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Question-213054

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Question Number 213054 by 281981 last updated on 29/Oct/24
Commented by Ghisom last updated on 30/Oct/24
in an equilateral triangle the ratio of the areas of the triangle and its circumcircle is 1:((4π(√3))/( 9)) ≈ 1:2.4 here the area of the circumcircle is ((4225π)/(64))≈207.4 the triangle is not equilateral ⇒ the ratio is smaller than 1:2.4 ⇒ max area =((12675(√3))/(256))≈85.8 ⇒ the answer should be 84 because the triangle still looks “almost equilateral”
$$\mathrm{in}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{areas}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{and}\:\mathrm{its}\:\mathrm{circumcircle} \\ $$$$\mathrm{is} \\ $$$$\mathrm{1}:\frac{\mathrm{4}\pi\sqrt{\mathrm{3}}}{\:\mathrm{9}}\:\approx\:\mathrm{1}:\mathrm{2}.\mathrm{4} \\ $$$$\mathrm{here}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{is} \\ $$$$\frac{\mathrm{4225}\pi}{\mathrm{64}}\approx\mathrm{207}.\mathrm{4} \\ $$$$\mathrm{the}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{not}\:\mathrm{equilateral}\:\Rightarrow\:\mathrm{the}\:\mathrm{ratio} \\ $$$$\mathrm{is}\:\mathrm{smaller}\:\mathrm{than}\:\mathrm{1}:\mathrm{2}.\mathrm{4}\:\Rightarrow \\ $$$$\mathrm{max}\:\mathrm{area}\:=\frac{\mathrm{12675}\sqrt{\mathrm{3}}}{\mathrm{256}}\approx\mathrm{85}.\mathrm{8} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{answer}\:\mathrm{should}\:\mathrm{be}\:\mathrm{84}\:\mathrm{because}\:\mathrm{the} \\ $$$$\mathrm{triangle}\:\mathrm{still}\:\mathrm{looks}\:“\mathrm{almost}\:\mathrm{equilateral}'' \\ $$
Answered by mr W last updated on 30/Oct/24
Commented by mr W last updated on 30/Oct/24
Δ=area of ABC s=((a+b+c)/2) r_1 =((rs)/(s−a))=((2Δ)/(−a+b+c)) ⇒−a+b+c=((2Δ)/r_1 ) ...(i) r_2 =((rs)/(s−b))=((2Δ)/(a−b+c)) ⇒a−b+c=((2Δ)/r_2 ) ...(ii) R=((abc)/(4Δ)) ⇒abc=4RΔ ...(iii) (i)+(ii): c=((1/r_1 )+(1/r_2 ))Δ=((2/(21))+(1/(12)))Δ=((5Δ)/(28)) (iii): ab=((4RΔ)/c)=((4×65×28)/(8×5))=182 (i)−(ii): b−a=((1/r_1 )−(1/r_2 ))Δ=((2/(21))−(1/(12)))Δ=(Δ/(84)) ((b−a)/c)=((28)/(5×84))=(1/(15)) ⇒b−a=(c/(15)) −a, b are roots of x^2 −(c/(15))x−182=0 a=−(c/(30))+(1/(30))(√(c^2 +182×30^2 )) b=(c/(30))+(1/(30))(√(c^2 +182×30^2 )) s=(c/2)+((√(c^2 +182×30^2 ))/(30)) Δ=(√(s(s−a)(s−b)(s−c))) =(√(((c/2)+((√(c^2 +182×30^2 ))/(30)))((c/2)−(c/(30)))((c/2)+(c/(30)))(((√(c^2 +182×30^2 ))/(30))−(c/2)))) =((2c)/(15))(√(14(((c^2 +182×30^2 )/(30^2 ))−(c^2 /4)))) =((28c)/(15))(√(13−((4c^2 )/(225)))) (iii): 182c=4×((65)/8)×((28c)/(15))(√(13−((4c^2 )/(225)))) 3=(√(13−((4c^2 )/(225)))) ⇒c=15 (⇒a=13, b=14) Δ=((abc)/(4R))=((182×8×15)/(4×65))=84 ✓
$$\Delta={area}\:{of}\:{ABC} \\ $$$${s}=\frac{{a}+{b}+{c}}{\mathrm{2}} \\ $$$${r}_{\mathrm{1}} =\frac{{rs}}{{s}−{a}}=\frac{\mathrm{2}\Delta}{−{a}+{b}+{c}}\:\Rightarrow−{a}+{b}+{c}=\frac{\mathrm{2}\Delta}{{r}_{\mathrm{1}} }\:\:\:…\left({i}\right) \\ $$$${r}_{\mathrm{2}} =\frac{{rs}}{{s}−{b}}=\frac{\mathrm{2}\Delta}{{a}−{b}+{c}}\:\Rightarrow{a}−{b}+{c}=\frac{\mathrm{2}\Delta}{{r}_{\mathrm{2}} }\:\:\:…\left({ii}\right) \\ $$$${R}=\frac{{abc}}{\mathrm{4}\Delta}\:\Rightarrow{abc}=\mathrm{4}{R}\Delta\:\:\:…\left({iii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$${c}=\left(\frac{\mathrm{1}}{{r}_{\mathrm{1}} }+\frac{\mathrm{1}}{{r}_{\mathrm{2}} }\right)\Delta=\left(\frac{\mathrm{2}}{\mathrm{21}}+\frac{\mathrm{1}}{\mathrm{12}}\right)\Delta=\frac{\mathrm{5}\Delta}{\mathrm{28}} \\ $$$$\left({iii}\right): \\ $$$${ab}=\frac{\mathrm{4}{R}\Delta}{{c}}=\frac{\mathrm{4}×\mathrm{65}×\mathrm{28}}{\mathrm{8}×\mathrm{5}}=\mathrm{182} \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$${b}−{a}=\left(\frac{\mathrm{1}}{{r}_{\mathrm{1}} }−\frac{\mathrm{1}}{{r}_{\mathrm{2}} }\right)\Delta=\left(\frac{\mathrm{2}}{\mathrm{21}}−\frac{\mathrm{1}}{\mathrm{12}}\right)\Delta=\frac{\Delta}{\mathrm{84}} \\ $$$$\frac{{b}−{a}}{{c}}=\frac{\mathrm{28}}{\mathrm{5}×\mathrm{84}}=\frac{\mathrm{1}}{\mathrm{15}} \\ $$$$\Rightarrow{b}−{a}=\frac{{c}}{\mathrm{15}} \\ $$$$−{a},\:{b}\:{are}\:{roots}\:{of} \\ $$$${x}^{\mathrm{2}} −\frac{{c}}{\mathrm{15}}{x}−\mathrm{182}=\mathrm{0} \\ $$$${a}=−\frac{{c}}{\mathrm{30}}+\frac{\mathrm{1}}{\mathrm{30}}\sqrt{{c}^{\mathrm{2}} +\mathrm{182}×\mathrm{30}^{\mathrm{2}} } \\ $$$${b}=\frac{{c}}{\mathrm{30}}+\frac{\mathrm{1}}{\mathrm{30}}\sqrt{{c}^{\mathrm{2}} +\mathrm{182}×\mathrm{30}^{\mathrm{2}} } \\ $$$${s}=\frac{{c}}{\mathrm{2}}+\frac{\sqrt{{c}^{\mathrm{2}} +\mathrm{182}×\mathrm{30}^{\mathrm{2}} }}{\mathrm{30}} \\ $$$$\Delta=\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)} \\ $$$$\:\:\:\:=\sqrt{\left(\frac{{c}}{\mathrm{2}}+\frac{\sqrt{{c}^{\mathrm{2}} +\mathrm{182}×\mathrm{30}^{\mathrm{2}} }}{\mathrm{30}}\right)\left(\frac{{c}}{\mathrm{2}}−\frac{{c}}{\mathrm{30}}\right)\left(\frac{{c}}{\mathrm{2}}+\frac{{c}}{\mathrm{30}}\right)\left(\frac{\sqrt{{c}^{\mathrm{2}} +\mathrm{182}×\mathrm{30}^{\mathrm{2}} }}{\mathrm{30}}−\frac{{c}}{\mathrm{2}}\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{2}{c}}{\mathrm{15}}\sqrt{\mathrm{14}\left(\frac{{c}^{\mathrm{2}} +\mathrm{182}×\mathrm{30}^{\mathrm{2}} }{\mathrm{30}^{\mathrm{2}} }−\frac{{c}^{\mathrm{2}} }{\mathrm{4}}\right)} \\ $$$$\:\:\:=\frac{\mathrm{28}{c}}{\mathrm{15}}\sqrt{\mathrm{13}−\frac{\mathrm{4}{c}^{\mathrm{2}} }{\mathrm{225}}} \\ $$$$\left({iii}\right): \\ $$$$\mathrm{182}{c}=\mathrm{4}×\frac{\mathrm{65}}{\mathrm{8}}×\frac{\mathrm{28}{c}}{\mathrm{15}}\sqrt{\mathrm{13}−\frac{\mathrm{4}{c}^{\mathrm{2}} }{\mathrm{225}}} \\ $$$$\mathrm{3}=\sqrt{\mathrm{13}−\frac{\mathrm{4}{c}^{\mathrm{2}} }{\mathrm{225}}} \\ $$$$\Rightarrow{c}=\mathrm{15}\:\:\left(\Rightarrow{a}=\mathrm{13},\:{b}=\mathrm{14}\right) \\ $$$$\Delta=\frac{{abc}}{\mathrm{4}{R}}=\frac{\mathrm{182}×\mathrm{8}×\mathrm{15}}{\mathrm{4}×\mathrm{65}}=\mathrm{84}\:\checkmark \\ $$

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