323. Easy, nice and interesting problems for high school.

by Virgil Nicula, Oct 17, 2011, 8:29 AM

ALGEBRA.

PP1. Eliminate the variables $x$ and $y$ between the relations $x+\frac{1}{x}=a\ \ \wedge\ \ y+\frac{1}{y}=$ $b\ \ \wedge\ \ xy+\frac{1}{xy}=c\ .$

Proof. Denote $\left\{\begin{array}{c} x+y=S\\\ xy=P\end{array}\right|$ . Observe that $xy+\frac{1}{xy}= c\iff \boxed{P^2+1=cP}\ \ (*)\ .$

$\blacktriangleright\ a+b=$ $\left(x+\frac 1x\right)+\left(y+\frac 1y\right)=$ $\frac {S(P+1)}{P}\implies$ $S^2\left[\left(P^2+1\right)+2P\right]=$ $(a+b)^2P^2\stackrel{(*)}{\implies}$ $\boxed{(c+2)S^2=(a+b)^2P}\ \ (1)\ .$

$\blacktriangleright\ ab=$ $\left(x+\frac 1x\right)\left(y+\frac 1y\right)=$ $\frac {S^2+P^2+1-2P}{P}$ $\stackrel{(*)}{\implies}$ $S^2+(c-2)P=abP\implies \boxed {S^2=(ab+2-c)P}\ \ (2)\ .$

From the relations $(1)\ \wedge\ (2)$ obtain that $(c+2)(ab+2-c)=(a+b)^2\iff$ $ab(c+2)+4-c^2=(a+b)^2\iff$ $\boxed{a^2+b^2+c^2=abc+4}$ .

PP2. Prove that $\left\{\begin{array}{c} a+b+c=1\\\\ a^3+b^3+c^3=1\end{array}\right|\ \implies\ (1-a)(1-b)(1-c)=0\ .$

Proof 1. $\sum a^3-3abc=\sum a\cdot\left(\sum a^2-\sum bc\right)\iff$ $\sum a^2-\sum bc+3abc=1\iff$

$\left(\sum a\right)^2-3\sum bc+3abc=1\iff$ $abc=\sum bc$ $\iff$ $1-\sum a+\sum bc-abc=0$ $\iff$ $(1-a)(1-b)(1-c)=0$ .

Proof 2. Denote $\left\{\begin{array}{ccc} s_1 & = & a+b+c\\\\ s_2 & = & ab+bc+ca\\\\ s_3\ & = & abc\end{array}\right|$ and $S_3=\sum a^3$ . Prove easily that $\left\{\begin{array}{cc} S_3=s_1^3-3s_1s_2+3s_3 & (1)\\\\ (1-a)(1-b)(1-c)=1-s_1+s_2-s_3 & (2)\end{array}\right|$ .

Therefore, $S_3=1\ \ \wedge\ \ s_1=1\stackrel{(1)}{\iff} s_2=s_3$ $\iff 1-s_1+s_2-s_3=0\stackrel{(2)}{\iff}\prod (1-a)=0$ .

PP3. Prove that for any $\{x,y\}\subset\mathbb R^*\ ,\ \left\{\begin{array}{cccc} (1) & (x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1 & \implies & x +y=0\\\\ (2) & (x+\sqrt{y^2+1})(y+\sqrt{x^2+1})=1 & \implies & x +y=0\end{array}\right|$ .

Proof 1. $(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$ $\implies$ $\begin{array}{c} y+\sqrt{y^2+1}=\sqrt{x^2+1}-x\iff x+y=\sqrt{x^2+1}-\sqrt{y^2+1}\\\\ x+\sqrt{x^2+1}=\sqrt{y^2+1}-y\iff x+y=\sqrt{y^2+1}-\sqrt{x^2+1}\end{array}$ $\implies$ $x+y=0$ .

Proof 2. The function $f:\mathbb R^*_+\rightarrow \mathbb R\ ,\ f(t)=\frac 12\cdot\left(t-\frac 1t\right)$ is surjectively $\implies (\forall )\{x,y\}\subset \mathbb R\ (\exists )\{u,v\}\subset R^*_+$ so that $x=f(u)\ ,\ y=f(v)$ .

Therefore, $\left(x+\sqrt{y^2+1}\right)\left(y+\sqrt{x^2+1}\right)=1\iff$ $\left(\frac {u^2-1}{2u}+\frac {v^2+1}{2v}\right)\cdot\left(\frac {v^2-1}{2v}+\frac {u^2+1}{2u}\right)=1\iff$

$\left[ uv(u+v)+(u-v)\right]\cdot\left[uv(u+v)-(u-v)\right]=1\iff$ $u^2v^2(u+v)^2-(u-v)^2=4u^2v^2\ (*)$ . Denote $uv=z$ . Thus, the relation $(*)$ becomes

$z^2\left(u^2+v^2+2z\right)-\left(u^2+v^2\right)+2z-4z^2=0\iff$ $P(z)\equiv 2\cdot \underline z^3+\left(u^2+v^2-4\right)\cdot \underline z^2+2\cdot \underline z-\left(u^2+v^2\right)=0$ . Observe that $P(1)=0$ .

Therefore, $P(z)=(z-1)\left[2z^2+\left(u^2+v^2-2\right)z+\left(u^2+v^2\right)\right]=0$ . Since $2z^2+\left(u^2+v^2-2\right)z+\left(u^2+v^2\right)=$

$2(uv)^2+uv\left(u^2+v^2\right)+(u-v)^2>0$ obtain that $uv=1$ . Hence $y=\frac {v^2-1}{2v}=\frac {\frac {1}{u^2}-1}{\frac 2u}=\frac {1-u^2}{2u}=-x$ . In conclusion, $x+y=0$ .

Proof 3. $\left\{\begin{array}{c} x+\sqrt {y^2+1}=a\\\\ y+\sqrt {x^2+1}=b\end{array}\right\|\iff$ $\left\{\begin{array}{c} a^2+x^2=2ax+y^2+1\\\\ b^2+y^2=2by+x^2+1\end{array}\right|\left|\begin{array}{c} \odot\ b\\\\ \odot\ a\end{array}\right\|\stackrel{(ab=1)}{\implies}$ $\left\{\begin{array}{c} a+bx^2=2x+by^2+b\\\\ b+ay^2=2y+ax^2+a\end{array}\right\|\ \bigoplus$ $\implies$

$(a-b)\left(y^2-x^2\right)=2(x+y)$ . Suppose against all reason that $x+y\ne 0$ . Thus, $(a-b)(x-y)=-2\ (*)$ . Since $a-b=$

$(x-y)+\frac {y^2-x^2}{\sqrt{x^2+1}+\sqrt{y^2+1}}=$ $(x-y)\cdot \frac {\left(\sqrt{x^2+1}-x\right)+\left(\sqrt{y^2+1}-y\right)}{\sqrt{x^2+1}+\sqrt{y^2+1}}$ and $\frac {\left(\sqrt{x^2+1}-x\right)+\left(\sqrt{y^2+1}-y\right)}{\sqrt{x^2+1}+\sqrt{y^2+1}}>0$

obtain that $(a-b)(x-y)\ge 0$ , what is absurd with $(*)$ . In conclusion, $x+y=0$ .

PP4. Solve the systems $\left\{\begin{array}{c} xy+x+y=1\\\ yz+y+z=5\\\ xz+x+z=2\end{array}\right\|\ ,\ \left\{\begin{array}{c} x+y+z=9\\\\ \frac{1}{x+1}+\frac{1}{y+2}+\frac{1}{z+3}=\frac{3}{5}\end{array}\right\|$ and prove that $\left\{\begin{array}{c} x^3+y^3+z^3=0\\\ x^5+y^5+z^5=0\end{array}\right\|\ \iff\ x=0\ \vee\ y=0\ \vee\ z=0$ .

Proof. $1\blacktriangleright$ This system is equivalent with $\left\{\begin{array}{c} (x+1)(y+1)=2\\\ (y+1)(z+1)=6\\\ (z+1)(x+1)=3\end{array}\right\|$ . Using the substitution

$\left\{\begin{array}{c} x+1=a\\\ y+1=b\\\ z+1=c\end{array}\right\|$ obtain the system $\left\{\begin{array}{c} ab=2\\\ bc=6\\\ ca=3\end{array}\right\|$ . Observe that $(abc)^2=36$ . Appear two cases :

$\odot\ abc=6\ \implies\ \left\{\begin{array}{c} a=1\\\ b=2\\\ c=3\end{array}\right\|\ \implies\ \left\{\begin{array}{c} x=0\\\ y=1\\\ z=2\end{array}\right\|$ .

$\odot\ abc=-6\ \implies\ \left\{\begin{array}{c} a=-1\\\ b=-2\\\ c=-3\end{array}\right\|\ \implies\ \left\{\begin{array}{c} x=-2\\\ y=-3\\\ z=-4\end{array}\right\|$ .

In conclusion, the solutions of the given system are $(0,1,2)$ and $(-2,-3,-4)$ .

$2\blacktriangleright$ Substitution $\left\{\begin{array}{c} x+1=a\\\ y+2=b\\\ z+3=c\end{array}\right\|\implies$ $\left\{\begin{array}{c} a+b+c=15\\\\ \frac 1a+\frac 1b+\frac 1c=\frac 35\end{array}\right\|$ . Observe that $9=15\cdot\frac 35=(a+b+c)\left(\frac 1a+\frac 1b+\frac 1c\right)\iff$

$\sum a\cdot\sum bc=9abc\iff$ $\sum\left(b^2+c^2\right)=6abc\iff$ $\sum (b-c)^2=0\iff$ $a=b=c=5\iff$ $\left\{\begin{array}{c} x=4\\\ y=3\\\ z=2\end{array}\right\|$ .

$3\blacktriangleright$ Consider the equation $t^3-mt^2+nt-p=0$ with the roots $\{x,y,z\}$ . Prove easily that $\left\{\begin{array}{ccc} S_3 & = & m^3-3mn+3p\\\\ S_5 & = & m^5-5m^3n+5m^2p+5mn^2-5np\end{array}\right\|$ .

Eliminate $p$ between $\left\{\begin{array}{ccc} m^3-3mn+3p & = & 0\\\\ m^5-5m^3n+5m^2p+5mn^2-5np & = & 0\end{array}\right\|$ and obtain $2m^5=5m^3n$ from where results the conclusion of our problem.

PP5. Ascertain $\min_{x^2y=k} (x^2+4xy)$ , where $x>0$ and $y>0$ (a box with open top).

Proof. $x^2+4xy=$ $x^2+\frac {4k}{x}=$ $x^2+\frac {2k}{x}+\frac {2k}{x}$ and $x^2\cdot \frac {2k}{x}\cdot\frac {2k}{x}=4k^2$ (constant). The sum $x^2+\frac {2k}{x}+\frac {2k}{x}$ is minimum iff $x^2=\frac {2k}{x}$ , i.e. $x=\sqrt[3]{2k}$ .

PP6. Prove that $a + b + c = 0\implies 6\cdot\left(a^5 + b^5 + c^5\right) = 5\cdot \left(a^2 + b^2 + c^2\right)\cdot\left( a^3 + b^3 + c^3\right)$ .

Proof. Let $f(x)=(x-a)(x-b)(x-c)=x^3-s_1x^2+s_2x-s_3=0$ be the equation with the roots $\{a,b,c\}$ , where $\left\{\begin{array}{c} s_1=a+b+c\\\ s_2=ab+bc+ca\\\ s_3=abc\end{array}\right|$ . Denote

$S_n=a^n+b^n+c^n$ , where $n\in\mathbb N$ . Observe that $\left\{\begin{array}{ccc} a^3-s_1a^2+s_2a-s_3=0 & \odot & a^n\\\\ b^3-s_1b^2+s_2b-s_3=0 & \odot & b^n\\\\ c^3-s_1c^2+s_2c-s_3=0 & \odot & c^n\end{array}\right|\ \bigoplus$ $\implies\ \boxed{S_{n+3}=s_1\cdot S_{n+2}-s_2\cdot S_{n+1}+s_3\cdot S_n}$ ,

where $n\in\mathbb N$ and $S_0=3\ ,\ S_1=s_1\ ,\ S_2=$ $s_1^2-2s_2\ ,\ S_3=s_1^3-3s_1s_2+3s_3$ . In the particular case when $s_1=0$ obtain that

$S_0=3\ ,\ S_1=0\ ,\ S_2=-2s_2$ and $\ \left\{\begin{array}{ccc} S_3=s_1S_2-s_2S_1+s_3S_0 & \implies & S_3=3s_3\\\\ S_4=s_1S_3-s_2S_2+s_3S_1 & \implies & S_4=2s_2^2\\\\ S_5=s_1S_4-s_2S_3+s_3S_2 & \implies & S_5=-5s_2s_3\end{array}\right|\ \implies\ 6\cdot S_5=5\cdot S_2\cdot S_3$ .

PP7. Find the number of common real roots of the polynomials $\left\{\begin{array}{c}x^5+3x^4-4x^3-8x^2+6x-1\\\ x^5-3x^4-2x^3+10x^2-6x+1\end{array}\right|$ .

Proof. $\left\{\begin{array}{ccc} x^5+3x^4-4x^3-8x^2+6x-1 & \odot & 1\\\ x^5-3x^4-2x^3+10x^2-6x+1 & \odot & (-1)\end{array}\right\|\ \oplus$

$\left\{\begin{array}{ccc} 3x^4-x^3-9x^2+6x-1 & \odot & \left(-x\right)\\\ x^5+3x^4-4x^3-8x^2+6x-1 & \odot & 3\end{array}\right\|\ \oplus$

$\left\{\begin{array}{ccc} 10x^4-3x^3-30x^2+19x-3 & \odot & 3\\\ 3x^4-x^3-9x^2+6x-1 & \odot & \left(-10\right)\end{array}\right\|\ \oplus$

$\left\{\begin{array}{ccc} x^3-3x+1 & \odot & (3x)\\\ 3x^4-x^3-9x^2+6x-1 & \odot & (-1)\end{array}\right\|\ \oplus$

$\left\{\begin{array}{ccc} x^3-3x+1 & \odot & 1\\\ x^3-3x+1 & \odot & (-1)\end{array}\right\|\ \oplus$

$\mathrm{STOP}$ . In conclusion, the common roots are $x^3-3x+1=0$ , i.e. the number of common real roots of these polynomials is $3$ (three).

PP8. Find the solution of the equation $\sqrt{\frac{x-7}{3}}+\sqrt{\frac{x-6}{4}}+\sqrt{\frac{x-8}{2}}=\sqrt{\frac{x-3}{7}}+\sqrt{\frac{x-4}{6}}+\sqrt{\frac{x-2}{8}}$ .

Proof. $\sqrt{\frac{x-7}{3}}+\sqrt{\frac{x-6}{4}}+\sqrt{\frac{x-8}{2}}=\sqrt{\frac{x-3}{7}}+\sqrt{\frac{x-4}{6}}+\sqrt{\frac{x-2}{8}}\Longleftrightarrow$

$\Longleftrightarrow(x-10)\left(\frac{\frac{4}{21}}{\sqrt{\frac{x-7}{3}}+\sqrt{\frac{x-3}{7}}}+\frac{\frac{1}{12}}{\sqrt{\frac{x-6}{4}}+\sqrt{\frac{x-4}{6}}}+\frac{\frac{3}{8}}{\sqrt{\frac{x-8}{2}}+\sqrt{\frac{x-2}{8}}}\right)=0$ $\Longleftrightarrow x=10$ .

GEOMETRY.

PP1. Ascertain the dimensions of a cylinder with the constant volume and with the minimum total surface.

Proof. Denote $x$ - the length of the radius and $y$ - the length of the altitude. Thus, $x^2y=k$ (constant) and the product $x^2+xy$ must be minimum. Therefore, can write

$x^2\cdot \frac {xy}{2}\cdot\frac {xy}{2}=$ $\frac {k^2}{4}$ (constant) and the sum $\left(x^2+\frac {xy}{2}+\frac {xy}{2}\right)$ must be minimum, i.e. $x^2=\frac {xy}{2}=\sqrt [3]{\frac {k^2}{4}}$ $\implies$ $\frac x1=\frac y2=\sqrt [3]{\frac k2}$ (the axial section is a square).

PP2. For the triangle $ABC$ denote $\left\|\ \begin{array}{c} A=(b+c)(c+a)(a+b)-8abc\\\\ B=abc-(b+c-a)(c+a-b)(a+b-c)\end{array}\ \right\|$ . Prove that

$A\ge B\ge 0$ , i.e. $\boxed{\prod (b+c)+\prod (b+c-a)\ge 9abc}$ (Virgil Nicula & Cosmin Pohoata, Mathematical Reflections).

Proof. Indeed, $A=\sum a\cdot\sum bc-9abc=2p(p^2+r^2+4Rpr)-36Rpr=2p(p^2+r^2-14Rr)\ge 0$ because

$p^2\ge 16Rr-5r^2\ge 14Rr-r^2$ . On other hands, $B=abc-8\prod (p-a)=4Rpr-8pr^2=4pr(R-2r)\ge 0$ .

In conclusion, $A-B=2p(p^2+r^2-14Rr-2Rr+4r^2)=2p(p^2+5r^2-16Rr)\ge 0$ .

PP3. Let $ABC$ be a triangle. Denote the midpoint $M$ of the side $[BC]$ . Suppose that

$\mathrm m\left(\widehat {MAB}\right)=90^{\circ}-C$ . Prove that $ABC$ is an $A$ -right triangle or $A$ -isosceles.

Proof. Observe that $\mathrm m\left(\widehat {MAC}\right)=90^{\circ}-B$ . Therefore, $1=\frac {MB}{MC}=\frac {AB}{AC}\cdot\frac {\sin\widehat {MAB}}{\sin\widehat{MAC}}=$ $\frac {\sin C}{\sin B}\cdot\frac {\cos C}{\cos B}=$ $\frac {\sin 2C}{\sin 2B}$ .

In conclusion, $\sin 2B=\sin 2C\iff$ $B=C\ \ \vee\ \ B+C=90^{\circ}$ $\iff$ $ABC$ is an $A$ -right triangle or $A$ -isosceles.

PP4. For $C\in (AB)$ , in the same semiplane w.r.t. the line $AB$ construct the $C$ -isosceles triangles $ACD$ and $BCE$

so that $\widehat{ACD} \equiv\widehat{BCE}$ . Denote $F\in AE \cap BD$ . Prove that the ray $[FC$ is the bisector of the angle $\widehat{AFB}$ .

Proof. Observe that $\widehat{CDA}\equiv\widehat{CEB}$ and $\left\{\begin{array}{c} AC=DC\\\\ CE=CB\\\\ \widehat {ACE}\equiv\widehat{DCB}\end{array}\right|\stackrel{(s.a.s)}{\implies}$ $\triangle ACE\equiv\triangle DCB\implies$ $\left\{\begin{array}{c} \widehat{CAE}\equiv\widehat{CDB}\\\\ \widehat {AEC}\equiv\widehat{DBC}\end{array}\right|\implies$ the quadrilaterals $ADFC$

and $BEFC$ are cyclically $\implies$ $\widehat{CFA}\equiv\widehat{CDA}\equiv \widehat{CEB}\equiv\widehat{CFB}$ , i.e. $\widehat{CFA}\equiv\widehat{CFB}\implies$ the ray $[FC$ is the bisector of $\widehat{AFB}$ . Nice problem !

PP5. Let $ABC$ be an $C$ -isosceles with the circumcircle $w$ . Let $P\in w$ be between $A$ and $B$ and on the opposite

side of the line $AB$ to $C$ . Denote $D\in PB$ so that $CD\perp PB$ . Show that $PA + PB = 2 \cdot PD$ .

Proof 1. Denote $CA=CB=a$ , $AB=c$ and the midpoint $M$ of $[AB]$ . Thus, $\widehat {CAM}$ $\equiv\widehat {CPD}$ $\Longrightarrow$ $\triangle CAM\sim\triangle CPD\Longrightarrow$ $\frac{CA}{AM}=\frac{CP}{PD}$ $\Longrightarrow$

$PC=\frac{2a}{c}\cdot PD$ . Apply Ptolemy's theorem $:\ a\cdot (PA+PB)=c\cdot PC$ $\Longrightarrow$ $PA+PB=\frac ca\cdot PC=\frac ca\cdot \frac{2a}{c}\cdot PD$ $\Longrightarrow$ $PA+PB=2PD$ .

Proof 2. Denote $E\in PB$ so that $B\in (PE)$ and $BE=PA$ and define $F\in w$ so that $\{C,F\}=CE\cap w$ . Observe that

$\widehat{CBE}\equiv$ $\widehat{CBF}+\widehat{FBE}\equiv$ $\widehat{CBF}+\widehat{PCF}=$ $\widehat{CAP}$ , i.e. $\widehat{CBE}\equiv\widehat{CAP}$ $\implies$ $\triangle CBE\stackrel{(s.a.s)}{\equiv}\triangle CAP\implies$

$CP=CE\implies$ $\triangle CPE$ is $C$ -isosceles, i.e. $PE=2\cdot PD$ . In conclusion, $PA+PB=BE+PB=PE=2\cdot PD$ .

An equivalent enunciation. Let $ABC$ be a triangle with the circumcircle $w$ . Denote the midpoint $M$ of the side $[BC]$ and the diameter

$[NS]$ of $w$ so that line $BC$ separates $A$ and $S$ . Denote $X\in AB$ so that $SX\perp AB$ . Prove that $MX\perp AS$ and $AX=\frac {b+c}{2}$ .

Proof 1. Apply Ptolemy's theorem to $ABSC\ :\ (b+c)\cdot SC=a\cdot SA$ . Prove easily that $\triangle AXS\sim\triangle CMS$ , i.e. $\frac {AX}{CM}=\frac {AS}{SC}$ . Thus, $\frac {b+c}{a}=\frac {SA}{SC}=$

$\frac 2a\cdot AX$ $\implies$ $AX=\frac {b+c}{2}$ . Since $MXBS$ is cyclically, obtain that $\widehat{MXS}\equiv$ $\widehat{MBS}\equiv$ $\widehat{SAX}$ , i.e.in the $X$ -right triangle $AXS$ we have $MX\perp AS$ .

Proof 2. Denote $D\in BC\cap BC$ . Using well-known relations $AD\cdot AS=bc$ and $AD=\frac {2bc\cos\frac A2}{b+c}$ obtain that $AX=AS\cdot \cos\frac A2=$ $\frac {bc\cdot\cos\frac A2}{AD}=$ $\frac {b+c}{2}$ .

PP6. Let $P$ be an interior point of $\triangle ABC$ . Denote $\left\{\begin{array}{c} D\in PA\cap BC\\\ E\in PB\cap CA\\\ F\in PC\cap AB\end{array}\right\|$ . Prove that $\frac{PA}{PD}\frac{PB}{PE}\frac{PC}{PF}=\frac{PA}{PD}+\frac{PB}{PE}+\frac{PC}{PF}+2$ (Euler's relation).

Proof. Denote $\left\{\begin{array}{c} \frac {DB}{DC}=x\\\\ \frac {EC}{EA}=y\\\\ \frac {FA}{FB}=z\end{array}\right\|$ . From the Ceva's theorem obtain that $xyz=1$ and from the Aubel's theorem obtain that $\left\{\begin{array}{c} \frac {PA}{PD}=z+\frac 1y\\\\ \frac {PB}{PE}=x+\frac 1z\\\\ \frac {PC}{PF}=y+\frac 1x\end{array}\right\|$ .

Thus, the relation from the conclusion becomes $\left(z+\frac 1y\right)+\left(x+\frac 1z\right)+\left(y+\frac 1x\right)+2=$ $\left(z+\frac 1y\right)\cdot\left(x+\frac 1z\right)\cdot \left(y+\frac 1x\right)\iff$

$z(1+x)+x(1+y)+y(1+z)+2=$ $z(1+x)\cdot x(1+y)\cdot y(1+z)\iff$ $(x+y+z)+$ $(xy+yz+zx)+2=$

$(1+x)(1+y)(1+z)\iff$ $1+(x+y+z)+$ $(xy+yz+zx)+xyz=$ $(1+x)(1+y)(1+z)$ , what is truly. I replaced $2=1+xyz$ .

PP7. Let $AC$ be a line segment in the plane and $B$ between $A$ and $C$ . Construct isosceles triangles $PAB$ and $QAC$ on one side of the segment $AC$ such that

$m(\angle APB) =m( \angle BQC) = 120^{\circ}$ and an isosceles triangle $RAC$ on the other side of $AC$ such that $m(\angle ARC) = 120^{\circ}.$ Show that $\triangle PQR$ is equilateral.

A equivalent enunciation. Let $ABCD$ be a rhombus with $m(\widehat {ABC})=60^{\circ}$ . For a point $M\in (BD)$ denote the

points $N\in (AB)$ , $P\in (AD)$ so that $MN\parallel AD$ and $MP\parallel AB$ . Prove that $CNP$ is an equilateral triangle.

Proof 1. $\left\{\begin{array}{c} CD=CA\\\\ DP=AN\\\\ \widehat{CDP}\equiv\widehat{CAN}\end{array}\right\|\implies$ $\triangle CDP\stackrel{(s.a.s.)}{\equiv}\triangle CAN\implies$ $\boxed{CP=CN}$ and $\widehat{DCP}\equiv\widehat{ACN}\implies$ $\widehat{DCA}\equiv\widehat{PCN}\implies$ $\boxed{m(\widehat{PCN})=60^{\circ}}$ .

Proof 2. Let the reflections $P_1$ and $N_1$ w.r.t. the line $BD$ of $P$ and $N$ respectively. Thus, $\implies\triangle CP_1P\equiv\triangle PAN\equiv\triangle NN_1C\implies \boxed{CP=PN=NC}$ .

Proof 3 (using complex numbers). Denote $\omega =\cos \frac{2\pi}{3}+i\sin \frac{2\pi}{3}$ , i.e. $\omega ^3=1$ , $\omega ^2=-1-\omega$ . Choose the origin $M(0)$ and

the line $BD$ as the axis $Ox$ . Thus, $B(-1)$ , $D(d)$ , $A(a)$ , $C(c)$ , $N(n)$ , $P(p)$ , where $d\in R$ , $d>0$ and $\{a,c,n,p\}\subset \Im\setminus \Re$ .

$\blacktriangleright NB=NM,\ m(\widehat {BNM})=120^{\circ}\Longrightarrow 0-n=(-1-n)\omega\Longrightarrow n=\frac{\omega}{1-\omega }\ .$
$\blacktriangleright PM=PD,\ m(\widehat {DPM})=120^{\circ}\Longrightarrow d-p=(0-p)\omega \Longrightarrow p=\frac{d}{1-\omega }\ .$
$\blacktriangleright CB=CD,\ m(\widehat {BCD})=120^{\circ}\Longrightarrow -1-c=(d-c)\omega\Longrightarrow c=\frac{-1-d\omega}{1-\omega}\ .$

$\bigstar \ \ \triangle CNP$ is equilateral $\Longleftrightarrow c^2+n^2+p^2=cn+np+pc\Longleftrightarrow$

$\omega^2 +d^2+(1+d\omega )^2=d\omega -\omega (1+d\omega )-d(1+d\omega )\Longleftrightarrow$

$(d^2+d+1)(\omega ^2+\omega +1)=0\iff$ $w^2+w+1=0$ , what is truly.

A generalization of the proposed problem. If $PA=PB\ ,\ QB=QC\ ,\ RA=$ $RC\ ,\ m(\widehat {APB})=m(\widehat {BQC})=m(\widehat {ARC})=\phi$ ,

then $PR=PQ\Longleftrightarrow RP=RQ\Longleftrightarrow QP=QR\Longleftrightarrow \phi=120^{\circ}$ and in this case $PQR$ is an equilateral triangle.

PP8. Let $P$ be an interior point w.r.t. the equlateral $\triangle ABC$ . Know that $AB$ . $\left\{\begin{array}{c} PA=x\\\ PB=y\\\ PC=z\end{array}\right\|$ . Prove that

$2\cdot AB^2=\sum x^2+4\sqrt 3\cdot S(x,y,z)$ , where $S(x,y,z)$ is area of the triangle with the lengths $\{x,y,z\}$ of its sides.

Proof. Denote $AB=l$ . Construct equilateral $\triangle APQ$ so that $AC$ separates $P$ and $Q$ . Observe that $\triangle ABP\stackrel{(s.a.s)}{\equiv}\triangle ACQ\implies$ $BP=CQ=y$ .

Apply the generalized Pytagoras' theorem in $\triangle APC$ , where $m\left(\widehat{APC}\right)=60^{\circ}+\phi$ and $\phi = m\left(\widehat{CPQ}\right)$ . Thus, $l^2=x^2+z^2-2xz\cdot\cos \left(\phi +60^{\circ}\right)=$

$x^2+z^2-xz\cdot\cos\phi+xz\sqrt 3\cdot \sin\phi =$ $x^2+z^2-\frac 12\cdot\left(x^2+z^2-y^2\right)+xz\sqrt 3\cdot\sin\phi =$ $\frac 12\cdot \left(x^2+y^2+z^2\right)+2\sqrt 3\cdot S(x,y,z)\implies$

$\boxed{2\cdot AB^2=x^2+y^2+z^2+4\sqrt 3\cdot S(x,y,z)}$ .

PP9. Let $M\ ,\ N$ be the midpoints of the sides $[AC]\ ,\ [BD]$ respectively in the cyclical quadrilateral $ABCD$ . Prove that $\frac{|AC-BD|}{2}\le MN\le \frac{AC+BD}{2}$ .

Proof. Denote $\left\{\begin{array}{c} AB=a\ ;\ BC=b\ ;\ CD=c\\\ DA=d\ ;\ AC=e\ ;\ BD=f\end{array}\right\|$ and the midpoints $X$ and $Y$ of the sides $AB$ and $BC$ respectively. I"ll use

the well-known relations $ef=ac+bd$ (Ptolemeu) and $e^2+f^2+4\cdot MN^2=$ $a^2+b^2+c^2+d^2$ (Euler). Since $XM=\frac b2$

and $XN=\frac d2$ obtain that $\frac {|b-d|}{2}\ \le\ MN\ \le\ \frac {b+d}{2}$ . From $YM=\frac a2$ and $YN=\frac c2$ obtain that $\frac {|a-c|}{2}\ \le\ MN\ \le\ \frac {a+c}{2}$ .

The relations Euler & Ptolemeu $\Longrightarrow$ $(e-f)^2+4\cdot MN^2=$ $(a-c)^2+(b-d)^2\ \le\ 8\cdot MN^2$ $\Longrightarrow$ $\frac {|e-f|}{2}\ \le\ MN$ .

The Relations Euler & Ptolemeu $\Longrightarrow$ $(e+f)^2+4\cdot MN^2=$ $(a+c)^2+(b+d)^2\ge$ $8\cdot MN^2\ \Longrightarrow$ $\frac {|e+f|}{2}\ \ge\ MN$ .

PP10. Consider three circles $w_k\ ,\ k\in\overline {1,3}$ with only a common point $D$ , i.e. $\{D\}=w_1\cap w_2\cap w_3$ . Prove that

$\left\|\begin{array}{ccc} \{D,A\}=w_2\cap w_3 & , & A_1\in (AD\cap w_1\\\\ \{D,B\}=w_3\cap w_1 & , & B_1\in (BD\cap w_2\\\\ \{D,C\}=w_1\cap w_2 & , & C_1\in (CD\cap w_3\end{array}\ \right\|\ \ \Longrightarrow\ \ \frac {AD}{AA_1}+\frac {BD}{BB_1}+\frac {CD}{CC_1}=1$ .

Proof. Denote $\begin{array}{c} \{B,C_2\}=A_1B\cap w_3\\\\ \{C,B_2\}=A_1C\cap w_2\end{array}$ . Prove easily that $A\in B_2C_2$ and $\begin{array}{c} B_1B_2\parallel A_1C_2\\\\ C_1C_2\parallel A_1B_2\end{array}$ . Denote $\begin{array}{c} B_0\in DB_2\cap A_1C_2\\\\ C_0\in DC_2\cap A_1B_2\end{array}$ .

Thus, $\left\|\begin{array}{c} B_1B_2\parallel A_1C_2\ \Longrightarrow\ \frac {BD}{BB_1}=\frac {B_0D} {B_0B_2}\\\\ C_1C_2\parallel A_1B_2\ \Longrightarrow\ \frac {CD}{CC_1}=\frac {C_0D}{C_0C_2}\end{array}\right\|$ $\Longrightarrow$ $\sum \frac {AD}{AA_1}=\frac {AD}{AA_1}+\frac {B_0D}{B_0B_2}+\frac {C_0D}{C_0C_2}=$ because $D\in AA_1\cap B_0B_2\cap C_0C_2$ in $\triangle A_1B_2C_2$ .

P11. The circles $(O_1)$ and $(O_2)$ are secant in $A$ and $B$ . A line $d$ which pass through the point $A$ cut again $(O_1)$ and $(O_2)$ in $C$ and

$D$ respectively. Let $M$ be a point of $[CD]$ . The line $BM$ cut again the circles $(O_1)$ and $(O_2)$ in $E$ and $F$ . Prove that $CE\parallel DF$ .

Proof. $\left\{\begin{array}{c} ME\cdot MB=MA\cdot MC\\\\ MF\cdot MB=MA\cdot MD\end{array}\right\|\ \implies\ \frac {ME}{MF}=$ $\frac {MC}{MD}\ \implies\ CE\parallel DF$ .

PP12. Let $ABC$ be an acute triangle with $b\ne c$ . Let $H$ be the orthocenter of $\triangle ABC$ . Denote the midpoint $M$ of $[BC]$ . Let $D\in (AB)$ and $E\in (AC)$ be two

points such that $AE=AD$ and $H\in ED$ . Prove that $HM$ is perpendicular to the common chord of the circumscribed circles of triangles $ABC$ and $ADE$ .

Lemma. Denote the circumcircle $c=C(O)$ and the orthocentre $H$ of $\triangle ABC$ , the midpoint $M$ of $[BC]$ , the intersection $N$ between $MH$ and the bisector

of the angle $\widehat {BAC}$ and $D\in AB\ ,\ E\in AC$ so that $H\in DE$ and $AD=AE$ . Then $N$ belongs to the circumcircle $w=C(O_a)$ of $\triangle ADE$ .

A metrical proof. $AH=2R\cos A$ and $P\in c\cap (AO_a\Longrightarrow$ $AP=2R\cos \frac{B-C}{2}\ ,$ $MP=R(1-\cos A)$ . Thus, $\frac{AN}{AP}=\frac{AH}{AH+MP}=$

$\frac{2\cos A}{1+\cos A}\Longrightarrow$ $\boxed {AN=\frac{2\cos A}{1+\cos A}\cdot AP}$ . Denote the point $N'\in AP\cap w$ , i.e. $DN'\perp AB$ . Thus, $\frac{AD}{\sin \left(B+\frac A2\right)}=\frac{AH}{\cos \frac A2}$ $\Longrightarrow$ $AD=\frac{\cos\frac{B-C}{2}}{\cos \frac A2}\cdot AH$

and $AN'=\frac{AD}{\cos \frac A2}$ $\Longrightarrow$ $AN'=\frac{\cos \frac{B-C}{2}}{\cos^2\frac A2}\cdot AH=$ $\frac{2\cos \frac{B-C}{2}}{1+\cos A}\cdot AH=$ $\frac{AP}{R}\cdot\frac{2R\cos A}{1+\cos A}=$ $\frac{2\cos A}{1+\cos A}\cdot AP$ $\Longrightarrow$ $AN'=AN$ $\Longrightarrow$ $N\equiv N'$ ,

i.e. the point $N$ belongs to the circumcircle of $\triangle ADE$ .

Proof of the proposed problem. Denote: $A'\in c\cap (AO$ ; the middlepoint $A_1$ of $[AH]$ ; $N\in w\cap (AO_a$ . From the above lemma results

$N\in MH$ . But $AA_1=HA_1=OM$ (the point $M$ is the middlepoint of $[HA']$ ) , $OA=OA'$ and $O_aA=O_aN$ . Thus, $A_1$ , $O$ , $O_a$ are

the middlepoints of $AH$ , $AA'$ , $AN$ respectively and $N\in HM\equiv HA'$ $\Longrightarrow$ $O_a\in OA_1$ and $OA_1\parallel MH$ . Therefore, $MH\parallel OO_a$ .

Remark. I denote ray $(XY$ without the point $X$ .

PP13 (Jayme). Let $ABCD$ be a square. For $E\in (AD)$ construct externally the square $AEFG$ .

Denote the middlepoint $I$ of $[DE]$ and $K\in BE\cap DG$ . Prove that $IK\cap CE\cap AB\ne\emptyset$ .

Proof (Yetti). Denote $J\in KI\cap AB$ and the midpoint $M$ of $[GB]$ . Thus, $\triangle ADG \sim \triangle ABE$ $\Longrightarrow$ $GK \perp KB$ and $\triangle KBG \sim \triangle ABE\Longrightarrow$

$\frac{GK}{KB} = \frac{GA}{AB}$ $\Longrightarrow$ $KA$ bisects $\angle BKG$ . From $\triangle KBG \sim \triangle KDE$ obtain that $KM \perp KI$ $\Longrightarrow$ $JKI$ is tangent to the circumcircle of $KBG$

and $JK = JA$ $\Longrightarrow$ $\triangle JKG \sim \triangle JBK$ $\Longrightarrow$ $\frac{JA}{JB} = \frac{JK}{JB} = \frac{GK}{KB} = \frac{EA}{CB}$ $\Longrightarrow$ $J \in CE$ .

An easy extension. Let $ABCD$ be a rhombus. For $E\in (AD)$ construct externally the rhombus $AEFG$ .

Denote the middlepoint $I$ of $[DE]$ and $K\in BE\cap DG$ . Prove that $IK\cap CE\cap AB\ne\emptyset$ .

Proof. Denote $L\in CE\cap AB$ . Observe that $\frac {LA}{LB}=\frac {EA}{BC}\ (1)$ and $GB=GA+AB=AE+AD=2\cdot AI\implies$ $\frac {IE}{IA}=\frac {2\cdot IE}{2\cdot IA}\implies$

$\frac {IE}{IA}=\frac {DE}{GB}\ (2)$ . Apply the Menelaus' theorem to the transversal $\overline {DKG}$ for $\triangle AEB\ :\ \frac {GA}{GB}\cdot\frac {KB}{KE}\cdot\frac {DE}{DA}=1$ $\implies$ $\frac {KB}{KE}=\frac {DA}{DE}\cdot\frac {GB}{GA}\ (3)$ .

Therefore, $\frac {LA}{LB}\cdot\frac {KB}{KE}\cdot\frac {IE}{IA}\ \stackrel{(1)\wedge (2)\wedge (3)}{=}\ \frac {EA}{BC}$ $\cdot \left(\frac {DA}{DE}\cdot\frac {GB}{GA}\right)\cdot\frac {DE}{GB}=1\implies$ $L\in IK$ . In conclusion, $IK\cap CE\cap AB\ne\emptyset$ .

PP14. Let $C$ and $D$ be intersection points of circles $w_1=C(O_1)$ and $w_2=C( O_2)$ . A line passing through $D$ intersects

again $O_1$ and $O_2$ at $A$ and $B$ respectively. For $P\in w_1$ and $Q\in w_2$ denote $H\in PD\cap AC$ and $M\in QD\cap BC$ .

Suppose that $O$ is the circumcenter of the triangle $ABC$ . Prove that $OD\perp MH\ \iff\ PQMH$ is cyclically.

Proof. I"ll use the power of a point. Thus, $PQMH$ is cyclically $\iff DP\cdot DH=DM\cdot DQ \iff$ $HD\cdot HP-HD^2=MD\cdot MQ-MD^2\iff$

$DM^2-DH^2=MD\cdot MQ-HP\cdot HD=MB\cdot MC-$ $HC\cdot HA=(OM^2-R^2)-(OH^2-R^2)=$ $OM^2-OH^2\iff$ $OD \perp MH$ .

PP15. Let $ABCD$ be a square and let $H\in (BC)$ and $G\in (CD)$ so that $\left(\widehat{HAG}\right)=45^{\circ}$ . Denote $Y\in AH\cap BD$ and

$Z\in AG\cap BD$ . Prove that $YZGH$ is a cyclical quadrilateral with its circumcenter on $GH$ and $GH^{2}=2\cdot\left(BY^{2}+CZ^{2}\right)$ .

Proof. $\left\{\begin{array}{ccccc} m\left(\widehat{HAZ}\right)=m\left(\widehat{HBZ}\right) & \iff & ABHZ\ \mathrm{is\ cyclically} & \iff & HZ\perp AG\\\\ m\left(\widehat{GAY}\right)=m\left(\widehat{GDY}\right) & \iff & ADGY\ \mathrm{is\ cyclically} & \iff & GY\perp AH\end{array}\right\|$ $\implies$ $YZGH$ is inscribed in the circle

with the diameter $[HG]$ . Remark that $X\in HZ\cap GY\implies$ $AX\perp HG$ , i.e. $X$ is the orthocenter of $\triangle AHG$ . Suppose w.l.o.g. $AB=1$ ,

i.e. $AC=\sqrt 2$ . Denote $BY=x$ and $DG=y$ . Observe that $\tan\widehat{BAH}=x\ ,\ \tan\widehat {CAG}=y$ and $m\left(\widehat{BAH}\right)+\left(\widehat {CAG}\right)=45^{\circ}$

$\implies$ $\frac {x+y}{1-xy}=1\implies$ $(x+y)+xy=1$ , i.e. $(x+1)(y+1)=2$ . Therefore, $\left\{\begin{array}{ccc} \frac {BY}{x}=\frac {YD}{1}=\frac {\sqrt 2}{x+1} & \implies & BY=\frac {x\sqrt 2}{x+1}\\\\ \frac {DZ}{y}=\frac {ZB}{1}=\frac {\sqrt 2}{y+1} & \implies & DZ=\frac {y\sqrt 2}{y+1}\end{array}\right\|$ .

Thus, $GH^2-2\cdot\left(BY^2+DZ^2\right)=$ $(1-x)^2+(1-y)^2-2\cdot\left[\frac {2x^2}{(x+1)^2}+\frac {2y^2}{(y+1)^2}\right]=$ $2(1-x-y)+x^2+y^2-x^2(y+1)^2-$

$y^2(y+1)^2=$ $(x+y)^2-(1-y)^2-(1-x)^2=0$ because $x(y+1)=1-y$ and $y(x+1)=1-x$ . Otherwise, $X\in AC\cap BD\implies$

$\left\{\begin{array}{ccccc} \triangle ABC\sim\triangle AYG\sim\triangle AXD & \implies & \frac{BY}{BX}=\frac {CG}{CD} & \implies & CG=BY\cdot\sqrt 2\\\\ \triangle ADC\sim\triangle AZH\sim\triangle AXB & \implies & \frac{DZ}{DX}=\frac {CH}{CB} & \implies & CH=DZ\cdot\sqrt 2\end{array}\right\|$ $\implies$ $GH^2=$ $CH^2+CG^2=$ $2\cdot\left(BY^2+DZ^2\right)$ .

Lemma. Let $\triangle ABC$ and $D\in BC$ , $E\in CA$ , $F\in AB$ . Denote $P\in AD\cap EF$ . Prove that $\boxed{\frac {PF}{PE} = \frac {DB}{DC}\cdot\frac {AF}{AE}\cdot \frac {AC}{AB}}$ .

Proof. Denote $\left\{\begin{array}{c} m\left(\widehat {DAB}\right)=x\\\\ m\left(\widehat {DAC}\right)=y\end{array}\right\|$ and apply the well-known relations $\left\{\begin{array}{c} \frac {PF}{PE}=\frac {AF}{AE}\cdot\frac {\sin x}{\sin y}\\\\ \frac {DC}{DB}=\frac {AC}{AB}\cdot\frac {\sin y}{\sin x}\end{array}\right\|\ \bigodot\ \implies \frac {PF}{PE} = \frac {DB}{DC}\cdot\frac {AF}{AE}\cdot \frac {AC}{AB}$ .

Application. Let $\triangle ABC$ with the incircle $w=C(I)$ and the points $D\in (BC)$ , $E\in (CA)$ , $F\in (AB)$ so

that $\frac {DB}{DC}=\frac {BF}{CE}$ . For a point $P\in AI$ denote $K\in DF\cap BP$ , $L\in DE\cap CP$ . Prove that $LK\parallel EF$ .

Proof. Denote $X\in AI\cap BC$ , $S\in BP\cap CA$ , $R\in CP\cap AB$ . Observe that $P\in AX\cap BS\cap CR\iff$

$\frac {XB}{XC}\cdot\frac {SC}{SA}\cdot\frac {RA}{RB}=1\iff$ $\frac {AB}{AC}\cdot\frac {SC}{SA}\cdot\frac {RA}{RB}=1\ (1)$ . Apply the upper lemma : $\left\{\begin{array}{c} \frac {LE}{LD}=\frac {RA}{RB}\cdot\frac {CE}{CD}\cdot\frac{CB}{CA}\\\\ \frac {KF}{KD}=\frac {SA}{SC}\cdot\frac {BF}{BD}\cdot\frac {BC}{BA}\end{array}\right\|$ .

Using the relations $\frac {DB}{DC}=\frac {BF}{CE}$ from the hypothesis and the relation $(1)$ obtain that $\frac {LE}{LD}=\frac {KF}{KD}$ , i.e. $LK\parallel EF$ .

PP16. Let $ABC$ be a triangle and let $D\in (BC)$ , $E\in (CA)$ and $F\in (AB)$ be three points such that $AD\cap BE\cap CF\ne\emptyset$ .

Let the points $M\in (EF)$ , $N\in (FD)$ and $P\in (DE)$ . Prove that $AM\cap BN\cap CP\ne\emptyset\iff$ $DM\cap EN\cap FP\ne\emptyset$ .

Proof. Denote $\left\{\begin{array}{c} X\in AM\cap BC\\\ Y\in BN\cap CA\\\ Z\in CP\cap AB\end{array}\right\|$ . Observe that $AD\cap BE\cap CF\ne\emptyset\iff$ $\frac {AF}{AE}\cdot \frac {CD}{CA}\cdot\frac {CD}{CA}=1$ . Apply the upper well-known

remarkable lemma : $\left\{\begin{array}{c} \frac {MF}{ME}=\frac {XB}{XC}\cdot\frac {AF}{AE}\cdot \frac {AC}{AB}\\\\ \frac {PE}{PD}=\frac {ZA}{ZB}\cdot \frac {CD}{CA}\cdot\frac {CA}{CB}\\\\ \frac {ND}{NF}=\frac {YC}{YA}\cdot\frac {CD}{CA}\cdot\frac {BC}{BA}\end{array}\right\|\ \bigodot\ \implies\ \frac {MF}{ME}\cdot\frac {PE}{PD}\cdot\frac {ND}{NF}=\frac {XB}{XC}\cdot\frac {ZA}{ZB}\cdot \frac {YC}{YA}$ . In conclusion,

$DM\cap EN\cap FP\ne\emptyset\iff\frac {MF}{ME}\cdot\frac {PE}{PD}\cdot\frac {ND}{NF}=1\iff$ $\frac {XB}{XC}\cdot\frac {ZA}{ZB}\cdot \frac {YC}{YA}=1\iff$ $AM\cap BN\cap CP\ne\emptyset$ .

PP17. Let $\triangle ABC$ with the incenter $I$ , the $A$ -exincenter $I_a$ and $c>b$ . Denote $D\in (AB)$ and $\{E,F\}\subset AA$ (the tangent in a point $A$ to the

circumcircle of $\triangle ABC$ ) so that $AD=AE=AF=b$ and the sideline $AB$ doesn't separate $C$ , $E$ . Prove that $I\in DE$ and $I_a\in FD$ .

Proof 1. $X\in AI\cap DE$ and $Y\in BC\cap AI\ \implies\ m(\angle BAE)=A+B$ , $YD=YC=\frac {ab}{b+c}$ , $m(\angle ADY)=C$ .

Thus, $DY\parallel AE\ \implies\ \frac {XA}{XY}=\frac {AE}{DY}=$ $\frac {b}{\frac {ab}{b+c}}\ \implies\ \frac {XA}{XY}=\frac {b+c}{a}$ $\stackrel{\mathrm{Van\ Aubel}}{\ \ \implies\ \ }X: =I$ . In conclusion, $I\in DE$ .

Proof 2. Denote $E_0\in DI\cap AA$ . Since $m(\angle BIC)=m(\angle BDC)=90^{\circ}+\frac A2$ it follows that $BDIC$ is a cyclical quadrilateral $\Longrightarrow$ $m(\angle E_0IC)=B\implies$

$m(\angle E_0AC)=B$ $\implies$ $AICE_0$ is a yclical quadrilateral. Since $m(\angle AID)=(\angle AIC)$ , then the chords $AC=AE_0$ are equally $\Longrightarrow$ $E \equiv E_0,$ as desired.

Remark. Similarly, we show that $DF$ passes through the $A$ -excenter of $\triangle ABC$ . Denote $F_0\in DI_a\cap AA$ . Since $m(\angle BI_aC)=m(\angle ADC)=90^{\circ}-\frac A2$

it follows that $BDCI_a$ is a cyclical quadrilateral $\Longrightarrow$ $m(\angle F_0I_aC)=B$ and $m(\angle F_0AC)=180^{\circ}-B$ $\implies$ $ACI_aF_0$ is a yclical quadrilateral.

Since $m(\angle AI_aD)=(\angle AI_aC)$ , then the chords $AC=AF_0$ are equally $\Longrightarrow$ $F \equiv F_0,$ as desired.

Proof 3. I"ll well-known equivalence that $\boxed{I\in DE\iff [ADE]=[ADI]+[AIE]}$ . Indeed, $[ADE]=[ADI]+[AIE]\iff$

$b^2\sin C=b\cdot AI\cdot\left[\sin\frac A2+\sin\left(B+\frac A2\right)\right]\iff$ $b\cdot\sin C=$ $\sqrt{\frac {bc(s-a)}{s}}\cdot 2\cos \frac B2\cos \frac C2\iff$

$b\cdot\sin\frac C2=\sqrt{\frac {bc(s-a)}{s}}\cdot\cos \frac B2\iff$ $b\cdot\sqrt {\frac {(s-a)(s-b)}{ab}}=\sqrt {\frac {bc(s-a)}{s}}\cdot\sqrt {\frac {s(s-b)}{ac}}$ , what is truly.

$\boxed{I_a\in DF\iff [AFI_a]=[AFD]+[ADI_a]}\iff$ $[AFD]=[AFI_a]-[ADI_a]\iff$

$b^2\sin C=b\cdot AI_a\cdot\left[\sin\left(\frac A2+C\right)-\sin \frac A2\right]\iff$ $b\cdot\sin C=\sqrt{\frac {bcs}{s-a}}\cdot 2\sin\frac B2\sin\frac C2\iff$

$b\cdot\cos\frac C2=\sqrt{\frac {bcs}{s-a}}\cdot\sin \frac B2\iff$ $b\cdot\sqrt {\frac {s(s-c)}{ab}}=\sqrt {\frac {bcs}{s-a}}\cdot\sqrt {\frac {(s-a)(s-c)}{ac}}$ , what is truly.

PP18. Let $ABCD$ be a trapezoid so that $AB\parallel CD$ and $AB\ne CD$ . Denote $\left\{\begin{array}{ccc} AB=a & ; & CD=b\\\ AC=e & ; & BD=f\\\ AD=m & ; & BC=n\end{array}\right\|$ . Prove that

$\frac {e^2-f^2}{m^2-n^2}=\frac {a+b}{a-b}$ and the intersection $I\in AC\cap BD$ belongs to the radical axis of the circles with the diameters $[AD]$ , $[BC]$ .

Proof 1. Denote the points $\{X,Y,Z\}\subset AB$ so that $CX\parallel AD$ , $CZ\parallel DB$ and $CY\perp AB$ . Observe that $CX=m$ , $CZ=f$ , $AX=BZ=b$ ,

$XB=a-b$ and $YX-YB=\left[(YX+XA)-(YB+BZ)\right]\implies$ $\boxed{YX-YB=YA-YZ}\ (*)$ . Therefore, $e^2-f^2=CA^2-CZ^2=$

$YA^2-YZ^2=(YA-YZ)(YA+YZ)$ $\implies$ $\boxed{e^2-f^2=(YA-YZ)(a+b)}\ (1)$ and $m^2-n^2=CX^2-CB^2=$ $YX^2-YB^2=$

$(YX-YB)(YX+YB)$ $\implies$ $\boxed{m^2-n^2=(YX-YB)(a-b)}\ (2)$ . Thus, using the relation $(*)$ obtain that $\boxed{\frac {e^2-f^2}{m^2-n^2}=\frac {a+b}{a-b}}\ (3)$ . Denote the

power $p_w(X)$ of $X$ w.r.t. the circle $w$ , $m\left(\widehat{AID}\right)=\phi$ , the midpoints $M$ , $N$ of $[AD]$ , $[BC]$ respectively and the circles $w_1$ , $w_2$ with the diameters $[AD]$ , $[BC]$

respectively. Thus, $\left\{\begin{array}{ccc} 4\cdot p_{w_1}(I)=4\cdot\left[IM^2-\left(\frac {AD}{2}\right)^2\right] & \implies & p_{w_1}(I)=\frac 12\cdot\left(IA^2+ID^2-AD^2\right)=IA\cdot ID\cdot\cos \phi\\\\ 4\cdot p_{w_2}(I)=4\cdot\left[IN^2-\left(\frac {BC}{2}\right)^2\right] & \implies & \cdot p_{w_2}(I)=\frac 12\cdot\left(IB^2+IC^2-BC^2\right)=IB\cdot IC\cdot \cos\phi\end{array}\right\|$ $\implies$ $p_{w_1}(I)=p_{w_2}(I)$ .

Proof 2. I"ll use same notations from the first proof. Denote $U\in AC\cap w_1$ and $V\in BD\cap w_2$ . Observe that $\frac {IA}{IC}=\frac {IB}{ID}\ (1)$ and $CUVD$ is inscribed in

the circle with the diameter $[CD]$ , i.e. $IU\cdot IC=IV\cdot ID\ (2)$ . The product of the relations $(1)$ and $(2)$ get $IU\cdot IA=IV\cdot IB$ , i.e. $p_{w_1}(I)=p_{w_2}(I)$ .

Observe that $AY-YZ=(\underline{AX}+XY)-(YB+\underline{BZ})=$ $XY-YB\implies \boxed{AY-YZ=XY-YB}\ (*)$ . Apply the generalized Pythagoras' theorem

to the triangles $\left\{\begin{array}{cccc} \triangle ACZ\ : & e^2=f^2+(a+b)^2-2(a+b)\cdot YZ & \implies & e^2-f^2=(a+b)(AY-YZ)\\\\ \triangle BCX\ : & m^2=n^2+(a-b)^2-2(a-b)\cdot YB &\implies & m^2-n^2=(a-b)(XY-YB)\end{array}\right\|$ $\stackrel{(*)}{\implies}\ \boxed{\ \frac {e^2-f^2}{m^2-n^2}=\frac {a+b}{a-b}\ }$ .

Otherwise. Apply the Stewart's theorem to the cevians $CX$ and $CZ$ in the triangles $ACB$ and $XCZ$ respectively :

$\left\{\begin{array}{ccc} CA^2\cdot XB+CB^2\cdot AX=AX\cdot XB\cdot AB+CX^2\cdot AB & \implies & e^2(a-b)+n^2b=ab(a-b)+m^2a\\\\ CX^2\cdot BZ+CZ^2\cdot XB=XB\cdot BZ\cdot XZ+ CB^2\cdot XZ & \implies & f^2(a-b)+m^2b=ab(a-b)+n^2a\end{array}\right\|\ \implies$

$\left\{\begin{array}{cccc} \ominus\ : & (e^2-f^2)(a-b)-b(m^2-n^2)=a(m^2-n^2) & \implies & \boxed{\frac {e^2-f^2}{m^2-n^2}=\frac {a+b}{a-b}}\\\\ \oplus\ : & (e^2+f^2)(a-b)+b(m^2+n^2)=2ab(a-b)+a(m^2+n^2) & \implies & e^2+f^2=2ab+m^2+n^2\ (\mathrm{\underline{Euler}'s \ relation})\end{array}\right\|$ .

Remark.. $\left\{\begin{array}{c} m^2+ab=\frac {a\cdot e^2+b\cdot f^2}{a+b}\\\\ n^2+ab=\frac {b\cdot e^2+a\cdot f^2}{b+a}\end{array}\right\|$ $\implies$ $\min\{e,f\}\le\min\left\{\sqrt {m^2+ab}\ ,\ \sqrt{n^2+ab}\right\}$ $\le\max\left\{\sqrt {m^2+ab}\ ,\ \sqrt{n^2+ab}\right\} \le \max\{e,f\}$ .

With other words, $\left[\sqrt {m^2+ab}\ ,\ \sqrt{n^2+ab}\right]\cup\left[\sqrt {n^2+ab}\ ,\ \sqrt{m^2+ab}\right]\subset [e,f]\cup [f,e]$ .

This post has been edited 379 times. Last edited by Virgil Nicula, Nov 19, 2015, 9:15 PM

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The element of Mathematics is dense here.
I love the style of your posts.
Thanks for writing the blog! I can learn Geometry from here too.
by epsilonist, Jun 27, 2011, 6:00 AM
Thank you very much for this blog.
by Affine, May 7, 2011, 6:20 PM
You seem to have solved every existing problem.
by Goutham, May 2, 2011, 10:59 AM
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!
by Caspar, Apr 11, 2011, 2:36 PM
Thank you a lot dear Virgil for an interesting and instructive blog.
by vittasko, Mar 12, 2011, 9:42 AM
Interesting Blog
by ShahinBJK, Mar 12, 2011, 5:07 AM
Interesant, d-le Nicula.
by silent_control, Mar 3, 2011, 6:00 PM
Yours is the largest blog full of math equations as far as I have known! Congrats!
by Goutham, Jan 30, 2011, 2:06 PM
15001th view.
by fries4guys, Jan 20, 2011, 4:57 PM
Thanks to all !
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

323. Easy, nice and interesting problems for high school.

by Virgil Nicula, Oct 17, 2011, 8:29 AM

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