Question Number 50352 by mr W last updated on 16/Dec/18
$${The}\:{distances}\:{from}\:{a}\:{point}\:{to}\:{the}\:{sides} \\ $$$${of}\:{a}\:{triangle}\:{are}\:{p},{q},{r}.\:{Find}\:{the}\: \\ $$$${maximum}\:\left({or}\:{minimum}\right)\:{area}\:{of}\:{the} \\ $$$${triangle},\:{if}\:{it}\:{exists}. \\ $$$${Assume}\:{r}\leqslant{q}\leqslant{p}. \\ $$
Commented by ajfour last updated on 16/Dec/18
Commented by ajfour last updated on 17/Dec/18
![D[rcos θ−pcos (θ+φ)]=Nsin φsin (2θ+φ) xcos θ−ysin θ = q eq. of BC xcos φ+ysin φ = r y_C = ((r−pcos φ)/(sin φ)) , y_A = ((pcos θ−q)/(sin θ)) x_B = ((qsin φ+rsin θ)/(sin (θ+φ))) △ = (((x_B −p)/2))(y_C −y_A ) = [((qsin φ+rsin θ−psin (θ+φ))/(2sin (θ+φ)))] ×[((qsin φ+rsin θ−psin (θ+φ))/(sin θsin φ))] ___________________________ △ = (([qsin φ+rsin θ−psin (θ+φ)]^2 )/(2sin (θ+φ)sin θsin φ)) ___________________________](https://www.tinkutara.com/question/Q50487.png)
$${D}\left[{r}\mathrm{cos}\:\theta−{p}\mathrm{cos}\:\left(\theta+\phi\right)\right]={N}\mathrm{sin}\:\phi\mathrm{sin}\:\left(\mathrm{2}\theta+\phi\right) \\ $$$${x}\mathrm{cos}\:\theta−{y}\mathrm{sin}\:\theta\:=\:{q} \\ $$$${eq}.\:{of}\:{BC} \\ $$$${x}\mathrm{cos}\:\phi+{y}\mathrm{sin}\:\phi\:=\:{r} \\ $$$${y}_{{C}} \:=\:\frac{{r}−{p}\mathrm{cos}\:\phi}{\mathrm{sin}\:\phi}\:\:\:\:,\:\:\:{y}_{{A}} \:=\:\frac{{p}\mathrm{cos}\:\theta−{q}}{\mathrm{sin}\:\theta} \\ $$$${x}_{{B}} \:=\:\frac{{q}\mathrm{sin}\:\phi+{r}\mathrm{sin}\:\theta}{\mathrm{sin}\:\left(\theta+\phi\right)} \\ $$$$\bigtriangleup\:=\:\left(\frac{{x}_{{B}} −{p}}{\mathrm{2}}\right)\left({y}_{{C}} −{y}_{{A}} \right) \\ $$$$\:\:\:\:=\:\left[\frac{{q}\mathrm{sin}\:\phi+{r}\mathrm{sin}\:\theta−{p}\mathrm{sin}\:\left(\theta+\phi\right)}{\mathrm{2sin}\:\left(\theta+\phi\right)}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:×\left[\frac{{q}\mathrm{sin}\:\phi+{r}\mathrm{sin}\:\theta−{p}\mathrm{sin}\:\left(\theta+\phi\right)}{\mathrm{sin}\:\theta\mathrm{sin}\:\phi}\right] \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\bigtriangleup\:=\:\frac{\left[{q}\mathrm{sin}\:\phi+{r}\mathrm{sin}\:\theta−{p}\mathrm{sin}\:\left(\theta+\phi\right)\right]^{\mathrm{2}} }{\mathrm{2sin}\:\left(\theta+\phi\right)\mathrm{sin}\:\theta\mathrm{sin}\:\phi} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:\: \\ $$
Commented by ajfour last updated on 16/Dec/18
$${yes}\:{Sir},\:{but}\:{still}\:{not}\:{an}\:{easy}\:{going}.. \\ $$
Commented by mr W last updated on 16/Dec/18
$${it}\:{is}\:{indeed}\:{not}\:{easy}. \\ $$
Commented by mr W last updated on 17/Dec/18
$${thank}\:{you}\:{sir}\:{for}\:{the}\:{working}! \\ $$$${i}'{m}\:{still}\:{thinking}\:{about}\:{this}\:{question}. \\ $$$${we}\:{can}\:{see}\:{in}\:{this}\:{diagram}: \\ $$$${when}\:\theta\:{and}\:\phi\:\rightarrow\pi/\mathrm{2},\:\Delta\rightarrow\infty \\ $$$${and}\:{when}\:\theta\:{and}\:\phi\:\rightarrow\mathrm{0},\:\Delta\rightarrow\infty \\ $$$${does}\:{it}\:{not}\:{mean}\:{that}\:{a}\:{mininmum} \\ $$$${for}\:\Delta\:{should}\:{exist}? \\ $$
Answered by mr W last updated on 16/Dec/18
Commented by mr W last updated on 16/Dec/18
![draw three circles from the same center with radii p,q,r. it is to find a triangle whose sides tangent these three circles and has maximum area (like ΔABC) or minimum area (like ΔA′B′C′). A(−a,0) and B(b,0) and C(h,k). Eqn. of AC: (y/k)=((x+a)/(h+a)) ⇒(k/(h+a))x−y+((ak)/(h+a))=0 Eqn. of circle with radius r: x^2 +(y−p)^2 =r^2 AC is tangent, ⇒((k/(h+a)))^2 r^2 +r^2 =(−p+((ak)/(h+a)))^2 =p^2 −2pa((k/(h+a)))+a^2 ((k/(h+a)))^2 ⇒(a^2 −r^2 )((k/(h+a)))^2 −2pa((k/(h+a)))+(p^2 −r^2 )=0 ⇒(k/(h+a))=((pa±(√(p^2 a^2 −(a^2 −r^2 )(p^2 −r^2 ))))/(a^2 −r^2 ))=((pa±r(√(p^2 −r^2 +a^2 )))/(a^2 −r^2 )) we consider at first the big triangle, ⇒(k/(h+a))=((pa+r(√(p^2 −r^2 +a^2 )))/(a^2 −r^2 )) ⇒h=((a^2 −r^2 )/(pa+r(√(p^2 −r^2 +a^2 ))))k−a Eqn. of BC: (y/k)=((x−b)/(h−b)) ⇒(k/(h−b))x−y−((bk)/(h−b))=0 Eqn. of circle with radius q: x^2 +(y−p)^2 =q^2 BC is tangent, ⇒((k/(h−b)))^2 q^2 +q^2 =(−p−((bk)/(h−b)))^2 =p^2 +2pb((k/(h−b)))+b^2 ((k/(h−b)))^2 ⇒(b^2 −q^2 )((k/(h−b)))^2 +2pb((k/(h−b)))+(p^2 −q^2 )=0 ⇒(k/(h−b))=((−pb+q(√(p^2 −q^2 +b^2 )))/(b^2 −q^2 )) ⇒h=((b^2 −q^2 )/(−pb+q(√(p^2 −q^2 +b^2 ))))k+b ⇒((a^2 −r^2 )/(pa+r(√(p^2 −r^2 +a^2 ))))k−a=((b^2 −q^2 )/(−pb+q(√(p^2 −q^2 +b^2 ))))k+b ⇒[((a^2 −r^2 )/(pa+r(√(p^2 −r^2 +a^2 ))))−((b^2 −q^2 )/(−pb+q(√(p^2 −q^2 +b^2 ))))]k=a+b ⇒k=((a+b)/(((a^2 −r^2 )/(pa+r(√(p^2 −r^2 +a^2 ))))−((b^2 −q^2 )/(−pb+q(√(p^2 −q^2 +b^2 )))))) area of ΔABC Δ=(1/2)(a+b)k ⇒Δ=(1/2)×(((a+b)^2 )/(((a^2 −r^2 )/(pa+r(√(p^2 −r^2 +a^2 ))))−((b^2 −q^2 )/(−pb+q(√(p^2 −q^2 +b^2 )))))) (∂Δ/∂a)=(∂Δ/∂b)=0 ......](https://www.tinkutara.com/question/Q50486.png)
$${draw}\:{three}\:{circles}\:{from}\:{the}\:{same}\:{center} \\ $$$${with}\:{radii}\:{p},{q},{r}. \\ $$$${it}\:{is}\:{to}\:{find}\:{a}\:{triangle}\:{whose}\:{sides} \\ $$$${tangent}\:{these}\:{three}\:{circles}\:{and}\:{has} \\ $$$${maximum}\:{area}\:\left({like}\:\Delta{ABC}\right)\:{or} \\ $$$${minimum}\:{area}\:\left({like}\:\Delta{A}'{B}'{C}'\right). \\ $$$${A}\left(−{a},\mathrm{0}\right)\:{and}\:{B}\left({b},\mathrm{0}\right)\:{and}\:{C}\left({h},{k}\right). \\ $$$${Eqn}.\:{of}\:{AC}: \\ $$$$\frac{{y}}{{k}}=\frac{{x}+{a}}{{h}+{a}} \\ $$$$\Rightarrow\frac{{k}}{{h}+{a}}{x}−{y}+\frac{{ak}}{{h}+{a}}=\mathrm{0} \\ $$$${Eqn}.\:{of}\:{circle}\:{with}\:{radius}\:{r}: \\ $$$${x}^{\mathrm{2}} +\left({y}−{p}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${AC}\:{is}\:{tangent}, \\ $$$$\Rightarrow\left(\frac{{k}}{{h}+{a}}\right)^{\mathrm{2}} {r}^{\mathrm{2}} +{r}^{\mathrm{2}} =\left(−{p}+\frac{{ak}}{{h}+{a}}\right)^{\mathrm{2}} ={p}^{\mathrm{2}} −\mathrm{2}{pa}\left(\frac{{k}}{{h}+{a}}\right)+{a}^{\mathrm{2}} \left(\frac{{k}}{{h}+{a}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left({a}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)\left(\frac{{k}}{{h}+{a}}\right)^{\mathrm{2}} −\mathrm{2}{pa}\left(\frac{{k}}{{h}+{a}}\right)+\left({p}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\frac{{k}}{{h}+{a}}=\frac{{pa}\pm\sqrt{{p}^{\mathrm{2}} {a}^{\mathrm{2}} −\left({a}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)\left({p}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)}}{{a}^{\mathrm{2}} −{r}^{\mathrm{2}} }=\frac{{pa}\pm{r}\sqrt{{p}^{\mathrm{2}} −{r}^{\mathrm{2}} +{a}^{\mathrm{2}} }}{{a}^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$${we}\:{consider}\:{at}\:{first}\:{the}\:{big}\:{triangle}, \\ $$$$\Rightarrow\frac{{k}}{{h}+{a}}=\frac{{pa}+{r}\sqrt{{p}^{\mathrm{2}} −{r}^{\mathrm{2}} +{a}^{\mathrm{2}} }}{{a}^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$$\Rightarrow{h}=\frac{{a}^{\mathrm{2}} −{r}^{\mathrm{2}} }{{pa}+{r}\sqrt{{p}^{\mathrm{2}} −{r}^{\mathrm{2}} +{a}^{\mathrm{2}} }}{k}−{a} \\ $$$$ \\ $$$${Eqn}.\:{of}\:{BC}: \\ $$$$\frac{{y}}{{k}}=\frac{{x}−{b}}{{h}−{b}} \\ $$$$\Rightarrow\frac{{k}}{{h}−{b}}{x}−{y}−\frac{{bk}}{{h}−{b}}=\mathrm{0} \\ $$$${Eqn}.\:{of}\:{circle}\:{with}\:{radius}\:{q}: \\ $$$${x}^{\mathrm{2}} +\left({y}−{p}\right)^{\mathrm{2}} ={q}^{\mathrm{2}} \\ $$$${BC}\:{is}\:{tangent}, \\ $$$$\Rightarrow\left(\frac{{k}}{{h}−{b}}\right)^{\mathrm{2}} {q}^{\mathrm{2}} +{q}^{\mathrm{2}} =\left(−{p}−\frac{{bk}}{{h}−{b}}\right)^{\mathrm{2}} ={p}^{\mathrm{2}} +\mathrm{2}{pb}\left(\frac{{k}}{{h}−{b}}\right)+{b}^{\mathrm{2}} \left(\frac{{k}}{{h}−{b}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left({b}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)\left(\frac{{k}}{{h}−{b}}\right)^{\mathrm{2}} +\mathrm{2}{pb}\left(\frac{{k}}{{h}−{b}}\right)+\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\frac{{k}}{{h}−{b}}=\frac{−{pb}+{q}\sqrt{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{b}^{\mathrm{2}} −{q}^{\mathrm{2}} } \\ $$$$\Rightarrow{h}=\frac{{b}^{\mathrm{2}} −{q}^{\mathrm{2}} }{−{pb}+{q}\sqrt{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{k}+{b} \\ $$$$\Rightarrow\frac{{a}^{\mathrm{2}} −{r}^{\mathrm{2}} }{{pa}+{r}\sqrt{{p}^{\mathrm{2}} −{r}^{\mathrm{2}} +{a}^{\mathrm{2}} }}{k}−{a}=\frac{{b}^{\mathrm{2}} −{q}^{\mathrm{2}} }{−{pb}+{q}\sqrt{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{k}+{b} \\ $$$$\Rightarrow\left[\frac{{a}^{\mathrm{2}} −{r}^{\mathrm{2}} }{{pa}+{r}\sqrt{{p}^{\mathrm{2}} −{r}^{\mathrm{2}} +{a}^{\mathrm{2}} }}−\frac{{b}^{\mathrm{2}} −{q}^{\mathrm{2}} }{−{pb}+{q}\sqrt{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right]{k}={a}+{b} \\ $$$$\Rightarrow{k}=\frac{{a}+{b}}{\frac{{a}^{\mathrm{2}} −{r}^{\mathrm{2}} }{{pa}+{r}\sqrt{{p}^{\mathrm{2}} −{r}^{\mathrm{2}} +{a}^{\mathrm{2}} }}−\frac{{b}^{\mathrm{2}} −{q}^{\mathrm{2}} }{−{pb}+{q}\sqrt{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} +{b}^{\mathrm{2}} }}} \\ $$$$ \\ $$$${area}\:{of}\:\Delta{ABC} \\ $$$$\Delta=\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{b}\right){k} \\ $$$$\Rightarrow\Delta=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\left({a}+{b}\right)^{\mathrm{2}} }{\frac{{a}^{\mathrm{2}} −{r}^{\mathrm{2}} }{{pa}+{r}\sqrt{{p}^{\mathrm{2}} −{r}^{\mathrm{2}} +{a}^{\mathrm{2}} }}−\frac{{b}^{\mathrm{2}} −{q}^{\mathrm{2}} }{−{pb}+{q}\sqrt{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} +{b}^{\mathrm{2}} }}} \\ $$$$\frac{\partial\Delta}{\partial{a}}=\frac{\partial\Delta}{\partial{b}}=\mathrm{0} \\ $$$$…… \\ $$
Commented by mr W last updated on 17/Dec/18
$${you}'{re}\:{right}. \\ $$$${x}\rightarrow\mathrm{0},{y}\rightarrow\mathrm{0}\:\Rightarrow\Delta\rightarrow\mathrm{0} \\ $$$${x}\rightarrow\infty,{y}\rightarrow\infty\Rightarrow\Delta\rightarrow\infty \\ $$$${I}\:{thought}\:{there}\:{is}\:{a}\:{minimum}\:{for} \\ $$$$\Delta{ABC}\:{and}\:{a}\:{maximum}\:{for}\:\Delta{A}'{B}'{C}'. \\ $$$${but}\:{this}\:{is}\:{not}\:{true}.\:{so}\:{this}\:{is}\:{not}\:{a} \\ $$$${good}\:{question}.\:{thanks}\:{for}\:{trying}\:{and} \\ $$$${for}\:{the}\:{correct}\:{conclusion}. \\ $$
Commented by ajfour last updated on 17/Dec/18
$${but}\:{i}\:{think}\:{minimum}\:{is}\:{zero} \\ $$$${and}\:{maximum}\:\rightarrow\:\infty. \\ $$