425. Geometrical extremity II.

by Virgil Nicula, May 14, 2015, 11:56 PM

PE0. Let $C(r,g)$ be a right circular cone with the lengths $r$ , $g$ of the radius and the generatrix respectively. Prove that if its

volume is constant, then $:$ its lateral surface is minimum $\iff g=r\sqrt 3\ ;$ its total surface is minimum $\iff g=3r\ .$

Proof. Let the circle $C(O,r)$ from base and the apex $A$ with $AO=h$ . Thus, $\boxed{g^2=r^2+h^2}$ and the volume $V=\frac {\pi r^2h}{3}$

is constant, i.e. $r^2h$ is constant. The lateral surface area is $\boxed{S_l=\pi rg}$ and the total surface area is $\boxed{S=\pi r(g+r)}$ .

$\blacktriangleright\ r^2h$ is constant $\iff r^8h^4$ is constant $\iff \frac {r^2h^2}{2}\cdot\frac {r^2h^2}{2}\cdot r^4$ is constant $(1)$ . Thus, $S_l-\min\iff rg-\min\iff$

$r^2g^2-\min\iff$ $r^2\left(r^2+h^2\right)-\min\iff$ $r^4+\frac {r^2h^2}2+\frac {r^2h^2}2-\min\ \stackrel{(1)}{\iff}$ $r^4=\frac {r^2h^2}2\iff h=r\sqrt 2\iff$ $\boxed{g=r\sqrt 3}\ .$

$\blacktriangleright\ r^2h$ is constant $\iff r^4h^2$ is constant $\iff r^4\left(g^2-r^2\right)$ is constant $\iff$ $r^2\cdot\frac {r(g+r)}4\cdot\frac {r(g-r)}2$ is constant $(2)$ . Therefore,

$S-\min\ \iff$ $\frac {3r(g+r)}4-\min\iff$ $r^2+\frac {r(g+r)}4+\frac {r(g-r)}2-\min\ \stackrel{(2)}{\iff}$ $r^2=\frac {r(g+r)}4=\frac {r(g-r)}2\iff$ $\boxed{\ g=3r\ }$ .

PE1.Sa se afle dimensiunile unui cilindru de rotatie fara un "capac" (de exemplu gen "cana","oala" etc), de volum constant si care are suprafata minima.

Dem. Fie $x$ - raza cercului din baza si $y$ - generatoarea cilindrului. Volumul cilindrului $V(x,y)=\pi x^2y$ este constant $\iff$ $\boxed{x^2y=k}\ (*)$ (constant) si suprafata fara un

"capac" este $A\equiv A(x,y)=2\pi xy+\pi x^2.$ Asadar $A$ este $\min\iff$ suma $S\equiv\left(2xy+x^2\right)$ este $\min\iff$ suma $\left(xy+xy+x^2\right)$ este $\min .$ Se observa ca

produsul $xy\cdot xy\cdot x^2=x^4y^2=\left(x^2y\right)^2\stackrel{*}{=}k^2$ (constant). Deci suma $S$ este $\min \iff xy=x^2=\sqrt[3]{k^2}.$ In concluzie, $\boxed{A(x,y)\ \mathrm{este}\ \min\iff x=y=\sqrt[3]k}\ .$

PE2. Sa se afle dimensiunile unui cilindru de rotatie cu volum constant si care are suprafata totala minima.

Dem. Fie $2x$ - diametrul cercului din baza si $y$ - generatoarea cilindrului. Volumul cilindrului $V(x,y)=\pi x^2y$ este constant $\iff$ $x^2y$ este constant $\iff$ $\frac {\left(x^2y\right)^2}{4}=x^2\cdot\frac {xy}{2}\cdot\frac {xy}{2}$

este constant. Aria totala este $A(x,y)=2\pi xy+2\pi x^2=$ $2\pi \left(x^2+xy\right)$ . Deci aria totala este minima $\iff$ $x^2+xy$ este minima $\iff$ suma $x^2+\frac {xy}{2}+\frac {xy}{2}$ este minima.

Produsul termenilor acestei sume este constant. Aria totala este minima $\iff x^2=\frac {xy}{2}$ , adica $\boxed{y=2x}$ , ceea ce inseamna ca in solutia de minim sectiunea axiala este un patrat.

PE3. Construim o fereastra $\mathcal F$ care este alcatuita dintr-un dreptunghi $ABCD$ si semicercul cu diametru $[CD]$ astfel incat suprafetele lor au in comun numai $[CD]$ . In ipoteza

ca perimetrul ferestrei este constant, sa se determine dimensiunile dreptunghiului $ABCD$ astfel incat aria ferestrei sa fie maxima, adica sa patrunda cat mai multa lumina pe timpul zilei.

Dem. Notam $AB=2x$ si $AD=y$ . Asadar, perimetrul ferestrei este $P(x,y)=(2+\pi )x+2y=k$ (constant) si aria ferestrei este $A(x,y)=2xy+\frac {\pi x^2}{2}$ . Deci, aria ferestrei

este maxima $\iff$ $\frac {\pi x^2}{2}+x\left[k-(2+\pi )x\right]$ este maxima $\iff$ $f(x)=-(4+\pi )x^2+2kx$ este maxim $\iff$ $f(x)=-(4+\pi )\left(x-\frac {k}{4+\pi}\right)^2+\frac {k^2}{4+\pi}$ este maxim $\iff$

$(4+\pi )\left(x-\frac {k}{4+\pi}\right)^2$ este minim $\iff x_{\mathrm{max}}=\frac {k}{4+\pi}$ si $2y_{\mathrm{max}}=k-(2+\pi )x_{\mathrm{max}}\iff$ $2y_{\mathrm{max}}=k-\frac {(2+\pi )k}{4+\pi}\implies$ $y_{\mathrm{max}}=\frac {k}{4+\pi}$ , adica $\boxed{x_{\mathrm{max}}=y_{\mathrm{max}}=\frac {k}{4+\pi}}$ .

PE4. Fie fix $F$ in interiorul unghiului $\widehat{XOY}$ si mobile $M\in (OX\ ,\ N\in (OY$ astfel incat $F\in MN$ . Sa se determine pozitia dreptei $MN$ astfel incat aria $\triangle MON$ sa fie minima.

Dem. Fie $\left\{\begin{array}{cc} A\in (OX\ ; & FA\parallel OY\\\\ B\in (OY\ ; & FB\parallel OX\end{array}\right\|$ si ariile $[AOF]=[BOF]=a\ ,\ [AFM]=$ $x\ ,\ [BFN]=y\implies$ $S\equiv [MON]=2a+x+y$ este minima

$\iff$ suma $x+y$ este minima. Se observa ca $\sqrt {\frac xS}+\sqrt {\frac yS}=\frac {FM}{MN}+\frac {FN}{MN}=1$ , i.e. $\sqrt x+\sqrt y=\sqrt S\iff$ $x+y+2\sqrt {xy}=2a+x+y\iff$ $xy=a^2$ (constant).

Astfel, deoarece $xy$ este constant, aria $\triangle MON$ este minima $\iff$ suma $x+y$ este minima $\iff x=y=a\iff$ $\left\{\begin{array}{c} AO=AM\\\ BO=BN\end{array}\right\|\iff$ $MN\parallel AB$ .

PE5. Sa se inscrie un dreptunghi $MNPQ$ de arie maxima in $\triangle ABC$ astfel incat $M\in (AB)\ ,\ N\in$ $(AC)\ ,\ \{P,Q\}\subset (BC)$ .

Dem. Fie $MN=x\ ,\ MQ=y$ si $D\in BC$ astfel ca $AD\perp BC$ . Se observa ca $\frac xa+\frac y{h_a}=$ $\frac {MN}{BC}+\frac {MQ}{AD}=$ $\frac {AM}{AB}+\frac {BM}{AB}=1\implies$ $\boxed{\frac xa+\frac y{h_a}=1}\ (*)$ (constant).

In concluzie, aria $S(x,y)=xy$ a dreptunghiului $ABCD$ este maxima $\iff$ produsul $\frac xa\cdot \frac y{h_a}$ este maxim $\stackrel{(*)}{\iff}$ $\frac xa=\frac y{h_a}=\frac 12\iff 2x=a$ $\iff MN$ este linie mijlocie in $\triangle ABC$ .

PE6. Fie un mobil $A$ pe semicercul de diametru $[BC]$ , unde $BC=a$ . Sa se determine maximul sumei $AC+AD$ , unde $D\in BC\ ,\ AD\perp BC$ .

Dem. Suma $AC+AD$ este max. $\iff$ $b+h_a$ este max. $\iff$ $b+\frac {bc}{a}$ este max. $\iff$ $b(a+c)$ este max. deoarece $a$ este constant $\iff$ $b^2(a+c)^2$ este max. $\iff$

$\left(a^2-c^2\right)(a+c)^2$ este max, $\iff$ $(a+c)^3(a-c)$ este max. $\iff$ $\frac {a+c}{3}\cdot\frac {a+c}{3}\cdot\frac {a+c}{3}\cdot (a-c)$ este max. , unde $\frac {a+c}{3}+\frac {a+c}{3}+\frac {a+c}{3}+(a-c)=2a$ (constant) $\iff$

$\frac {a+c}{3}=a-c=\frac a2\iff$ $c=\frac a2\ ,\ b=\frac {a\sqrt 3}{2}$ . In concluzie, $b+h_a=b+\frac {bc}{a}\le \frac {a\sqrt 3}{2}+\frac {a\sqrt 3}{4}=\frac {3a\sqrt 3}{4}\implies$ $\boxed{b+h_a\le \frac {3a\sqrt 3}{4}}$ .

Observatie. Fie $E$ a doua intersectie intre $AD$ si cercul circumscris al $\triangle ABC$ . Aplicam inegalitatea $p\le \frac {3R\sqrt 3}{2}$ in $\triangle ACE.$ Deci $2h_a+2b\le 3\sqrt 3\cdot \frac a2\ \implies\ b+h_a\le \frac {3a\sqrt 3}{4}$ .

PE7. Sa se arate ca intr-un triunghi $A$ -dreptunghic exista inegalitatea $\frac cb+\frac {2b}{a+c}\ge \sqrt 3$ .

Dem. $b^2=a^2-c^2$ $\implies$ $\left\{\begin{array}{c} E=\frac cb+\frac {2b}{a+c}=\frac cb+\frac {2(a-c)}{b}=\frac {2a-c}{b}\\\\ F=\frac ab+\frac {b}{a+c}=\frac ab+\frac {a-c}{b}=\frac {2a-c}{b}\end{array}\right\|\implies$ $E=F\implies$ $2E=E+F=\frac {a+c}{b}+\frac {3b}{a+c}\ge 2\sqrt 3$

cu egalitate $\iff \frac {a+c}{b}=\frac {3b}{a+c}=\sqrt 3\ \iff\ \frac a2=\frac b{\sqrt 3}=\frac c1$ , adica $B=60^{\circ}$ .

PE8. Fie $\triangle ABC$ , unde $b\ge c$ . Sa se arate ca $\frac ba+\frac {a}{b+c}\ge \sqrt {2+\frac {b^2-c^2}{a^2}}$ cu egalitate $\iff A=2C$ .

Dem. Daca $b=c$ , atunci $\frac ba+\frac {a}{2b}\ge$ $\sqrt 2\iff$ $\left(a-b\sqrt 2\right)^2\ge 0$ cu egalitate $\iff a=b\sqrt 2$ , adica $\frac a{\sqrt 2}=\frac b1=$ $\frac c1\iff$ $A=90^{\circ}\ ,\ B=C=45^{\circ}$ $\implies A=2C$ . Fie $b>c$ si

$b^2=c^2+\lambda a^2$ , unde $\lambda >0$ . Deci $\frac ba+\frac {b-c}{\lambda a}\ge$ $\sqrt {2+\lambda}\iff$ $(\lambda +1)^2\left(\lambda a^2+c^2\right)\ge c^2+$ $2ac\lambda\sqrt {\lambda +2}+\lambda^2a^2(\lambda +2)\iff$ $a^2-2ac\sqrt {\lambda +2}+(\lambda +2)c^2\ge 0\iff$

$\left(a-c\sqrt {\lambda +2}\right)^2\ge 0$ , adevarat. Avem egaltate $\iff a=c\sqrt {\lambda +2}$ , adica $\left|a^2-c^2\right|=bc\iff$ $|A-C|=C\iff$ $A=2C$ .

Obs. Daca $\lambda =1$ , adica $B=90^{\circ}$ , atunci inegalitatea devine $\frac ba+\frac {a}{b+c}\ge\sqrt 3$ , cu egalitate $\iff A=60^{\circ}$ (vezi problema propusa PE6).

PE9. Pe laturile $\triangle ABC$ consideram $\left\{\begin{array}{cc} M\in (BC)\ : & MB=m\cdot MC\\\\ N\in (CA)\ : & NC=n\cdot NA\\\\ P\in (AB)\ : & PA=p\cdot PB\end{array}\right\|$ astfel incat $AM\cap BN\cap CP\ne \emptyset$ , adica $mnp=1$ . Cercul circumscris $w$ al $\triangle ABC$

intalneste a doua oara $AM$ , $BN$ si $CP$ in $X$ , $Y$ si $Z$ respectiv. Sa se arate ca $\frac {MA}{MX}+\frac {NB}{NY}+\frac {PC}{PZ}\ge -3+2\sum\sqrt {\frac {(m+1)(n+1)}{n}}\ge$ $-3+4\sum\sqrt {\frac {m+1}{n+1}}\ge 9$ .

Dem. $\frac {MB}{m}=\frac {MC}{1}=\frac a{m+1}$ . Aplicand relatia Stewart la $AM$ in $\triangle ABC$ se obtine $(m+1)^2\cdot AM^2=(m+1)\left(mb^2+c^2\right)-ma^2$ si din puterea lui $M$ in raport cu $w$

obtinem $MA\cdot MX=MB\cdot MC$ . Asadar, $\sum \frac {MA}{MX}=\sum\frac {MA^2}{MA\cdot MX}=$ $\sum\frac {MA^2}{MB\cdot MC}=$ $\sum\frac {(m+1)\left(mb^2+c^2\right)-ma^2}{(m+1)^2\cdot \frac {ma}{m+1}\cdot \frac a{m+1}}=$ $\sum\frac {(m+1)\left(mb^2+c^2\right)-ma^2}{ma^2}=$

$-3+\sum\frac {(m+1)\left(mb^2+c^2\right)}{ma^2}=$ $-3+\sum\left[\frac {(m+1)b^2}{a^2}+\frac {(n+1)a^2}{nb^2}\right]\ge$ $-3+2\sum\sqrt {\frac {(m+1)(n+1)}{n}}\ge$ $-3+2\sum\sqrt {\frac {m+1}{n+1}\cdot\frac {(n+1)^2}{n}}\ge$ $-3+4\sum\sqrt {\frac {m+1}{n+1}}\ge 9$

because $\frac {(n+1)^2}{n}\ge 4$ and $\sum\sqrt {\frac {m+1}{n+1}}\ge 3\prod\sqrt {\frac {m+1}{n+1}}=3$ . Se arata usor ca avem egalitate $\iff m=n=p=1$ , adica centrul de greutate $G\in AM\cap BN\cap CP$ .

PE10. Let $ABCDE$ be a convex pentagon inscribed in a given circle $C(O,R)$ and for which $AC\perp BD$ . Ascertain the its maximum area $[ABCDE]$ .

Proof. Suppose w.l.o.g. $DE = AE$ . Let $\left\|\ \begin{array}{ccc} AB = u & ; & CD = v \\ \\ BC = x & ; & AD = y\end{array}\ \right\|$ and the distance $\delta_d(X)$ of $X$ to $d$ . Since $AC\perp BD$ obtain $x^2 + y^2 = u^2 + v^2 = 4R^2\ (*)$ and

$\delta_{AD}(O) = \frac x2$ . Apply the Ptolemy's theorem in the cyclic $ABCD\ \ : \ \ AC\cdot BD = xy + uv$ . Thus, $[ABCDE] = [ABCD] + [AED] =$ $\frac 12\cdot AC\cdot BD + \frac 12\cdot AD\cdot \delta_{AD}(E) =$

$\frac {xy + uv}{2} + \frac y2\cdot\left(R - \frac x2\right)$ . Thus, the area $[ABCDE]$ is maximum $\Longleftrightarrow$ $\left[2uv + y(2R + x)\right]$ is maximum, where exists $(*)$ .

$\blacktriangleright\ \ \ uv\mathrm {\ - \ max.}$ $\Longleftrightarrow$ $\left\|\begin{array}{c} u^2\cdot v^2\mathrm {\ - \ max.} \\ \\ u^2 + v^2 = 4R^2\mathrm{\ - \ constant}\end{array}\right\|$ $\Longleftrightarrow$ $u^2 = v^2 = 2R^2$ $\Longleftrightarrow$ $u = v = R\sqrt 2$ .

$\blacktriangleright\ \ \ y(2R + x)\mathrm {\ - \ max.}$ $\Longleftrightarrow$ $y^2\cdot (2R + x)^2\mathrm {\ - \ max.}$ $\Longleftrightarrow$ $\left(4R^2 - x^2\right)\left(2R + x\right)^2\mathrm {\ - \ max.}\ \Longleftrightarrow$ $\left\|\begin{array}{c} (2R - x)\cdot\left(\frac {2R + x}{3}\right)^3\ \mathrm {\ - \ max.} \\ \\ (2R - x) + 3\cdot\frac {2R + x}{3} = 4R\mathrm{\ - \ constant}\end{array}\right\|$ $\Longleftrightarrow$

$2R - x = \frac {2R + x}{3} = R$ and $x^2+y^2=4R^2$ . Thus, $x = R$ si $y = R\sqrt 3$ , i.e. $AB = CD = R\sqrt 2$ , $BC = R$ and $AD = R\sqrt 3$ $\Longrightarrow$ $\boxed {\ \begin{array}{c} AB = CD = R\sqrt 2 \\ \\ BC = DE = EA = R\\\\ (\ 2\cdot 90^{\circ}\ +\ 3\cdot 60^{\circ}\ =\ 360^{\circ}\ )\end{array}\ }$ .

PE11. Let $\triangle ABC$ and the mobile point $M\in [BC]$ . A circle which passes through $A$ and is tangent to $[BC]$ in $M$ cut again $[AB]$ , $[AC]$ in $X$ , $Y$ respectively.

Let the area $\sigma$ of the cyclic $AXMY$ and the area $S$ of $\triangle ABC$ . Prove that exists the inequality $\frac {\sigma}{S} + \left(\frac {a}{b + c}\right)^2\le 1$ with equality if and only if $\frac {MB}{MC} = \frac {AB}{AC}$ .

Proof. Let $\left\{\begin{array}{c} MB=x\\\\ MC=y\\\\ (x+y=a)\end{array}\right\|$ . Thus, $\left\{\begin{array}{ccc} BX\cdot BA=BM^2 & \implies & BX=\frac {x^2}c\\\\ CY\cdot CA=CM^2 & \implies & CY=\frac {y^2}b\end{array}\right\|$ . Thus, $[AXMY]$ is $\max\iff$ $[MBX]+[MCY]$ is $\min\iff$

$\frac 12\cdot\left(BM\cdot BX\cdot\sin B+CM\cdot CY\cdot\sin C\right)$ is $\min\iff$ $\frac 12\cdot \left(x\cdot \frac {x^2}c\cdot \frac b{2R}+y\cdot \frac {y^2}b\cdot \frac c{2R}\right)$ is $\min\iff$ $\frac {b^2x^3+c^2y^3}{4Rbc}$ is $\min\iff$ $\min_{\left|\begin{array}{c} x+y=a\\\ x>0\ ,\ y>0\end{array}\right|}\left(b^2x^3+c^2y^3\right)$ .

Let $f(x)=b^2x^3+c^2(a-x)^3$ , where $x\in(0,a)$ . Thus, $f'(x)=3b^2x^2-3c^2(a-x)^2\ .s.s.\ (bx)^2-$ $[c(a-x)]^2\ .s.s.\ bx-c(a-x)=(b+c)x-ac$ .

So $f'(x)\ .s.s.\ x-x_{\min}$ , where $x_{\min}=\frac {ac}{b+c}$ and $y_{\min}=\frac {ab}{b+c}$ , i.e. $\frac xc=\frac yb=\frac a{b+c}\iff$ the ray $[AM$ is the $A$ -bisector of $\triangle ABC$ . In this case,

$S-\sigma =\frac {b^2x^3+c^2y^3}{4Rbc}\ge \frac {f\left(x_{\min}\right)}{4Rbc}\ge$ $\left(\frac a{b+c}\right)^2\cdot\frac {abc}{4R}=S\cdot\left(\frac a{b+c}\right)^2\implies$ $S-\sigma\ge S\cdot\left(\frac a{b+c}\right)^2\implies$ $1-\frac {\sigma}{S}\ge \left(\frac a{b+c}\right)^2\implies$ $\boxed{\frac {\sigma}{S} + \left(\frac {a}{b + c}\right)^2\le 1}\ (*)$ .

Remark. The inequality $(*)$ is equivalently with $\boxed{\left\{\begin{array}{c} x+y=a\\\\ x>0\ ;\ y>0\end{array}\right\}\ \implies\ \frac {x^3}{c^2}+\frac {y^3}{b^2}\ \ge\ a\cdot\left(\frac a{b+c}\right)^2}$ , what is Holder's inequality $\left(\frac {x^3}{c^2}+\frac {y^3}{b^2}\right)(c+b)(c+b) \geq (x+y)^3$ .

Consequence. In the case $AM\perp BC$ , i.e. $\left\{\begin{array}{c} x:=c\cdot\cos B\\\\ y:=b\cdot \cos C\end{array}\right\|$ obtain that in an acute $\triangle ABC$ exists the inequality $\frac {(a^2 + c^2 - b^2)^3}{c^2} + \frac {(a^2 + b^2 - c^2)^3}{b^2}\ge \frac {8a^6}{(b + c)^2}$ .

Extension. Let $\triangle ABC$ and let $M\ ,\ N$ be two mobile points which belong to the side $[BC]$ so that $MN = k < a$ (constant) . The circumcircle of the triangle $MAN$ cut again

the sides $[AB]$ , $[AC]$ in the points $X$ , $Y$ respectively. Find the maximum of the area of the cyclic pentagon $AXMNY$ and the position of the points $M$ , $N$ where it is touched.

PE12. Let $ABCD$ be an convex quadrilateral with the fixed $\{A,B,C\}$ and the mobile $D$ . Find the maximum value of the area $S=[ABCD]$ , where $AB=BC=CD=1$ .

Proof. Let $AD=x$ . Bretschneider's formula $\implies 16\cdot S^2=(1+x)^3(3-x)-16x\cdot \cos^2\frac {A+C}{2}$ . Thus, $x\cdot \cos^2\frac {A+C}{2}$ is absolute $\min\iff A+C=\pi$ , i.e. $ABCD$ is cyclic. In this case $S$ is $\max\iff$ $(1+x)^3(3-x)$ is $\max\iff$ $\left(\frac {1+x}{3}\right)^3(3-x)$ is $\max\iff$ $\frac {1+x}{3}=3-x=\frac 44=1$ because $3\cdot\frac {1+x}{3}+(3-x)=4$ (constant) and $S$ is $\max\iff x=2$ and $S_{\mathrm{max}}=\frac {3\sqrt 3}{4}$ . Remark. Prove easily that $27+(x-3)(x+1)^3=(x-2)^2(x^2+4x+6)\ge 0$ with equality iff $x=2$ .

Therefore $(x+1)^3(3-x)\le 27$ $\implies$ $16\cdot S^2\le 27$ $\implies$ $S\le\frac {3\sqrt 3}{4}$ with equality iff $x=2$ .

PE13. Peter and Jane begin running along a straight track starting from point $O$ while Lily looks on at the side. The diagram shows a particular instant when Peter, Lily and Jane are at

$A$ , $B$ and $C$ respectively. Given that Peter runs $3$ times faster than Jane and $m\left(\widehat{AOB}\right) = 90^{\circ}$ , find the maximum value of Lily's angle of sight, i.e. $\widehat{ABC}$ between the two runners.

http://i944.photobucket.com/albums/ad288/GemenLeu/acc5b34d-5e52-4c6e-9ee5-f26b1af78e37_zpsft5pui2z.png

Proof 1 (analytic). Let $\left\{\begin{array}{cc} O(0,0)\ ; & A(0,3x)\\\\ B(d,0)\ ; & C(0,x)\end{array}\right\|\ ,\ \left\{\begin{array}{c} m\left(\widehat {OBA}\right)=\alpha\\\ m\left(\widehat {OBC}\right)=\beta\\\ m\left(\widehat {ABC}\right)=\phi\end{array}\right\|$ and the slopes $\left\{\begin{array}{c} s(BA)=-\tan\alpha =\frac {3x}d\\\\ s(BC)=-\tan\beta =\frac xd\end{array}\right\|$ where $\phi =\alpha -\beta >0\implies$ $\tan\phi =\tan (\alpha -\beta)=$

$\frac {\tan\alpha -\tan\beta}{1+\tan\alpha\cdot\tan\beta}\implies$ $\tan\phi =\frac {2dx}{3x^2+d^2}$ . Let $t=\frac dx\ ,\ x>0$ . Thus, $\tan\phi =f(t)\equiv \frac {2t}{t^2+3}$ . Prove easily that $\phi$ is $\max \iff f(t)$ is $\min\iff t=\sqrt 3\iff \phi =30^{\circ}$ .

Proof 2. Let $OB=d$ (constant) and $OC=x\ ,\ OA=3x$ , i.e. $AC=2x$ and $m\left(\widehat{ABC}\right)=\phi <90^{\circ}$ . Thus, $\left\{\begin{array}{ccc} BC^2 & = & x^2+d^2\\\\ BA^2 & = & 9x^2+d^2\end{array}\right\|$ .I"ll apply the identity

$4[ABC]=\left(b^2+c^2-a^2\right)\tan\phi$ in any $\triangle ABC$ . Therefore, $2\cdot AC\cdot BO=\left(BA^2+BC^2-AC^2\right)\tan\phi\iff$ $2dx=\left(3x^2+d^2\right)\tan\phi\iff$

$\boxed{\tan\phi =\frac {2dx}{3x^2+d^2}}\ (*)$ . In conclusion, $\phi$ is $\max$ . $\iff$ $\tan\phi$ is $\max$ . $\iff$ $\frac {3x^2+d^2}{2dx}$ is $\min$ . $\iff$ $\frac{3x^2+d^2}{dx}$ is $\min$ . $\iff$ $\frac {3x}d+\frac dx$ is $\min$ .

Observe that the product $\frac {3x}d\cdot\frac dx=3$ (constant).Thus, $\phi$ is $\max$ . $\iff$ $\frac {3x}d=\frac dx=\sqrt 3$ and $\cos\phi =\frac {\sqrt 3}2$ , i.e. $\boxed{d=x\sqrt3\ \wedge\ \phi_{\mathrm{max}} =30^{\circ}}$ .

Proof 3. I"ll use same notations from the upper proof. Thus, $\left\{\begin{array}{ccc} BC^2 & = & x^2+d^2\\\\ BA^2 & = & 9x^2+d^2\end{array}\right\|$ . Apply the generalized Pythagoras's theorem $:$

$AC^2=BA^2+BC^2-2\cdot BA\cdot BC\cdot\cos\phi\iff$ $4x^2=\left(x^2+d^2\right)+\left(9x^2+d^2\right)-2\cos\phi \sqrt{\left(x^2+d^2\right)\left(9x^2+d^2\right)}\iff$ $\cos\phi =\frac {3x^2+d^2}{\sqrt{\left(x^2+d^2\right)\left(9x^2+d^2\right)}}$ . In conclusion,

$\phi$ is $\max .\iff$ $\cos \phi$ is $\min .\iff$ $\cos^2\phi$ is $\min .\iff$ $\frac {\left(3x^2+d^2\right)^2}{\left(x^2+d^2\right)\left(9x^2+d^2\right)}$ is $\min .\iff$ $\frac {9x^4+6d^2x^2+d^4}{9x^4+10d^2x^2+d^4}$ is $\min .\iff$ $1-\frac {4d^2x^2}{9x^4+10d^2x^2+d^4}$ is $\min .\iff$

$\frac {4d^2x^2}{9x^4+10d^2x^2+d^4}$ is $\max .\iff$ $\frac {9x^4+10d^2x^2+d^4}{4d^2x^2}$ is $\min .\iff$ $\frac {9x^4+10d^2x^2+d^4}{d^2x^2}$ is $\min .\iff$ $\left(\frac {3x}d\right)^2+10+\left(\frac dx\right)^2$ is $\min .\iff$ $\left(\frac {3x}d\right)^2+\left(\frac dx\right)^2$ is $\min$ $.$

Observe that $\left(\frac {3x}d\right)^2\cdot \left(\frac dx\right)^2=9$ (constant). Therefore, $\phi$ is $\max$ $.$ $\iff$ $\left(\frac {3x}d\right)^2=\left(\frac dx\right)^2=3$ and $\tan\phi =\frac 1{\sqrt 3}$ , i.e. $\boxed{d=x\sqrt3\ \wedge\ \phi_{\mathrm{max}} =30^{\circ}}$ .

Lemma. Let $\{M,S\}$ be a pair of isogonal points on the side $[BC]$ , i.e. $\widehat{SAB}\equiv\widehat{MAC}$ . Then exists the relation $\boxed{\frac sm=\frac {bc(\lambda +1)}{\lambda b^2+c^2}}\ (*)\ ,$ where $\frac {MB}{MC}=\lambda$ .

Proof. From $\frac {MB}{MC}=\lambda$ obtain that $\frac {MB}{\lambda}=\frac {MC}1=\frac a{\lambda +1}\implies$ $\boxed{MC=\frac a{\lambda +1}}\ (3)$ . Apply the Steiner's relation $\frac {SB}{SC}\cdot\frac {MB}{MC}=\frac {c^2}{b^2}$ , i.e. $\frac {SB}{SC}=\frac {c^2}{\lambda b^2}\iff$

$\frac {SB}{c^2}=\frac{SC}{\lambda b^2}=\frac a{\lambda b^2+c^2}\implies$ $\boxed{SB=\frac {ac^2}{\lambda b^2+c^2}}\ (4)$ . Denote $AM=m$ , $AS=s$ and $\left(\widehat{SAB}\right)=\left(\widehat{MAC}\right)=\phi$ . Apply the theorem of Sines to the triangles $:$

$\left\{\begin{array}{cccccccc} \triangle ACM\ : & \frac {AM}{\sin C} & = & \frac {MC}{\sin\phi} & \implies & m\cdot\sin\phi & \stackrel {(3)}{=} & \frac a{\lambda +1}\cdot \sin C\\\\ \triangle ABS\ : & \frac {AS}{\sin B} & = & \frac {BS}{\sin\phi} & \implies & s\cdot\sin\phi & \stackrel {(4)}{=} & \frac {ac^2}{\lambda b^2+c^2}\cdot \sin B\end{array}\right\|\ (:)\ \implies$ $\frac ms=\frac a{\lambda +1}\cdot\frac {\lambda b^2+c^2}{ac^2}\cdot\frac {\sin C}{\sin B}\implies$ $\frac ms= \frac a{\lambda +1}\cdot\frac {\lambda b^2+c^2}{ac^2}\cdot\frac cb\implies$ $\frac sm=\frac {bc(\lambda +1)}{\lambda b^2+c^2}$

PE14. Prove that there is the inequality $\sum\frac {b^2+c^2}{m_a}\le 12R$ , where $R$ is the length of the circumradius and $m_a$ is the length of

the $A$ -median of $\triangle ABC$ (author: Dmitry Tereshin from Moscow. This problem was proposed in "All Russian Olympiad 1994").

Proof. Prove easily or is well-known the relation $\boxed{bc=2Rh_a}\ (1)$ and from upper lemma for $\lambda :=1$ obtain that $\boxed{\frac {b^2+c^2}{m_a}=\frac {2bc}{s_a}}\ (2)\ ,$ where $h_a$ and

$s_a$ are the lengths of the $A$ -altitude and the $A$ -symmedian respectively. Therefore, $\frac {b^2+c^2}{m_a}\ \stackrel{1\wedge 2}{=}\ \frac {4Rh_a}{s_a}\le 4R\ .$ In conclusion, $\sum\frac {b^2+c^2}{m_a}\le 12R\ .$

PE15. What is the maximum volume of a cylinder what is inscribed within a cone given with radius $r$ of the base and height $h$ ?

Proof. Denote the lengths $x$ and $y$ of radius and height for the required cylinder. Observe that $\frac xr+\frac yh=1$ and its volume is $V=V(x,y)=\pi x^2y$ . Thus, $V$ is $\max\iff$

$x^2y$ is $\max\iff$ $\frac {x}{2r}\cdot\frac {x}{2r}\cdot \frac yh$ is $\max$ , where $\frac {x}{2r}+\frac {x}{2r}+\frac yh=1$ (constant) $\iff$ $\frac {x}{2r}=\frac yh=\frac 13\iff$ $\left\{\begin{array}{c} x=\frac {2r}{3}\\\\ h=\frac h3\end{array}\right\|$ and in this case $V_{\max .}=V\left(\frac {2r}{3},\frac h3\right)=\frac {4\pi r^2h}{27}$ .

PE16. What is the maximum of whole surface for a cylinder what is inscribed within a cone given with radius $r$ of the base and height $h$ ?

Proof. Denote the lengths $x$ and $y$ of radius and height for the required cylinder. Observe that $\frac xr+\frac yh=1$ , i.e. $\boxed{y=\frac {h(r-x)}r}$ and its total surface is

$S_t=S_t(x,y)=2\pi x(x+y)$ . Thus, $S_t$ is $\max\iff$ $x(x+y)$ is $\max\iff$ $x\left[x+\frac {h(r-x)}r\right]$ is $\max\iff$ $x\left[rx+h(r-x)\right]$ is $\max\iff$

$(r-h)x^2+hrx$ is $\max$ . Appear two cases. If If $h>2r$ , then $S_t$ is $\max\iff$ $x_{\max}=\frac {hr}{2(h-r)}$ and in this case $h_{\max}=\frac {h(h-2r)}{2(h-r)}$ . If $h\le 2r$ , then $x=r$ .

PE17. Find the maximum value of $f(x)=\sin^2\frac x2\cos^6\frac x2$ , where $x\in\mathbb R$ ..

Proof 1. Let $t = \sin^2\frac x2$ , where $t\in [0,1]$ . Our function becomes $g(t)=t(1 - t)^3$ . By AM-GM follows: $3t(1 - t)^3\le \left[\frac{3t + (1 - t) + (1 - t) + (1 - t)}{4}\right]^4 = \frac{81}{256}$ .

Dividing both sides by $3$ gives $\boxed{f(x)\le\frac {27}{256}}\ (*)$ . Equality holds iff $3t = 1 - t$ , i.e. $t = \frac{1}{4}$ . Get $\sin\frac{x}{2} = \pm \frac{1}{2}$ . which holds for $x\in \left\{\frac{6k\pi\pm 1}3\right\}$ where $k\in\mathbb Z$ .

Proof 2. $16f(x)=2\sin^2\frac x2\left(2\cos^2\frac x2\right)^3=(1-\cos x)(1+\cos x)^3\implies$ $\frac {16f(x)}{27}=(1-\cos x)\left(\frac {1+\cos x}3\right)^3\le\left[\frac {(1-\cos x)+3\frac {1+\cos x}3}4\right]^4=\frac 1{16}\implies$ $f(x)\le\frac {27}{256}$ .

PE18. Suppose we have a (right, circular) truncated cone $T\equiv F(h,R,r)$ , where: its height $h$ and its volume are constants; its upper radius $R$ and

its lower radius $r$ are variables, where $r<R$ . Find the ratio $\frac Rr$ when the sum between its lateral surface and surface of lower base is minimum.

Proof.

PE19. Find $\max\left(u^4v^3\right)$ where $\{u,v\}\subset (0,1)$ and $u+v+uv=1.$

Proof 1 (without derivatives). Observe that the sum $\frac u3+\frac u3+\frac u3+\frac v2+\frac v2+uv=3\cdot\frac u3+2\cdot\frac v2+uv=u+v+uv=1$ (constant) and $u^4v^3$ is $\max\iff$

$\frac {u^4v^3}{3^3\cdot 2^2}$ is $\max$ $\iff$ $uv\cdot\frac u3\cdot\frac u3\cdot\frac u3\cdot\frac v2\cdot\frac v2=$ $uv\cdot \left(\frac u3\right)^3\cdot\left(\frac v2\right)^2$ is $\max$ $\iff$ $uv=\frac u3=\frac v2=\frac 1{3+2+1}=\frac 16,$ i.e. $u=\frac 12,$ $v=\frac 13$ and $uv=\frac 16,$ what is compatibly.

Proof 2 (with derivatives) Observe that $u+v+uv=1$ $\implies$ $v(u)=\frac {1-u}{1+u},$ $v'(u)=$ $-\frac 2{(1+u)^2}$ and $f(u,v)=u^4v^3$ becomes $g(u)=\frac {u^4(1-u)^3}{(1+u)^3}$ and

$g'(u)=4u^3v^3+3u^4v^2v'=$ $4u^3v^3-\frac {6u^4v^2}{(1+u)^2}=2u^3v^2\cdot\left[2v-\frac {3u}{(1+u)^2}\right]=$ $\frac {2u^3v^2}{(1+u)^2}\cdot\left[2v(1+u)^2-3u\right]=$ $\frac {2u^3v^2}{(1+u)^2}\cdot\left[2(1-u)(1+u)-3u\right]=$

$-\frac {2u^3v^2}{(1+u)^2}\cdot\left (2u^2+3u-2\right)\implies$ $\boxed{g'(u)=-\frac { 2u^3v^2(u+2)(2u-1)}{(1+u)^2}}\ (1).$ Prove easily that $\max_{u\in (0,1)}g(u)=g\left(\frac 12\right).$

Example (Israel Diaz Acha). Find the maximum of $f(x,y)=\tan^4x\tan^3y,$ where $\{x,y\}\subset \left(0,\frac {\pi}4\right)$ and $x+y=\frac {\pi}4.$

Proof. I"ll use the notations $\boxed{\tan x=u}$ and $\boxed{\tan y=v}.$ Thus, $\{u,v\}\subset (0,1)$ and $1=\tan\frac {\pi}4=\tan (x+y)=$ $\frac {\tan x+\tan y}{1-\tan x\tan y}\iff$ $1=\frac {\tan x+\tan y}{1-\tan x\tan y}\iff$

$1=\frac{u+v}{1-uv}\iff$ $1-uv=u+v,$ i.e. $u+v+uv=1.$ Using upper proposed problem P18 obtain that the function $f(x,y)$ touches its maximum for $2\cdot \tan x=3\cdot\tan y=1.$

PE20. Let the $C$ -right $\triangle ABC,$ the midpoint $M$ of $[AC]$ and the point $D\in (MC)$ so that $\widehat{MBA}\equiv\widehat{MBD}$ and $\widehat{DBA}\equiv\widehat{DBC}.$ Prove that $\frac 52<\frac ca<3.$

Proof 1 (with derivatives). $CB\perp CA\iff \boxed{b^2=c^2-a^2}\ (1)$ and $\widehat{DBA}\equiv\widehat{DBC}\iff$ $\frac {DA}c=\frac {DC}a=\frac b{a+c}.$ Hence $\boxed{DC=\frac {ab}{a+c}}\ (2)$ and $DM=DA-MA=$

$\frac {bc}{a+c}-\frac b2=$ $\frac {b(c-a)}{2(a+c)}$ $\implies$ $\boxed{DM=\frac {b(c-a)}{2(a+c)}}\ (3)\ .$ Thus, $\widehat{MBD}\equiv\widehat{MBA}\iff$ $\frac {BD}c=\frac {BD}{BA}=\frac {MD}{MA}=\frac {\frac {b(c-a)}{2(a+c)}}{\frac b2}=$ $\frac {c-a}{a+c}\implies$ $\boxed{BD=\frac {c(c-a)}{a+c}}\ (4)\ .$ Therefore,

$CB^2+CD^2=BD^2\iff$ $a^2+\left(\frac {ab}{a+c}\right)^2=\left[\frac {c(c-a)}{a+c}\right]^2\iff$ $a^2(a+c)^2+a^2\left(c^2-a^2\right)=c^2(c-a)^2\iff$ $\boxed{2a^3+a^2c=c^3-2ac^2}\ (5)\ .$ Denote $\boxed{\frac ac=t}\ ,$

where $t\in (0,1)\ .$ In conclusion, the relation $(5)$ becomes the equation $\boxed{f(t)\equiv 2t^3+t^2+2t-1=0}\ (6)$ where $0<t<1\ .$ Observe that $f'(t)=6t^2+2t+2>0,$

i.e. the function $f$ is increasing ( $\uparrow$ ) on $(0,1)$ (only one real root). Observe that $f\left(\frac 13\right)=-\frac 4{27}<0<\frac {11}{125}=f\left(\frac 25\right) \iff$ $\frac 13<t<\frac 25\iff$ $\frac 13<\frac ac<\frac 25$ $\iff$ $\frac 52<\frac ca<3.$

See here and here.

http://i944.photobucket.com/albums/ad288/GemenLeu/crystinel_zpsxdhp4p33.png

PE21. Find the maximum and the minimum of the function $f(x,y)=3x+4y$ over $\mathbb R^2,$ where $x^2+2y^2=17\ .$

Proof. See its extension P14 from here.

This post has been edited 209 times. Last edited by Virgil Nicula, Feb 25, 2019, 7:23 PM

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El_Ectric:
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425. Geometrical extremity II.

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