426. Problems from and for baccalaureate.

by Virgil Nicula, Jun 25, 2015, 8:01 PM

PP0. Let $\triangle ABC$ with orthocenter $H\ ,$ $E\in BH\cap AC\ ,\ F\in CH\cap AB$ and midpoint $M$ of $[BC]\ .$ Prove that $m\left(\widehat{ EMF}\right)=\left|180^{\circ}-2A\right|$ (without proof).

PP1. Let $\triangle ABC$ with incircle $w=\mathbb C(I,r)$ and excircles $\left\{\begin{array}{ccc} w_b & = & \mathbb C\left(I_b,r_b\right)\\\\ w_c & = & \mathbb C\left(I_c,r_c\right)\end{array}\right\|$ . If $BC=5$ , $r_b=3$ , $r_c=2\ ,$ then find $\{A,b,c,\underline r,S\}$ (standard notations).

Answer: $A=90^{\circ}\ ;\ b=4\ ;\ c=3\ ;\ ;\ r=1\ ;\ S=6$ . Generally $r=f\left(a,r_b,r_c\right)=\frac {-a^2+a\sqrt {a^2+4r_br_c}}{2\left(r_b+r_c\right)}$ . See PP16 from here.

PP2. Let $\triangle ABC$ with incircle $w=\mathbb C(I,r)$ and the $A$ -excircle $w_a=\mathbb C\left(I_a,r_a\right)$ . If $AB=5$ , $AC=8$ and $r_a=\frac {10}{\sqrt 3}\ ,$ then find $\{A,a,\underline r,S\}$ (standard notations).

Answer: $A=60^{\circ}\ ;\ a=7\ ;\ r=\sqrt 3\ ;\ S=10\sqrt 3$ . Generally $(b+c)^2\cdot\left(r_a-r\right)^2=$ $\left(r_a+r\right)^2\left[(b-c)^2+4rr_a\right]\implies r$ (explicit). See PP17 from here.

PP3. Let $\overarc[]{BC}\subset w$ (circle) be an arc with the midpoint $A$ and a mobile point $M\in\overarc[]{BC}$ . Prove that there is the relation $MA^2+MB\cdot MC=AB^2$ (constant).

Proof 1. Suppose w.l.o.g. that $M\in\overarc[]{AC}$ and let $M^{\prime}\in \overarc[]{AB}$ so that $AM^{\prime}\parallel BM$ . Thus, $\left\{\begin{array}{ccc} M^{\prime}B& = & MA\\\\ M^{\prime}A & = & MC\\\\ M^{\prime}M & = & AB\end{array}\right\|$ . Apply Ptolemy's

theorem to the isosceles trapezoid $AMBM^{\prime}\ :\ M^{\prime}B\cdot MA+MB\cdot M^{\prime}A=M^{\prime}M\cdot AB\iff$ $MA^2+MB\cdot MC=AB^2$ .

Proof 2. Let the diameter $[AS]$ of the circle $w$ . Thus, $MA\perp MS$ and $\widehat{SMB}\equiv\widehat{SMC}$ . Therefore,

$\left\{\begin{array}{cccccc} \underline{MD}^2 & = & MB\cdot MC & - & \underline{\underline{\underline{DB\cdot DC}}} & \mathrm{(theorem\ of\ bisector)}\\\\ AB^2 & = & \underline{\underline{AD}}^2 & + & \underline{\underline{\underline{DB\cdot DC}}} & \mathrm{(Stewart's\ theorem)}\\\\ \underline{\underline{AD}}^2 & = & \underline{MD}^2 & + & MA^2 & \mathrm{(Pythagoras'\ theorem)}\end{array}\right\|$ $\bigoplus$ $\implies$ $AB^2=MB\cdot MC+MA^2$ .

PP4. Let $ABC$ be an isosceles triangle with $AB=AC$ and $M\in (BC)$ . Then $AB^2=MA^2+MB\cdot MC$ (particular case of the Stewart's relation) .

Remark. If $M$ is the midpoint of $[BC]$ , then is evidently. Therefore, can suppose w.l.o.g. that $M$ isn't the midpoint of $[BC]$ .

Method 1. Denote the midpoint $D$ of $[BC]$ and suppose w.l.o.g. that $M\in (BD)$ . Apply the well-known property $\boxed{AD\perp BM\iff AB^2-AM^2=DB^2-DM^2}$

(can prove easily). Therefore, $AB^2-AM^2=(DB-DM)(DB+DM)=$ $MB(DA+DM)=MB\cdot MC\implies$ $AB^2=AM^2+MB\cdot MC$ .

Method 2. Apply the Stewart's relation to $AM$ in $\triangle ABC\ :\ AB^2\cdot MC$ $+AC^2\cdot MB=AM^2\cdot BC+MB\cdot MC\cdot BC$ $\iff$ $AB^2=AM^2+MB\cdot MC$ .

Method 3. Let the circumcircle $w$ of $\triangle ABC$ and $\{A,N\}=\{A,M\}\cap w$ . From the power $p_w(M)$ of $M$ w.r.t. $w$ obtain that

$\boxed{MB\cdot MC=MA\cdot MN}\ (*)$ . Thus, $\widehat{ACM}\equiv\widehat{ACB}\equiv$ $\widehat{ABC}\equiv\widehat{ANC}\implies$ $\widehat{ACM}\equiv\widehat{ANC}\implies$ $\triangle ACM\sim\triangle ANC\implies$

$\frac {AC}{AN}=\frac {AM}{AC}\implies$ $AC^2=AM\cdot AN\implies$ $AB^2=AM(AM+MN)=$ $AM^2+MA\cot MN\ \stackrel{(*)}{\implies}\ AB^2=AM^2+MB\cdot MC$ .

Method 4. Denote $\left\{\begin{array}{ccc} AB=AC=l & ; & AM=x\\\\ MB=a & ; & MC=b\end{array}\right\|$ . I"ll apply the well-known $4S=\left(b^2+c^2-a^2\right)\tan A$ in any $\triangle ABC$ where $S$ is its area. Thus, $B=C\implies$

$\tan B=\tan C\implies$ $\frac {MB}{MC}=\frac {[ABM]}{[ACM]}\implies$ $\frac ab=\frac {l^2+a^2-x^2}{l^2+b^2-x^2}\implies$ $b\left(l^2+a^2-x^2\right)=a\left(l^2+b^2-x^2\right)\implies$ $l^2(a-b)+ab(b-a)=x^2(a-b)\iff$ $l^2=x^2+ab$ .

Particular case. For $\triangle ABE$ and $F\in (AE)$ with $BA=BE=1$ , $BF=x$ and $FA=a$ , $FE=b$ obtain that $\boxed{1=x^2+ab}\ (*)$ .

PP5. Let $ABCD$ be a square and $F\in (CD)\cap (BE)$ so that $CE=AB$ and $BF=a$ , $FE=b$ . Prove that $\tan\phi =\sqrt {\frac {a-b}{a+b}}$ , where $\phi =m\left(\widehat{CBF}\right)$ .

Proof 1. Denote w.l.o.g. $AB=1$ and $CF=x\in (0,1)$ . Thus, $\tan\phi =\frac {CF}{CB}=x$ and $\left\{\begin{array}{ccc} 1+x^2 & = & a^2\\\\ 1-x^2 & \stackrel{(*)}{=} & ab\end{array}\right\|$ $\implies$ $\boxed{2x^2=a(a-b)}\ (1)$ . Thus, $\widehat{BDF}\equiv\widehat{BED}$

$($ same value $45^{\circ})$ $\iff$ $\triangle BDF\sim\triangle BED\iff$ $\frac {BD}{BE}=\frac {BF}{BD}$ i.e. $BD^2=BF\cdot BE\iff$ $2=a(a+b)\ \stackrel{(1)}{\implies}\ \frac {2x^2}{2}=\frac {a(a-b)}{a(a+b)}\implies$ $x^2=\frac {a(a-b)}{a(a+b)}\iff$ $x=\sqrt{\frac {a-b}{a+b}}$ .

Proof 2. Denote w.l.o.g. $AB=1$ and $CF=x\in (0,1)$ . Thus, $\frac {CF}{CB}=\tan\phi =x$ . Apply an well-known relation $:$

$\frac {FB}{FE}=\frac {CB}{CE}\cdot\frac {\sin\widehat {FCB}}{\sin\widehat{FCE}}\implies$ $\frac ab=\frac 1{\cos 2\phi}\implies$ $\cos 2\phi =\frac ba\implies$ $\frac {1-x^2}{1+x^2}=\frac ba\implies$ $x^2=\frac {a-b}{a+b}\implies$ $x=\sqrt{\frac {a-b}{a+b}}$ .

PP6. Let $\triangle ABC$ with the circumcircle $w$ and the incentre $I$ . Denote $\{A,L\}=\{A,I\}\cap w$ . Prove that $LA^2=LI^2+AB\cdot AC$ .

Proof. Prove easily that $\boxed{LB=LC=LI}\ (*)$ . Thus, $\left\{\begin{array}{ccccccc} \triangle LBD\sim\triangle LAB & \implies & \frac {LB}{LA}=\frac {LD}{LB} & \stackrel{(*)}{\implies} & LA\cdot LD & = & LI^2\\\\ \triangle LAB\sim\triangle CAD & \implies & \frac {LA}{CA}=\frac {AB}{AD} & \implies & LA\cdot AD & = & AB\cdot AC\end{array}\right\|\ \bigoplus\implies$ $LA^2=LI^2+AB\cdot AC$ .

PP7. Let $ABCD$ be a trapezoid so that $AD\parallel CD$ and $\widehat{CAD}\equiv\widehat{BDC}$ . Denote $\left\{\begin{array}{cccccc} AB=a\ ; & CD=b\\\ AD=c\ ; & BC=d\\\ AC=e\ ; & BD=f\end{array}\right\|$ Prove that $\frac {d^2-b^2}a=\frac {c^2-b^2}b$ .

Proof. Let $E\in AB$ be the point so that $A\in (BE)$ and $AE=CD$ . Thus, $ACDE$ is a parallelogram, i.e. $AE=b\ ,\ DE=e$ and $\widehat{ADE}\equiv\widehat{CAD}\equiv\widehat{BDC}\equiv\widehat {DBE}\implies$

$\triangle ADE\sim \triangle DBE\iff$ $\frac {AD}{DB}=\frac {DE}{BE}=\frac {AE}{DE}\iff$ $\frac cf=\frac e{a+b}=\frac be\iff$ $\odot\begin{array}{ccccc} \nearrow & e^2 & = & b(a+b) & \searrow\\\\ \searrow & ef & = & c(a+b) &\nearrow\end{array}\odot$ . Apply the well-known (or prove easily) Euler's relation $:$

$\boxed{e^2+f^2=2ab+c^2+d^2}\ (*)$ . Thus, $b(a+b)+\frac {c^2(a+b)}b=2ab+c^2+d^2\iff$ $b^2+\frac {ac^2}b=ab+d^2\iff$ $d^2-b^2=a\cdot\left(\frac {c^2}b-b\right)\iff$ $\frac {d^2-b^2}a=\frac {c^2-b^2}b$ .

PP8. Let a rectangle $ABCD$ with $O\in AC\cap BD$ and the midpoints $M$ , $N$ of $[AB]$ , $[OD]$ respectively. Prove that $ABCD$ is a square $\iff \triangle MNC$ is isosceles and $NM\perp NC$ .

Proof 1. Denote $AB=a$ , $AD=b$ and $E\in CN\cap AD$ , $F\in MN\cap AD$ . Thus, $NM\perp NC\iff$ $CNDF$ is cyclically $\iff$ the power of $E$ w.r.t.

the circumcircle of $CNDF$ is $ED\cdot EF=EN\cdot EC\iff$ $\frac b3\cdot \frac {5b}6=\frac 14\cdot \left(\frac {b^2}9+a^2\right)\iff$ $10b^2=b^2+9a^2\iff a=b$ and in this case $NM=NC$ .

Proof 2. Denote $AB=a$ , $AD=b$ and $E\in CN\cap AD$ , $F\in MN\cap AD$ . Thus, $NM\perp NC\iff$ $AENM$ is cyclically $\iff$

$\widehat{FEN}\equiv\widehat{FMA}\iff$ $\tan\widehat{FEN}=\tan\widehat{FMA}\iff$ $\frac {DC}{DE}=\frac {AF}{AM}\iff$ $\frac {a}{\frac b3}=\frac {\frac {3b}2}{\frac a2}\iff$ $a^2=b^2\iff$ $a=b$ and in this case $NM=NC$ .

Proof 3. Denote $AB=a$ , $AD=b$ and $E\in CN\cap AD$ and $F\in MN\cap AD$ and $\left\{\begin{array}{ccc} m\left(\widehat{DCE}\right) & = & x\\\\ m\left(\widehat{AFM}\right) & = & y\end{array}\right\|$ . Therefore,

$NM\perp NC\iff$ $x=y\iff$ $\tan x=\tan y\iff$ $\frac {DE}{DC}=\frac {AM}{AF}\iff$ $\frac {\frac b3}a=\frac {\frac a2}{\frac {3b}2}\iff$ $\frac {b^2}2=\frac {a^2}2\iff$ $a=b$ a.s.o.

Proof 4. I"ll use the analytical geometry. Thus, for $\left\{\begin{array}{ccc} A(0,0) & ; & B(a,0)\\\\ C(a,b) & ; & D(0,b)\end{array}\right\|$ obtain that $M\left(\frac a2,0\right)$ , $O\left(\frac a2,\frac b2\right)$ and $N\left(\frac a4,\frac {3b}4\right)$ . The slopes

of $NM$ , $NC$ are $:\ \left\{\begin{array}{ccccc} s(NM) & = & \frac {\frac {3b}4-0}{\frac a4-\frac a2} & \implies & s(NM)=-\frac {3b}a\\\\ s(NC) & = & \frac {b-\frac {3b}4}{a-\frac a4} & \implies & s(NC)=\frac b{3a}\end{array}\right\|$ $\implies$ $s(NM)\cdot s(NC)=-1\implies NM\perp NC$ a.s.o.

PP9. Let an acute $\triangle ABC$ and denote $\left\{\begin{array}{cccccccc} E\in AC & , & BE\perp AC & ; & R\in EF & , & CR\perp EF\\\\ F\in AB & , & CF\perp AB & ; & P\in EF & , & BP\perp EF\end{array}\right\|$ . Prove that $EP=FR$ .

Proof. $\left\{\begin{array}{ccc} \triangle BCE\sim\triangle FCR & \implies & \frac {BC}{FC}=\frac {BE}{FR}\\\\ \triangle BEP\sim\triangle BCF & \implies & \frac {BE}{BC}=\frac {EP}{CF}\end{array}\right\|\bigodot\implies EP=FR$ .

PP10. Let $\triangle ABC$ with the orthocenter $H$ and the orthic $\triangle DEF$ , where $\left\{\begin{array}{c} D\in BC\\\ E\in CA\\\ F\in AB\end{array}\right\|$ . Prove that $DH\cdot DA=DB\cdot DC=DE\cdot DF$ .

Proof. $\left\{\begin{array}{ccccc} \triangle DBF\sim \triangle DEC & \iff & \frac {DB}{DE}=\frac {DF}{DC} & \iff & DE\cdot DF=DB\cdot DC\\\\ \triangle DEH\sim \triangle DAF & \iff & \frac {DE}{DA}=\frac {DH}{DF} & \iff & DE\cdot DF=DH\cdot DA\end{array}\right\|$ $\implies$ $DH\cdot DA=DB\cdot DC=DE\cdot DF$ . Remark $:\ \frac {DA}{DH}=\tan B\tan C$ .

PP11. Let $\triangle ABC$ with the circumcircle $w=\mathbb C\left(O,R\right)$ , the incircle $\mathbb C(I,r)$ and the $A$ -excircle $\mathbb C\left(I_a,r_a\right)$ . Denote $:\ D\in BC\ ,\ AD\perp BC$

with $AD=h_a\ ;\ L\in AI\cap BC\ ;\ \{A,S\}=\{A,I\}\cap w\ ;\ \{A,A'\}=\{A,O\}\cap w$ . Prove that $AL\cdot AS=AI\cdot AI_a=AI_b\cdot AI_c=2Rh_a=bc$ .

Proof. $\left\{\begin{array}{ccccccc} \triangle ABS\sim \triangle ALC & \iff & \frac {AB}{AL}=\frac {AS}{AC} & \iff & AL\cdot AS & = & bc\\\\ \triangle ABI_a\sim \triangle AIC & \iff & \frac {AB}{AI}=\frac {AI_a}{AC} & \iff & AI\cdot AI_a & = & bc\\\\ \triangle ABD\sim \triangle AA'C & \iff & \frac {AB}{AA'}=\frac {AD}{AC} & \iff & 2Rh_a & = & bc\\\\ \triangle I_bAC\sim \triangle BAI_c & \iff & \frac {I_bA}{BA}=\frac {AC}{AI_c} & \iff & AI_b\cdot AI_c & = & bc\end{array}\right\|$ $\implies$ $AL\cdot AS=AI\cdot AI_a=AI_b\cdot AI_c=2Rh_a=bc$ .

PP12. Solve the system $\left\{\begin{array}{ccc} x^2-y^2 & = & 4x-5y\\\\ x^2+y^2 & = & 2x+y\end{array}\right\|$ over $\mathbb R^*$ .

Proof. Denote $\boxed{\ x\ =\ ty\ }\ (*)$ . Thus, $\left\{\begin{array}{ccc} x^2-y^2 & = & 4x-5y\\\\ x^2+y^2 & = & 2x+y\end{array}\right\|\iff$ $\left\{\begin{array}{cccc} x^2 & = & 3x-2y & (1)\\\\ y^2 & = & -x+3y & (2)\end{array}\right\|\implies$ $\left(\frac xy\right)^2=\frac {3x-2y}{-x+3y}\implies$ $t^2=\frac {3t-2}{-t+3}\implies$

$t^3-3t^2+3t-2=0\iff$ $(t-2)\left(t^2-t+1\right)=0\iff$ $t=2\ \stackrel{(*)}{\iff}\ x=$ $2y\ \stackrel{(1)}{\implies}\ 4y^2=$ $6y-2y\implies$ $y=1\implies$ $\left\{\begin{array}{ccc} x & = & 2\\\ y & = & 1\end{array}\right\|$ .

PP13. Let $\{x,y,z\}\subset\mathbb R$ so that $\left\{\begin{array}{ccc} x+y+z & = & 2\\\ xy+yz+zx & = & 1\end{array}\right\|$ . Prove that $\{\ x\ ,\ y\ ,\ z\ \}\subset\left[\ 0\ ,\ \frac 43\ \right]$ and $xyz\le\frac 4{27}$ .

Proof. Denote $\left\{\begin{array}{ccc} y+z & = & S\\\ yz & = & P\end{array}\right\|$ . Thus, $\left\{\begin{array}{ccc} x+y+z & = & 2\\\ xy+yz+zx & = & 1\end{array}\right\|\implies$ $\left\{\begin{array}{ccc} S+x & = & 2\\\\ xS+P & = & 1\end{array}\right\|\implies$ $\left\{\begin{array}{ccc} S & = & 2-x\\\\ P & = & (x-1)^2\end{array}\right\|$ . Therefore, $S^2\ge 4P\iff$

$(2-x)^2\ge 4(x-1)^2\iff$ $[2(x-1)+(x-2)][2(x-1)-(x-2)]\le 0\iff$ $x(3x-4)\le 0\iff x\in\left[0,\frac 43\right]$ . Hence $\{x,y,z\}\subset\left[0,\frac 43\right]\ .$ Therefore,

$xyz\le \frac 4{27}\iff$ $xP\le \frac 4{27}\iff$ $27x(x-1)^2-4\le 0\iff$ $27x^3-54x^2+27x-4\le 0\iff$ $(3x-1)^2(3x-4)\le 0$ , what is true because $x\in\left[0,\frac 43\right]$ .

PP14. Let $\{x,y,z\}\subset\mathbb R$ so that $\left\{\begin{array}{ccc} x+y+z & = & 6\\\\ xy+yz+zx & = & 9\end{array}\right\|$ . Prove that $\{\ x\ ,\ y\ ,\ z\ ,\ xyz\ \}\subset\left[\ 0\ ,\ 4\ \right]$ .

Proof. Denote $\left\{\begin{array}{ccc} y+z & = & S\\\ yz & = & P\end{array}\right\|$ . Thus, $\left\{\begin{array}{ccc} S+x & = & 6\\\\ xS+P & = & 9\end{array}\right\|\implies$ $\left\{\begin{array}{ccc} S & = & 6-x\\\\ P & = & (x-3)^2\end{array}\right\|$ . Therefore, $S^2\ge 4P\iff$ $(6-x)^2\ge 4(x-3)^2\iff$

$[2(x-3)]^2-(x-6)^2\le 0\iff$ $[2(x-3)+(x-6)][2(x-3)-(x-6)]\le 0\iff$ $x(3x-12)\le 0\iff x\in\left[0,4\right]$ . Hence, $\{x,y,z\}\subset[0,4]$ .

Observe that $xyz\le 4\iff$ $xP\le 4\iff$ $x(x-3)^2-4\le 0\iff$ $x^3-6x^2+9x-4\le 0\iff$ $(x-1)^2(x-4)\le 0$ , what is truly because $x\in\left[0,4\right]$ .

PP15. Ascertain $z\in\mathbb C$ so that $y\equiv z^2+2iz+3\in \mathbb R$ and $y<0$ .

Proof 1. $y\in\mathbb R\iff$ $y=\overline y\iff$ $z^2+2iz=\overline z^2-2i\overline z\iff$ $(z+\overline z)(z-\overline z+2i)=0\iff$ $z+\overline z=0\ \vee\ z-\overline z=-2i\iff$ exist $\{a,b\}\subset\mathbb R$ so that $z=bi\ \vee\ z=a-i$ .

If $z=bi$ , then $-b^2-2b+3<0\iff$ $b\not\in [-3,1]$ and if $z=a-i$ , then $a^2+4<0\iff a\in\emptyset$ . In conclusion, the all solutions are $z=bi$ , where $b\in (-\infty ,-3)\cup (1,\infty )$ .

Proof 2 (A.Mot). $\left(\exists\right)\ m<0$ so that $z^2+2iz+3-m=0$ . Observe that $\Delta'(m)=m-4<0$ and $z^2=\Delta'(m)\iff$

$z\in\left\{\pm i\sqrt{4-m}\right\}$ . Therefore, the solution of our problem is the reunion of $\mathbb S_m$ , where $m<0$ and $\mathbb S_m=\left\{i\left(-1\pm\sqrt{4-m}\right)\right\}$ .

Remark. If denote $b\in\left\{-1\pm\sqrt{4-m}\right\}$ for $m<0$ , then $(b+1)^2=4-m\iff$ $m=-\left(b^2+2b-3\right)$ . Thus, $m<0\iff$

$b^2+2b-3>0\iff$ $b\in (-\infty ,-3)\cup (1,\infty )$ and find again the solution from first proof, i.e. $z=bi$ , where $b\in (-\infty ,-3)\cup (1,\infty )$ .

PP16. Solve the inequation $1<\frac {5x-1}{x+3}<2$ , where $x\in\mathbb R$ and $x\not =-3$ .

Proof 1. I"ll use the equivalence $x\in (a,b) \iff (x-a)(x-b)<0$ . Indeed, $1<\frac {5x-1}{x+3}<2\iff$

$\left(\frac {5x-1}{x+3}-1\right)\left(\frac {5x-1}{x+3}-2\right)<0\iff$ $(x-1)(3x-7)<0\iff$ $x\in\left(1,\frac 73\right)$ .

Proof 2. I"ll use the equivalence $x\in (a,b) \iff \left|x-\frac {a+b}2\right|<\left|\frac {b-a}2\right|$ . Indeed, $1<\frac {5x-1}{x+3} <2\iff$ $\left|\frac {5x-1}{x+3}-\frac 32\right|<\frac 12\iff$ $\left|\frac {7x-11}{x+3}\right|<1\iff$

$|7x-11|<|x+3|\iff$ $(7x-11)^2-(x+3)^2<0\iff$ $[(7x-11)+(x+3)]\cdot [(7x-11)-(x+3)]<0\iff$

$8(x-1)\cdot 2(3x-7)<0\iff$ $(x-1)(3x-7)<0\iff$ $x\in\left(1,\frac 73\right)$ .

Proof 3 (A.Mot). The inequality becomes $\frac {5x-1}{x+3}=1+a=2-b$ , where $a+b=1$ and $0<a<1$ or $0<b<1$ . Also obtain that

$x=\frac {4+3a}{4-a}=\frac {7-3b}{3+b}$ from where results: if $a\searrow 0$ or $b\nearrow 1$ , then $x\searrow 1\ ;$ if $a\nearrow 1$ or $b\searrow 0$ , then atunci $x\nearrow \frac 73$ . Thus, $1<x<\frac 73$ . Nice!

Remark. $x\in[a,b]\cup[b,a]\iff$ $(x-a)(x-b)\le 0\iff$ $\left|x-\frac {a+b}2\right|\le\left|\frac{a-b}2\right|\iff$ $|x-a|+|x-b|=|a-b|$ .

Rezolvarea ecuatiei de gradul doi cu coeficienti complecsi.

Preliminary. Let $az^2+bz+c=0$ be a second degree polynomial equation with complex coefficients, i.e. $\{a,b,c\}\subset\mathbb C$ and $a\ne 0$ . Suppose w.l.o.g. that exists at least a nonreal number

in the set $\{a,b,c\}$ and $\Delta=b^2-4ac=u+iv$ , where $\{u,v\}\subset\mathbb R$ and $v\ne 0$ . Observe that $az^2+bz+c=0\iff (2az+b)^2=\Delta\ (*)$ . Therefore, at first solve the equation

$y^2=u+iv$ . Prove easily that $y_{1,2}=\pm w$ , where $\boxed{w=\sqrt{\frac {r+u}{2}}+i\cdot\mathrm{sgn}(v)\cdot\sqrt {\frac {r-u}{2}}}$ and $r=|\Delta|$ . Indeed, $w^2=\frac {r+u}2-\frac {r-u}2+2i\cdot\mathrm{sgn}(v)\cdot\sqrt {\frac {r^2-u^2}{4}}=$

$u+2i\cdot\mathrm{sgn}(v)\cdot\frac{|v|}2=u+iv\implies w^2=u+iv\ .$ At second from the relation $(*)$ obtain that $z_{1,2}=\frac {-b\pm w}{2a}$ .

Example. Solve the equation $x^4-8(1-i)x^2+(63-16i)=0$ . With the substitution $\boxed{x^2=z}$ our equation becomes $\boxed{z^2-8(1-i)z+(63-16i)=0}\ (*)\ .$

$\Delta^{\prime}=16(1- i)^2-(63-16i)\implies$ $\Delta^{\prime}=-63-16i=y^2\implies\ \left\{\begin{array}{c} r=\sqrt {63^2+16^2}=65\ \iff\ 65^2-63^2=(65-63)(65+63)=16^2\\\\\ y_{1,2}=\pm\left(\sqrt {\frac {65-63}{2}}-i\sqrt {\frac {65+63}{2}}\right)=\pm(1-8i)\end{array}\right\|\ .$ Therefore,

$z_{1,2}=4(1- i)\pm (1-8i)\ \implies\ z_1=5-12i$ and $z_2=3+4i\ .$ Verify Viete's relations $:\ S=z_1+z_2=-\frac ba=8(1-i)$ and $P=z_1z_2=\frac ca=63-16i\ .$

$\odot\begin{array}{cccccccc} \nearrow & x^2=z_1=5-12i & ; & \left|z_1\right|=\sqrt{5^2+12^2}=13 & \implies & x_{1,2}=\pm\left(\sqrt{\frac {13+5}2}-i\sqrt{\frac {13-5}2}\right) & \implies & \odot\begin{array}{ccc} \nearrow & x_1=3-2i & \searrow\\\\ \searrow & x_2=-3+2i & \nearrow\end{array}\odot\\\\ \searrow & x^2=z_2=3+4i & ; & \left|z_2\right|=\sqrt{3^2+4^2}=5 & \implies & x_{3,4}=\pm\left(\sqrt{\frac {5+3}2} +i\sqrt{\frac {5-3}2}\right) & \implies & \odot\begin{array}{ccc} \nearrow & x_3=2+i & \searrow\\\\ \searrow & x_4=-2-i & \nearrow\end{array}\odot\end{array}$

PP17. Let $\boxed{\ z_{1,2}\in\mathbb C\ }$ be the roots of the equation with complex coefficients $\boxed{\ az^2+bz+c=0\ ,\ a\ne 0\ }$ . Prove that $\boxed{\left|z_1\right|+\left|z_2\right|=\sqrt {\frac {|b|^2+|b^2-4ac|+4|ac|}{2|a|^2}}}$ .

Proof.

$\blacktriangleright\ \left|z_1-z_2\right|^2=$ $\left|\left(z_1+z_2\right)^2-4z_1z_2\right|\implies$ $\left|z_1-z_2\right|^2=$ $\left|\frac {b^2-4ac}{a^2}\right|\ (1)\ .$

$\blacktriangleright\ \left|z_1+z_2\right|^2+\left|z_1-z_2\right|^2=$ $2\left(\left|z_1\right|^2+\left|z_2\right|^2\right)\stackrel{(1)}{\implies}$ $\left|z_1\right|^2+\left|z_2\right|^2=$ $\frac 12\cdot \left(\left|\frac ba\right|^2+\frac {\left|b^2-4ac\right|}{|a|^2}\right)\ (2)\ .$

$\blacktriangleright\ \left(\left|z_1\right|+\left|z_2\right|\right)^2=$ $\left|z_1\right|^2+\left|z_2\right|^2+2\left|z_1z_2\right| \stackrel{(2)}{\implies}$ $\left(\left|z_1\right|+\left|z_2\right|\right)^2=$ $\frac 12\cdot\left(\left|\frac ba\right|^2+\frac {\left|b^2-4ac\right|}{|a|^2}\right)+2\cdot\left|\frac ca\right|\implies$ $\left|z_1\right|+\left|z_2\right|=$ $\sqrt {\frac {|b|^2+|b^2-4ac|+4|ac|}{2|a|^2}}$ .

Application. Let $\left\{z_1,z_2\right\}\subset\mathbb C$ be the roots of the equation $z^2-2iz+m=0$ , where $m\in\mathbb C$ is a parameter. Ascertain the values of $m$ for which $\left|z_1\right|+\left|z_2\right|=2$ .

Proof. For $a:=1\ ,\ b:=-2i$ and $c:=m\in\mathbb C$ obtain that $\left|z_1\right|+\left|z_2\right|=2\iff$ $\sqrt {\frac {4+4|1+m|+4|m|}{2}}=2\iff$ $|m|+|m+1|=1$ . Denote the

points $M(m)\ ,\ A(0)\ ,\ B(-1)$ . Therefore, $|m|+|m+1|=1\iff$ $MA+MB=AB\iff$ $M\in[AB]\iff m\in\mathbb R\ ,\ m\in [-1,0]$ . Nice exercise !

$\boxed{\begin{array}{c} \underline{\mathrm{PROPOSED\ PROBLEM}}.\ \mathrm{Let}\ \{a,b,c\}\subset\mathbb C\ \mathrm{where}\ a\ne 0\ \mathrm{and}\ az^2+bz+c=0\begin{array}{ccc} \nearrow & z_1 & \searrow\\\\ \searrow & z_2 & \nearrow\end{array}\odot\implies\left|z_1\right|+\left|z_2\right|=\sqrt {\frac {|b|^2+|b^2-4ac|+4|ac|}{2|a|^2}}\\\\ \underline{\mathrm{PROOF}}.\ \left|z_1-z_2\right|^2=\left|\left(z_1+z_2\right)^2-4z_1z_2\right|\implies\left|z_1-z_2\right|^2=\left|\frac {b^2-4ac}{a^2}\right|\ (1)\ .\\\\ \left|z_1+z_2\right|^2+\left|z_1-z_2\right|^2=2\left(\left|z_1\right|^2+\left|z_2\right|^2\right)\stackrel{(1)}{\implies}\left|z_1\right|^2+\left|z_2\right|^2=\frac 12\cdot \left(\left|\frac ba\right|^2+\frac {\left|b^2-4ac\right|}{|a|^2}\right)\ (2)\ .\\\\ \left(\left|z_1\right|+\left|z_2\right|\right)^2=\left|z_1\right|^2+\left|z_2\right|^2+2\left|z_1z_2\right| \stackrel{(2)}{\implies}\left(\left|z_1\right|+\left|z_2\right|\right)^2=\frac 12\cdot\left(\left|\frac ba\right|^2+\frac {\left|b^2-4ac\right|}{|a|^2}\right)+2\cdot\left|\frac ca\right|\implies\left|z_1\right|+\left|z_2\right|=\sqrt {\frac {|b|^2+|b^2-4ac|+4|ac|}{2|a|^2}}\ .\end{array}}$

PP18. Ascertain the roots of the equation $f(z)\equiv 2z^3-(7+2i)z^2+3(2-i)z+(5-i)=0\ (*)\ ,$ where $z\in\mathbb C$ and exists at least $x\in\mathbb R$ so that $f(x)=0\ .$

Proof. $(\exists )\ x\in\mathbb R$ so that $f(x)=\left(2x^3-7x^2+6x+5\right)-i\left(2x^2+3x+1\right)=0\iff$ $\left\{\begin{array}{ccc} 2x^3-7x^2+6x+5 & = & 0\\\\ 2x^2+3x+1 & = & 0\end{array}\right\|$ $\iff$ $\left\{\begin{array}{ccc} (2x+1)\left(x^2-4x+5\right) & = & 0\\\\ (2x+1)(x+1) & = & 0\end{array}\right\|\iff$

$x=-\frac 12$ . In conclusion, $f(z)=(2z+1)\left[\left(z^2-4z+5\right)-i(z+1)\right]=0\iff$ $z_1=-\frac 12$ and $z^2-(4+i)z+(5-i)=0$ . Observe that $\Delta =(4+i)^2-4(5-i)\implies$

$\boxed{\Delta=-5+12i}$ . The equation $z^2=\Delta\iff$ $z^2=-5+12i$ has the roots $z'_{1,2}=\pm\left(\sqrt{\frac {13-5}2}+\sqrt{\frac {13+5}2}\right)$ , where $r=|-5+12i|=13$ , i.e. $z^2=-5+12i\iff$

$z'_{1,2}=\boxed{\pm(2+3i)}$ . In conclusion, $z_{1,2}=\frac {(4+i)\pm(2+3i)}2\begin{array}{cccc} \nearrow & z_2 & = & 1-i\\\\ \searrow & z_3 & = & 3+2i\end{array}\ .$

PP19. Let $BCQP$ be the trapezoid with $BP\parallel CQ\perp BC$ , where $BP=p$ and $CQ=q$ . Suppose that exists an interior point $A$

so that $AP=AQ$ and $AP\perp AQ$ . Prove that $\left\{\begin{array}{ccc} 2ap & = & a^2+b^2-c^2+4S\\\\\ 2aq & = & a^2-b^2+c^2+4S\end{array}\right\|$ , where $S=[ABC]$ and $\left\{\begin{array}{ccc} BC & = & a\\\\ CA & = & b\\\\ AB & = & c\end{array}\right\|$ .

Proof (metric). Denote $:\ D\in (BC)$ so that $AD\perp BC\ ;\ M\in (BP)$ and $N\in (CP)$ so that $A\in MN\ ,\ MN\parallel BC$ and $\left\{\begin{array}{ccccc} DB & = & AM & = & m\\\\ DC & = & AN & = & n\end{array}\right\|$ .

Apply the generalized Pythagoras' theorem in the triangle $ABC\ :\ \left\{\begin{array}{cccc} b^2 & = & a^2+c^2-2am & (1)\\\\ c^2 & = & a^2+b^2-2an & (2)\end{array}\right\|$ . Observe that $\triangle AMP\equiv\triangle QNA$ $\iff$ $\left\{\begin{array}{ccc} MP=NA=DC=n\\\\ MA=NQ=DB=m\end{array}\right\|$ ,

i.e. $MB=AD=NC\iff \boxed{p-n=h_a=q-m}\ .$ In conclusion, $\left\{\begin{array}{ccccc} 2S=ah_a=a(q-m) & \implies & 2aq=4S+2am & \stackrel{(1)}{=} & 4S+a^2+c^2-b^2\\\\ 2S=ah_a=a(p-n) & \implies & 2ap=4S+2an & \stackrel{(2)}{=} & 4S+a^2+b^2-c^2\end{array}\right\| \ .$

PP20. Let $\triangle ABC$ with the orthocenter $H$ , $A=45^{\circ}$ , $D\in (BC)$ so that $AD\perp BC$ and $AD=c$ , $DB=a$ and $DC=b$ . Prove that $\frac 1{c-(a+b)}=\frac 1c+\frac 1a+\frac 1b$ .

Proof 1. Denote the symmetrical $S$ of $H$ w.r.t. $D$ . Observe that $AH=2R\cos A=2R\sin A=BC\iff AH=a+b$ . Is well known that $S$ belongs to the circumcircle $w$

of $\triangle ABC$ , $DH=DS=c-(a+b)$ and (from the power of $D$ w.r.t. $w$ ) $DA\cdot DS=DB\cdot DC$ , i.e. $c(c-a-b)=ab$ , what is equivalently with the required relation. Indeed,

prove easily that $\frac 1{c-(a+b)}=\frac 1c+\frac 1a+\frac 1b\iff$ $\frac 1{c-a-b}-\frac 1c =\frac 1a+\frac 1b\iff$ $\frac{a+b}{c(c-a-b)} =$ $\frac{a+b}{ab}\iff$ $c(c-a-b)=ab\iff$ $\boxed{(c+a)(c+b)=2c^2}$ .

Proof 2. I"ll apply the well-known identity $4S=\left(AC^2+AB^2-BC^2\right)\tan A\ (*)$ in any $\triangle ABC$ . Using the notations from the previous method, obtain that the relation $(*)$

for $A=45^{\circ}$ is equivalently with $2c(a+b)=\left(a^2+c^2\right)+$ $\left(b^2+c^2\right)-(a+b)^2\iff$ $c(a+b)=c^2-ab\iff$ $c(c-a-b)=ab \iff$ $\boxed{(c+a)(c+b)=2c^2}$ .

Remark. Verify for the values $c=6k\ ,\ a=2k\ ,\ b=3k$ where $k\in\mathbb R^*_{+}\ ,$ i.e. $\frac 16+\frac12+\frac 13=1$ .

PP21 (F.J. Garcia Capitan). Let an equilateral $\triangle ABC$ with $AB=a$ and circles $\beta =\mathbb C(B,a)$ , $\gamma =\mathbb C(C,a)$ . Construct the circle $w=\mathbb C(I,r)$

which is tangent to $BC$ , interior tangent to $\beta$ and exterior tangent to $\gamma$ . Prove that $2r=h$ , where $h=\frac {a\sqrt 3}2$ is the length of altitude for $\triangle ABC$ .

http://i944.photobucket.com/albums/ad288/GemenLeu/89991a3c-8d39-4d2a-bcf9-8eb347e24844_zpszheyyx4l.jpg

Proof 1. Denote the projection $K$ of $I$ on $BC$ . Prove easily that $\left\{\begin{array}{ccc} IB & = & a-r\\\ IC & = & a+r\end{array}\right\|$ and $KB^2=IB^2-IK^2=(a-r)^2-r^2=a^2-2ar\implies$ $KB=\sqrt {a^2-2ar}\implies$

$\boxed{KC=a+\sqrt {a^2-2ar}}\ (*)$ . Apply the Pythagoras' theorem in the $K$ -right triangle $IKC\ :\ IC^2=KI^2+KC^2\iff$ $(a+r)^2\ \stackrel{(*)}{=}\ r^2+$ $\left(a+\sqrt {a^2-2ar}\right)^2\iff$

$a^2+r^2+2ar=r^2+a^2+a^2-2ar+2a\sqrt{a^2-2ar}\iff$ $4r-a=2\sqrt{a^2-2ar}\iff$ $16r^2+a^2-8ar=4a^2-8ar\iff$ $2r=\frac {a\sqrt 3}2\iff$ $2r=h$ .

Remark. Denote the diameter $[KL]$ of the left circle and the midpoint $M$ of $[BC]$ . Prove easily that $AMKL$ is a square! Indeed, $MK=MA\iff$

$KB+\frac a2=h\iff$ $2\sqrt{a^2-2ar}=a\left(\sqrt 3-1\right)\iff$ $4a^2-8ar=2a^2(2-\sqrt 3)\iff$ $4r=a\sqrt 3\iff$ $2r=h$ , what is truly.

Proof 2. Let $K$ be the projection of $I$ on $BC$ . Prove easily that $\left\{\begin{array}{ccc} IB & = & a-r\\\ IC & = & a+r\end{array}\right\|$ .Thus, $IK\perp BC\iff$ $IC^2-IB^2=KC^2-KB^2\iff$ $(a+r)^2-(a-r)^2=$

$a(2\cdot KB+a)\iff$ $4r=2\sqrt{a^2-2ar}+a\iff$ $(4r-a)^2=4\left(a^2-2ar\right)\iff$ $16r^2+a^2-8ar=4a^2-8ar\iff$ $2r=\frac {a\sqrt 3}2$ . In conclusion, $2r=h$ .

PP22 (F.J. Garcia Capitan). Let a line $d$ and four points $\{A,C,B,D\}\subset d$ in this order so that $AB=CD=2R$ . Denote the midpoints $K$ , $L$ of $[AB]$ , $[CD]$ respectively and construct

the semicircles $\alpha =\mathbb C(K,R)$ , $\beta =\mathbb C(L,R)$ and the circles $w=\mathbb C(I,r)$ , $w_1=\mathbb C(I_1,r)$ , $w_2=\mathbb C(I_2,r)$ which are tangent to the line $d$ so that $:$ the circle $w$ is interior tangent to the

circles $\alpha$ and $\beta\ ;$ the circle $w_1$ (left) is interior tangent to $\alpha$ and exterior tangent to $\beta\ ;$ the circle $w_2$ (right) is exterior tangent to $\alpha$ and interior tangent to $\beta\ .$ Prove that $5r=2R$ .

http://i944.photobucket.com/albums/ad288/GemenLeu/0545f130-bf94-43bf-bb61-78e627f70aed_zpsaebrthqt.jpg

Proof. Denote the projections $T$ and $T_1$ of $I$ and $I_1$ on $d$ . Thus, $KI=R-r\implies KT=TL=\sqrt {R^2-2Rr}$ . Therefore, $KI_1=R+r$

and $KT_1=3\sqrt{R^2-2Rr}$ . In conclusion, $KI_1^2=T_1I_1^2+T_1K^2\iff$ $(R+r)^2=r^2+9\left(R^2-2Rr\right)\iff$ $8R^2=20Rr\iff$ $\frac rR=\frac 25$ .

PP23. ascertain the sum of the real numbers $a$ , $b$ which verify the relations $\left\|\ \begin{array}{ccc} a^3+3a^2+4a+1 & = & 0\\\\ b^3+3b^2+4b+3 & = & 0\end{array}\ \right\|$ .

Proof. Let the function $f(x)=x^3+3x^2+4x+5\ ,\ x\in\mathbb R$ . Observe that $f(a)=4$ and $f(b)=2$ . Prove easily that $f$ is strict increasing $(\nearrow )$

and its graph $G_f$ has the symmetry point $S(-1,3)$ , i.e. $f(-2-x)=6-f(x)$ . For $x:=a$ get $f(-2-a)=6-f(a)=2=f(b)$ , i.e.

$f(-2-a)=f(b)$ . Since function $f$ is strict increasing $\implies$ $f$ is injective obtain that $f(-2-a)=f(b)\implies$ $-2-a=b$ , i.e. $a+b=-2$ .

PP24. Let the "triangle" of all odd natural numbers $\begin{array}{ccc} L_1 & : & 1\\\ L_2 & : & 3\ \ 5\\\ L_3 & : & 7\ \ 9\ 11\\\ L_4 & : & 13\ 15\ 17\ 19\\\ \ldots & : & \ldots\ \ldots\ \ldots\ \ldots\end{array}$ . Ascertain $:$

$\blacktriangleright$ The first term and the last term of the $\mathrm{n^{th}}$ line, where $n\in \mathbb N^*\ ;$

$\blacktriangleright$ The sum of the terms from the $\mathrm{n^{th}}$ line $\ ;$

$\blacktriangleright$ Prove that $1^3+2^3+\ \ldots\ +(n-1)^3+n^3=\left[1+2+3+\ \ldots\ +(n-1)+n\right]^2\ ;$

$\blacktriangleright$ Find the "coordinates" of the number $1001$ , i.e. find the line that belongs to and where is number the $1001$ in this line.

PP25. Let $\triangle ABC$ with the circumcircle $\alpha=C(O,R)$ and the incircle $w=C(I,r)$ for which denote $E\in AC\cap w$ ,

$F\in AB\cap w$ and $\{A,S\}=\{A,I\}\cap \alpha$ . Prove that $\widehat{SEC}\equiv\widehat{SFB}$ and $\boxed{SF^2+s(s-a)=SA^2}\ (*)\ .$

Proof. Prove easily that $AESF$ is a deltoid, i.e. $AS$ is a symmetry axis. Suppose w.l.o.g. $c<b$ and denote $\left\{\begin{array}{ccc} P\in AB & ; & SP\perp AB\\\\ R\in AC & ; & SR\perp AC\end{array}\right\|$ . Observe that $B\in (AP)\ ,\ R\in (AC)$

and $\triangle SBP\ \stackrel{s.s}{\equiv}\ \triangle SCR$ , i.e. $PB=RC=x$ and $c+x=AP=AR=b-x\iff$ $c+x=b-x\iff$ $\left\{\begin{array}{cccc} x & = & \frac {b-c}2 & (1)\\\\ PA & = & \frac {b+c}2 & (2)\end{array}\right\|$ . Thus, $SP\perp AF\iff$ $SA^2-SF^2=$

$PA^2-PF^2=$ $(PA-PF)(PA+PF)=$ $AF\cdot (2\cdot PA-AF)=$ $(s-a)[(b+c)-(s-a)]=s(s-a)$ $\implies$ $SA^2-SF^2=s(s-a)\iff$ $SA^2=SF^2+s(s-a)\ .$

Remark 1. Let $K\in AC$ so that $I_aK\perp AB$ , where $I_a$ is the $A$ -excenter of $\triangle ABC$ . Thus, $\left\{\begin{array}{ccc} SK & = & SF\\\\ PK & = & PF\end{array}\right\|$ . So $SP\perp AK\iff$ $SA^2-SF^2=SA^2-SK^2=$

$PA^2-PK^2=$ $(PA+PK)(PA-PK)=AK\cdot (PA-PF)=$ $AK\cdot AF=s(s-a)\implies$ $SA^2-SF^2=s(s-a)$ , i.e. $SA^2=SF^2+s(s-a)\ .$

Remark 2. $AS\cos\frac A2=2R\sin\left(B+\frac A2\right)\cos\frac A2=$ $R(\sin C+\sin B)=\frac {b+c}2\implies \boxed{AS\cos\frac A2=\frac {b+c}2}\ (1)\ .$ Apply the generalized Pythagoras' theorem to $\triangle ASF\ :$

$SF^2=AS^2+AF^2-2\cdot AF\cdot AS\cos\frac A2\ \stackrel{(1)}{=}\ AS^2+AF^2-(b+c)\cdot AF=$ $AS^2-AF(b+c-AF)=(s-a)[b+c)-(s-a)]\implies$ $SF^2=SA^2-s(s-a)$ .

PP26. Let the parabola $w$ with the equation $y=x^2$ and the line $d$ with the equation $y=2x-3$ . For a mobile point $P(a,b)\in d$ , where $a^2>b$ denote

$\{M,N\}\subset w$ so that $P\in MM\cap NN$ (for $X\in w$ let $XX$ - the tangent line to $w$ ). Find the minimum value of the area $[PMN]$ , letting $P$ moves on $d$ .

Proof. $P\in d\iff \boxed{b=2a-3}\ (*)$ and the equations of $MM$ , $NN$ are $:\ \left\{\begin{array}{ccccc} M\left(m,m^2\right) & \implies & MM\ :\ y-m^2=2m(x-m) & \implies & y=2mx-m^2\\\\ N\left(n,n^2\right) & \implies & NN\ :\ y-n^2=2n(x-n) & \implies & y=2nx-n^2\end{array}\right\|$ ,

where $m\ne n\ .$ Thus, $\left\{\begin{array}{ccc} P\in MM & \implies & b=2ma-m^2\\\\ P\in NN & \implies & b=2na-n^2\end{array}\right\|$ $\implies$ $\left\{\begin{array}{cccccccc} 2ma-m^2=2na-n^2 & \implies & 2a(m-n)=m^2-n^2 & \implies & m+n & = & 2a & (1)\\\\ n\left(b+m^2\right)=m\left(b+n^2\right) & \implies & b(m-n)=mn(m-n) & \implies & mn & = & b & (2)\end{array}\right\|\ .$

The area $S=[PMN]=\frac 12\cdot|\Delta|$ , where $\Delta =\left|\begin{array}{ccc} a & b & 1\\\\ m & m^2 & 1\\\\ n & n^2 & 1\end{array}\right|\ \stackrel{1\wedge 2}{=}\ \left|\begin{array}{ccc} \frac {m+n}2 & mn & 1\\\\ m & m^2 & 1\\\\ n & n^2 & 1\end{array}\right|\ \stackrel{\begin{array}{c} L_2:=L_2-L_1\\\ L_3:=L_3-L_1\end{array}}{=}\ \left|\begin{array}{ccc} \frac {m+n}2 & mn & 1\\\\ \frac{m-n}2 & m(m-n) & 0\\\\ \frac{n-m}2 & n(n-m) & 0\end{array}\right|=$ $\left|\begin{array}{cc} \frac{m-n}2 & m(m-n)\\\\ \frac{n-m}2 & n(n-m)\end{array}\right|=$

$\frac {(m-n)^2}2\cdot \left|\begin{array}{cc} 1 & m\\\\ -1 & -n\end{array}\right|\implies$ $\Delta=\frac {(m-n)^3}2$ and $\boxed{S=\frac 14\cdot |m-n|^3}\ (3)\ .$ In conclusion, $S$ is $\min\iff$ $|m-n|$ is $\min\iff$ $|m-n|^2=(m-n)^2$ is $\min\iff$

$(m+n)^2-4mn$ is $\min\iff$ $4a^2-4b$ is $\min\iff$ $a^2-b$ is $\min\stackrel{*}{\iff} a^2-2a+3$ is $\min\iff$ $\boxed{a=1\ \wedge\ b=-1}$ , i.e. $P(1,-1)$ $\iff$ $m+n=2\ \wedge\ mn=-1\iff$

$|m-n|=2\sqrt 2\ \stackrel{3}{\iff}\ 4S=16\sqrt 2\iff$ $\boxed{S=4\sqrt 2}\ .$ I denoted $L_k$ - the line $k\in \overline{1,3}$ in the determinant $\Delta$ .

PP27. Let $\triangle ABC$ with the incircle $w=\mathbb C(I,r)$ and the circumcircle $\alpha =\mathbb C(O,R)$ . Denote $D\in BC\cap AI$ ,

$F\in AB\cap w$ , $\{A,S\}=\{A,I\}\cap \alpha$ and $X\in BC\cap FS$ . Prove that $\frac {SA}{SD}=\left(\frac {b+c}2\right)^2$ and find the ratio $\frac {XB}{XD}$ .

Proof. I"ll use the well known properties $:\ \left\|\begin{array}{cccc} DS\cdot DA & = & DB\cdot DC & (0)\\\\ AD^2 & = & bc-DB\cdot DC & (2)\end{array}\right\|\ \ \wedge\ \ \left\|\begin{array}{cccc} \frac {IA}{ID} & = & \frac {b+c}a & (1)\\\\ \frac {DA}{DS} & = & \frac {BA\cdot CA}{SB\cdot SC} & (3)\end{array}\right\|$ . Thus, $\frac {DA}{DS}=\frac {DA^2}{DS\cdot DA}\ \stackrel{0\wedge 2}{=}\ \frac {bc-DB\cdot DC}{DB\cdot DC}=$

$\frac {bc}{DB\cdot DC}-1$ $\implies$ $1+\frac {DA}{DS}=\frac {bc}{DB\cdot DC}$ $\implies$ $\frac {SA}{SD}=$ $\frac {CA}{CD}\cdot \frac {BA}{BD}=\frac {IA}{ID}\cdot\frac {IA}{ID}=$ $\left(\frac {IA}{ID}\right)^2\ \stackrel{1}{\implies}\ \boxed{\frac {SA}{SD}=\left(\frac {b+c}a\right)^2}\ (4)\ .$ Apply the Menelaus' theorem to the

transversal $\overline{SXF}/\triangle ABD\ :\ \frac {SD}{SA}\cdot \frac {FA}{FB}\cdot\frac {XB}{XD}=1\iff$ $\frac {XB}{XD}=\frac {FB}{FA}\cdot\frac {SA}{SD}\ \stackrel{4}{\implies}\ \boxed{\frac {XB}{XD}=\frac {s-b}{s-a}\cdot \left(\frac {b+c}a\right)^2}\ .$ See and the proposed problem PP25 from here.

Remark. If $I_a$ is the $A$ -excenter of $\triangle ABC$ , then $\left(A,I,D,I_a\right)$ is an harmonic division and $S$ is the midpoint of $[I_aI]$ . From well-known property get that $\frac {SA}{SD}=\left(\frac {IA}{ID}\right)^2=\left(\frac {b+c}a\right)^2$ .

PP28. Prove the identity $C_{2k+3}^{k+1}-C_{2k+1}^k=\frac {3k+4}{k+2}\cdot C_{2k+1}^{k+1}$ and calculate the sum $\sum_{k=0}^n\frac {3k+4}{k+2}\cdot C_{2k+1}^{k+1}$ .

Proof 1.1 $C_{2k+3}^{k+1}-C_{2k+1}^k=\frac {(2k+3)!}{(k+1)!(k+2)!}-\frac {(2k+1)!}{k!(k+1)!}=$ $\left[\frac {2(k+1)(2k+3)}{(k+1)(k+2)}-1\right]\cdot\frac {(2k+1)!}{k!(k+1)!}=$ $\left[\frac {2(2k+3)}{k+2}-1\right]\cdot\mathrm C_{2k+1}^{k}\implies$ $\boxed{C_{2k+3}^{k+1}-C_{2k+1}^k=\frac {3k+4}{k+2}\cdot C_{2k+1}^{k+1}}$

Proof 1.2 $C_{2k+3}^{k+1}=$ $C_{2k+2}^{k+1}+C_{2k+2}^{k}=$ $\left(C_{2k+1}^{k+1}+C_{2k+1}^{k}\right)+\left(C_{2k+1}^{k}+C_{2k+1}^{k-1}\right)=$ $3C_{2k+1}^{k}+C_{2k+1}^{k-1}$ $\implies$ $C_{2k+3}^{k+1}-C_{2k+1}^{k}=2C_{2k+1}^{k}+C_{2k+1}^{k-1}=C_{2k+1}^{k}+C_{2k+2}^{k}=$

$\frac {(2k+1)!}{k!(k+1)!}+\frac {(2k+2)!}{k!(k+2)!}=$ $\left(1+\frac {2k+2}{k+2}\right)\cdot \frac {(2k+1)!}{k!(k+1)!}=$ $\frac {3k+4}{k+2}\cdot\mathrm C_{2k+1}^k$ $\implies$ $\boxed{C_{2k+3}^{k+1}-C_{2k+1}^{k}=\frac {3k+4}{k+2}\cdot\mathrm C_{2k+1}^k}$

Proof 2. $\sum_{k=0}^n\frac {3k+4}{k+2}\cdot C_{2k+1}^{k+1}=$ $\sum_{k=0}^n\left(C_{2k+3}^{k+1}-C_{2k+1}^k\right)=$ $\left(C_{3}^{1}-C_{1}^0\right)+\left(C_{5}^{2}-C_{3}^1\right)+\left(C_{7}^{3}-C_{5}^2\right)+\ \ldots\ +\left(C_{2n+1}^{n}-C_{2n-1}^{n-1}\right)+\left(C_{2n+3}^{n+1}-C_{2n+1}^n\right)=\mathrm C_{2n+3}^{n+1}-1\ .$

PP29. Let the equation $x^2+x+1=0$ with $x_{1,2}\in\mathbb C\ .$ Denote $S_n=x_1^n+x_2^n\ ,\ n\in\mathbb N\ .$ Calculate "instant" $S_{1001}$ without complex numbers. Hint: $1001=7\cdot 11\cdot 13\ .$

Proof 1.

PP30. Let $\triangle ABC$ with $A$ -excircle $w_a=\mathbb C\left(I_a,r_a\right)$ for what denote $\left\{\begin{array}{ccccccc} U & \in & AB\cap w_a & ; & M & \in & I_aB\cap UV\\\\ V & \in & AC\cap w_a & ; & N & \in & I_aC\cap UV\end{array}\right\|\ .$ Prove that $\boxed{MN=BC\cdot\sin\frac A2}\ (*)\ .$

Proof. Let incenter $I\ ,$ $W\in BC\cap w_a$ and $L\in UV\cap AI_a\ .$ Thus, $\overline{AII_a}_a\perp UV$ and $IBI_aC$ is inscribed in circle with the diameter $\left[II_a\right]$ $\implies$

$\left\{\begin{array}{ccccc} m\left(\widehat{BI_aI}\right)=m\left(\widehat{BCI}\right) & \implies & m\left(\widehat{MI_aL}\right)=\frac C2 & \implies & m\left(\widehat{NMI_a}\right)=90^{\circ}-\frac C2\\\\ m\left(\widehat{CI_aI}\right)=m\left(\widehat{CBI}\right) & \implies & m\left(\widehat{NI_aL}\right)=\frac B2& \implies & m\left(\widehat{MNI_a}\right)=90^{\circ}-\frac B2\end{array}\right\|\ .$ $\implies$ $\triangle MI_aN\sim\triangle CI_aB\ ,$ i.e. $MN$ is antiparallel

to $BC$ in the triangle $BI_aC\ .$ In conclusion, $\frac {MN}{BC}=\frac {I_aL}{I_aW}\implies$ $MN=BC\cdot \frac {I_aL}{I_aU}=BC\cdot\sin\widehat {LUI_a}\implies$ $MN=BC\cdot\sin \frac A2\ ,$ i.e. the relation $(*)\ .$

$\mathrm{END}$

This post has been edited 432 times. Last edited by Virgil Nicula, Feb 19, 2018, 9:29 PM

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long blog
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372554 views!
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369302 views!

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Thank you very much for this blog.
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You seem to have solved every existing problem.
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I love your posts, man (those which I understand)! It's interesting how you punish integrals with recurrence relations and sometimes make seemingly unrelated integrals evaluate each other!
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Thank you a lot dear Virgil for an interesting and instructive blog.
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Interesting Blog
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Interesant, d-le Nicula.
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15001th view.
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Thanks to all !
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El_Ectric:
You can either copy some parts from the blog post, paste it at "Search Blogs" (which is lower on the page, after "Blog Stats") and find the whole post.

You're welcome.
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Thanks Thomas_.
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El_Ectric: Try to open this blog with Internet Explorer (browser). If it still doesn't work, just copy the post you're interested in, and than paste it into Word Document, or almost any kind of reading documents and you'll have all the details.

Nice blog!
by Thomas_, Nov 24, 2010, 4:59 PM
Dear Virgil !

Congratulations for this blog !
Babis stergiou - Greece
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72 shouts

My (math)blog

426. Problems from and for baccalaureate.

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