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Question Number 211058 by mr W last updated on 27/Aug/24

Commented by mr W last updated on 27/Aug/24

find the radius R of the largest sphere  which can be placed between the red  and blue planes in the first octant.  assume 0<p≤a, 0<q≤b, 0<r≤c.

findtheradiusRofthelargestspherewhichcanbeplacedbetweentheredandblueplanesinthefirstoctant.assume0<pa,0<qb,0<rc.

Answered by mr W last updated on 29/Aug/24

Commented by mr W last updated on 28/Aug/24

the largest sphere above and touching  the lower plane (x/p)+(y/q)+(z/r)=1:  say its radius is R_o , then its center  is at (R_o ,R_o ,R_o ).  we have  (((R_o /p)+(R_o /q)+(R_o /r)−1)/( (√((1/p^2 )+(1/q^2 )+(1/r^2 )))))=R_o   ⇒R_o =(1/((1/p)+(1/q)+(1/r)−(√((1/p^2 )+(1/q^2 )+(1/r^2 )))))

thelargestsphereaboveandtouchingthelowerplanexp+yq+zr=1:sayitsradiusisRo,thenitscenterisat(Ro,Ro,Ro).wehaveRop+Roq+Ror11p2+1q2+1r2=RoRo=11p+1q+1r1p2+1q2+1r2

Commented by mr W last updated on 29/Aug/24

Commented by mr W last updated on 28/Aug/24

the largest sphere under the  upper plane (x/a)+(y/b)+(z/c)=1:  say its radius is R_i , then its center  is at (R_i ,R_i ,R_i ).  we have  (((R_i /a)+(R_i /b)+(R_i /c)−1)/( (√((1/a^2 )+(1/b^2 )+(1/c^2 )))))=−R_i   ⇒R_i =(1/((1/a)+(1/b)+(1/c)+(√((1/a^2 )+(1/b^2 )+(1/c^2 )))))

thelargestsphereundertheupperplanexa+yb+zc=1:sayitsradiusisRi,thenitscenterisat(Ri,Ri,Ri).wehaveRia+Rib+Ric11a2+1b2+1c2=RiRi=11a+1b+1c+1a2+1b2+1c2

Commented by mr W last updated on 28/Aug/24

Commented by mr W last updated on 29/Aug/24

let′s ignore the lower plane at first.  as shown previously, the largest   sphere under the upper plane has   the radius R_i  and and its center is   at (R_i ,R_i ,R_i ).  R_i =(1/((1/a)+(1/b)+(1/c)+(√((1/a^2 )+(1/b^2 )+(1/c^2 )))))  if the distance from (R_i ,R_i ,R_i ) to  the lower plane is equal to or larger  than R_i , i.e. if  (((R_i /p)+(R_i /q)+(R_i /r)−1)/( (√((1/p^2 )+(1/q^2 )+(1/r^2 )))))≥R_i , i.e. if  (1/((1/p)+(1/q)+(1/r)−(√((1/p^2 )+(1/q^2 )+(1/r^2 )))))≤R_i   i.e. if R_o ≤R_i ,  that means if   (1/p)+(1/q)+(1/r)−(√((1/p^2 )+(1/q^2 )+(1/r^2 )))  ≥(1/a)+(1/b)+(1/c)+(√((1/a^2 )+(1/b^2 )+(1/c^2 )))  then the lower plane is beneath  the sphere with radius R_i  and this   sphere is also the largest sphere  between both planes. that means  in this case  R=R_i =(1/((1/a)+(1/b)+(1/c)+(√((1/a^2 )+(1/b^2 )+(1/c^2 )))))    if R_o >R_i , i.e. if  (1/p)+(1/q)+(1/r)−(√((1/p^2 )+(1/q^2 )+(1/r^2 )))  <(1/a)+(1/b)+(1/c)+(√((1/a^2 )+(1/b^2 )+(1/c^2 )))  then the largest sphere between  both planes is the maximum sphere  from following three cases:    case 1:   center at (u, R_1 ,R_1 ), radius R_1   (((u/a)+(R_1 /b)+(R_1 /c)−1)/( (√((1/a^2 )+(1/b^2 )+(1/c^2 )))))=−R_1   ⇒u=a−aR_1 ((1/b)+(1/c)+(√((1/a^2 )+(1/b^2 )+(1/c^2 ))))  ⇒u=a−aR_1 ((1/a)+(1/b)+(1/c)+(√((1/a^2 )+(1/b^2 )+(1/c^2 ))))+R_1   ⇒u=a−((aR_1 )/R_i )+R_1   (((u/p)+(R_1 /q)+(R_1 /r)−1)/( (√((1/p^2 )+(1/q^2 )+(1/r^2 )))))=R_1   ⇒u=p−pR_1 ((1/q)+(1/r)−(√((1/p^2 )+(1/q^2 )+(1/r^2 ))))  ⇒u=p−pR_1 ((1/p)+(1/q)+(1/r)−(√((1/p^2 )+(1/q^2 )+(1/r^2 ))))+R_1   ⇒u=p−((pR_1 )/R_o )+R_1   p−((pR_1 )/R_o )+R_1 =a−((aR_1 )/R_i )+R_1   (R_1 /R_i )(1−(p/a)×(R_i /R_o ))=1−(p/a)  ⇒R_1 =((1−(p/a))/(1−(p/a)×(R_i /R_o )))×R_i   u=a(1−(R_1 /R_i ))+R_1   ⇒u=a(1−((1−(p/a))/(1−(p/a)×(R_i /R_o ))))+R_1  > R_1     similarly  case 2:   center at (R_2 ,v,R_2 ), radius R_2   ⇒R_2 =((1−(q/b))/(1−(q/b)×(R_i /R_o )))×R_i   ⇒v=b(1−((1−(q/b))/(1−(q/b)×(R_i /R_o ))))+R_2  > R_2     case 3:   center at (R_3 ,R_3 ,w), radius R_3   ⇒R_3 =((1−(r/c))/(1−(r/c)×(R_i /R_o )))×R_i   ⇒w=c(1−((1−(r/c))/(1−(r/c)×(R_i /R_o ))))+R_3  > R_3     since (R_i /R_o )<1 and 0<(p/a), (q/b), (r/c)≤1  and f(x)=((1−x)/(1−kx)) is a strctly   decreasing function for 0<k<1 and  0≤x≤1, we get  R=max(R_1 ,R_2 ,R_3 )     =((1−min((p/a),(q/b),(r/c)))/(1−min((p/a),(q/b),(r/c))×(R_i /R_o )))×R_i

letsignorethelowerplaneatfirst.asshownpreviously,thelargestsphereundertheupperplanehastheradiusRiandanditscenterisat(Ri,Ri,Ri).Ri=11a+1b+1c+1a2+1b2+1c2ifthedistancefrom(Ri,Ri,Ri)tothelowerplaneisequaltoorlargerthanRi,i.e.ifRip+Riq+Rir11p2+1q2+1r2Ri,i.e.if11p+1q+1r1p2+1q2+1r2Rii.e.ifRoRi,thatmeansif1p+1q+1r1p2+1q2+1r21a+1b+1c+1a2+1b2+1c2thenthelowerplaneisbeneaththespherewithradiusRiandthissphereisalsothelargestspherebetweenbothplanes.thatmeansinthiscaseR=Ri=11a+1b+1c+1a2+1b2+1c2ifRo>Ri,i.e.if1p+1q+1r1p2+1q2+1r2<1a+1b+1c+1a2+1b2+1c2thenthelargestspherebetweenbothplanesisthemaximumspherefromfollowingthreecases:case1:centerat(u,R1,R1),radiusR1ua+R1b+R1c11a2+1b2+1c2=R1u=aaR1(1b+1c+1a2+1b2+1c2)u=aaR1(1a+1b+1c+1a2+1b2+1c2)+R1u=aaR1Ri+R1up+R1q+R1r11p2+1q2+1r2=R1u=ppR1(1q+1r1p2+1q2+1r2)u=ppR1(1p+1q+1r1p2+1q2+1r2)+R1u=ppR1Ro+R1ppR1Ro+R1=aaR1Ri+R1R1Ri(1pa×RiRo)=1paR1=1pa1pa×RiRo×Riu=a(1R1Ri)+R1u=a(11pa1pa×RiRo)+R1>R1similarlycase2:centerat(R2,v,R2),radiusR2R2=1qb1qb×RiRo×Riv=b(11qb1qb×RiRo)+R2>R2case3:centerat(R3,R3,w),radiusR3R3=1rc1rc×RiRo×Riw=c(11rc1rc×RiRo)+R3>R3sinceRiRo<1and0<pa,qb,rc1andf(x)=1x1kxisastrctlydecreasingfunctionfor0<k<1and0x1,wegetR=max(R1,R2,R3)=1min(pa,qb,rc)1min(pa,qb,rc)×RiRo×Ri

Commented by mr W last updated on 27/Aug/24

summary:  with   R_i =(1/((1/a)+(1/b)+(1/c)+(√((1/a^2 )+(1/b^2 )+(1/c^2 )))))  R_o =(1/((1/p)+(1/q)+(1/r)−(√((1/p^2 )+(1/q^2 )+(1/r^2 )))))  if R_o ≤R_i , then  R=R_i   if R_o >R_i , then  R=((1−min((p/a),(q/b),(r/c)))/(1−min((p/a),(q/b),(r/c))∙(R_i /R_o )))∙R_i

summary:withRi=11a+1b+1c+1a2+1b2+1c2Ro=11p+1q+1r1p2+1q2+1r2ifRoRi,thenR=RiifRo>Ri,thenR=1min(pa,qb,rc)1min(pa,qb,rc)RiRoRi

Commented by mr W last updated on 27/Aug/24

formula used:  distance from point to plane in space

formulaused:distancefrompointtoplaneinspace

Commented by mr W last updated on 27/Aug/24

Commented by ajfour last updated on 27/Aug/24

immense ! i will look carefully. get  the idea very well. great thing for  question treasure.

immense!iwilllookcarefully.gettheideaverywell.greatthingforquestiontreasure.

Commented by mr W last updated on 27/Aug/24

thanks for checking sir!

thanksforcheckingsir!

Commented by mr W last updated on 27/Aug/24

example 1:  upper plane (x/6)+(y/9)+(z/(10))=1  lower plane (x/5)+(y/3)+(z/4)=1  a=6, b=9, c=10  p=5, q=3, r=4  (p/a)=(5/6), (q/b)=(3/9)=(1/3), (r/c)=(4/(10))=(2/5)   ⇒min=(q/b)=(1/3)  R_i =(1/((1/6)+(1/9)+(1/(10))+(√((1/6^2 )+(1/9^2 )+(1/(10^2 ))))))        =((102−3(√(406)))/(25))≈1.662  R_o =(1/((1/5)+(1/3)+(1/4)−(√((1/5^2 )+(1/3^2 )+(1/4^2 )))))       =((47+(√(769)))/(24))≈3.114  R_o >R_i    R=((1−(1/3))/(1−(1/3)×((24×(102−3(√(406))))/(25×(47+(√(769)))))))×((102−3(√(406)))/(25))      ≈1.348

example1:upperplanex6+y9+z10=1lowerplanex5+y3+z4=1a=6,b=9,c=10p=5,q=3,r=4pa=56,qb=39=13,rc=410=25min=qb=13Ri=116+19+110+162+192+1102=1023406251.662Ro=115+13+14152+132+142=47+769243.114Ro>RiR=113113×24×(1023406)25×(47+769)×1023406251.348

Commented by behi834171 last updated on 27/Aug/24

mathmatic is very beautiful.  your Q′s & A′s ,are more than   beautiful!  They are amazing sir mrW.  Thanks god for being of you in this forum.

mathmaticisverybeautiful.yourQs&As,aremorethanbeautiful!TheyareamazingsirmrW.Thanksgodforbeingofyouinthisforum.

Commented by mr W last updated on 27/Aug/24

thanks back to you, father from Behi,   for your appreciation and support!

thanksbacktoyou,fatherfromBehi,foryourappreciationandsupport!

Commented by necx122 last updated on 27/Aug/24

Mr. Behi its been a really long time. Feels  good to see your post here again.

Mr.Behiitsbeenareallylongtime.Feelsgoodtoseeyourposthereagain.

Commented by ajfour last updated on 27/Aug/24

yes so glad for your remark bsck on this forum.

yessogladforyourremarkbsckonthisforum.

Commented by behi834171 last updated on 28/Aug/24

Red heart to you dear mr.W

Redhearttoyoudearmr.W

Commented by behi834171 last updated on 28/Aug/24

Thank you so much sir with red herat!

Thankyousomuchsirwithredherat!

Commented by behi834171 last updated on 28/Aug/24

so happy for see you  again sir ajfour   and red heart to you!

sohappyforseeyouagainsirajfourandredhearttoyou!

Commented by BHOOPENDRA last updated on 01/Sep/24

great Mr.W

greatMr.W

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