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Question Number 61861 by mr W last updated on 10/Jun/19

Commented by mr W last updated on 10/Jun/19

$$HistheorthocenteroftriangleABC.$$$$Itsdistancetothevertexesis\alpha ,\beta ,\gamma$$$$respectively.$$$$Findthesidelengthesa,b,cofthe$$$$triangle.$$

Commented by MJS last updated on 11/Jun/19

$$\mathrm{there}\mathrm{should}\mathrm{be}2\mathrm{triangles}(\mathrm{at}\mathrm{least}\mathrm{if}\mathrm{the}$$$$\mathrm{triangle}\mathrm{is}\mathrm{not}\mathrm{rectangular}).H\mathrm{might}\mathrm{be}$$$$\mathrm{outside}\mathrm{the}\mathrm{triangle}...$$

Commented by MJS last updated on 11/Jun/19

$$\mathrm{the}\mathrm{distances}\mathrm{are}$$$$\left(1\right)\alpha =\frac{a(-a^2+b^2+c^2)}{\delta }$$$$\left(2\right)\beta =\frac{b(a^2-b^2+c^2)}{\delta }$$$$\left(3\right)\gamma =\frac{c(a^2+b^2-c^2)}{\delta }$$$$[\mathrm{with}\delta =\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}]$$$$\mathrm{we}\mathrm{can}\mathrm{solve}\mathrm{this}\mathrm{for}\mathrm{given}\alpha ,\beta ,\gamma$$$$\mathrm{for}\alpha =5,\beta =6,\gamma =7\mathrm{I}\mathrm{get}$$$$\left(\begin{array}{c}a\\ b\\ c\end{array}\right)=\left(\begin{array}{c}5.142053\\ 3.929466\\ 1.562276\end{array}\right)\vee \left(\begin{array}{c}a\\ b\\ c\end{array}\right)=\left(\begin{array}{c}10.94952\\ 10.43514\\ 9.792451\end{array}\right)$$$$\mathrm{the}{3}^{\mathrm{rd}}\mathrm{solution}\mathrm{is}\mathrm{not}\mathrm{valid}\mathrm{because}a,b,c\notin \mathbb{R}$$

Commented by mr W last updated on 12/Jun/19

$$thankyousir!canwegettheformula$$$$fora,b,cintermsof\alpha ,\beta ,\gamma ?$$

Commented by MJS last updated on 12/Jun/19

$$\mathrm{transformed}\mathrm{equations}:$$$$\left(1\right)a^6+a^4({\alpha }^2-2(b^2+c^2))-a^2(2{\alpha }^2(b^2+c^2)-{(b^2+c^2)}^2)+{\alpha }^2{(b^2-c^2)}^2=0$$$$\left(2\right)b^6+b^4({\beta }^2-2(a^2+c^2))-b^2(2{\beta }^2(a^2+c^2)-{(a^2+c^2)}^2)+{\beta }^2{(a^2-c^2)}^2=0$$$$\left(3\right)c^6+c^4({\gamma }^2-2(a^2+b^2))-c^2(2{\gamma }^2(a^2+b^2)-{(a^2+b^2)}^2)+{\gamma }^2{(a^2-b^2)}^2=0$$$$\mathrm{these}\mathrm{are}\mathrm{of}\mathrm{degree}6\mathrm{for}\mathrm{the}\mathrm{leading}\mathrm{variable}$$$$\mathrm{but}\mathrm{only}\mathrm{of}\mathrm{degree}4\mathrm{for}\mathrm{the}2\mathrm{others}.\mathrm{we}\mathrm{can}$$$$\mathrm{omit}{a}^4\mathrm{from}\left(2\right)\mathrm{and}\left(3\right)$$$$\left(2\right){Aa}^4+...$$$$\left(3\right){Ba}^4+...$$$$B\times \left(2\right)-A\times \left(3\right)\mathrm{leads}\mathrm{to}$$$$-4a^2{b}^2{c}^2(b^2-c^2+{\beta }^2-{\gamma }^2)=0$$$$\Rightarrow c^2=b^2+{\beta }^2-{\gamma }^2$$$$\mathrm{now}\left(2\right)=\left(3\right)\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}{Pb}^2+Q=0$$$$\Rightarrow b^2=\frac{{\beta }^2{(a^2-{\beta }^2+{\gamma }^2)}^2}{(a+\beta +\gamma )(-a+\beta +\gamma )(a-\beta +\gamma )(a+\beta -\gamma )}$$$$\mathrm{we}\mathrm{are}\mathrm{left}\mathrm{with}\left(1\right)\mathrm{which}\mathrm{leads}\mathrm{to}$$$$a^{14}+\mathcal{B}{a}^{12}+\mathcal{C}{a}^{10}+\mathcal{D}{a}^8+\mathcal{E}{a}^6+\mathcal{F}{a}^4+\mathcal{G}{a}^2+\mathcal{H}=0$$$$a=\sqrt{s}$$$$s^7+\mathcal{B}{s}^6+\mathcal{C}{s}^5+\mathcal{D}{s}^4+\mathcal{E}{s}^3+\mathcal{F}{s}^2+\mathcal{G}s+\mathcal{H}=0$$$$\mathrm{with}\mathcal{H}={\alpha }^2{(\beta -\gamma )}^6{(\beta +\gamma )}^6.\mathrm{luckily}\mathrm{we}\mathrm{find}$$$$\mathrm{that}\pm ({\beta }^2-{\gamma }^2)\mathrm{are}\mathrm{double}\mathrm{zeros}\mathrm{each}\Rightarrow$$$$s^3+({\alpha }^2-2({\beta }^2+{\gamma }^2))s^2-(2{\alpha }^2({\beta }^2+{\gamma }^2)-{({\beta }^2+{\gamma }^2)}^2)s+{\alpha }^2{({\beta }^2-{\gamma }^2)}^2=0$$$$\left(\mathrm{which}\mathrm{btw}\mathrm{looks}\mathrm{exactly}\mathrm{like}\mathrm{the}\mathrm{above}\mathrm{equations}\right)$$$$s=t-\frac{{\alpha }^2-2({\beta }^2+{\gamma }^2)}{3}$$$$t^3-\frac{{({\alpha }^2+{\beta }^2+{\gamma }^2)}^2}{3}t+\frac{2{({\alpha }^2+{\beta }^2+{\gamma }^2)}^3}{27}-4{\alpha }^2{\beta }^2{\gamma }^2=0$$$$\mathrm{this}\mathrm{has}\mathrm{got}3\mathrm{real}\mathrm{solutions}\mathrm{for}\alpha ,\beta ,\gamma \in \mathbb{R}$$$$\Rightarrow \mathrm{we}\mathrm{need}\mathrm{the}\mathrm{trigonometric}\mathrm{method}\mathrm{which}$$$$\mathrm{gives}\mathrm{no}\mathrm{useable}\mathrm{general}\mathrm{solution}\mathrm{in}\mathrm{this}\mathrm{case}$$

Answered by ajfour last updated on 10/Jun/19

Commented by ajfour last updated on 10/Jun/19

$$A(\alpha ,0);B(-x,\sqrt{{\beta }^2-x^2});$$$$C(-x,-\sqrt{{\gamma }^2-x^2})$$$$BH\mathrm{\perp }AC$$$$\Rightarrow \frac{\sqrt{{\gamma }^2-x^2}}{\alpha +x}=\frac{x}{\sqrt{{\beta }^2-x^2}}$$$$\Rightarrow ({\beta }^2-x^2)({\gamma }^2-x^2)=x^2{(\alpha +x)}^2$$$$2\alpha x^3+({\alpha }^2+{\beta }^2+{\gamma }^2)x^2-{\beta }^2{\gamma }^2=0$$$$(xisobtainedfromthiseq.)$$$$Now$$$$\mathit{a}=\mid \sqrt{{\beta }^2-x^2}\pm \sqrt{{\gamma }^2-x^2}\mid$$$$\left(oneofthetwosigns\right)$$$$\mathit{b}=\sqrt{{(\alpha +x)}^2+{\gamma }^2-x^2}$$$$\mathit{c}=\sqrt{{(\alpha +x)}^2+{\beta }^2-x^2}.$$

Commented by mr W last updated on 10/Jun/19

$$thankyousir!itriedthesameway.$$$$butthecubicequationhasusually3$$$$solutions.howcanweseewhichone$$$$istherightone?oristhetriangle$$$$unique?$$

Answered by mr W last updated on 12/Jun/19

Commented by mr W last updated on 21/Jun/19

$$assume\alpha ⩾\beta ⩾\gamma .$$$$letHD=u=\lambda \alpha$$$$BD=\sqrt{{\beta }^2-u^2}$$$$DC=\sqrt{{\gamma }^2-u^2}$$$$\frac{DC}{HD}=\frac{AD}{BD}$$$$\frac{\sqrt{{\gamma }^2-u^2}}{u}=\frac{u+\alpha }{\sqrt{{\beta }^2-u^2}}$$$$\Rightarrow \sqrt{({\beta }^2-u^2)({\gamma }^2-u^2)}=u(u+\alpha )$$$$\Rightarrow ({\beta }^2-u^2)({\gamma }^2-u^2)=u^2(u^2+2\alpha u+{\alpha }^2)$$$$\Rightarrow 2\alpha u^3+({\alpha }^2+{\beta }^2+{\gamma }^2)u^2-{\beta }^2{\gamma }^2=0$$$$\Rightarrow {\lambda }^3+\frac{{\alpha }^2+{\beta }^2+{\gamma }^2}{2{\alpha }^2}{\lambda }^2-\frac{{\beta }^2{\gamma }^2}{2{\alpha }^4}=0$$$$withq=\frac{{\alpha }^2+{\beta }^2+{\gamma }^2}{6{\alpha }^2},r=\frac{{\beta }^2{\gamma }^2}{4{\alpha }^4}$$$$letp=\frac{r}{q^3}=\frac{54{\alpha }^2{\beta }^2{\gamma }^2}{{({\alpha }^2+{\beta }^2+{\gamma }^2)}^3}⩽2$$$$\Rightarrow {\lambda }^3+3q{\lambda }^2-2r=0$$$$with\lambda =s-q$$$$\Rightarrow s^3-3q^{2}s+2(q^3-r)=0$$$$\mathrm{\Delta }={(-q^2)}^3+{(q^3-r)}^2=-(2-\frac{r}{q^3}){rq}^3<0$$$$\Rightarrow s=2q\sin \left\{\frac{1}{3}[{\sin }^{-1}(1-p)+2k\pi ]\right\},k=0,1,2$$$$\Rightarrow \lambda =s-q=q\{2\sin [\frac{1}{3}{\sin }^{-1}(1-p)+\frac{2}{3}k\pi ]-1\},k=0,1,2$$$$twosolutionsaresuitable:k=0and1$$$$\Rightarrow \lambda =q[2\sin \left\{\frac{1}{3}{\sin }^{-1}(1-p)\right\}-1]<0or$$$$\Rightarrow \lambda =q[2\sin \{\frac{1}{3}{\sin }^{-1}(1-p)+\frac{2\pi }{3}\}-1]>0$$$$\Rightarrow a=\sqrt{{\beta }^2-{\lambda }^2{\alpha }^2}\pm \sqrt{{\gamma }^2-{\lambda }^2{\alpha }^2}(-vefor\lambda <0)$$$$\Rightarrow b=\sqrt{(1+2\lambda ){\alpha }^2+{\gamma }^2}$$$$\Rightarrow c=\sqrt{(1+2\lambda ){\alpha }^2+{\beta }^2}$$$$$$$$example:\alpha =7,\beta =6,\gamma =5$$$$\Rightarrow \lambda =0.35604/-0.597544$$$$\Rightarrow a=9.792451/1.562276$$$$\Rightarrow b=10.435137/3.929466$$$$\Rightarrow c=10.949525/5.142053$$

Commented by MJS last updated on 12/Jun/19

$$\mathrm{yes}{\mathrm{you}}^{\prime }\mathrm{re}\mathrm{right}$$

Answered by ajfour last updated on 12/Jun/19

Commented by mr W last updated on 13/Jun/19

$$thankyouforthisnewwaysir!$$$$itseemsthatallmethodsleadfinally$$$$toacubicequation.$$

Commented by ajfour last updated on 13/Jun/19

$$letBorigin.$$$$H(\beta \cos \theta ,\beta \sin \theta )....\left(i\right)$$$$A(\beta \cos \theta ,\beta \sin \theta \pm \alpha )...\left(ii\right)$$$$eq.ofAB:$$$$y=x\cot \varphi$$$$eq.ofAC:$$$$y=-(x-a)\cot \theta$$$$y_A=x_A\cot \varphi =(a-x_A)\cot \theta$$$$\Rightarrow x_A=\frac{a\cot \theta }{\cot \theta +\cot \varphi }=\beta \cos \theta \left[see\left(i\right)\right]$$$$and{y}_A=x_A\cot \varphi =\beta \sin \theta \pm \alpha \left[see\left(ii\right)\right]$$$$Also\beta \sin \theta =\gamma \sin \varphi$$$$\Rightarrow \beta \cos \theta \cot \varphi =\beta \sin \theta \pm \alpha$$$$\left(\beta \cos \theta \right)\left(\frac{\pm \sqrt{1-\frac{{\beta }^2\sin {}^2\theta }{{\gamma }^2}}}{\frac{\beta \sin \theta }{\gamma }}\right)=\beta \sin \theta \pm \alpha$$$$let\sin {}^2\theta =t$$$$\Rightarrow (1-t)({\gamma }^2-{\beta }^{2}t)=t({\beta }^{2}t+{\alpha }^2\pm 2\alpha \beta \sqrt{t})$$$$\Rightarrow {\gamma }^2-{\beta }^{2}t-{\gamma }^{2}t+{\beta }^2{t}^2={\beta }^2{t}^2+{\alpha }^{2}t\pm 2\alpha \beta t\sqrt{t}$$$$\Rightarrow 4{\alpha }^2{\beta }^2{t}^3={[{\gamma }^2-({\alpha }^2+{\beta }^2+{\gamma }^2)t]}^2$$$$or$$$$4{\alpha }^2{\beta }^2{t}^3-{({\alpha }^2+{\beta }^2+{\gamma }^2)}^2{t}^2$$$$+2{\gamma }^2{({\alpha }^2+{\beta }^2+{\gamma }^2)}^{2}t-{\gamma }^4=0$$$$for\alpha =7,\beta =6,\gamma =5$$$$....$$

Commented by ajfour last updated on 13/Jun/19

Commented by ajfour last updated on 13/Jun/19

$$x=\sin {}^2\theta ,ais\alpha ,b=\beta ,c=\gamma$$$$\left(justforthecalculatorresults\right)$$$$\sin {}^2\theta =1.05631goesinvalid$$$$hencetwotrianglesarepossible.$$

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