Question Number 217080 by hardmath last updated on 28/Feb/25
Commented by mr W last updated on 01/Mar/25
$${also}\:{Area}\left({ABCD}\right)\leqslant\sqrt{{abcd}} \\ $$
Answered by mr W last updated on 01/Mar/25
Commented by mr W last updated on 01/Mar/25
![say the diagonals are e and f with angle θ between them. according to euler′s quadrilateral theorem: a^2 +b^2 +c^2 +d^2 =e^2 +f^2 +4g^2 ⇒e^2 +f^2 ≤a^2 +b^2 +c^2 +d^2 [ABCD]=(([A′B′C′D′])/2) =((ef sin θ)/2) ≤((ef)/2) ≤((e^2 +f^2 )/4) ≤((a^2 +b^2 +c^2 +d^2 )/4) ✓](Q217117.png)
$${say}\:{the}\:{diagonals}\:{are}\:{e}\:{and}\:{f}\:{with} \\ $$$${angle}\:\theta\:{between}\:{them}. \\ $$$${according}\:{to}\:{euler}'{s}\:{quadrilateral} \\ $$$${theorem}: \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} ={e}^{\mathrm{2}} +{f}^{\mathrm{2}} +\mathrm{4}{g}^{\mathrm{2}} \\ $$$$\Rightarrow{e}^{\mathrm{2}} +{f}^{\mathrm{2}} \leqslant{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \\ $$$$\left[{ABCD}\right]=\frac{\left[{A}'{B}'{C}'{D}'\right]}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{ef}\:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leqslant\frac{{ef}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leqslant\frac{{e}^{\mathrm{2}} +{f}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leqslant\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }{\mathrm{4}}\:\:\:\checkmark \\ $$
Answered by mr W last updated on 02/Mar/25
$${a}\:{quadrilateral}\:{with}\:{sides}\:{a},\:{b},\:{c},\:{d} \\ $$$${has}\:{the}\:{maximum}\:{area}\:{when}\:{it}\:{is} \\ $$$${cyclic}.\:\:{this}\:{maximum}\:{area}\:{is} \\ $$$$\sqrt{\left({s}−{a}\right)\left({a}−{b}\right)\left({s}−{c}\right)\left({s}−{d}\right)}\:{with} \\ $$$${s}=\frac{{a}+{b}+{c}+{d}}{\mathrm{2}}.\:{my}\:{proof}\:{see}\:{Q}\mathrm{30233}. \\ $$$${that}\:{means} \\ $$$${Area}\:\left({ABCD}\right)_{{arbitrary}} \\ $$$$\:\:\:\:\:\leqslant{Area}\:\left({ABCD}\right)_{{cyclic}} \\ $$$$\:\:\:\:\:=\sqrt{\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)\left({s}−{d}\right)}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:=\sqrt[{\mathrm{4}}]{\left({s}−{a}\right)^{\mathrm{2}} \left({s}−{b}\right)^{\mathrm{2}} \left({s}−{c}\right)^{\mathrm{2}} \left({s}−{d}\right)^{\mathrm{2}} }\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\leqslant\frac{\left({s}−{a}\right)^{\mathrm{2}} +\left({s}−{b}\right)^{\mathrm{2}} +\left({s}−{c}\right)^{\mathrm{2}} +\left({s}−{d}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{4}{s}^{\mathrm{2}} −\mathrm{2}{s}\left({a}+{b}+{c}+{d}\right)+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{4}{s}^{\mathrm{2}} −\mathrm{4}{s}^{\mathrm{2}} +{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }{\mathrm{4}}\:\checkmark \\ $$
Answered by mr W last updated on 04/Mar/25
$${Area}\left({ABCD}\right) \\ $$$$\:\:\:\:\:\:\:\:=\Delta\left({ABC}\right)+\Delta\left({CDA}\right) \\ $$$$\:\:\:\:\:\:\:\:=\frac{{ab}\:\mathrm{sin}\:\angle{B}}{\mathrm{2}}+\frac{{cd}\:\mathrm{sin}\:\angle{D}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\leqslant\frac{{ab}}{\mathrm{2}}+\frac{{cd}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\leqslant\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{4}}+\frac{{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }{\mathrm{4}}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }{\mathrm{4}}\:\checkmark \\ $$
Commented by hardmath last updated on 03/Mar/25
$$\mathrm{cool}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$