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Question Number 218580 by mr W last updated on 12/Apr/25

Commented by vnm last updated on 12/Apr/25

$$$$$$\mathrm{Let}\mathrm{triangle}\mathrm{ABC}\mathrm{be}\mathrm{inscribed}\mathrm{in}\mathrm{equilateral}\mathrm{triangle}\mathrm{DEF}.$$$$\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{DEF}\mathrm{is}\mathrm{maximal}$$$$\mathrm{if}\mathrm{there}\mathrm{exists}\mathrm{a}\mathrm{point}\mathrm{G}\mathrm{inside}\mathrm{ABC}\mathrm{such}\mathrm{that}\mathrm{GA},\mathrm{GB},\mathrm{GC}\mathrm{are}\mathrm{perpendicular}$$$$\mathrm{to}\mathrm{EF},\mathrm{FD},\mathrm{DE}.$$$$\mathrm{let}BC=a,CA=b,AB=c$$$$GA=x,GB=y,GC=z$$$$\mathrm{\angle }BGC=\mathrm{\angle }CGA=\mathrm{\angle }AGB=\frac{2\pi }{3}$$$$\{\begin{array}{l}{y}^2+z^2+yz=a^2\\ z^2+x^2+zx=b^2\\ x^2+y^2+xy=c^2\end{array}$$$$y^2-x^2+z(y-x)=a^2-b^2$$$$(x+y+z)(y-x)=a^2-b^2$$$$(x+y+z)(z-x)=a^2-c^2$$$$x+y+z=w$$$$x-y=\frac{b^2-a^2}{w},x-z=\frac{c^2-a^2}{w}$$$$x=\frac{1}{3}(w+\frac{b^2+c^2-2a^2}{w})$$$$y=\frac{1}{3}(w+\frac{c^2+a^2-2b^2}{w})$$$$z=\frac{1}{3}(w+\frac{a^2+b^2-2c^2}{w})$$$$x^2+y^2+xy=c^2$$$$.................................$$$${\left(w^2\right)}^2-w^2(a^2+b^2+c^2)+$$$$(a^4+b^4+c^4-b^2{c}^2-c^2{a}^2-a^2{b}^2)=0$$$$w^2=\frac{a^2+b^2+c^2+\sqrt{3(2(b^2{c}^2+c^2{a}^2+a^2{b}^2)-a^4-b^4-c^4)}}{2}$$$$\mathrm{In}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}\mathrm{the}$$$$\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{distances}\mathrm{from}\mathrm{any}$$$$\mathrm{point}\mathrm{to}\mathrm{the}\mathrm{sides}\mathrm{is}\mathrm{a}\mathrm{constant}$$$$\mathrm{value}\mathrm{and}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{altitude}\mathrm{of}\mathrm{the}\mathrm{triangle},w\mathrm{is}\mathrm{the}\mathrm{altitude}.$$$$S_{△DEF}=\frac{w^2}{\sqrt{3}}=\frac{a^2+b^2+c^2+\sqrt{3(2(b^2{c}^2+c^2{a}^2+a^2{b}^2)-a^4-b^4-c^4)}}{2\sqrt{3}}$$

Commented by mr W last updated on 12/Apr/25

$$foragiventrianglewithsidesa,b,c,$$$$1)findtheareaofthelargest$$$$equilateraltrianglewhichcan$$$$circumscribeit.\left(\mathit{r}\mathit{e}\mathit{d}infigure\right)$$$$2)findthesmallestequilateral$$$$trianglewhichcanbeinscribed$$$$init.\left(\mathit{b}\mathit{l}\mathit{u}\mathit{e}infigure\right)$$

Commented by mr W last updated on 13/Apr/25

$$thansk!$$$$perfectlysolved!$$$$considering{heron}^{\prime }sformula,the$$$$resultcanalsobeexpressedas$$$$\frac{a^2+b^2+c^2}{2\sqrt{3}}+2\mathrm{\Delta }$$$$with\mathrm{\Delta }=areaoftriangleABC$$

Answered by vnm last updated on 18/Apr/25

$$$$$$Analgorithmforfindingthe$$$$areaofthebluetriangle.$$$$TriangleABC:a=BC,b=CA,c=AB$$$$\alpha =\arccos \frac{b^2+c^2-a^2}{2bc},\beta =\arccos \frac{c^2+a^2-b^2}{2ca},\gamma =\arccos \frac{a^2+b^2-c^2}{2ab}$$$${\alpha }_1=\pi -\alpha ,{\beta }_1=\pi -\beta ,{\gamma }_1=\pi -\gamma$$$$PQRisanequilateraltrianglewithsidelength=1$$$$pointMinsidePQR,suchthat$$$${\alpha }_1=\mathrm{\angle }QMR,{\beta }_1=\mathrm{\angle }RMP,{\gamma }_1=\mathrm{\angle }PMQ$$$$p=MP,q=MQ,r=MR$$$${\phi }_{\alpha }=\mathrm{\angle }MQR=\frac{\pi -{\alpha }_1}{2}-\delta ,{\theta }_{\alpha }=\mathrm{\angle }MRQ=\frac{\pi -{\alpha }_1}{2}+\delta$$$$\frac{r}{\sin {\phi }_{\alpha }}=\frac{1}{\sin {\alpha }_1}=\frac{q}{\sin {\theta }_{\alpha }}$$$$q=\frac{\sin {\theta }_{\alpha }}{\sin {\alpha }_1},r=\frac{\sin {\phi }_{\alpha }}{\sin {\alpha }_1}n$$$${\theta }_{\gamma }=\mathrm{\angle }MQP=\frac{\pi }{3}-{\phi }_{\alpha }=$$$$\frac{\pi }{3}-(\frac{\pi -{\alpha }_1}{2}-\delta )=\frac{{\alpha }_1}{2}-\frac{\pi }{6}+\delta$$$${\phi }_{\beta }=\mathrm{\angle }MRP=\frac{\pi }{3}-{\theta }_{\alpha }=$$$$\frac{\pi }{3}-(\frac{\pi -{\alpha }_1}{2}+\delta )=\frac{{\alpha }_1}{2}-\frac{\pi }{6}-\delta$$$$\frac{p}{\sin {\theta }_{\gamma }}=\frac{1}{\sin {\gamma }_1},p=\frac{\sin {\theta }_{\gamma }}{\sin {\gamma }_1}$$$$\frac{p}{\sin {\phi }_{\beta }}=\frac{1}{\sin {\beta }_1},p=\frac{\sin {\phi }_{\beta }}{\sin {\beta }_1}$$$$\frac{\sin {\theta }_{\gamma }}{\sin {\gamma }_1}=\frac{\sin {\phi }_{\beta }}{\sin {\beta }_1},\sin {\beta }_1\cdot \sin {\theta }_{\gamma }=\sin {\gamma }_1\cdot \sin {\phi }_{\beta }$$$$\sin {\beta }_1\cdot \sin (\frac{{\alpha }_1}{2}-\frac{\pi }{6}+\delta )=\sin {\gamma }_1\cdot \sin (\frac{{\alpha }_1}{2}-\frac{\pi }{6}-\delta )$$$$\sin {\beta }_1(\sin \frac{3{\alpha }_1-\pi }{6}\cos \delta +\cos \frac{3{\alpha }_1-\pi }{6}\sin \delta )=$$$$\sin {\gamma }_1(\sin \frac{3{\alpha }_1-\pi }{6}\cos \delta -\cos \frac{3{\alpha }_1-\pi }{6}\sin \delta$$$$(\sin {\gamma }_1-\sin {\beta }_1)\tan \frac{3{\alpha }_1-\pi }{6}=(\sin {\gamma }_1+\sin {\beta }_1)\tan \delta$$$$\delta =\arctan \left(\frac{\sin {\gamma }_1-\sin {\beta }_1}{\sin {\gamma }_1+\sin {\beta }_1}\tan \frac{3{\alpha }_1-\pi }{6}\right)$$$$Nowwecancalculatep,q,r.$$$$DrawlinesperpendiculartoMP,MQ,MR$$$$throughpointsP,Q,R.$$$$ThepointsF,G,Hwherethese$$$$linesintersectarethevertices$$$$ofatrianglesimilartotriangle$$$$ABC.$$$$f=GH=\frac{1}{2}((p+q)\tan \frac{{\gamma }_1}{2}+(q-p)\cot \frac{{\gamma }_1}{2})+\frac{1}{2}((p+r)\tan \frac{{\beta }_1}{2}+(r-p)\cot \frac{{\beta }_1}{2})$$$$g=HF=\frac{1}{2}((q+r)\tan \frac{{\alpha }_1}{2}+(r-q)\cot \frac{{\alpha }_1}{2})+\frac{1}{2}((q+p)\tan \frac{{\gamma }_1}{2}+(p-q)\cot \frac{{\gamma }_1}{2})$$$$h=FG=\frac{1}{2}((r+p)\tan \frac{{\beta }_1}{2}+(p-r)\cot \frac{{\beta }_1}{2})+\frac{1}{2}((r+q)\tan \frac{{\alpha }_1}{2}+(q-r)\cot \frac{{\alpha }_1}{2})$$$$△PQRistheminimalequilateraltriangleinscribedin△FGH.$$$$Answer:$$$$S_{△min.equilat}=S_{△ABC}\frac{S_{△PQR}}{S_{△FGH}}=\frac{\sqrt{3}{S}_{△ABC}}{4S_{△FGH}}$$$$forexample$$$$a=17,b=13,c=11$$$$S_{△ABC}=71.4995629357271$$$$\mathrm{p}=0.368961052327$$$$q=0.689772071985$$$$r=0.721618806573$$$$\mathrm{f}=2.755346135575$$$$g=2.107029397793$$$$h=1.782871028902$$$$\frac{f}{a}=\frac{g}{b}=\frac{h}{c}=0.162079184446$$$$S_{△min.equilat}=16.483375438503$$

Commented by mr W last updated on 20/Apr/25

$$thankssir!$$

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