_1^2)}^2+{(x_2-x_1)}^2=s^2$$$$[{(x_2+x_1)}^2+1][{(x_2+x_1)}^2-4x_1{x}_2]=s^2$$$$4(4h^2+1)(h^2+k-\frac{h(a-h)}{k})=s^2...\left(iii\right)$$$$from\left(ii\right):$$$$k=\frac{a-h}{2h}...\left(I\right)$$$$put\left(I\right)into\left(i\right):$$$$\frac{{(a-h)}^2}{4h^2}+{(a-h)}^2=\frac{3s^2}{4}$$$$(1+4h^2){(a-h)}^2=3s^2{h}^2$$$$\Rightarrow a=h(1+s\sqrt{\frac{3}{1+4h^2}})...\left(II\right)$$$$putinto\left(iii\right):$$$$\Rightarrow 2(1+4h^2)(s\sqrt{\frac{3}{1+4h^2}}-2h^2)=s^2...\left(III\right)$$$$$$$$foragivens(⩽4.73565)wegethfrom$$$$\left(III\right)andthenafrom\left(II\right).$$

Commented by mr W last updated on 12/Dec/20

$$weseefor0<s<3.4641thereisone$$$$possibletriangle.$$$$for3.4641⩽s⩽4.73565therearetwo$$$$possibletriangles.$$$$$$$$example:s=4$$

Commented by mr W last updated on 12/Dec/20

Commented by mr W last updated on 12/Dec/20

Commented by mr W last updated on 12/Dec/20

Commented by mr W last updated on 12/Dec/20

Commented by mr W last updated on 12/Dec/20

Answered by ajfour last updated on 12/Dec/20

$$leteq.ofBCbe$$$$y=2mx+c$$$$\tan \theta =2m$$$$Intersectionwithparabola$$$$x^2-2mx-c=0$$$$x=m\pm \sqrt{m^2+c}$$$$midpointD(m,2m^2+c)$$$$(x_2-x_1)\sqrt{1+4m^2}=s\Rightarrow$$$$4(m^2+c)(1+4m^2)=s^2...\left(i\right)$$$$\frac{s\sqrt{3}}{2}\left(\frac{1}{\sqrt{1+4m^2}}\right)=2m^2+c$$$$\frac{s\sqrt{3}}{2\sqrt{1+4m^2}}=m^2+\frac{s^2}{4(1+4m^2)}...\left(ii\right)$$$$x_A=a=m+\frac{s\sqrt{3}}{2}\left(\frac{2m}{\sqrt{1+4m^2}}\right)...\left(iii\right)$$$$let1+4m^2=t^2$$$$nowfrom\left(ii\right)$$$$2\sqrt{3}st=(t^2-1)t^2+s^2$$$$\Rightarrow t^4-t^2-2\sqrt{3}st+s^2=0$$$${(s-\sqrt{3}t)}^2=4t^2-t^4$$$$s=\sqrt{3}t\pm \sqrt{3t^2+t^2-t^4}$$$$s=t(\sqrt{3}+\sqrt{4-t^2})$$$$s_{max}\approx 4.7356$$$$Discriminant$$$$D=12t^2-4(t^4-t^2)⩾0$$$$\Rightarrow t^4⩽4t^2$$$$\Rightarrow t^2=1+4m^2⩽4$$$$\Rightarrow \tan \theta =2m⩽\sqrt{3}$$

Commented by ajfour last updated on 12/Dec/20

$$Sir,pleasehelpcheckingmy$$$$solution,ouranswersfor{s}_{max}$$$$differ.$$$$Yoursolutionevenseemserrorless$$$$tome.$$

Commented by ajfour last updated on 12/Dec/20

Commented by mr W last updated on 12/Dec/20

$$veryniceapproachsir!$$$$$$$$s=t(\sqrt{3}\pm \sqrt{4-t^2})$$$$\Rightarrow s_{max}=4.73565$$$$s_{max}isnotatt=2,butatt=1.7047!$$$$$$$$x_A=a=m+\frac{s\sqrt{3}}{2}\left(\frac{2m}{\sqrt{1+4m^2}}\right)...\left(iii\right)$$

Commented by ajfour last updated on 12/Dec/20

$$thankyouSir.$$

Commented by mr W last updated on 12/Dec/20

Commented by ajfour last updated on 12/Dec/20

$$s=f\left(t\right)$$

Commented by mr W last updated on 12/Dec/20

$$exactvaluefor{s}_{max}:$$$$s=t(\sqrt{3}+\sqrt{4-t^2})$$$$\frac{ds}{dt}=\sqrt{3}+2\sqrt{4-t^2}-\frac{4}{\sqrt{4-t^2}}=0$$$$letx=\sqrt{4-t^2}$$$$2x^2+\sqrt{3}x-4=0$$$$\Rightarrow x=\frac{\sqrt{35}-\sqrt{3}}{4}=\sqrt{4-t^2}$$$$\Rightarrow t=\sqrt{4-{\left(\frac{\sqrt{35}-\sqrt{3}}{4}\right)}^2}$$$$=\frac{\sqrt{2(\sqrt{105}+13)}}{4}\approx 1.70466$$$$s_{max}=\sqrt{4-{\left(\frac{\sqrt{35}-\sqrt{3}}{4}\right)}^2}(\sqrt{3}+\frac{\sqrt{35}-\sqrt{3}}{4})$$$$=\frac{\sqrt{2(35\sqrt{105}+359)}}{8}\approx 4.73565$$

Commented by ajfour last updated on 12/Dec/20

$$Wonderful,Sir,goodanalysis!$$

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