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Question Number 60313 by ajfour last updated on 19/May/19
Commented by ajfour last updated on 19/May/19
Find side length l of largest equilateral  triangle circumscribing a given  △ABC (sides a,b,c).
$$\mathrm{Find}\:\mathrm{side}\:\mathrm{length}\:\boldsymbol{{l}}\:\mathrm{of}\:\mathrm{largest}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}\:\mathrm{circumscribing}\:\mathrm{a}\:\mathrm{given} \\ $$$$\bigtriangleup\mathrm{ABC}\:\left(\mathrm{sides}\:\mathrm{a},\mathrm{b},\mathrm{c}\right). \\ $$
Answered by ajfour last updated on 19/May/19
  This question wasn′t solved  satisfactorily before, i just now  find   l_(max) =(√((2/3)(a^2 +b^2 +c^2 )+(8/( (√3)))(△_(ABC) )))  .
$$\:\:\mathrm{This}\:\mathrm{question}\:\mathrm{wasn}'\mathrm{t}\:\mathrm{solved} \\ $$$$\mathrm{satisfactorily}\:\mathrm{before},\:\mathrm{i}\:\mathrm{just}\:\mathrm{now} \\ $$$$\mathrm{find}\:\:\:{l}_{{max}} =\sqrt{\frac{\mathrm{2}}{\mathrm{3}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}}\left(\bigtriangleup_{\mathrm{ABC}} \right)}\:\:. \\ $$
Commented by ajfour last updated on 21/May/19
dont you like this solution mrW Sir,  it is to meet your previous objection  to a previous coordinate geometry  based solution of mine which wasn′t  as symmetric in a,b,c as this; if you  remember it at all..
$$\mathrm{dont}\:\mathrm{you}\:\mathrm{like}\:\mathrm{this}\:\mathrm{solution}\:\mathrm{mrW}\:\mathrm{Sir}, \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{to}\:\mathrm{meet}\:\mathrm{your}\:\mathrm{previous}\:\mathrm{objection} \\ $$$$\mathrm{to}\:\mathrm{a}\:\mathrm{previous}\:\mathrm{coordinate}\:\mathrm{geometry} \\ $$$$\mathrm{based}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{mine}\:\mathrm{which}\:\mathrm{wasn}'\mathrm{t} \\ $$$$\mathrm{as}\:\mathrm{symmetric}\:\mathrm{in}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{as}\:\mathrm{this};\:\mathrm{if}\:\mathrm{you} \\ $$$$\mathrm{remember}\:\mathrm{it}\:\mathrm{at}\:\mathrm{all}.. \\ $$
Commented by mr W last updated on 22/May/19
your solution is wonderful sir. the  result is in pefectform. i don′t  know how you came to taking the points  M and N. i would try to take ∠ABR=θ  as variable and get l=f(θ) and (dl/dθ)=0.
$${your}\:{solution}\:{is}\:{wonderful}\:{sir}.\:{the} \\ $$$${result}\:{is}\:{in}\:{pefectform}.\:{i}\:{don}'{t} \\ $$$${know}\:{how}\:{you}\:{came}\:{to}\:{taking}\:{the}\:{points} \\ $$$${M}\:{and}\:{N}.\:{i}\:{would}\:{try}\:{to}\:{take}\:\angle{ABR}=\theta \\ $$$${as}\:{variable}\:{and}\:{get}\:{l}={f}\left(\theta\right)\:{and}\:\frac{{dl}}{{d}\theta}=\mathrm{0}. \\ $$
Commented by ajfour last updated on 21/May/19
constancy of the vertex angles at  P,Q,R  led me into thinking this   Sir!
$$\mathrm{constancy}\:\mathrm{of}\:\mathrm{the}\:\mathrm{vertex}\:\mathrm{angles}\:\mathrm{at} \\ $$$$\mathrm{P},\mathrm{Q},\mathrm{R}\:\:\mathrm{led}\:\mathrm{me}\:\mathrm{into}\:\mathrm{thinking}\:\mathrm{this}\: \\ $$$$\mathrm{Sir}! \\ $$
Commented by mr W last updated on 21/May/19
great thought sir! this gives the  answer to my question how to determine  the max. circumscribing equilateral  triangle geometrically,  i.e. without using calculus.  i have been thinking about  this for two days. now i know how.  it answers also the question that  the searched triangle is a maximum,  not a minimum.
$${great}\:{thought}\:{sir}!\:{this}\:{gives}\:{the} \\ $$$${answer}\:{to}\:{my}\:{question}\:{how}\:{to}\:{determine} \\ $$$${the}\:{max}.\:{circumscribing}\:{equilateral} \\ $$$${triangle}\:{geometrically}, \\ $$$${i}.{e}.\:{without}\:{using}\:{calculus}. \\ $$$${i}\:{have}\:{been}\:{thinking}\:{about} \\ $$$${this}\:{for}\:{two}\:{days}.\:{now}\:{i}\:{know}\:{how}. \\ $$$${it}\:{answers}\:{also}\:{the}\:{question}\:{that} \\ $$$${the}\:{searched}\:{triangle}\:{is}\:{a}\:{maximum}, \\ $$$${not}\:{a}\:{minimum}. \\ $$
Commented by mr W last updated on 21/May/19
my result using geometrical method  is the same as your method using  calculus, see below.
$${my}\:{result}\:{using}\:{geometrical}\:{method} \\ $$$${is}\:{the}\:{same}\:{as}\:{your}\:{method}\:{using} \\ $$$${calculus},\:{see}\:{below}. \\ $$
Answered by ajfour last updated on 20/May/19
Commented by ajfour last updated on 20/May/19
cos 30° = ((BC/2)/(CM))   ⇒  ((√3)/2)=(a/2)×(1/(CM))  ⇒   PM=CM=(a/( (√3)))       (similarly  BN=(c/( (√3)))  )   .....(ii)  ∠BMP=180°−60°−θ  BP=2PMsin (((∠BMP)/2))  BP=((2a)/( (√3)))sin (60°−(θ/2))        ......(i)  let ∠B=β  and since ∠PBR=180°,  φ+30°+β+30°+(30°+(θ/2))=180°  ⇒   φ=90°−β−(θ/2)    BR = 2BNcos φ             =((2c)/( (√3)))sin (β+(θ/2))  let side of equilateral △PQR be l.   l= BP+BR      = ((2a)/( (√3)))sin (60°−(θ/2))+((2c)/( (√3)))sin (β+(θ/2))  ⇒l(√(3 ))=a((√3)cos (θ/2)−sin (θ/2))             +2c(sin βcos (θ/2)+cos βsin (θ/2))      l(√3) =(2ccos β−a)sin (θ/2)              +(2csin β+a(√3))cos (θ/2)    ((d(l(√3)))/dθ)=0   ⇒        tan (θ/2)=((2ccos β−a)/(2csin β+a(√3)))    l(√3)= (2ccos β−a){sin (θ/2)+((cos^2 (θ/2))/(sin (θ/2)))}         l(√3)  =  ((2ccos β−a)/(sin (θ/2)))   l_(max) (√3) = (√((2ccos β−a)^2 +(2csin β+a(√3))^2 ))       =(√(4a^2 +4c^2 +4ac((√3)sin β−cos β)))       = (√(4a^2 +4c^2 +8(√3)((1/2)acsin β)−4ac(((a^2 +c^2 −b^2 )/(2ac)))))    ⇒l(√3) = (√(2(a^2 +b^2 +c^2 )+8(√3)△_(ABC) ))    ___________________________     l_(max) =(√((2/3)(a^2 +b^2 +c^2 )+(8/( (√3)))△_(ABC) ))    ___________________________.
$$\mathrm{cos}\:\mathrm{30}°\:=\:\frac{\mathrm{BC}/\mathrm{2}}{\mathrm{CM}}\:\:\:\Rightarrow\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{a}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{CM}} \\ $$$$\Rightarrow\:\:\:\mathrm{PM}=\mathrm{CM}=\frac{\mathrm{a}}{\:\sqrt{\mathrm{3}}} \\ $$$$\:\:\:\:\:\left(\mathrm{similarly}\:\:\mathrm{BN}=\frac{\mathrm{c}}{\:\sqrt{\mathrm{3}}}\:\:\right)\:\:\:…..\left(\mathrm{ii}\right) \\ $$$$\angle\mathrm{BMP}=\mathrm{180}°−\mathrm{60}°−\theta \\ $$$$\mathrm{BP}=\mathrm{2PMsin}\:\left(\frac{\angle\mathrm{BMP}}{\mathrm{2}}\right) \\ $$$$\mathrm{BP}=\frac{\mathrm{2a}}{\:\sqrt{\mathrm{3}}}\mathrm{sin}\:\left(\mathrm{60}°−\frac{\theta}{\mathrm{2}}\right)\:\:\:\:\:\:\:\:……\left(\mathrm{i}\right) \\ $$$$\mathrm{let}\:\angle\mathrm{B}=\beta\:\:\mathrm{and}\:\mathrm{since}\:\angle\mathrm{PBR}=\mathrm{180}°, \\ $$$$\phi+\mathrm{30}°+\beta+\mathrm{30}°+\left(\mathrm{30}°+\frac{\theta}{\mathrm{2}}\right)=\mathrm{180}° \\ $$$$\Rightarrow\:\:\:\phi=\mathrm{90}°−\beta−\frac{\theta}{\mathrm{2}} \\ $$$$\:\:\mathrm{BR}\:=\:\mathrm{2BNcos}\:\phi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2c}}{\:\sqrt{\mathrm{3}}}\mathrm{sin}\:\left(\beta+\frac{\theta}{\mathrm{2}}\right) \\ $$$$\mathrm{let}\:\mathrm{side}\:\mathrm{of}\:\mathrm{equilateral}\:\bigtriangleup\mathrm{PQR}\:\mathrm{be}\:\boldsymbol{{l}}. \\ $$$$\:{l}=\:\mathrm{BP}+\mathrm{BR} \\ $$$$\:\:\:\:=\:\frac{\mathrm{2a}}{\:\sqrt{\mathrm{3}}}\mathrm{sin}\:\left(\mathrm{60}°−\frac{\theta}{\mathrm{2}}\right)+\frac{\mathrm{2c}}{\:\sqrt{\mathrm{3}}}\mathrm{sin}\:\left(\beta+\frac{\theta}{\mathrm{2}}\right) \\ $$$$\Rightarrow{l}\sqrt{\mathrm{3}\:}={a}\left(\sqrt{\mathrm{3}}\mathrm{cos}\:\frac{\theta}{\mathrm{2}}−\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2c}\left(\mathrm{sin}\:\beta\mathrm{cos}\:\frac{\theta}{\mathrm{2}}+\mathrm{cos}\:\beta\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\right) \\ $$$$\:\:\:\:{l}\sqrt{\mathrm{3}}\:=\left(\mathrm{2ccos}\:\beta−\mathrm{a}\right)\mathrm{sin}\:\frac{\theta}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:+\left(\mathrm{2csin}\:\beta+\mathrm{a}\sqrt{\mathrm{3}}\right)\mathrm{cos}\:\frac{\theta}{\mathrm{2}} \\ $$$$\:\:\frac{{d}\left({l}\sqrt{\mathrm{3}}\right)}{{d}\theta}=\mathrm{0}\:\:\:\Rightarrow \\ $$$$\:\:\:\:\:\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}=\frac{\mathrm{2ccos}\:\beta−\mathrm{a}}{\mathrm{2csin}\:\beta+\mathrm{a}\sqrt{\mathrm{3}}} \\ $$$$\:\:{l}\sqrt{\mathrm{3}}=\:\left(\mathrm{2}{c}\mathrm{cos}\:\beta−\mathrm{a}\right)\left\{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}+\frac{\mathrm{cos}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}\right\} \\ $$$$\:\:\:\:\:\:\:{l}\sqrt{\mathrm{3}}\:\:=\:\:\frac{\mathrm{2ccos}\:\beta−\mathrm{a}}{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}} \\ $$$$\:{l}_{{max}} \sqrt{\mathrm{3}}\:=\:\sqrt{\left(\mathrm{2}{c}\mathrm{cos}\:\beta−\mathrm{a}\right)^{\mathrm{2}} +\left(\mathrm{2csin}\:\beta+\mathrm{a}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:=\sqrt{\mathrm{4a}^{\mathrm{2}} +\mathrm{4c}^{\mathrm{2}} +\mathrm{4ac}\left(\sqrt{\mathrm{3}}\mathrm{sin}\:\beta−\mathrm{cos}\:\beta\right)} \\ $$$$\:\:\:\:\:=\:\sqrt{\mathrm{4a}^{\mathrm{2}} +\mathrm{4c}^{\mathrm{2}} +\mathrm{8}\sqrt{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{acsin}\:\beta\right)−\mathrm{4ac}\left(\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }{\mathrm{2ac}}\right)} \\ $$$$\:\:\Rightarrow\boldsymbol{{l}}\sqrt{\mathrm{3}}\:=\:\sqrt{\mathrm{2}\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} \right)+\mathrm{8}\sqrt{\mathrm{3}}\bigtriangleup_{\mathrm{ABC}} } \\ $$$$\:\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:\:\boldsymbol{{l}}_{\boldsymbol{{max}}} =\sqrt{\frac{\mathrm{2}}{\mathrm{3}}\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} \right)+\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}}\bigtriangleup_{\mathrm{ABC}} } \\ $$$$\:\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_. \\ $$$$ \\ $$
Commented by ajfour last updated on 20/May/19
much symmetric this time !                say if a=b=c  l=(√((2/3)(3a^2 )+(8/( (√3)))×((√3)/4)a^2 )) = 2a .  if c=0  ⇒ a=b  l=(√((2/3)(2a^2 )+0)) = ((2a)/( (√3))) .
$$\mathrm{much}\:\mathrm{symmetric}\:\mathrm{this}\:\mathrm{time}\:!\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\mathrm{say}\:\mathrm{if}\:\mathrm{a}=\mathrm{b}=\mathrm{c} \\ $$$${l}=\sqrt{\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{3}{a}^{\mathrm{2}} \right)+\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{a}^{\mathrm{2}} }\:=\:\mathrm{2}{a}\:. \\ $$$${if}\:{c}=\mathrm{0}\:\:\Rightarrow\:\mathrm{a}=\mathrm{b} \\ $$$${l}=\sqrt{\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}{a}^{\mathrm{2}} \right)+\mathrm{0}}\:=\:\frac{\mathrm{2}{a}}{\:\sqrt{\mathrm{3}}}\:. \\ $$
Answered by mr W last updated on 21/May/19
Commented by mr W last updated on 03/Feb/24
PQ_(max) =2AB, i.e. when CP and CQ are  diameters of the circles. PQ//AB.
$${PQ}_{{max}} =\mathrm{2}{AB},\:{i}.{e}.\:{when}\:{CP}\:{and}\:{CQ}\:{are} \\ $$$${diameters}\:{of}\:{the}\:{circles}.\:{PQ}//{AB}. \\ $$
Commented by mr W last updated on 21/May/19
this is the geometrical solution:  construct point M and point N such  that ∠BMC=120° and ∠ANB=120°.  construct circle with point M as  center and through B and C.  construct circle with point N as  center and through A and B.  both circles intersect at point D.  construct diameter DP and DR.  PR is the side length of the maximum  equilateral triangle PQR which we  are searching.
$${this}\:{is}\:{the}\:{geometrical}\:{solution}: \\ $$$${construct}\:{point}\:{M}\:{and}\:{point}\:{N}\:{such} \\ $$$${that}\:\angle{BMC}=\mathrm{120}°\:{and}\:\angle{ANB}=\mathrm{120}°. \\ $$$${construct}\:{circle}\:{with}\:{point}\:{M}\:{as} \\ $$$${center}\:{and}\:{through}\:{B}\:{and}\:{C}. \\ $$$${construct}\:{circle}\:{with}\:{point}\:{N}\:{as} \\ $$$${center}\:{and}\:{through}\:{A}\:{and}\:{B}. \\ $$$${both}\:{circles}\:{intersect}\:{at}\:{point}\:{D}. \\ $$$${construct}\:{diameter}\:{DP}\:{and}\:{DR}. \\ $$$${PR}\:{is}\:{the}\:{side}\:{length}\:{of}\:{the}\:{maximum} \\ $$$${equilateral}\:{triangle}\:{PQR}\:{which}\:{we} \\ $$$${are}\:{searching}. \\ $$
Commented by mr W last updated on 21/May/19
MB=MC=(a/2)×(2/( (√3)))=(a/( (√3)))  NB=NA=(c/2)×(2/( (√3)))=(c/( (√3)))  ∠MBN=∠B+60°  MN^2 =BN^2 +BM^2 −2×BM×BN×cos (B+60°)  =(a^2 /3)+(c^2 /3)−((2ac )/3)((1/2)cos B−((√3)/2) sin B)  =((a^2 +c^2 )/3)−((ac cos B )/3)+(1/( (√3))) ac sin B  =((a^2 +c^2 )/3)−((ac(a^2 +c^2 −b^2 ) )/(6ac))+(1/( (√3))) ac sin B  =((a^2 +b^2 +c^2  )/6)+(1/( (√3))) ac sin B  l=PR=2MN=2(√(((a^2 +b^2 +c^2  )/6)+(1/( (√3))) ac sin B))  =(√(((2(a^2 +b^2 +c^2 ) )/3)+((4 ac sin B)/( (√3)))))  since Δ=((ac sin B)/2)=area of ΔABC  ⇒l=(√(((2(a^2 +b^2 +c^2 ) )/3)+((8Δ)/( (√3)))))
$${MB}={MC}=\frac{{a}}{\mathrm{2}}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}=\frac{{a}}{\:\sqrt{\mathrm{3}}} \\ $$$${NB}={NA}=\frac{{c}}{\mathrm{2}}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}=\frac{{c}}{\:\sqrt{\mathrm{3}}} \\ $$$$\angle{MBN}=\angle{B}+\mathrm{60}° \\ $$$${MN}^{\mathrm{2}} ={BN}^{\mathrm{2}} +{BM}^{\mathrm{2}} −\mathrm{2}×{BM}×{BN}×\mathrm{cos}\:\left({B}+\mathrm{60}°\right) \\ $$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{3}}+\frac{{c}^{\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{2}{ac}\:}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:{B}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{sin}\:{B}\right) \\ $$$$=\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{3}}−\frac{{ac}\:\mathrm{cos}\:{B}\:}{\mathrm{3}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:{ac}\:\mathrm{sin}\:{B} \\ $$$$=\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{3}}−\frac{{ac}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\:}{\mathrm{6}{ac}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:{ac}\:\mathrm{sin}\:{B} \\ $$$$=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:}{\mathrm{6}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:{ac}\:\mathrm{sin}\:{B} \\ $$$${l}={PR}=\mathrm{2}{MN}=\mathrm{2}\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:}{\mathrm{6}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:{ac}\:\mathrm{sin}\:{B}} \\ $$$$=\sqrt{\frac{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\:}{\mathrm{3}}+\frac{\mathrm{4}\:{ac}\:\mathrm{sin}\:{B}}{\:\sqrt{\mathrm{3}}}} \\ $$$${since}\:\Delta=\frac{{ac}\:\mathrm{sin}\:{B}}{\mathrm{2}}={area}\:{of}\:\Delta{ABC} \\ $$$$\Rightarrow{l}=\sqrt{\frac{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\:}{\mathrm{3}}+\frac{\mathrm{8}\Delta}{\:\sqrt{\mathrm{3}}}} \\ $$
Commented by ajfour last updated on 22/May/19
Why this way we get the maximum  size △PQR , i fail to follow the reason,  Sir; please enlighten..
$$\mathrm{Why}\:\mathrm{this}\:\mathrm{way}\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{maximum} \\ $$$$\mathrm{size}\:\bigtriangleup\mathrm{PQR}\:,\:\mathrm{i}\:\mathrm{fail}\:\mathrm{to}\:\mathrm{follow}\:\mathrm{the}\:\mathrm{reason}, \\ $$$$\mathrm{Sir};\:\mathrm{please}\:\mathrm{enlighten}.. \\ $$
Commented by mr W last updated on 22/May/19
we can see that R must be on the circle  with center N and P on the circle with  center M.  now the question is to find  the positions of P and R such that PR  reaches the maximum.  we can see ∠DPB and ∠DRB are both  constant, therefore ∠PDR is also   constant. in this case the PR reaches  its msximum if both DP and DR  reach their maximum, i.e. when  DP and DR are the diameters of the  circles. i can remember once i put  this question in an earlier post, and  you have solved it using calculus  method. unfortunately i can not  find this old post, but maybe you can  also remember it, see image below.
$${we}\:{can}\:{see}\:{that}\:{R}\:{must}\:{be}\:{on}\:{the}\:{circle} \\ $$$${with}\:{center}\:{N}\:{and}\:{P}\:{on}\:{the}\:{circle}\:{with} \\ $$$${center}\:{M}.\:\:{now}\:{the}\:{question}\:{is}\:{to}\:{find} \\ $$$${the}\:{positions}\:{of}\:{P}\:{and}\:{R}\:{such}\:{that}\:{PR} \\ $$$${reaches}\:{the}\:{maximum}. \\ $$$${we}\:{can}\:{see}\:\angle{DPB}\:{and}\:\angle{DRB}\:{are}\:{both} \\ $$$${constant},\:{therefore}\:\angle{PDR}\:{is}\:{also}\: \\ $$$${constant}.\:{in}\:{this}\:{case}\:{the}\:{PR}\:{reaches} \\ $$$${its}\:{msximum}\:{if}\:{both}\:{DP}\:{and}\:{DR} \\ $$$${reach}\:{their}\:{maximum},\:{i}.{e}.\:{when} \\ $$$${DP}\:{and}\:{DR}\:{are}\:{the}\:{diameters}\:{of}\:{the} \\ $$$${circles}.\:{i}\:{can}\:{remember}\:{once}\:{i}\:{put} \\ $$$${this}\:{question}\:{in}\:{an}\:{earlier}\:{post},\:{and} \\ $$$${you}\:{have}\:{solved}\:{it}\:{using}\:{calculus} \\ $$$${method}.\:{unfortunately}\:{i}\:{can}\:{not} \\ $$$${find}\:{this}\:{old}\:{post},\:{but}\:{maybe}\:{you}\:{can} \\ $$$${also}\:{remember}\:{it},\:{see}\:{image}\:{below}. \\ $$
Commented by mr W last updated on 22/May/19
Commented by ajfour last updated on 22/May/19
Thanks Sir, for the geometric way!
$$\mathrm{Thanks}\:\mathrm{Sir},\:\mathrm{for}\:\mathrm{the}\:\mathrm{geometric}\:\mathrm{way}! \\ $$
Commented by mr W last updated on 22/May/19
this is a very interesting question.  i spent a long time to find a way to  understand and solve it geometrically.  i′d never come to this solution without  your image. therefore thanks alot!
$${this}\:{is}\:{a}\:{very}\:{interesting}\:{question}. \\ $$$${i}\:{spent}\:{a}\:{long}\:{time}\:{to}\:{find}\:{a}\:{way}\:{to} \\ $$$${understand}\:{and}\:{solve}\:{it}\:{geometrically}. \\ $$$${i}'{d}\:{never}\:{come}\:{to}\:{this}\:{solution}\:{without} \\ $$$${your}\:{image}.\:{therefore}\:{thanks}\:{alot}! \\ $$

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