Question Number 183951 by Acem last updated on 01/Jan/23
Commented by Acem last updated on 01/Jan/23
$$\:\bullet{c}\:\:{Are}\:{the}\:{points}\:{A},\:{F},\:{N},\:{M}\:\in\:{a}\:{circle}? \\ $$$$\:\:\:\:\:\:\:\:\:{if}\:{they}\:{are}\:{then}\:{find}\:{it}\:{center}\:{position}\:{and}\:{it}\:{radius} \\ $$$$\:\ast\:{AN}\bot{BC},\:{BA}\bot{CA} \\ $$$$\: \\ $$
Answered by HeferH last updated on 01/Jan/23
$$\:{a}.\:{Since}\:\angle{BAC}\:=\:\mathrm{90}°\:{and}\:{AM}\:{is}\:{median}\:\Rightarrow \\ $$$$\angle{NAM}\:=\:\mathrm{30}° \\ $$$$\:{b}.\:{Yes},\:\:{the}\:{ratio}\:{of}\:{their}\:{sides}\:{is}\:\frac{\mathrm{1}}{\mathrm{2}}, \\ $$$$\:{there}\:{is}\:{a}\:{specific}\:{theorem},\:{if}\:{you}\:{join} \\ $$$$\:{the}\:{midpoints}\:{of}\:{any}\:{two}\:{sides}\:{of}\:{a}\:{triangle} \\ $$$$\:{the}\:{triangle}\:{that}\:{you}\:{get}\:{is}\:{going}\:{to}\:{be}\:{similar} \\ $$$$\:{to}\:{the}\:{first}\:{one}.\: \\ $$$$\:{c}.\:{Yes},\:{the}\:{quadrilateral}\:{AFNM}\:{would}\:{need}\:{to} \\ $$$$\:{be}\:{cyclic},\:{for}\:{that}\:\angle{F}\:+\:\angle{M}\:=\:\mathrm{180}°\:\vee\: \\ $$$$\angle{A}\:+\:\angle{N}\:=\:\mathrm{180}°.\:{In}\:{this}\:{case} \\ $$$$\:\:\mathrm{60}°\left(\angle{MAF}\:\right)+\mathrm{120}°\:\left({MNF}\right)=\:\mathrm{180}° \\ $$$$\left.\:{I}\:{will}\:{send}\:{an}\:{image}\:{later}\::\right)\:{and}\:{the}\:{radius} \\ $$$$\:\:{stuff} \\ $$$$\: \\ $$
Commented by HeferH last updated on 01/Jan/23
Commented by HeferH last updated on 01/Jan/23
$${AMNF}\:{end}\:{up}\:{being}\:{an}\:{isosceles}\:{trapezoid}. \\ $$$$\Rightarrow\:{r}\:=\:\mathrm{1} \\ $$$$\:{Also}: \\ $$$$\:\left({x},\:{y}\right)\:=\:\left(\frac{\mathrm{5}}{\mathrm{2}},\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:\:\left({According}\:{to}\:{the}\:{first}\:{diagram}\right)\: \\ $$$$\:{but}\:{I}\:{may}\:{have}\:{a}\:{mistake}\:{somewhere} \\ $$
Commented by Acem last updated on 01/Jan/23
$${Big}\:{Thanks}\:{for}\:{your}\:{efforts}\:{dear}\:{friend}! \\ $$$$\:{I}\:{only}\:{didn}'{t}\:{get}\:{the}\:{point}\:\left({x},\:{y}\right)\:,\:{if}\:{the}\:{origin}\:{was} \\ $$$$\:{M},\:{the}\:{center}\:{should}\:{be}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\:,\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\:{P}.{s}:\:{You}\:{can}\:{check}\:{my}\:{method} \\ $$$$ \\ $$
Commented by HeferH last updated on 01/Jan/23
$${I}\:{tought}\:{B}\:{was}\:{the}\:{origin},\:{so}\:{thats}\:{why}\: \\ $$$$\:{i}\:{got}\:{something}\:{else}\:{thank}\:{you}\:{for}\:{the} \\ $$$$\:{interesting}\:{problem}\:{friend}!\: \\ $$
Commented by Acem last updated on 01/Jan/23
$$\:{There}\:{is}\:{not}\:{any}\:{problem}\:{at}\:{your}\:{solution} \\ $$$$\:{If}\:{you}\:{deem}\:{B}\:{as}\:{the}\:{origin}\:{so}\:{the}\:{point}\:\left(\frac{\mathrm{5}}{\mathrm{2}}\:,\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\:{is}\:\mathrm{100\%}\:\:{correct}. \\ $$$$\:{Your}\:{solution}\:{is}\:{completly}\:{correct} \\ $$$$\:{Thank}\:{you}\:{very}\:{much}! \\ $$
Answered by Acem last updated on 01/Jan/23
Commented by Acem last updated on 01/Jan/23
Commented by Acem last updated on 01/Jan/23
