Question-222958

Question Number 222958 by mr W last updated on 11/Jul/25

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

find the area of the smallest inscribed equilateral triangle of a given triangle with sides of lengthes 3, 5, 7 respectively.

$${find}\:{the}\:{area}\:{of}\:{the}\:{smallest}\: \\ $$$${inscribed}\:{equilateral}\:{triangle}\:{of}\:{a} \\ $$$${given}\:{triangle}\:{with}\:{sides}\:{of}\:{lengthes} \\ $$$$\mathrm{3},\:\mathrm{5},\:\mathrm{7}\:{respectively}. \\ $$

Commented by Ghisom last updated on 12/Jul/25

do you have a solution? I get sidelength of △ ((15(√3))/(16))≈1.62 ⇒ area is ((675(√3))/(1024))≈1.14 ...there seems to be only one △ can you confirm this? I made some assumptions which could be wrong... I′ll post my path later.

$$\mathrm{do}\:\mathrm{you}\:\mathrm{have}\:\mathrm{a}\:\mathrm{solution}? \\ $$$$\mathrm{I}\:\mathrm{get}\:\mathrm{sidelength}\:\mathrm{of}\:\bigtriangleup\:\frac{\mathrm{15}\sqrt{\mathrm{3}}}{\mathrm{16}}\approx\mathrm{1}.\mathrm{62} \\ $$$$\Rightarrow\:\mathrm{area}\:\mathrm{is}\:\frac{\mathrm{675}\sqrt{\mathrm{3}}}{\mathrm{1024}}\approx\mathrm{1}.\mathrm{14} \\ $$$$…\mathrm{there}\:\mathrm{seems}\:\mathrm{to}\:\mathrm{be}\:\mathrm{only}\:\mathrm{one}\:\bigtriangleup \\ $$$$\mathrm{can}\:\mathrm{you}\:\mathrm{confirm}\:\mathrm{this}?\:\mathrm{I}\:\mathrm{made}\:\mathrm{some}\:\mathrm{assumptions} \\ $$$$\mathrm{which}\:\mathrm{could}\:\mathrm{be}\:\mathrm{wrong}…\:\mathrm{I}'\mathrm{ll}\:\mathrm{post}\:\mathrm{my}\:\mathrm{path} \\ $$$$\mathrm{later}. \\ $$

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

$${yes},\:{that}'{s}\:{correct}. \\ $$$${there}\:{are}\:{infinite}\:{such}\:{equilateral} \\ $$$${triangles}.\:{but}\:{there}\:{is}\:{a}\:{minimum}. \\ $$

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

$${please}\:{show}\:{your}\:{path}.\:{thanks}. \\ $$

Commented by mr W last updated on 17/Jul/25

Commented by mr W last updated on 17/Jul/25

say the smallest inscribed equilateral triangle is s_(min) . all triangles with angles α,β,γ are similar. for an equilateral triangle with given side length 1 we can find its largest circumtriangle with given angles α,β,γ. say the sides of this triangle are a′,b′,c′ respectively. we know (s_(min) /1)=(a/(a′))=(b/(b′))=(c/(c′)) in this way we can find s_(min) =((2(√2)Δ)/( (√(a^2 +b^2 +c^2 +4(√3)Δ)))) with Δ=area of triangle ABC.

$${say}\:{the}\:{smallest}\:{inscribed}\:{equilateral} \\ $$$${triangle}\:{is}\:{s}_{{min}} . \\ $$$${all}\:{triangles}\:{with}\:{angles}\:\alpha,\beta,\gamma\:{are} \\ $$$${similar}. \\ $$$${for}\:{an}\:{equilateral}\:{triangle}\:{with}\: \\ $$$${given}\:{side}\:{length}\:\mathrm{1}\:{we}\:{can}\:{find}\:{its} \\ $$$${largest}\:{circumtriangle}\:{with}\:{given} \\ $$$${angles}\:\alpha,\beta,\gamma.\:{say}\:{the}\:{sides}\:{of}\:{this} \\ $$$${triangle}\:{are}\:{a}',{b}',{c}'\:{respectively}. \\ $$$${we}\:{know} \\ $$$$\frac{{s}_{{min}} }{\mathrm{1}}=\frac{{a}}{{a}'}=\frac{{b}}{{b}'}=\frac{{c}}{{c}'} \\ $$$${in}\:{this}\:{way}\:{we}\:{can}\:{find} \\ $$$${s}_{{min}} =\frac{\mathrm{2}\sqrt{\mathrm{2}}\Delta}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{3}}\Delta}} \\ $$$${with}\:\Delta={area}\:{of}\:{triangle}\:{ABC}. \\ $$

Answered by gabthemathguy25 last updated on 12/Jul/25

s=((3+5+7)/2)=7.5 (√(7.5(7.5−3)(7.5−5)(7.5−7))) ⇒(√(7.5(4.5)(2.5)(0.5))) ⇒≈(√(42.1875))≈6.5 0.2 × 6.5 ≈ 1.3

$${s}=\frac{\mathrm{3}+\mathrm{5}+\mathrm{7}}{\mathrm{2}}=\mathrm{7}.\mathrm{5} \\ $$$$\sqrt{\mathrm{7}.\mathrm{5}\left(\mathrm{7}.\mathrm{5}−\mathrm{3}\right)\left(\mathrm{7}.\mathrm{5}−\mathrm{5}\right)\left(\mathrm{7}.\mathrm{5}−\mathrm{7}\right)} \\ $$$$\Rightarrow\sqrt{\mathrm{7}.\mathrm{5}\left(\mathrm{4}.\mathrm{5}\right)\left(\mathrm{2}.\mathrm{5}\right)\left(\mathrm{0}.\mathrm{5}\right)} \\ $$$$\Rightarrow\approx\sqrt{\mathrm{42}.\mathrm{1875}}\approx\mathrm{6}.\mathrm{5} \\ $$$$\mathrm{0}.\mathrm{2}\:×\:\mathrm{6}.\mathrm{5}\:\approx\:\mathrm{1}.\mathrm{3} \\ $$

Commented by Hanuda354 last updated on 12/Jul/25

$$\mathrm{What}\:\mathrm{do}\:\mathrm{you}\:\mathrm{mean},\:\:''\mathrm{0}.\mathrm{2}\:×\:\mathrm{6}.\mathrm{5}''\:? \\ $$

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

$${how}\:{can}\:{you}\:{say}\:{this}? \\ $$$${this}\:{question}\:{is}\:{quite}\:{hard}. \\ $$

Commented by gabthemathguy25 last updated on 12/Jul/25

Theres no formula for an equilateral triangle inscribes in a triangle, so i approximated. 0.2 because empirically, the smallest inscribed equilateral triangle often has an area about 20%-25% of the triangle's area (from geometric results).

Commented by AntonCWX8 last updated on 12/Jul/25

$${No}\:{worry}\:{sir}. \\ $$$${He}\:{is}\:{very}\:{good}\:{in}\:{approximating}\:{stuffs}. \\ $$$${Just}\:{look}\:{at}\:{his}\:{attempts}\:{at}\:{those}\:{hard}\:{integrals}\:{and}\:{you}'{ll}\:{see}. \\ $$

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

$${in}\:{current}\:{case}\:{he}\:{is}\:{even}\:{not} \\ $$$${approximating},\:{but}\:{just}\:{guessing}. \\ $$$${sorry},\:{here}\:{we}\:{are}\:{talking}\:{about}\: \\ $$$$“{mathematics}'',\:{such}\:{guessing}\:{or} \\ $$$${approximating}\:{don}'{t}\:{make}\:{any} \\ $$$${sense},\:{they}\:{are}\:{actually}\:{just}\:{spam}. \\ $$$${if}\:{we}\:{can}'{t}\:{solve}\:{a}\:{question}, \\ $$$${we}\:{can}\:{just}\:{leave}\:{it}\:{unsolved}.\:{we} \\ $$$${don}'{t}\:{need}\:{to}\:{post}\:{a}\:{naked}\:{value} \\ $$$${or}\:{approximation}\:{or}\:{something} \\ $$$${obtained}\:{by}\:{using}\:{a}\:{calculator}\:{or}\: \\ $$$${an}\:{AI}\:{app}\:{or}\:{WolframAlpha}\:{or}\: \\ $$$${whatever}\:{without}\:{any}\:{path}.\:{because}\: \\ $$$${other}\:{people}\:{have}\:{asked}\:{questions},\: \\ $$$${not}\:{because}\:{they}\:{don}'{t}\:{have}\:{a}\: \\ $$$${calculator}\:{or}\:{AI}\:{app}\:{or}\:{whatever},\: \\ $$$${but}\:{they}\:{want}\:{an}\:{accurate} \\ $$$$“{mathematic}''\:{solution}\:{with}\:{path}. \\ $$

Answered by AntonCWX8 last updated on 12/Jul/25

Denote P = point touching base BC L=length of equilateral triangle α=∠ABC x = BP PC = 7−x (L/(sin(21.8°)))=((7−x)/(sin(180°−α)))⇒L=(((7−x)sin(21.8°))/(sin(180°−α))) (L/(sin(38.2°)))=(x/(sin(α)))⇒L=((xsin(38.2°))/(sin(α))) (((7−x)sin(21.8°))/(sin(180°−α)))=((xsin(38.2))/(sin(α))), 0<x<7 Only solution: x≈2.626 but α does not exist... not sure why...

$${Denote}\: \\ $$$${P}\:=\:{point}\:{touching}\:{base}\:{BC} \\ $$$${L}={length}\:{of}\:{equilateral}\:{triangle} \\ $$$$\alpha=\angle{ABC} \\ $$$${x}\:=\:{BP} \\ $$$${PC}\:=\:\mathrm{7}−{x} \\ $$$$ \\ $$$$\frac{{L}}{{sin}\left(\mathrm{21}.\mathrm{8}°\right)}=\frac{\mathrm{7}−{x}}{{sin}\left(\mathrm{180}°−\alpha\right)}\Rightarrow{L}=\frac{\left(\mathrm{7}−{x}\right){sin}\left(\mathrm{21}.\mathrm{8}°\right)}{{sin}\left(\mathrm{180}°−\alpha\right)} \\ $$$$\frac{{L}}{{sin}\left(\mathrm{38}.\mathrm{2}°\right)}=\frac{{x}}{{sin}\left(\alpha\right)}\Rightarrow{L}=\frac{{xsin}\left(\mathrm{38}.\mathrm{2}°\right)}{{sin}\left(\alpha\right)} \\ $$$$ \\ $$$$\frac{\left(\mathrm{7}−{x}\right){sin}\left(\mathrm{21}.\mathrm{8}°\right)}{{sin}\left(\mathrm{180}°−\alpha\right)}=\frac{{xsin}\left(\mathrm{38}.\mathrm{2}\right)}{{sin}\left(\alpha\right)},\:\mathrm{0}<{x}<\mathrm{7} \\ $$$$ \\ $$$${Only}\:{solution}: \\ $$$${x}\approx\mathrm{2}.\mathrm{626} \\ $$$${but}\:\alpha\:{does}\:{not}\:{exist}… \\ $$$${not}\:{sure}\:{why}… \\ $$

Commented by AntonCWX8 last updated on 12/Jul/25