Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 129244 by ajfour last updated on 14/Jan/21

Commented by ajfour last updated on 14/Jan/21

In terms of the sides of △ABC,  find largest radius sphere that  can be placed against the room  corner and triangle.

$${In}\:{terms}\:{of}\:{the}\:{sides}\:{of}\:\bigtriangleup{ABC}, \\ $$$${find}\:{largest}\:{radius}\:{sphere}\:{that} \\ $$$${can}\:{be}\:{placed}\:{against}\:{the}\:{room} \\ $$$${corner}\:{and}\:{triangle}.\:\:\:\:\: \\ $$

Answered by mr W last updated on 22/Feb/21

Commented by mr W last updated on 22/Feb/21

Commented by mr W last updated on 22/Feb/21

Commented by mr W last updated on 22/Feb/21

say the plane containing the triangle  ABC intersects the coordinate axes  at P, Q, R.  ΔPQR is a circumtriangle of the  given triangle ΔABC.  say the sides of ΔABC are a, b, c.  its area is Δ_(ABC) .  Δ_(ABC) =(√(s(s−a)(s−b)(s−c)))  with s=((a+b+c)/2)    say the radius of the small sphere  under the plane is R and   its center is G(R,R,R). the distance  from G to the plane ΔPQR is R.    say P(p,0,0), Q(0,q,0), R(0,0,r).  the equation of the plane containing  ΔABC as well as ΔPQR is  (x/p)+(y/q)+(z/r)=1  we have  R=((∣(R/p)+(R/q)+(R/r)−1∣)/( (√((1/p^2 )+(1/q^2 )+(1/r^2 )))))=((1−((R/p)+(R/q)+(R/r)))/( (√((1/p^2 )+(1/q^2 )+(1/r^2 )))))  ⇒R=(1/((1/p)+(1/q)+(1/r)+(√((1/p^2 )+(1/q^2 )+(1/r^2 )))))  due to symmetry of r w.r.t. p,q,r we  know R is maximum when p=q=r.    PQ=(√(p^2 +q^2 ))  QR=(√(q^2 +r^2 ))  RP=(√(r^2 +p^2 ))  p=q=r means PQ=QR=RP, i.e.  ΔPQR is equilateral.    the largest equilateral triangle ΔPQR  circumscribing ΔABC has side length  PQ=QR=RP=l=(√(((2(a^2 +b^2 +c^2 ))/3)+((8Δ_(ABC) )/( (√3)))))  with Δ_(ABC) =area of ΔABC.  (see Q60313 for more details)  in this case: p=q=r=(l/( (√2)))  R_(max) =(1/((3/p)+((√3)/p)))=(p/(3+(√3)))=(l/((3+(√3))(√2)))  R_(max) =(1/((3+(√3))(√2)))×(√(((2(a^2 +b^2 +c^2 ))/3)+((8Δ_(ABC) )/( (√3)))))  R_(max) =((√(a^2 +b^2 +c^2 +4(√3)Δ_(ABC) ))/(3((√3)+1)))    btw,for the big sphere over the plane  containing ΔABC we have  R=((∣(R/p)+(R/q)+(R/r)−1∣)/( (√((1/p^2 )+(1/q^2 )+(1/r^2 )))))=((((R/p)+(R/q)+(R/r))−1)/( (√((1/p^2 )+(1/q^2 )+(1/r^2 )))))  ⇒R_(max) =((√(a^2 +b^2 +c^2 +4(√3)Δ_(ABC) ))/(3((√3)−1)))

$${say}\:{the}\:{plane}\:{containing}\:{the}\:{triangle} \\ $$$${ABC}\:{intersects}\:{the}\:{coordinate}\:{axes} \\ $$$${at}\:{P},\:{Q},\:{R}. \\ $$$$\Delta{PQR}\:{is}\:{a}\:{circumtriangle}\:{of}\:{the} \\ $$$${given}\:{triangle}\:\Delta{ABC}. \\ $$$${say}\:{the}\:{sides}\:{of}\:\Delta{ABC}\:{are}\:{a},\:{b},\:{c}. \\ $$$${its}\:{area}\:{is}\:\Delta_{{ABC}} . \\ $$$$\Delta_{{ABC}} =\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)} \\ $$$${with}\:{s}=\frac{{a}+{b}+{c}}{\mathrm{2}} \\ $$$$ \\ $$$${say}\:{the}\:{radius}\:{of}\:{the}\:{small}\:{sphere} \\ $$$${under}\:{the}\:{plane}\:{is}\:{R}\:{and}\: \\ $$$${its}\:{center}\:{is}\:{G}\left({R},{R},{R}\right).\:{the}\:{distance} \\ $$$${from}\:{G}\:{to}\:{the}\:{plane}\:\Delta{PQR}\:{is}\:{R}. \\ $$$$ \\ $$$${say}\:{P}\left({p},\mathrm{0},\mathrm{0}\right),\:{Q}\left(\mathrm{0},{q},\mathrm{0}\right),\:{R}\left(\mathrm{0},\mathrm{0},{r}\right). \\ $$$${the}\:{equation}\:{of}\:{the}\:{plane}\:{containing} \\ $$$$\Delta{ABC}\:{as}\:{well}\:{as}\:\Delta{PQR}\:{is} \\ $$$$\frac{{x}}{{p}}+\frac{{y}}{{q}}+\frac{{z}}{{r}}=\mathrm{1} \\ $$$${we}\:{have} \\ $$$${R}=\frac{\mid\frac{{R}}{{p}}+\frac{{R}}{{q}}+\frac{{R}}{{r}}−\mathrm{1}\mid}{\:\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}}=\frac{\mathrm{1}−\left(\frac{{R}}{{p}}+\frac{{R}}{{q}}+\frac{{R}}{{r}}\right)}{\:\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}} \\ $$$$\Rightarrow{R}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{r}}+\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}} \\ $$$${due}\:{to}\:{symmetry}\:{of}\:{r}\:{w}.{r}.{t}.\:{p},{q},{r}\:{we} \\ $$$${know}\:{R}\:{is}\:{maximum}\:{when}\:{p}={q}={r}. \\ $$$$ \\ $$$${PQ}=\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} } \\ $$$${QR}=\sqrt{{q}^{\mathrm{2}} +{r}^{\mathrm{2}} } \\ $$$${RP}=\sqrt{{r}^{\mathrm{2}} +{p}^{\mathrm{2}} } \\ $$$${p}={q}={r}\:{means}\:{PQ}={QR}={RP},\:{i}.{e}. \\ $$$$\Delta{PQR}\:{is}\:{equilateral}. \\ $$$$ \\ $$$${the}\:{largest}\:{equilateral}\:{triangle}\:\Delta{PQR} \\ $$$${circumscribing}\:\Delta{ABC}\:{has}\:{side}\:{length} \\ $$$${PQ}={QR}={RP}={l}=\sqrt{\frac{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}{\mathrm{3}}+\frac{\mathrm{8}\Delta_{{ABC}} }{\:\sqrt{\mathrm{3}}}} \\ $$$${with}\:\Delta_{{ABC}} ={area}\:{of}\:\Delta{ABC}. \\ $$$$\left({see}\:{Q}\mathrm{60313}\:{for}\:{more}\:{details}\right) \\ $$$${in}\:{this}\:{case}:\:{p}={q}={r}=\frac{{l}}{\:\sqrt{\mathrm{2}}} \\ $$$${R}_{{max}} =\frac{\mathrm{1}}{\frac{\mathrm{3}}{{p}}+\frac{\sqrt{\mathrm{3}}}{{p}}}=\frac{{p}}{\mathrm{3}+\sqrt{\mathrm{3}}}=\frac{{l}}{\left(\mathrm{3}+\sqrt{\mathrm{3}}\right)\sqrt{\mathrm{2}}} \\ $$$${R}_{{max}} =\frac{\mathrm{1}}{\left(\mathrm{3}+\sqrt{\mathrm{3}}\right)\sqrt{\mathrm{2}}}×\sqrt{\frac{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}{\mathrm{3}}+\frac{\mathrm{8}\Delta_{{ABC}} }{\:\sqrt{\mathrm{3}}}} \\ $$$${R}_{{max}} =\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{3}}\Delta_{{ABC}} }}{\mathrm{3}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)} \\ $$$$ \\ $$$${btw},{for}\:{the}\:{big}\:{sphere}\:{over}\:{the}\:{plane} \\ $$$${containing}\:\Delta{ABC}\:{we}\:{have} \\ $$$${R}=\frac{\mid\frac{{R}}{{p}}+\frac{{R}}{{q}}+\frac{{R}}{{r}}−\mathrm{1}\mid}{\:\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}}=\frac{\left(\frac{{R}}{{p}}+\frac{{R}}{{q}}+\frac{{R}}{{r}}\right)−\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}} \\ $$$$\Rightarrow{R}_{{max}} =\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{3}}\Delta_{{ABC}} }}{\mathrm{3}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com