313. For middle school - some nice problems.

by Virgil Nicula, Aug 24, 2011, 9:43 PM

PP0 Solve the equation $f_a(x)\equiv x^4 + a^4 - 3ax^3+3a^3x = 0$ , where $a\ne 0$ .

Proof. $x^4 + a^4 - 3ax^3+3a^3x=$ $x^4+a^4-3ax\left(x^2-a^2\right)=$ $\left(x^2-a^2\right)^2+2a^2x^2-3ax\left(x^2-a^2\right)=$ $a^2x^2\cdot\left(\frac {x^2-a^2}{ax}-1\right)\left(\frac {x^2-a^2}{ax}-2\right)$ .

In conclusion, $x^4 + a^4 - 3ax^3+3a^3x=\left(x^2-ax-a^2\right)\left(x^2-2ax-a^2\right)$ . Thus, $f_a(x)=0\iff \frac xa\in\left\{\ \frac {1\pm\sqrt 5}{2}\ ,\ 1\pm\sqrt 2\ \right\}$ .

PP1. Prove the following implication :

$\boxed{f(x,y,z)\equiv 2x^2 + 2y^2 + 5z^2 - 2xy - 4yz - 4x - 2z + 15\ \implies\ \min_{\{x,y,z\}\subset\mathbb R}\ f(x,y,z)=f(2,2,1)=10}$ .

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Proof.

$\blacktriangleright\ h_{y,z}(x)=2\cdot \underline x^2-2(y+2)\cdot \underline x+(2y^2+5z^2-4yz-2z+15)\implies$ $\min _x(y,z)=\frac 12\cdot \left[3y^2-4y-8yz+10z^2-4z+26\right]$ and $\boxed{x_m=\frac {y_m+2}{2}}$ .

$\blacktriangleright\ g_{z}(y)=\frac 12\cdot \left[3\cdot \underline y^2-4(1+2z)\cdot\underline y+(10z^2-4z+26)\right]\implies$ $\min_{(x,y)}(z)=\frac 13\cdot \left(7z^2-14z+37\right)$ and $\boxed{y_m=\frac {2(1+2z_m)}{3}}$ .

$\blacktriangleright\ l(z)=$ $\frac 13\cdot \left(7\cdot\underline {z}^2-14\cdot\underline {z}+37\right)$ $\implies$ $\min_{(x,y,z)} =-\frac {1}{21}\cdot\left(49-7\cdot 37\right)=\frac {30}{3}=10$ , $\boxed{z_m=1}\implies$ $\boxed{\ \min_{\{x,y,z\}\subset\mathbb R}\ f(x,y,z)=f(2,2,1)=10\ }$ , where $\left\{\begin{array}{c} x_m=2\\\\ y_m=2\\\\ z_m=1\end{array}\right\|$

Remark. I used $f(x)=ax^2+bx+c\ ,\ x\in \mathbb R$ , where $a>0\ \implies\ \min_{x\in\mathbb R} f(x)=$ $-\frac {\Delta}{4a}=-\frac {\Delta '}{a}$ ,

where $\Delta =b^2-4ac$ and $b=2b_1\ \implies\ \Delta '=b_1^2-ac$ . Indeed, can write $f(x)=a\cdot \left(x^2+\frac ba\cdot x+\frac ca\right)=$

$a\cdot\left[\left(x+\frac {b}{2a}\right)^2-\frac {b^2}{4a^2}+\frac ca\right]=$ $a\cdot\left[\left(x+\frac {b}{2a}\right)^2-\frac {b^2-4ac}{4a^2}\right]\implies$ $\boxed{\ f(x)=a\cdot \left[\left(x+\frac {b}{2a}\right)^2-\frac {\Delta}{4a}\right]\ }$ ,

where $\Delta =b^2-4ac$ . If $a>0$ , then $(\forall)\ x\in\mathbb R\ \ ,\ \ f(x)\ge \frac {-\Delta}{4a}$ , with the equality if and only if $x=-\frac {b}{2a}$ .

For example, here is another proof of the upper proposed problem :

$f(x,y,z)=2\cdot\underline x^2-2(y+2)\cdot\underline x+\left(2y^2+5z^2-4yz-2z+15\right)=$

$2\cdot\left(x-\frac {y+2}{2}\right)^2-\frac {(y+2)^2}{2} +2y^2+5z^2-4yz-2z+15=$

$2\cdot\left(x-\frac {y+2}{2}\right)^2+\frac {-y^2-4y-4+4y^2+10z^2-8yz-4z+30}{2}=$

$\frac {(y+2-2x)^2}{2}+\frac {3\cdot\underline y^2-4(1+2z)\cdot\underline y+10z^2-4z+26}{2}=$

$\frac {(y+2-2x)^2}{2}+\frac 32\cdot\left\{\left[y-\frac 23\cdot (1+2z)\right]^2-\frac 49\cdot (1+2z)^2+\frac {10z^2-4z+26}{3}\right\}=$

$\frac {(-2x+y+2)^2}{2}+\frac 16\cdot\left[(4z-3y+2)^2-4(2z+1)^2+30z^2-12z+78\right]=$

$\frac {(-2x+y+2)^2}{2}+\frac {(4z+2-3y)^2}{6}+\frac {14z^2-28z+74}{6}=$ $\frac {(-2x+y+2)^2}{2}+\frac {(4z+2-3y)^2}{6}+\frac {7(z-1)^2+30}{3}\implies$

$\boxed{\ f(x,y,z)=\frac {(-2x+y+2)^2}{2}+\frac {(4z+2-3y)^2}{6}+\frac {7(z-1)^2}{3}+10\ \ge\ 10\ }$ with $f(x,y,z)=10\ \iff\ \left\{\begin{array}{c} x_m=2\\\\ y_m=2\\\\ z_m=1\end{array}\right\|$ .

PP2. Let $ABCDEF$ be a regular hexagon. The points $M\in (AC)$ and $N\in (CE)$ satisfy $B\in MN$ and $AM=CN$ . Find $r=\frac {AC}{AM}$ .

Proof 1. Consider the regular $12$ -gon $AA_1BB_1CC_1DD_1EE_1FF_1$ . Suppose w.l.o.g. that $AB=1$ . Denote $M_1\in \ BD_1\cap AC$

and $N_1\in BD_1\cap CE$ . Observe that $AC=CE=\sqrt 3$ , $m(\angle CBD_1)=45^\circ$ , $(\angle BCE)=90^\circ$ , $m(\angle CN_1B)=45^\circ$

and $CN_1=CB$ . Similarly, $AM_1=AF$ , $\frac {AC}{AM_1} =\frac {CE}{CN_1}=\sqrt 3$ . In conclusion, $M_1\equiv M$ and $N_1\equiv N$ , $r=\sqrt 3$ .

Proof 2. Denote $\left\{\begin{array}{c} m(\angle CBN)=x\\\\ m(\angle EBN)=y\\\\ \left(x+y=60^\circ\right)\end{array}\right|$ . Apply an well-known relation to the cevians : $\left\{\begin{array}{cc} BM/\triangle ABC\ : & \frac {MC}{MA}=\frac {\sin x}{\sin (y+60^\circ)}\\\\ BN/\triangle CBE\ : & \frac {NE}{NC}=\frac {2\sin y}{\sin x}\end{array}\right|\implies$

$\sin^2x=2\sin y\sin (y+60^\circ )\iff$ $\sin^2x=\cos 60^\circ -\cos (2y+60^\circ )\iff$ $1-\cos 2x=$ $1-2\cos (2y+60^\circ )\iff$

$\cos 2x=\cos (2y+60^\circ )\iff$ $\left\{\begin{array}{c} x=y+30^\circ\\\ x+y=60^\circ\end{array}\right|\iff$ $\left\{\begin{array}{c} x=45^\circ\\\ y=15^\circ\end{array}\right|$ . In conclusion, $MA=1$ and $r=\sqrt 3$ .

PP3. Let $ABC$ be a triangle. The $A$ -bisector meet $[BC]$ în $D$ and the circumcircle again in $M$ .

Denote the midpoint $S$ of $[BC]$ and the projection $P$ of $S$ on $AM$ . Prove that $AP\cdot DM=\frac{a^2}{4}$ .

Proof. Suppose w.l.o.g. $b>c$ and note the diameter $[MN]$ of the circumcircle, $m(\angle ADB)=\phi$ and $L\in AN\cap BC$ .

Since $\triangle SDM\sim\triangle SNL$ obtain that $\frac {SD}{SM}=\frac {SN}{SL}$ and $BS^2=SM\cdot SN\implies$ $SD\cdot SL=BS^2$ . Observe that

$AP=SL\cos\phi$ and $DM\cos\phi =SD$ . In conclusion, $AP\cdot DM=SL\cdot SD=SB^2\implies AP\cdot DM=\frac {a^2}{4}$ .

Remark. Know that $\frac {DB}{c}=\frac {DC}{b}=\frac {a}{b+c}$ and $AD^2=bc-DB\cdot DC=\frac {2bc\cdot\cos\frac A2}{b+c}$ . Therefore, $DB\cdot DC=DA\cdot DM\implies$

$DM=\frac {a^2bc}{(b+c)^2}\cdot\frac {b+c}{2bc\cos\frac A2}\implies$ $\boxed{\ DM=\frac {a^2}{2(b+c)\cos\frac A2}\ }\ (1)$ . Denote $\left\{\begin{array}{c} X\in SP\cap AB\\\ Y\in SP\cap AC\end{array}\right|$ and projections $U$ , $V$ of $B$ , $C$ on $AD$ respectively.

Prove easily that $AX=AY=\frac {b+c}{2}$ and $AP=\frac {BU+CV}{2}\implies$ $\boxed{\ AP=\frac{b+c}{2}\cdot\cos\frac A2\ }\ (2)$ . From the relations $(1)$ , $(2)$ obtain that $AP\cdot DM=\frac {a^2}{4}$ .

PP4. Let $ABC$ be a triangle with the circumcircle $w=C(O,R)$ . Let $AD$ , $BE$ be the $A$ -bisector and the $B$ -bisector, where $D\in BC$ and

$E\in CA$ . Let $\{F,G\}\subset w$ be two points so that $AF\parallel DE$ , $FG\parallel BC$ and $AC$ separates $B$ and $G$ . Prove that $GA=GB+GC$ .

Proof. Denote $x=m(\angle BAF)$ and suppose w.l.o.g. that $2R=1$ . Prove easily that $\left\{\begin{array}{c} m(\angle EDA)=\frac A2+x\\\\ m(\angle EDC)=B-x\end{array}\right|$ and $\left\{\begin{array}{c} GA=\sin (B-x)\\\ GB=\sin (A+x)\\\ GC=\sin x\end{array}\right|$ .

Apply an well-known property to $DE$ in $\triangle ADC\ : \frac ca=\frac {EA}{EC}=\frac {DA}{DC}\cdot\frac {\sin\widehat{EDA}}{\sin\widehat{EDC}}\implies$ $\frac ca=\frac {\sin C}{\sin\frac A2}\cdot \frac {\sin\left(\frac A2+x\right)}{\sin (B-x)}\implies$ $\frac {\sin(B-x)}{\sin\left(\frac A2+x\right)}=$

$\frac {a\sin C}{c\sin\frac A2}=$ $\frac {c\sin A}{c\sin \frac A2}\implies$ $\boxed{\ 2\cos\frac A2=\frac {\sin(B-x)}{\sin\left(\frac A2+x\right)}\ }\ \ (*)$ . Therefore, $GA=GB+GC\iff$ $\sin(B-x)=\sin (A+x)+\sin x\iff$

$\sin (B-x)=2\sin\left(\frac A2+x\right)\cos\frac A2\iff$ $2\cos\frac A2=\frac {\sin (B-x)}{\sin\left(\frac A2+x\right)}$ , what is the true relation $(*)$ . In conclusion, $GA=GB+GC$ .

PP5. Let $r_1$ and $r_2$ be the radii of two circles through $A$ and which touch $BC$ at $B$ , $C$ respectively. Prove that $r_1r_2= R^2$ .

Proof. Denote $\{A,P\}=w_1\cap w_2$ and $S\in AP\cap BC$ . Observe that $m(\angle SPB)=B$ and $m(\angle SPC)=C$ . Therefore,

$\left\{\begin{array}{c} 2R\sin C= c=2r_1\sin B\\\\ 2R\sin B=b=2r_2\sin C\end{array}\right|\implies$ $\left\{\begin{array}{c} R\sin C=r_1\sin B\\\\ R\sin B=r_2\sin C\end{array}\right|\ \bigodot\implies$ $R^2=r_1r_2$ . Remark that $\frac {r_1}{r_2}=\left(\frac cb\right)^2$ and

$\frac {r_1}{c^2}=\frac {r_2}{b^2}=$ $\frac {\sqrt{r_1r_2}}{bc}=\frac {R}{bc}=\frac {1}{2h_a}=\frac {a}{4S}$ . Thus, $r_1+r_2=\frac {a\left(b^2+c^2\right)}{4S}\ge \frac {abc}{2S}=2R\implies$ $\boxed{\frac {r_1+r_2}{2}\ge R=\frac {br_1+cr_2}{b+c}}$ .

PP6. $\{x,y,z\}\subset\mathbb R\implies x^2 + y^2 + z^2 \ge \sqrt 2\cdot (xy + yz)$ .

Method 1. $\left(y-x\sqrt 2\right)^2+\left(y-z\sqrt 2\right)^2\ge 0$ .

Method 2. $f_{x,z}(y)=\underline y^2-(x+z)\sqrt 2\cdot\underline y+\left(x^2+z^2\right)$ with the discrimination $\Delta (x,z)=2(x+z)^2-4\left(x^2+z^2\right)=$ $-2(x-z)^2\le 0\implies$ $f_{x,z}(y)\ge 0$ .

Method 3 (Uzbekistan). $x^2+y^2+z^2=$ $\left(x^2+\frac {y^2}{2}\right)+\left(\frac {y^2}{2}+z^2\right)\ge xy\sqrt 2+yz\sqrt 2=x(y+z)\sqrt 2$ .

An easy and nice extension. $\boxed{\left(x^2+y^2+z^2\right)\cdot \sqrt {p^2+q^2}\ \ge\ 2y\cdot(px+qz)}$ .

PP7. $K$ , $L$ , $M$ and $N$ are points on sides $AB$ , $BC$ , $CD$ and $DA$ respectively of the unit square $ABCD$ such that $KM$

is parallel to $BC$ and $LN$ is parallel to $AB$ . The perimeter of triangle $KBL$ is equal to $1$ . What is the area of triangle $MND$ ?

Proof. Denote the midpoints $X$ , $Y$ of the sides $[AB]$ , $[BC]$ respectively. Since $BX=BY=\frac 12$ - semiperimeter of $\triangle KBL$ obtain that the incircle $w$

of the square is the $B$ -exincircle of $\triangle KBL$ , i.e. $KL$ is tangent to $w$ . Denote $T\in KL\cap w$ and $KT=KX=x$ , $LT=LY=y$ . Therefore,

$BK+x=BL+y=\frac 12$ and $KL=x+y$ . Since $KL^2=BK^2+BL^2\iff$ $\frac 12-(x+y)=2xy\iff$ $4xy+2(x+y)+1=2$ , i.e.

$\boxed{(1+2x)(1+2y)=2}\ (*)$ . On other hand, $AK=DM=\frac 12+x$ , $CL=DN=\frac 12+y$ and $[MDN]=\frac 12\cdot\left(\frac 12+x\right)\left(\frac 12+y\right)$ , i.e.

$[MDN]=\frac 18\cdot (1+2x)(1+2y)$ . In conclusion, using the relation $(*)$ obtain that $[MDN]=\frac 14$ .

An easy and nice extension. Let $ABCD$ be a rhombus with the incircle $w(O,r)$ and $m\left(\widehat{ABC}\right)\le 90^{\circ}$ . The circle $w$ touches $AB$ , $BC$

at $E$ , $F$ respectively and denote $BE=u$ , $AE=v$ . Consider two mobile points $K\in (BE)$ , $L\in (BF)$ so that the semiperimeter of

$\triangle KBL$ is equal to $BE$ . Denote $M\in (CD)$ , $N\in (DA)$ for which $KM\parallel BC$ and $LN\parallel CD$ . Ascertain the area of the triangle $MND$ .

Proof. Denote $\phi =m\left(\widehat{ABD}\right)$ . Observe that $u>v$ , $r^2=uv$ , $\tan\phi =\sqrt{\frac vu}$ and $\cos \phi =\frac {u-v}{u+v}$ , $\sin \phi =\frac {2\sqrt{uv}}{u+v}$ . Since the semiperimeter of $\triangle KBL$

is equally to $BE$ obtain that the circle $w$ is the $B$ -exincircle of $\triangle KBL$ . Denote $T\in KL\cap w$ . Therefore, $KT=KE=x$ and $LT=LF=y$ .

Thus, $BK=u-x$ , $BL=u-y$ and $KL=x+y$ . Apply the generalized Pytagoras' theorem to $\triangle KBL\ :$

$\ (u-x)^2+(u-y)^2-2(u-x)(u-y)\cdot$ $\frac {u-v}{u+v}=(x+y)^2\iff$ $\boxed{v(x+y)+xy=uv}\ \ (*)$ . On other hand,

$[MDN]=\frac 12\cdot (v+x)(v+y)\cdot\sin 2\phi =$ $\frac 12\cdot\left[v^2+v(u+v)+xy\right]\cdot$ $\frac {2\sqrt {uv}}{u+v}\ \stackrel{(*)}{=}\ \frac 12$ $\cdot v(u+v)\cdot \frac {2\sqrt {uv}}{u+v}\iff$ $\boxed{\ [MDN]=v\sqrt{uv}\ }$ .

PP8 (1995 - AIME). Circles $w_1=C(O_1,r_1)$ , $w_2=C(O_2,r_2)$ are externally tangent to each other and are internally tangent to a circle $w=C(O,r)$

so that $r_1+r_2=r$ . The circle $w$ has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord.

Proof. Suppose w.l.o.g. $r_1\le r_2$ . Denote the chord $[MN]$ of circle $w$ that is a common external tangent of circles $w_1$ and $w_2$ . Denote the projections $T_1$ , $T$ , $T_2$

of $O_1$ , $O$ , $O_2$ on the line $MN$ . Thus, $\left\{\begin{array}{c} O_1T_1=OO_2=r_1\\\\ O_2T_2=OO_1=r_2\\\\ OM=ON=r=r_1+r_2\end{array}\right|$ . Denote $U\in (OT)$ and $V\in O_2T_2$ so that $O_1\in UV$ and $UV\parallel T_1T_2$ .

In the right trapezoid $O_1O_2T_2T_1$ , where $OT\parallel O_1T_1\parallel O_2T_2\perp T_1T_2$ and $OO_1=r_2$ , $OO_2=r_1$ obtain easily $OT=OU+UT$ ,

where $UT=r_1$ and $\frac {OU}{O_2V}=\frac {O_1O}{O_1O_2}\iff$ $\frac {OU}{r_2-r_1}=\frac {r_2}{r_1+r_2}$ . Therefore, $OT=r_1+\frac {r_2(r_2-r_1)}{r_1+r_2}$ , i.e. $\boxed{OT=\frac {r_1^2+r_2^2}{r_1+r_2}}$ .

In conclusion, $MN^2=4\cdot TM^2=$ $4\cdot\left(OM^2-OT^2\right)=$ $4\cdot\left[(r_1+r_2)^2-\left(\frac {r_1^2+r_2^2}{r_1+r_2}\right)^2\right]\implies$ $\boxed{\ MN^2=\frac {16r_1r_2\left(r_1^2+r_1r_2+r_2^2\right)}{(r_1+r_2)^2}\ }$ .

In the particular case $r_1=3$ and $r_2=6$ obtain that $MN^2=224$ (the proposed problem 4 at AIME - 1995).

Extension. Circles $w_1=C(O_1,r_1)$ , $w_2=C(O_2,r_2)$ are externally tangent to each other and are internally tangent to a circle $w=C(O,r)$ so that $O\in O_1O_2$

and $r_1+r_2< 2r$ . The circle $w$ has a chord that is a common external tangent $[MN]$ of the other two circles. Prove that $MN^2=\frac {16(r-r_1)(r-r_2)(r^2-r_1r_2)}{\left[2r-\left(r_1+r_2\right)\right]^2}$ .

PP9. Let $ABC$ be a triangle with the centroid $G$ and let $M$ be an arbitrary interior point. The line

$MG$ cut $AB$ , $BC$ , $CA$ in $Z$ , $X$ , $Y$ respectively. Prove that $\frac{\overline {XM}}{\overline {XG}}+\frac{\overline {YM}}{\overline {YG}}+\frac{\overline {ZM}}{\overline {ZG}}=3$ .

Proof. Denote the midpoints $D$ , $E$ , $F$ of the sides $[BC]$ , $[CA]$ , $[AB]$ respectively and $U\in AM\cap BC$ , $V\in BM\cap CA$ ,

$W\in CM\cap AB$ . Prove easily that $\frac {\overline{UM}}{\overline {UA}}+\frac {\overline{VM}}{\overline {VB}}+\frac {\overline{WM}}{\overline {WC}}=\sum\frac {[BMC]}{[BAC]}=1\ \ (*)$ . Apply the Menelaus' theorem to

transversals : $\left\{\begin{array}{cccc} \overline{XDU}/\triangle AGM\ : & \frac {\overline{XM}}{\overline{XG}}\cdot \frac {\overline{DG}}{\overline{DA}}\cdot \frac {\overline{UA}}{\overline{UM}}=1 & \implies & \frac {\overline{XM}}{\overline {XG}}=3\cdot\frac {\overline{UM}}{\overline{UA}}\\\\ \overline{YEV}/\triangle BGM\ : & \frac {\overline{YM}}{\overline{YG}}\cdot \frac {\overline{EG}}{\overline{EB}}\cdot \frac {\overline{VB}}{\overline{VM}}=1 & \implies & \frac {\overline{YM}}{\overline {YG}}=3\cdot\frac {\overline{VM}}{\overline{VB}}\\\\ \overline{ZFW}/\triangle CGM\ : & \frac {\overline{ZM}}{\overline{ZG}}\cdot \frac {\overline{FG}}{\overline{FC}}\cdot \frac {\overline{WC}}{\overline{WM}}=1 & \implies & \frac {\overline{ZM}}{\overline {ZG}}=3\cdot\frac {\overline{WM}}{\overline{WC}}\end{array}\right|\ \stackrel{(*)}{\implies}\ \sum$ $\frac{\overline {XM}}{\overline {XG}}=3$ .

PP10 In the $x$ - $y$ plane, given are circles $C_1$ with radius $r$ , $C_2$ with radius $R$ . Their centers lie on the first quadrant. Suppose that $C_1$ touches

the $x$ axis at $(1,\ 0)$ , $C_2$ touches the $y$ axis at $(0,\ 2)$ and $C_1$ touches externally $C_2$ . When $R,\ r$ vary, find the minimum value of $R+r$ .

Proof. Observe that $(R-1)^2+(r-2)^2=(R+r)^2$ , i.e. $R+2r+Rr=\frac 52\iff (R+2)(r+1)=\frac 92$ . Therefore, $R+r$

is max. $\iff$ $(R+2)+(r+1)$ is max. $\iff$ $R+2=r+1=\sqrt {\frac 92}=\frac {3\sqrt 2}{2}$ . In conclusion, $R+r\ge 3\sqrt 2-3=3\left(\sqrt 2-1\right)$ .

An easy extension In the $x$ - $y$ plane, given are circles $w_1=C(O_1,r)$ , $w_2=C_2(O_2,R)$ . Their centers lie on the first quadrant. Suppose that $w_1$ touches

the $x$ axis at $(a,\ 0)$ , $w_2$ touches the $y$ axis at $(0,\ b)$ and $w_1$ touches externally $w_2$ . When $R,\ r$ vary, find the minimum value of $R+r$ .

Answer. In the minimum case, the slope of the line $O_1O_2$ is equally to $-1$ and $R+b=r+a=\frac {(a+b)\sqrt 2}{2}$ , i.e. $R+r\ge (a+b)\cdot \left(\sqrt 2-1\right)$ .

PP11. Let $ABCD$ be a square and $M$ be an interior point so that $MA=AB$ and $MC\perp MD$ . Prove that the midpoint of $[BC]$ belongs to $DM$ .

Proof 1. $\left\{\begin{array}{c} E\in DM\cap BC\\\ F\in CM\cap AB\end{array}\right|\implies$ $ADMF$ is cyclically and $AM=AD$ . Thus, $\widehat {BFC}\equiv$ $\widehat {ADM}\equiv$ $\widehat {AMD}\equiv$ $\widehat{AFD}\implies$

$\widehat {BFC}\equiv\widehat{AFD}\iff$ $FA=FB$ . Since $\triangle CBF\equiv\triangle DCE$ obtain that $CE=BF$ , i.e. $E$ is the midpoint of the side $[BC]$ .

Proof 2. Denote $E\in DM\cap BC$ . Observe that $EC^2=EM\cdot ED=EB^2\ \implies\ EB=EC$ . With other words, the line $DM$ is the radical axis between

the circle with diameter $[CD]$ and the circle with the center $A$ and the radius $[AB]$ . Thus, $DM$ cut the common tangent $BC$ of these circles in the midpoint $E$ of $[BC]$ .

Proof 3. Denote $\left\{\begin{array}{c} E\in DM\cap BC\\\ F\in CM\cap AB\end{array}\right|$ . Thus, $M\in C(A,AB)\implies$ $m\left(\widehat {BMD}\right)=135^{\circ}$ $\iff$ $m\left(\widehat {BME}\right)=45^{\circ}$ and $BEMF$ is cyclically $\iff$

$m\left(\widehat {BFE}\right)=45^{\circ}$ $\implies$ $\boxed{BE=BF}$ . Since $\triangle BCF\equiv\triangle CED$ obtain that $\boxed{BF=CE}$ . In conclusion, $EB=EC$ , i.e. $E$ is the midpoint of $[BC]$ .

Extension. Let $M$ be a point in the square $ABCD$ so that $MA=AB$ and $m\left(\widehat {CMD}\right)=x\in \left (45^{\circ},180^{\circ}\right)$ .

Denote $E\in DM\cap BC$ . Prove that $\frac {EB}{EC}=1-\cot x$ . For example, $\left\{\begin{array}{ccc} x=90^{\circ} & \implies & EB=EC\\\\ x=135^{\circ} & \implies & EB=2\cdot EC\end{array}\right|$ .

Proof. Suppose w.l.o.g. $AB=1$ . Denote $m\left(\widehat{DAM}\right)=2y$ .Thus, $m\left(\widehat{CDM}\right)=y$ and $MD=2\cdot\sin y$ . Therefore, $\frac {MD}{\sin\widehat {DCM}}=\frac {CD}{\sin\widehat{CMD}}\iff$

$\frac {2\cdot\sin y}{\sin (x+y)}=\frac {1}{\sin x}\iff$ $\boxed{\cot x+\cot y=2}$ . In conclusion, $1+\frac {EB}{EC}=$ $\frac {BC}{EC}=$ $\frac {CD}{CE}=\cot y\implies$ $\frac {EB}{EC}=\cot y-1\implies$ $\frac {EB}{EC}=1-\cot x$ .

PP12. Let $ABC$ be an equilateral triangle and $\left\{\begin{array}{cc} M\in (BC)\ :\ MB=MC\\\\ N\in (Ca)\ :\ NC=3\cdot NA\\\\ P\in (AB)\ :\ PA=2\cdot PB\end{array}\right|$ . Ascertain $m\left(\widehat {NMP}\right)$ .

Proof 1. Suppose w.l.o.g. that $AB=12$ . Thus, $\left\{\begin{array}{ccc} MB=6 & ; & MC=6\\\ NC=9 & ; & NA=3\\\ PA=8 & ; & PB=4\end{array}\right|$ . Apply the Pytagoras' theorem in the triangles

$\left\{\begin{array}{ccc} \triangle ANP & \implies & NP=7\\\\ \triangle BPM & \implies & PM=2\sqrt 7\\\\ \triangle CMN & \implies & MN=3\sqrt 7\end{array}\right|$ and in the triangle $MNP\ \implies\ \cos \widehat{NMP}=\frac 12$ , i.e. $m\left(\widehat {NMP}\right)=60^{\circ}$ .

Proof 2. Observe that $\frac{BM}{BP}=\frac{CN}{CM}$ and $\widehat{MBP}\equiv\widehat {MCN}$ . Hence $\triangle BMP\sim\triangle CNM$ , i.e.

$\left\{\begin{array}{c} \widehat{BMP}\equiv\widehat{CNM}\\\\ \widehat{BPM}\equiv\widehat{CMN}\end{array}\right|\implies$ $m\left(\widehat{NMP}\right)=$ $180^{\circ}-\left[m\left(\widehat{BMP}\right)+m\left(\widehat{CMN}\right)\right]=60^{\circ}$ .

PP13, Let $x,y\in\Bbb{R},\ ;\ x^2+y^2=1$ . Ascertain the maximum value of the function $f(x,y)=\mid x-y\mid +\mid x^3-y^3\mid$ .

Proof. Can use the substitution $|x-y|=t\in \left[0,\sqrt 2\right]$ because $x^2+y^2=1\ .$ In conclusion, the problem

becomes to find the maximum of the function $f(t)=\frac 12\cdot\left(5t-t^3\right)$ $\implies$ $f(t)\le f\left(\sqrt{\frac 53}\right)=\left(\frac 53\right)^{\frac 32}$ .

Remark. $\frac 12\cdot\left(5t-t^3\right)=$ $\frac{1}{2}\cdot t(5-t^2)=$ $\frac{1}{2\sqrt{2}}\cdot \sqrt{2t^2(5-t^2)(5-t^2)}\le$ $\frac{1}{2\sqrt{2}}\sqrt{\left(\frac{10}{3}\right)^3}=$ $\left(\frac{5}{3}\right)^{\frac{3}{2}}$ with equality iff $2t^2=5-t^2\iff$ $t=\sqrt{\frac 53}$ .

PP14. $\{m,p,q\}\subset\mathbb N^*$ and $\{x,y\}\subset \mathbb R^*_+$ . Ascertain $\max_{x^m+y^m=1}\ x^py^q$ without derivatives.

Example. Ascertain the maximum area of an isosceles trapezoid $ABCD$ which is inscribed in a semicircle $w=C(O,R)$ with the diameter $[AB]$ .

Proof. $x^py^q$ is $\max\iff$ $\left(\frac {x^m}{p}\right)^p\cdot \left(\frac {y^m}{q}\right)^q$ is $\max$ . Since $p\cdot\frac {x^m}{p}+q\cdot\frac {y^m}{q}=1$ (constant), then $x^py^q$ is $\max\iff$ $\frac {x^m}{p}=\frac {y^m}{q}=\frac {1}{p+q}$ .

Proof of given example. Denote $m\left(\widehat{DAB}\right)=m\left(\widehat{CBA}\right)=\alpha$ and the projections $E$ and $F$ of $D$ and $C$ on the diameter $[AB]$ . Observe

that $m\left(\widehat{AOD}\right)=m\left(\widehat{BOC}\right)=180^{\circ}-2\alpha$ and $m\left(\widehat{COD}\right)=4\alpha -180^{\circ}$ . Thus, $[ABCD]=2\cdot [ADE]+[CDEF]=$

$2\cdot\frac12\cdot R^2\cdot \sin(180^\circ-2\alpha)+\frac12\cdot R^2\cdot \sin(4\alpha-180^\circ)=$ $R^2\cdot\left(\sin 2\alpha-\frac12\cdot \sin 4\alpha\right)=$ $R^2\cdot (\sin 2\alpha-\sin 2\alpha\cos 2\alpha )=$

$R^2\cdot \sin 2\alpha (1-\cos2\alpha )=$ $R^2\cdot 2\sin\alpha\cos\alpha\cdot 2 \sin^2\alpha =$ $4R^2\cdot \sin^3\alpha\cos\alpha\implies$ $\boxed{[ABCD]=4R^2\cdot \sin^3\alpha\cos\alpha}$ . Thus, $[ABCD]$ is $\max\iff$

the product $\sin^3\alpha\cos\alpha$ is $\max\ (m=2\ ,\ p=3\ ,\ q=1)\iff$ $\frac {\sin^2\alpha}{3}=\frac {\cos^2\alpha}{1}=\frac14\iff$ $\tan\alpha =\sqrt 3\iff$ $\alpha =60^{\circ}\iff$ $BC=CD=DA$ .

PP15. Solve the system of equations $x^m+y^n=2z^p\ ,\ y^m+z^n=2x^p\ ,\ z^m+x^n=2y^p$ , where $\{m,n,p\}\subset \mathbb N^*$ and $\left\{x,y,z\right\}\subset (0,\infty )$ . .

Proof. Suppose w.l.o.g. $x=\max \{x,y,z\}$ . Observe that

$\blacktriangleright\ x\ge y\ge z \implies 2(z^p-x^p)=\left(x^m-y^m\right)+$ $\left(y^n-z^n\right)\ge 0 \implies z\ge x\implies x\ge y\ge z\ge x \implies x=y=z$

$\blacktriangleright\ x\ge z\ge y \implies 2(y^p-x^p)=\left(z^m-y^m\right)+$ $\left(x^n-z^n\right)\ge 0 \implies y\ge x\implies x\ge z\ge y\ge x\implies x=y=z$ .

In conclusion, $x=y=z\in \left\{0,\pm 1,\pm i\sqrt 2\right\}$ .

PP16. Let $ABC$ be an acute triangle with the circumcircle $w=C(O,R)$ . Denote $K\in BC\cap AO$

and the points $L\in AB$ , $M\in AC$ so that $KL = KB$ and $KM = KC$ . Prove that $LM \parallel BC$ .

Proof 1 $.\ \frac {KB}{KC}=\frac {AB}{AC}\cdot\frac {\sin\widehat{KAB}}{\sin\widehat{KAC}}=$ $\frac cb\cdot\frac{\sin \left(90^{\circ} -C\right)}{\sin \left(90^{\circ}-B\right)}\implies$ $\boxed{\frac {KB}{KC}=\frac {c\cdot \cos C}{b\cdot\cos B}}\ (*)$ . Thus, $\left\{\begin{array}{c} AL=AB-LB=c-2KB\cdot\cos B\\\\ AM=AC-MC=b-2KC\cdot\cos C\end{array}\right\|$ . Hence

$LM\parallel BC\iff$ $\frac {AL}{AB}=\frac {AM}{AC}\iff$ $\frac {c-2\cdot KB\cdot\cos B}{c}=\frac {b-2\cdot KC\cdot\cos C}{b}\iff$ $\frac {KB\cdot\cos B}{c}=\frac {KC\cdot\cos C}{b}\iff$ $\frac {KB}{KC}=\frac {c\cdot \cos C}{b\cdot\cos B}\ (*)$ .

Proof 2 Denote the midpoints $U$ , $V$ of $[BL]$ , $[CM]$ respectively and $D\in BC$ so that $AD\perp BC$ . Since $\widehat{DAB}\equiv\widehat {KAC}$ , i.e. the rays $[AD$ ,

$[AK$ are $A$ -isogonal obtain from the Steiner's theorem that $\boxed{\frac {DB}{DC}\cdot\frac {KB}{KC}=\frac {c^2}{b^2}}\ (*)$ . Since the quadrilaterals $AUDK$ , $AVKD$ are cyclically

and $\left\{\begin{array}{c} BL=2\cdot BU\\\\ CM=2\cdot CV\end{array}\right\|$ obtain that $\left\{\begin{array}{ccc} BA\cdot BU=BK\cdot BD\\\\ CA\cdot CV=CK\cdot CD\end{array}\right\|\implies$ $\frac cb\cdot \frac {BL}{CM}=\frac {KB}{KC}\cdot\frac {DB}{DC}\stackrel{(*)}{\implies}$ $\frac {BL}{CM}=\frac cb\implies LM\parallel BC$ .

PP17. Find $a\in\mathbb R$ for which the equation $\sqrt[3]{{(x + 1)^2 }} - \sqrt[3]{{1 - x^2 }} + \sqrt[3]{{(x - 1)^2 }} = a\sqrt[3]{{x^2 - 1}}$ has an unique solution in $\mathbb R$ .

Proof. We can observe that $x=\pm 1$ is not a solution. Divide by $\sqrt[3]{x^2-1}\ :\ \sqrt[3]{\frac{x+1}{x-1}}+1+\sqrt[3]{\frac{x-1}{x+1}}=a$ . Denote $t=\sqrt[3]{\frac{x+1}{x-1}}$ .

Then our equation becomes $t^2+(1-a)t+1=0$ . For $x$ to be unique, we need $t$ to be unique, i.e. $\Delta =0\Longrightarrow a\in\{-1\ ;\ 3\}$ .

PP18. Let $ABCD$ be a quadrilateral with $AC = 6$ , $BD = 8$ . Prove that $\max\left\{AB;BC;CD;DA\right\}\ge 5\ (*)$ .

Proof. Denote $m=\max\left\{AB;BC;CD;DA\right\}$ . Thus, $4m^2\ge AB^2+BC^2+CD^2+DA^2\stackrel{(1)}{\ge} AC^2+BD^2=100$ $\implies m\ge 5$ .

Remark. I used the Euler's relation $AB^2+BC^2+CD^2+DA^2=AC^2+BD^2+4\cdot MN^2$ , where $M$ and $N$ are midpoints of $[AC]$

and $[BD]$ , from where obtain the previous inequality $(1)$ . In conclusion, the relation $(*)$ becomes equality if and only if $ABCD$ is a paralellogram.

PP19. Let $\triangle ABC$ with $b>c$ the circumcircle $w$ . Let $P\in BB\cap CC$ , $M\in (AC)$ so that $m\left(\widehat{ABM}\right)=C$ and midpoint $N$ of $[BM]$ . Prove that $N\in AP$ .

Proof. It is well-known that the ray $[AP$ is the $A$ -symmedian in $\triangle ABC$ . Since $BM$ is a anti-parallel to $BC$ and the

geometrical locus of the midpoint of the anti-parallel to $BC$ is the $A$ -symmedian of $\triangle ABC$ , results that $N\in AP$ .

Remark. I used $XX$ - the tangent to the circle $w$ in the point $X\in w$ . If denote $S\in AP\cap BC$ and $\{A,D\}=AP\cap w$ , then $\frac {SB}{SC}=\frac {BA\cdot BD}{CA\cdot CD}\ (*)$

and $\left\{\begin{array}{cc} \triangle ABD\ : & \frac {PA}{PD}=\left(\frac {BA}{BD}\right)^2\\\\ \triangle ACD\ : & \frac {PA}{PD}=\left(\frac {CA}{CD}\right)^2\end{array}\right\|\ \implies$ $\frac {BA}{BD}=\frac {CA}{CD}\implies$ $\frac {BD}{CD}=\frac {BA}{CA}\stackrel{(*)}{\implies}$ $\frac {SB}{SC}=\left(\frac {BA}{CA}\right)^2\implies$ $AP$ is $A$ -symmedian in $\triangle ABC$ .

PP20. Let $A$ , $B$ be two fixed points. For a mobile point $M$ define the points $\left\{\begin{array}{cc} C\in (MA\ ; & MC=p\cdot MA\\\\ D\in (MB\ ; & MD=q\cdot MB\\\\ L\in (CD)\ ; & LC=r\cdot LD\end{array}\right\|$ ,

where $p,q,r$ are positive real numbers. Denote the midpoint $L$ of $[CD]$ . Prove that the point $F\in AB\cap ML$ is fixed.

Proof. $\boxed{\frac {FA}{FB}=\frac {LC}{LD}\cdot \frac {MA}{MB}\cdot\frac{MD}{MC}}=$ $r\cdot \frac {MA}{MB}\cdot\frac {q\cdot MB}{p\cdot MA}=r\cdot \frac qp$ (constant).

PP21. Solve the equation $\sqrt{4-3\sqrt{10-3x}}=x-2$ .

Proof 1. Observe that $2\le \frac {74}{27}<x<\frac {10}{3}<4$ and $\sqrt{4-3\sqrt{10-3x}}=x-2\iff$ $3\sqrt{10-3x}=-x^2+4x\iff$ $3\left(1-\sqrt{10-3x}\right)=$

$x^2-4x+3\iff$ $\frac {9(x-3)}{1+\sqrt{10-3x}}=(x-1)(x-3)$ , i.e. $x=3$ is a zero. Suppose $x\ne 3$ . In this case our equation becomes $\frac {9}{1+\sqrt{10-3x}}=x-1$

$\iff$ $\sqrt {10-3x}=\frac {10-x}{x-1}\iff$ $\sqrt {10-3x}=-1+\frac {9}{x-1}$ . Since for $2<x<4$ we have $\sqrt {10-3x}<2<-1+\frac {9}{x-1}$ obtain that $x\in \emptyset$ .

In conclusion, the equation has only the zero $\boxed{x=3}$ .

Proof 2. Observe that $2\le \frac {74}{27}<x<\frac {10}{3}<4$ and by squaring both sides we get $4-3\sqrt{10-3x}=(x-2)^2\iff$ $3\sqrt{10-3x}=-x^2+4x$ .

Squaring again we get $x^4-8x^3+16x^2+27x-90=0$ . Factorizing we get $(x+2)(x-3)(x^2-7x+15)=0$ from here we get only $x=3$ .

Proof 3 . From the given equation we deduce that $\frac{74}{27}\leq x\leq \frac{10}{3}$ . Squaring, we get $4-3\sqrt{10-3x}=x^2-4x+4\iff$ $-3\sqrt{10-3x}=x^2-4x$ . Introduce

another variable $y$ such that $\sqrt{10-3x}=ay+b$ for some real constants $a$ and $b$ . Then we have two system of equations $\left\{\begin{array}{c} x^2-4x+3ay+3b=0\\\ a^2y^2+2aby+3x+b^2-10=0 \end{array}\right\|$ .

We need to find $a$ and $b$ in such a way that the corresponding coefficients of $x^2$ and $y^2$ are both equal and the constant terms are both equal. From this we set

$a^2=1$ , $b^2-10=3b\iff a=1$ , $b=-2$ . Thus, we have an equivalent system of equations $\left\{\begin{array}{c} x^2-4x+3y-6=0\\\ y^2-4y+3x-6=0 \end{array}\right\|$ .

Subtracting leaves $x^2-y^2-7x+7y=0\iff$ $(x-y)(x+y-7)=0$ . We need to consider two cases.

Case 1. If $x=y$ , then $\sqrt{10-3x}=x-2$ or $x^2-4x+4=10-3x\iff$ $x^2-x-6=(x-3)(x+2)=0$ . As $x\geq\frac{74}{27}$ , we get $x=3$ .

Case 2. If $y=7-x$ , then $\sqrt{10-3x}=7-x-2$ or $\sqrt{10-3x}=5-x\iff$ $x^2-10x+25=10-3x\iff$ $x^2-7x+15=0\implies x\in\emptyset$ .

Therefore, the only real solution is $\boxed{x=3}$ (thugzmath10).

PP22. Triangle $ABC$ has $AB=6$ , $BC=8$ , and $AC=10$ . A semicircle is inscribed in $ABC$ such that it is tangent to $AC$ in $M\in (AC)$ and the endpoints

$E$ and $F$ of its diameter lie on $AB$ and $BC$ respectively. Let $X$ be the midpoint of segment $EF$ . If $XA=XC$ , what is the length of the radius of the semicircle ?

Proof. Denote $EF=2r$ . Thus, $XA=XC\iff$ $M$ is the midpoint of $[AC]\iff$ $BEMF$ is cyclically $\iff$

$\left\{\begin{array}{ccccc} AM^2=AE\cdot AB & \iff & AE=\frac {25}{6} & \iff & BE=\frac {11}{6}\\\\ CM^2=CF\cdot CB & \iff & CF=\frac {25}{8} & \iff & BF=\frac {39}{8}\end{array}\right\|$ . In conclusion, $EF^2=BE^2+BF^2\iff$

$4r^2=\left(\frac {11}{6}\right)^2+\left(\frac {39}{8}\right)^2\iff$ $4\cdot 16\cdot 36\cdot r^2=16\cdot 121+9\cdot 1521=15625=125^2\iff$ $r=\frac {125}{48}$ .

PP23. $ABCD$ is a parallelogram. Let $\left\{\begin{array}{cc} P\in (CD)\ ;& \frac {PD}{PC}=p\\\\ Q\in (BC)\ ; & \frac {QB}{QC}=q\end{array}\right\|$ . Let $T\in AP\cap DQ$ . Find $\frac {TA}{TP}$ and $\frac {TD}{TQ}$ .

Proof. Let $\left\{\begin{array}{c} AD=BC=a\\\\ AB=DC=b\end{array}\right\|$ and $\left\{\begin{array}{c} m\left(\widehat {TDA}\right) =m\left(\widehat {CQD}\right)=x\\\\ m\left(\widehat {TDP}\right) =m\left(\widehat {CDQ}\right)=y\end{array}\right\|$ . Thus, $\left\{\begin{array}{ccc} \frac {PD}{p}=\frac {PC}{1}=\frac {b}{p+1}\\\\ \frac {QB}{q}=\frac {QC}{1}=\frac {a}{q+1}\end{array}\right\|$

and $\frac {TA}{TP}=\frac {DA}{DP}\cdot\frac {\sin x}{\sin y}=$ $\frac {DA}{DP}\cdot\frac {CD}{CQ}=$ $\frac {a}{\frac {bp}{p+1}}\cdot\frac {b}{\frac {a}{q+1}}=\frac {(p+1)(q+1)}{p}\implies$ $\boxed{\frac {TA}{TP}=\frac{(p+1)(q+1)}{p}}$ . Let $R\in AP\cap BC$ .

Thus, $\frac {TQ}{TD}=\frac {QR}{AD}=$ $\frac {QC+CR}{AD}=\frac {QC}{BC}+\frac {CR}{AD}=$ $\frac {QC}{BC}+\frac {PC}{PD}=$ $\frac {1}{q+1}+\frac 1p\implies$ $\boxed{\frac {TD}{TQ}=\frac {p(q+1)}{p+q+1}}$ .

Remark. If $p=2$ and $q=1$ obtain that $TA=3\cdot TP$ and $TD=TQ$ .

PP24. On $[AB]$ of $\triangle ABC$ is built a equilateral triangle $ABD$ . Show that with $[AC]$ , $[CB]$ and $[CD]$ can construct a triangle.

Proof. Suppose $AB$ separates $C, D$ . Construct the equilateral triangle $\triangle BCE$ , $BC$ separates $A$ , $E$ . See that a $60^\circ$ rotation about $B$ maps $\triangle BCD$ and $\triangle BEA$ .

Hence $AE=CD$ and $\triangle ACE$ is the one constructed with the required segments. Similarly when $C$ , $D$ are on the same side of $AB$ and $A$ , $E$ on the same side of $BC$

PP25. Let $ABM$ , $BCN$ , $CDP$ , $d$ be equilateral triangles and a line so that $\{A,B,C,D\}\subset d$ (in this order) and $\left\{\begin{array}{c} AB=a\\\ BC=b\\\ CD=c\end{array}\right\|$ and $M$ , $N$ , $P$

are situated in the same side of $d$ . Show that $\boxed{\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}\ge\sqrt{a^2+b^2+c^2+ab-ac+bc}}\ (*)$ (Lajos Longaver).

Proof, Let $R\in AM\cap DP$ . Thus, $\left\{\begin{array}{c} RM=b+c\\\ RP=a+b\end{array}\right\|\implies$ $MP^2=(a+b)^2+(b+c)^2-(a+b)(b+c)\implies$

$MP^2=a^2+b^2+c^2+ab+bc-ac$ . Hence $\left\{\begin{array}{c} MN^2=a^2+b^2-ab\\\\ PN^2=b^2+c^2-bc\end{array}\right\|$ . Thus, $NM+NP\ge MP\implies\ (*)$ .

An easy extension. Prove that for $a_k\in\mathbb R^*_+$ , where $k\in\overline{0,n}$ , exists the inequality $\sum_{k=0}^{n-1}\sqrt{a_k^2-a_ka_{k+1}+a_{k+1}^2}\ge \sqrt{a_0^2+a_n^2+\left(\sum_{k=1}^{n-1}\right)^2-a_0a_n+(a_0+a_n)\sum_{k=1}^{n-1}a_k}$ .

PP26. Let $\triangle ABC$ with $b>c$ , the midpoints $M$ , $N$ , $P$ of $BC$ , $[CA]$ , $[AB]$ respectively and $\left\{\begin{array}{cc} D\in (BC)\ ; & \widehat{DAB}\equiv\widehat{DAC}\\\\ E\in AD\ :& BE\perp AD\\\\ F\in AD\ ;& CF\perp AD\end{array}\right\|$ . Prove that $NF\cap PE\cap BC\ne\emptyset$ .

Proof. $\left\{\begin{array}{ccccccccc} EA\perp ED & \iff & PA=PE & \iff & \widehat{PEA}\equiv\widehat{PAE}\equiv\widehat{CAE} & \iff & PE\parallel AC & \iff & M\in PE\\\ FA\perp FD & \iff & NA=NF & \iff & \widehat{NFA}\equiv\widehat{NAF}\equiv\widehat{BAF} & \iff & NF\parallel AB & \iff & M\in NF\end{array}\right\|$ $\implies$ $M\in PE\cap NF$ .

PP27. Let $ABCD$ be a square and let $F\in (CD)$ , $E\in (BC)$ be two points so that $m(\angle EAF)=45^{\circ}$ . Prove that the ray $[FA$ is the bisector of $\widehat{DFE}$ .

Proof. Let $G\in CD$ so that $D\in (FG)$ and $DG=BE$ . Thus, $\triangle ADG\equiv\triangle ABFE\implies$ $AE=AG\ \wedge\ \widehat{FAG}\equiv\widehat{FAE}\implies$

$\triangle FAE\equiv\triangle FAG\implies$ $\widehat{AFE}\equiv\widehat{AFG}\implies$ the ray $[FA$ is the bisector of $\widehat{DFE}$ .

Remark. If $\left\{\begin{array}{c} m(\angle BAE)=x\\\\ m(\angle DAF)=y\end{array}\right\|$ , then prove easily that $\tan\widehat{AFE}=\frac {1-\tan x\tan y}{1+\tan^2y-(\tan x+\tan y)}$ . Indeed, denote $\left\{\begin{array}{c} m\left(\widehat{AFE}\right)=\phi\\\ m\left(\widehat{CFE}\right)=z\end{array}\right\|$ and can

suppose w.l.o.g. that $AB=1$ . Thus, $\left\{\begin{array}{c} BE=\tan x\ ;\ CE=1-\tan x\\\ DF=\tan y\ ;\ CF=1-\tan y\end{array}\right\|\implies$ $\tan z=\frac {1-\tan x}{1-\tan y}$ and $\tan\phi=\cot(z-y)=\frac {1+\tan z\tan y}{\tan z-\tan y}$ a.s.o.

PP28 Let $\{x,y,z\}\subset\mathbb C$ such that $\left\{\begin{array}{ccc} x^2 & = & 2(2y-9)\\\\ y^2 & = & 2(z-4)\\\\ z^2 & = & 2(3x-1)\end{array}\right\|$ . Find $x^4 + y^4 + z^4$ .

Proof. $x^4+y^4+z^4=$ $4(2y-9)^2+4(z-4)^2+4(3x-1)^2=$ $4\cdot\left[\left(4y^2-36y+81\right)+\left(z^2-8z+16\right)+\left(9x^2-6x+1\right)\right]=$

$4\cdot\left(9x^2+4y^2+z^2-6x-36y-8z+98\right)=$ $4\cdot\left[18(2y-9)+8(z-4)+2(3x-1)-6x-36y-8z+98\right]=-4\cdot 98=-392$ .

An easy extension. Let $\{x,y,z,a,b,c\}\subset\mathbb C$ such that $\left\{\begin{array}{ccc} x^2 & = & ay-\frac {c^2}{2}\\\\ y^2 & = & bz-\frac {a^2}{2}\\\\ z^2 & = & cx-\frac {b^2}{2}\end{array}\right\|$ . Prove that $4\cdot\left(x^4+y^4+z^4\right)+\left(a^4+b^4+c^4\right)=0$ .

Remark. For $a:=4\ ,\ b:=2\ ,\ c:=6$ obtain the proposed problem PP28.

PP29. Let $\{a,b\}\subset (0,1)$ and $\{m,n\}\subset \mathbb R^*_+$ so that $m+n>1$ and $ma+nb=1$ . Prove that $\boxed{\frac {1-ab}{(1-a)(1-b)}\ge \frac {1+2\sqrt{mn}}{m+n-1}}\ (*)$ .

Proof. Apply C.B.S. inequality $\left(\frac {1}{1-a}+\frac {1}{1-b}\right)\cdot\left(\frac {m}{\frac {1}{1-a}}+\frac {n}{\frac {1}{1-b}}\right)\ge \left(\sqrt m+\sqrt n\right)^2\iff$ $\left(\frac {1}{1-a}+\frac {1}{1-b}\right)\cdot\left[m(1-a)+n(1-b)\right]\ge \left(\sqrt m+\sqrt n\right)^2\iff$

$\frac {1}{1-a}+\frac {1}{1-b}\ge \frac {\left(\sqrt m+\sqrt n\right)^2}{m+n-1}$ . Observe that $\frac {1-ab}{(1-a)(1-b)}=\frac {1}{1-a}+\frac {1}{1-b}-1\ge\frac {m+n+2\sqrt{mn}}{m+n-1}-1$ . In conclusion, $\frac {1-ab}{(1-a)(1-b)}\ge \frac {1+2\sqrt{mn}}{m+n-1}$ .

PP30. Let a rectangle $ABCD$ , $\{E,M\}\subset AC$ and $N\in CD$ so that $BE\perp AC$ and $\frac {MA}{ME}=\frac {ND}{NC}$ . Prove that $MN\perp MB$ .

Proof 1. Let $P\in AB$ so that $NP\parallel BC$ . Thus, $\frac {MA}{ME}=\frac{ND}{NC}=\frac {PA}{PB}\implies$ $MP\parallel BE\implies$ $ME\perp AC\implies$ $BCNMP$ is inscribed in the circle with the diameter $[CP]$ or $[BN]\implies$ $MN\perp MB$ .

Proof 2. $\triangle ABE\sim\triangle DBC\implies$ $M\in (AE)$ and $N\in (DC)$ are omologously $\implies\widehat{BME}\equiv\widehat{BNC}\implies$ $BCMN$ is cyclically $\implies$ $MN\perp MB$ .

An easy extension. Let a convex cyclical $ABCD$ , $\{E,M\}\subset (AC)$ , $N\in (CD)$ so that $\widehat{AEB}\equiv\widehat {DCB}$ and $\frac {MA}{ME}=\frac {ND}{NC}$ . Prove that $BCNM$ is a cyclically.

This post has been edited 278 times. Last edited by Virgil Nicula, Nov 20, 2015, 8:57 AM

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I hav easy solution PP6
$(x^2+\frac{y^2}{2})+(\frac{y^2}{2}+z^2)\geq \sqrt{2}xy+\sqrt{2}yz=\sqrt{2}(xy+yz)$

by Uzbekistan, Jun 10, 2012, 12:14 PM

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by Virgil Nicula, Dec 11, 2010, 8:32 PM
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.
by Thomas_, Dec 8, 2010, 6:42 PM
Thanks Thomas_.
by El_Ectric, Dec 8, 2010, 4:21 PM
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
by stergiu, Nov 12, 2010, 6:15 AM

72 shouts

My (math)blog

313. For middle school - some nice problems.

by Virgil Nicula, Aug 24, 2011, 9:43 PM

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