382. Peruan "jewels" (Israel Diaz, M. O. Sanchez a.s.o.)

by Virgil Nicula, Jul 30, 2013, 2:00 PM

PP1 (Israel Diaz Acha). Let an acute $\triangle ABC$ with circumcircle $w=C(O,R)$ . Denote the midpoints $M$ , $N$ ,

$P$ of $[BC]$ , $[CA]$ , $[AB]$ respectively. Suppose that $OM=\sqrt 2$ , $ON=1$ , $OP=\sqrt 5$ . Find $R$ and $\widehat{ACB}$ .

Proof 1. Apply the Ptolemy's relation to $:\ \left\{\begin{array}{cc} ONAP\ : & -aR+b\sqrt 5+c=0\\\\ OMBP\ : & a\sqrt 5-bR+c\sqrt 2=0\\\\ OMCN\ : & a+b\sqrt 2-cR=0\end{array}\right\|$ . Thus, $\left|\begin{array}{ccc} -R & \sqrt 5 & 1\\\\ \sqrt 5 & -R & \sqrt 2\\\\ 1 & \sqrt 2 & -R\end{array}\right|=0\ \iff\ R^3$ $-8R-2\sqrt {10}=0\iff$

$\left(R-\sqrt {10}\right)\left(R^2+R\sqrt {10}+2\right)=0\iff \boxed{R=\sqrt{10}}\ (*)$ . Therefore, $\left\{\begin{array}{c} -aR+b\sqrt 5+c=0\\\\ a\sqrt 5-bR+c\sqrt 2=0\end{array}\right\|\iff$ $\frac {a}{\left|\begin{array}{cc} \sqrt 5 & 1\\\ -R & \sqrt 2\end{array}\right|}=$ $\frac {b}{\left|\begin{array}{cc} 1 & -R\\\ \sqrt 2 & \sqrt 5\end{array}\right|}=$ $\frac {c}{\left|\begin{array}{cc} -R & \sqrt 5\\\ \sqrt 5 & -R\end{array}\right|}\stackrel{(*)}{\iff}$

$\frac a{2\sqrt {10}}=\frac b{3\sqrt 5}=\frac c5$ $\iff$ $\boxed{\frac a{2\sqrt 2}=\frac b3=\frac c{\sqrt 5}}$ . In conclusion, $\cos C=\frac {a^2+b^2-c^2}{2ab}=$ $\frac {8+9-5}{12\sqrt 2}=\frac 1{\sqrt 2}\implies$ $\cos C=\frac {\sqrt 2}{2}\implies$ $\boxed{C=45^{\circ}}$ .

Proof 2. Apply the Ptolemy's relation in the quadrilateral $ANOP\ :\ c+b\sqrt 5=aR$ . Prove easily that $\boxed{a^2+8=b^2+4=c^2+20=4R^2}\ (1)$ . Therefore,

$c+b\sqrt 5=aR\ \stackrel{(1)}{\iff}\ \sqrt {R^2-5}+$ $\sqrt {5\left(R^2-1\right)}=R\sqrt{R^2-2}\iff$ $6R^2-10+2\sqrt {5\left(r^4-6R^2+5\right)}=R^4-2R^2\iff$ $R^4-8R^2+10=2\sqrt{5\left(R^4-6R^2+5\right)}\iff$

$\left(t^2-8t+10\right)^2=20\left(t^2-6t+5\right)\iff$ $t^4-16t^3+64t^2-40t=0\iff$ $9(t-10)\left(t^2-6t+4\right)=0$ , where $\boxed{R^2=t>5}$ . Thus, $\boxed{R=\sqrt {10}}$ and from the relation $(1)$

obtain that $\left\{\begin{array}{ccc} a & = & 4\sqrt 2\\\\ b & = & 6\\\\ c & = & 2\sqrt 5\end{array}\right\|$ , i.e. $\boxed{\frac a{2\sqrt2}=\frac b3=\frac c{\sqrt 5}=2}$ . In conclusion, $\cos C=\frac {a^2+b^2-c^2}{2ab}=$ $\frac {8+9-5}{12\sqrt 2}=\frac 1{\sqrt 2}\implies$ $\cos C=\frac {\sqrt 2}{2}\implies$ $\boxed{C=45^{\circ}}$ .

Proof 3. Prove easily $\boxed{\frac {a^2}4+2=\frac {b^2}4+1=\frac {c^2}4+5=R^2}\ (1)$ and Pythagoras' theorem $:\ \left\{\begin{array}{ccc} PN^2=\frac {a^2}{4}=5+1+2\sqrt 5\cos A & \stackrel{(1)}{\implies} & R^2=8+2\sqrt 5\cos A\\\\ PM^2=\frac {b^2}{4}=5+2+2\sqrt {10}\cos B & \stackrel{(1)}{\implies} & R^2=8+2\sqrt {10}\cos B\\\\ MN^2=\frac {c^2}{4}=2+1+2\sqrt 2\cos C & \stackrel{(1)}{\implies} & R^2=8+2\sqrt 2\cos C\end{array}\right\|$ $\implies$

$R^2-8=2\sqrt 5\cos A=2\sqrt {10}\cos B=2\sqrt 2\cos C\implies$ $\frac {\cos A}{\sqrt 2}=\frac {\cos B}{1}=\frac {\cos C}{\sqrt 5}=\lambda$ , where $\lambda=\frac {R^2-8}{2\sqrt{10}}$ . Apply identity $\boxed{\ \sum\cos A+2\prod\cos A=1\ }\ :$

$2\lambda^2+\lambda^2+5\lambda^2+2\lambda^3\sqrt {10}=1\iff$ $2\lambda^3\sqrt {10}+8\lambda^2-1=0\iff$ $\left(\lambda\sqrt {10}-1\right)\left(2\lambda^2+\lambda\sqrt {10}+1\right)=0\iff$ $\lambda=\frac 1{\sqrt {10}}\iff$ $\left\{\begin{array}{c} R=\sqrt{10}\\\\ \cos C=\frac {\sqrt 2}{2}\end{array}\right\|\implies$ $\boxed{C=45^{\circ}}$ .

PP2 (Miquel Ochoa Sanchez). Let an equilateral $\triangle ABC$ and $\left\{\begin{array}{ccc} E\in (AC) & ; & F\in (AF)\\\\ G\in BE\cap CF & ; & L\in EF\cap AG\end{array}\right\|$ . Prove that $[AEGF]=[BGC]\implies \frac {LF}{LE}=\frac {GB}{GC}$ .

Proof. $[AEGF]=[BGC]\implies [ABE]=[BCF]\implies$ $AB\cdot AE=BC\cdot BF\implies$ $AE=BF\implies$ $ABE=BCF\implies \widehat{BFC}\equiv\widehat{BEA}\implies$ $AEGF$ is cyclic $\implies$

$\triangle CEG\sim\triangle CFA\implies$ $\boxed{\frac {EG}{FA}=\frac {CG}{CA}}\ (*)$ . Apply the Menelaus' theorem to $\overline{ALG}/\triangle BEF\ :\ \frac {AF}{AB}\cdot\frac {GB}{GE}\cdot\frac {LE}{LF}=1\stackrel{(*)}{\implies}$ $\frac {EG}{GC}\cdot\frac {GB}{GE}\cdot\frac {LE}{LF}=1\implies$ $\frac {LF}{LE}=\frac {GB}{GC}$ .

An easy extension. Let an $A$ -isosceles $\triangle ABC$ be and $\left\{\begin{array}{ccc} E\in (AC) & ; & F\in (AF)\\\\ G\in BE\cap CF & ; & L\in EF\cap AG\end{array}\right\|$ so that $AEGF$ is cyclic. Prove that $\frac {LF}{LE}=\frac {GB}{GC}$ .

Proof 1. cyclic $AEGF\implies$ $\triangle CEG\sim\triangle CFA\implies$ $\boxed{\frac {EG}{FA}=\frac {CG}{CA}}\ (*)$ . Menelaus' to $\overline{ALG}/\triangle BEF\ :\ \frac {AF}{AB}$ $\cdot\frac {GB}{GE}\cdot\frac {LE}{LF}=1\stackrel{(*)}{\implies}$ $\frac {EG}{GC}\cdot\frac {GB}{GE}\cdot\frac {LE}{LF}=1\implies$ $\frac {LF}{LE}=\frac {GB}{GC}$ .

Proof 2. $AEGF$ is cyclic $\implies$ $\left\{\begin{array}{ccc} \triangle CEG\sim\triangle CFA & \implies & \frac {FA}{EG}=\frac {CA}{CG}\\\\ \triangle BFG\sim\triangle BEA & \implies & \frac {FG}{EA}=\frac {BG}{BA}\end{array}\right\|\ (*)$ . Apply $\frac {LF}{LE}=\frac {FA\cdot FG}{EA\cdot EG}\implies$ $\frac {LF}{LE}=\frac {FA}{EG}\cdot\frac {FG}{EA}\stackrel{(*)}{\implies}$ $\frac {LF}{LE}=\frac {CA}{CG}\cdot\frac {BG}{BA}\implies$ $\frac {LF}{LE}=\frac {GB}{GC}$

Lemma. Let $ABC$ be a triangle so that $c<a$ , i.e. exists $D\in (BC)$ so that $CD=AB$ . Suppose that $C<\frac {\pi}{4}$ . Then $\boxed{B=2C\iff DA=DC}$ .

Proof. Denote $m\left(\widehat{CAD}\right)=x$ and observe that $CD=AB\iff$ $\frac {CD}{AD}=\frac {AB}{AD}\iff$ $\frac{\sin x}{\sin C}=\frac {\sin (x+C)}{\sin B}$ . Thus, $B=2C\iff$ $\frac{\sin x}{\sin C}=\frac {\sin (x+C)}{\sin 2C}\iff$

$\sin x\sin 2C=\sin C\sin (x+C)\iff$ $\cos (2C-x)-\cos (2C+x)=\cos x-\cos (2C+x)\iff$ $\cos (2C-x)=\cos x\iff x=C\iff$ $DA=DC$ .

Now suppose that $DA=DC$ . Then $DA=AB$ and $x=C\ ,\ B=m\left(\widehat{ADB}\right)\implies$ $B=x+C=2C\implies$ $B=2C$ .

PP3 (Miquel Ochoa Sanchez). Let $ABC$ be a triangle with $B=40^{\circ}$ and $C=20^{\circ}$ . Denote $\left\{\begin{array}{cc} D\in (BC)\ : & DC=AB\\\\ E\in (AC)\ : & CE=BD\end{array}\right\|$ . Prove that $m\left(\widehat{CED}\right)=30^{\circ}$ .

Proof 1. Denote $m\left(\widehat{CED}\right)=x$ , Thus, $DC=c$ , $CE=a-c$ and $\frac {CD}{\sin \widehat{CED}}=\frac {CE}{\sin\widehat{CDE}}\iff$ $\frac {c}{\sin x}=\frac {a-c}{\sin\left(x+20^{\circ}\right)}=\frac {a}{\sin x+\sin\left(x+20^{\circ}\right)}$ . Since

$\frac {c}{\sin 20^{\circ}}=\frac a{\sin 60^{\circ}}$ obtain that $\frac {\sin 20^{\circ}}{\sin x}=\frac {\sin 60^{\circ} }{\sin x+\sin\left(x+20^{\circ}\right)}\iff$ $\frac {\sin 20^{\circ}}{\sin x}=\frac {\cos 30^{\circ}}{2\cos 10^{\circ}\sin\left(x+10^{\circ}\right)}\iff$ $\sin \left(x+10^{\circ}\right)\left(\sin 30^{\circ}+\sin 10^{\circ}\right)=\sin x\cos 30^{\circ}\iff$

$\sin \left(x+10^{\circ}\right)\left(1+2\sin 10^{\circ}\right)=2\sin x\cos 30^{\circ}\iff$ $\sin \left(x+10^{\circ}\right)+\cos x-\cos \left(x+20^{\circ}\right)=\sin \left(x+30^{\circ}\right)+\sin \left(x-30^{\circ}\right)\iff$

$\sin \left(x-30^{\circ}\right)+\sin \left(x+30^{\circ}\right)-\sin \left(90^{\circ}-x\right)+$ $\cos \left(x+20^{\circ}\right)-\cos \left(80^{\circ}-x\right)=0\iff$ $\sin \left(x-30^{\circ}\right)+2\sin \left(x-30^{\circ}\right)\cos 60^{\circ}+2\sin 50^{\circ}\sin \left(30^{\circ}-x\right)=0\iff$

$\sin \left(x-30^{\circ}\right)\left(1-\sin 50^{\circ}\right)=0\iff$ $x=30^{\circ}\iff$ $m\left(\widehat{CED}\right)=30^{\circ}$ .

Proof 2. Construct $\triangle BDF$ so that $BC$ separates $A$ , $F$ and $\triangle BDF\sim\triangle CED$ . Hence $m\left(\widehat{BDF}\right)=m\left(\widehat{CED}\right)=x$ , $m\left(\widehat{ABF}\right)=60^{\circ}$ , $DF=ED$ , $BF=CD=AB$ ,

i.e. $\triangle ABF$ is equilateral. Using upper lemma obtain $DA=DC$ , i.e. $m\left(\widehat{CAD}\right)=20^{\circ}$ , $m\left(\widehat{DAF}\right)=m\left(\widehat{CDA}\right)=40^{\circ}$ and $ADF$ is an $A$ -isosceles triangle, i.e.

$m\left(\widehat{AFD}\right)=m\left(\widehat{ADF}\right)=70^{\circ}\implies$ $x+40^{\circ}=70^{\circ}\iff$ $x=30^{\circ}\iff$ $m\left(\widehat{CED}\right)=30^{\circ}$ .

An easy extension. Let $\triangle ABC$ with $D\in (BC)$ , $DC=c$ . Let $E\in (AC)$ for which $CE=a-c$ and $m\left(\widehat{CED}\right)=x$ . Prove that $B=2C\iff \tan x=\frac {\sin C}{2\cos 2C-\cos C}$ .

Proof. Apply the theorem of Sines in $\triangle CED\ :\ \frac {c}{\sin x}=$ $\frac {a-c}{\sin (x+C)}=\frac a{\sin x+\sin (x+C)}$ . Therefore, $B=2C\iff$ $\frac {\sin C}{\sin x}=\frac {\sin 3C}{\sin x+\sin (x+C)}\iff$

$\sin x(1+2\cos 2C)=\sin x+\sin (x+C)\iff$ $2\sin x\cos 2C=\sin (x+C)\iff$ $2\tan x\cos 2C=\tan x\cos C+\sin C\iff$ $\tan x=\frac {\sin C}{2\cos 2C-\cos C}$ .

Particular case. $C=20^{\circ}\implies$ $\frac {\sin C}{2\cos 2C-\cos C}=$ $\frac {\sin 20^{\circ}}{2\cos 40^{\circ}-\cos 20^{\circ}}=$ $\frac{\sin 20^{\circ}}{2\cos (60^{\circ}-20^{\circ})-\cos 20^{\circ}}=$ $\frac {\sin 20^{\circ}}{\cos 20^{\circ}+\sin 20^{\circ}\sqrt 3-\cos 20^{\circ}}=$

$\frac {\sin 20^{\circ}}{\sin 20^{\circ}\sqrt 3}=\frac 1{\sqrt 3}$ . Thus, $B=40^{\circ}\ \wedge\ C=20^{\circ}\iff$ $\tan x=\frac 1{\sqrt 3}$ , i.e. $m\left(\widehat{CED}\right)=30^{\circ}$ .

PP4. Let an equilateral $\triangle ABC$ with incircle $w$ for which denote $E\in AC\cap w$ , $F\in AB\cap w$ . Let $X\in (AF)$ , $Y\in (AE)$ so that $XY$ is tangent to $w$ . Prove that $\frac {XA}{XB}+\frac {YA}{YC}=1$ .

Proof. Let $T\in XY\cap w$ , $\left\{\begin{array}{c} XF=x\\\\ YE=y\end{array}\right\|$ . Suppose w.l.o.g. $AB=2$ . Thus, $XY=x+y$ , $\left\{\begin{array}{c} XA=1-x\ ;\ YA=1-y\\\\ XB=1+x\ ;\ YC=1+y\end{array}\right\|$ . Apply Pythagoras' theorem to $\triangle XAY\ :$

$(1-x)^2$ $+(1-y)^2=(1-x)(1-y)+(x+y)^2\iff$ $\boxed{(x+y)+3xy=1}\ (*)$ . So $\frac {XA}{XB}+\frac {YA}{YC}=\frac {1-x}{1+x}+\frac {1-y}{1+y}=$ $\frac {2(1-xy)}{1+(x+y)+xy}\ \stackrel{(*)}{\implies}\ \boxed{\frac {XA}{XB}+\frac {YA}{YC}=1}$ .

Remark (Ruben Auqui). $\frac {1}{XB}+\frac {1}{YC}=\frac {1}{1+x}+\frac 1{1+y}=\frac {2+x+y}{1+x+y+xy}\stackrel{(*)}{=}\frac 32\implies$ $\boxed{\frac {1}{XB}+\frac {1}{YC}=\frac 3{AB}}$ .

PP5. Let a trapezoid $ABCD$ with $AB\parallel CD$ , $P\in CB\cap AD$ . For $X\in (AB)$ let $Y\in CD\cap PX$ , $R\in AY\cap BD$ , $T\in PR\cap AB$ . Prove that $\frac 1{AT}=\frac 1{AB}+\frac 1{AX}$ .

Proof. Menelaus' to $\overline{PRT}/\triangle ABD\ :\ \frac {PD}{PA}\cdot$ $\frac {TA}{TB}\cdot\frac{RB}{RD}=1\iff$ $\frac {DY}{AX}\cdot \frac {TA}{TB}\cdot\frac {AB}{DY}=1\iff$ $\frac {TB}{TA}=\frac {AB}{AX}\iff$ $\frac {AB}{AT}=1+\frac {TB}{AT}=1+\frac {AB}{AX}\iff$ $\frac 1{AT}=\frac 1{AB}+\frac 1{AX}$ .

PP6. Let $\triangle ABC$ and $A$ -bisector $AD$ , where $D\in (BC)$ . For $P\in (AD)$ denote $\left\{\begin{array}{c} R\in AB\cap CP\\\\ Q\in AC\cap BP\end{array}\right\|$ . Prove that $[ARPQ]=[BPC]\ \iff\ \frac {PA}{PD}=\frac {b+c}{\sqrt{bc}}$ .

Proof. Ceva's theorem to $P\ :\ \frac {DB}{DC}\cdot \frac {QC}{QA}\cdot \frac {RA}{RB}=1\iff$ $\boxed{\frac {QA}{QC}=\frac cb\cdot\frac {RA}{RB}}\ (*)$ . Thus, $[ARPQ]=[BPC]\iff$ $[ABQ]=[BCR]\iff$ $\frac {[ABQ]}{[ABC]}=\frac {[BCR]}{[ACB]}\iff$

$\frac {AQ}{AC}=\frac{BR}{AB}\iff$ $\frac {QA}{QC}=\frac {RB}{RA}\ \stackrel{(*)}{\iff}\ \boxed{\frac {QA}{QC}=\frac {RB}{RA}=\sqrt {\frac cb}}\ (1)$ . Van Aubel's $:\ \frac {PA}{PD}=\frac {RA}{RB}+\frac {QA}{QC}=$ $\sqrt{\frac bc}+\sqrt {\frac cb}\implies$ $\frac {PA}{PD}=\frac {b+c}{\sqrt{bc}}$ .

An easy extension. Let $\triangle ABC$ and $D\in (BC)$ so that $\frac {DB}{DC}=m\in\mathbb R^*_+$ . For $P\in (AD)$ denote $\left\{\begin{array}{c} R\in AB\cap CP\\\\ Q\in AC\cap BP\end{array}\right\|$ . Prove that $[ARPQ]=[BPC]\ \iff\ \frac {PA}{PD}=\frac {m+1}{\sqrt m}$ .

Particular case. If $P:=L$ - Lemoine's point, then $\frac {DB}{DC}=m=\frac {c^2}{b^2}$ and $[ARPQ]=[BPC]\iff \frac {PA}{PD}=\frac {b^2+c^2}{bc}$ .

PP7 (Miguel Ochoa Sanchez). Let an $A$ -isosceles $\triangle ABC$ and the interior point $P$ so that $\left\{\begin{array}{c} m\left(\widehat{PAB}\right)=12^{\circ}\\\\ m\left(\widehat{PAC}\right)=24^{\circ}\\\\ m\left(\widehat{PBA}\right)=18^{\circ}\end{array}\right\|$ . Prove that $m\left(\widehat{PCA}\right)=30^{\circ}$ .

Proof 1 (trigonometric). Denote $m\left(\widehat{PCA}\right)=x$ . Observe that $m\left(\widehat{PBC}\right)=54^{\circ}$ and $m\left(\widehat{PCB}\right)=72^{\circ}-x$ . Apply trigonometric form of Ceva's theorem:

$\sin \widehat{PAB}\sin\widehat{PBC}\sin \widehat{PCA}=\sin\widehat{PAC}\sin\widehat{PBA}\sin\widehat{PCB}\iff$ $\sin 12^{\circ}\sin 54^{\circ}\sin x=\sin 24^{\circ}\sin 18^{\circ}\sin (18^{\circ}+x)\iff$

$\cos 36^{\circ}\sin x=2\cos 12^{\circ}\sin 18^{\circ}\cos (18^{\circ}+x)\iff$ $(2\cos 24^{\circ}-1)\sin x=2\sin 18^{\circ}\cos (18^{\circ}+x)\iff$ $(\cos 24^{\circ}-\cos 60^{\circ})\sin x=\sin 18^{\circ}\cos (18^{\circ}+x)\iff$

$2\sin 42^{\circ}\sin x=\cos (18^{\circ}+x)\iff$ $\cos (42^{\circ}-x)-\cos (42^{\circ}+x)=\cos (18^{\circ}+x)\iff$ $\cos (42^{\circ}-x)-\cos (18^{\circ}+x)=\cos (42^{\circ}+x)\iff$

$2\sin 30^{\circ}\sin (x-12^{\circ})=\cos (42^{\circ}+x)\iff$ $\sin (x-12^{\circ})=\cos (42^{\circ}+x)\iff$ $\cos (102^{\circ}-x)=\cos (42^{\circ}+x)\iff$ $102^{\circ}-x=42^{\circ}+x\iff$ $m\left(\widehat{PCA}\right)=30^{\circ}$ .

Proof 2 (synthetic). Take $D$ the reflection of $A$ in $BP$ , see that $\Delta ADP$ is equilateral, while $\Delta BAD\cong\Delta ABC(s.a.s.)$ ; consequently $ABCD$ is an isosceles trapezoid, i.e. cyclic.

Since $BD$ is the bisector of $\angle ABC$ , we get $CD=AD$ , hence $AD=CD=PD$ , or $D$ is the circumcenter of $\Delta APC\implies\angle ACP=\frac{\widehat{ADP}}{2}=30^\circ$ (Sunken Rock).

Proof 3 (synthetic). Take $D$ as above and $E\in AC\cap BP$ , see that $\Delta BCE$ is isosceles, so $CE=BC$ and $m(\widehat{BDE})=m(\widehat{BAC}) = 36^\circ=\frac{m\widehat{ACB}}{2}$ , hence $C$ is

the circumcenter of $\Delta BDE$ and $CD=BC$ ; but, as shown above, $AD=BC$ , consequently $D$ is the circumcenter of $\Delta APC\implies\angle ACP=\frac{\widehat{ADP}}{2}=30^\circ$ . (Sunken Rock)

Lemma. Prove that equation $\boxed{\sin 3x\sin 5x=\sin 2x\sin 8x}\ ,\ x\in\left(0,\frac {\pi}{10}\right)\ (*)$ has only one zero $x=6^{\circ}$ , i.e. $x=\frac {\pi}{30}$ .

Proof 1. $\sin 3x\sin 5x=\sin 2x\sin 8x\ \|\ \odot\ 2\cos 3x\ne 0\iff$ $\sin 6x\sin 5x=\sin 2x(\sin 11x+\sin 5x)\iff$

$\sin 5x(\sin 6x-\sin 2x)=$ $\sin 2x\sin 11x\iff$ $2\sin 5x\sin 2x\cos 4x=$ $\sin 2x\sin 11x\iff$ $2\sin 5x\cos 4x=$

$\sin 11x\iff$ $\sin 9x+\sin x=\sin 11x\iff$ $\sin x=\sin 11x-\sin 9x\iff$ $\sin x=2\sin x\cos 10x\iff$ $x=6^{\circ}$

Remark. Let $0<b<a<\frac {\pi}{2}$ so that $\frac a5=\frac b3=x$ . Thus, $\sin b\sin a=\sin (a-b)\sin (a+b)\ \odot\ 2\cos b\ \implies\ \sin a$ $\sin 2b=\sin (a-b)[\sin (a+2b)+\sin a]\implies$

$\sin a[\sin 2b-\sin (a-b)]=\sin (a-b)\sin (a+2b)\implies$ $2\sin a\sin\frac {3b-a}{2}\cos\frac {a+b}{2}=\sin (a-b)\sin (a+2b)$ . Since $\frac{3b-a}{2}=a-b$ get

$2\sin a\cos \frac {a+b}{2}=\sin (a+2b)\implies$ $\sin\frac {3a+b}{2}+\sin\frac {a-b}{2}=\sin (a+2b)\implies$ $\sin \frac {a-b}{2}=\sin (a+2b)-\sin\frac {3a+b}{2}\implies$ $\sin\frac {a-b}{2}=2\sin\frac {3b-a}{4}\cos\frac {5(a+b)}{4}$ .

Since $2(a-b)=3b-a$ get $\cos\frac {5(a+b)}{4}=\frac 12\implies$ $5(a+b)=240^{\circ}\implies$ $a+b=48^{\circ}\implies$ $x=\frac a5=\frac b3=\frac {48^{\circ}}{8}=6^{\circ}\implies$ $x=6^{\circ}$ .

Proof 2. Let $0<b<a<\frac {\pi}{2}$ . Thus, $\sin a\sin b=\sin (a-b)\sin(a+b)\iff \sin a\sin b=\sin^2a-\sin^2b\iff$ $\left(\frac {\sin a}{\sin b}\right)^2-\left(\frac {\sin a}{\sin b}\right)-1=0\iff$ $\frac {\sin a}{\sin b}=\frac {1+\sqrt 5}{2}=$

$\frac {{\frac 12}}{\frac {-1+\sqrt 5}{4}}=\frac {\sin 30}{\sin 18}$ . In the particular case $a=5x\ ,\ b=3x$ obtain that $\frac {\sin 5x}{\sin 3x}=\frac {\sin (5\cdot 6)}{\sin (3\cdot 6)}\implies x=6$ because the function $f(x)=\frac {\sin 5x}{\sin 3x}\ ,\ x\in\left(0,\frac {\pi}{10}\right)$ is injectively. Indeed,

$f'(x)\ .s.$ $s.\ (5\cos 5x\sin 3x-3\cos 3x\sin 5x)\ .a.$ $s.\ (5\tan3x-3\tan 5x)\ .s.$ $s.\ \left(\frac{\tan 3x}{3}-\frac {\tan 5x}{5}\right)=g(x)$ , where $g'(x)=\left(\tan^23x-\tan^25x\right)\ .s.s.$

$\left(\tan 3x-\tan 5x\right)\ .s.s.$ $(3x-5x)<0$ . Hence $g'(x)<0\implies$ $g\searrow$ . Since $g(0)=0$ get $g(x)<0$ , i.e. $f'(x)<0\implies$ $f\searrow$ , i.e. $f$ is strict decreasing $\implies$ $f$ is injectively.

Remark. I defined the equivalence relation $X\ .s.s.\ Y\ \iff\ X=Y=0\ \vee\ XY>0$ , i.e. $\mathrm{sign}(X)=\mathrm{sign}(Y)$ what means that $X$ and $Y$ have same sign.

PP8. Let $\triangle ABC$ and $D\in (BC)$ so that exists $x\in \left(0,\frac {\pi}{8}\right)$ such that $\left\{\begin{array}{c} m\left(\widehat{DAB}\right)=x\\\\ B=2x\ ;\ C=5x\end{array}\right\|$ . Prove that $DC=AB\ \iff\ x=6^{\circ}$ .

Proof. $m\left(\widehat{DAC}\right)=\pi -8x>0$ , $m\left(\widehat{ADC}\right)=3x$ and $DC=AB\iff$ $\frac {DC}{DA}=\frac {AB}{DA}\iff$ $\frac {\sin 8x}{\sin 5x}=\frac {\sin 3x}{\sin 2x}\iff$ $\sin 3x\sin 5x=\sin 2x\sin 8x\ \stackrel{(\mathrm{lemma})}{\iff}\ x=6^{\circ}$ .

PP9 (Miquel Ochoa Sanchez). Let $\triangle ABC$ and its interior point $P$ so that $\left\{\begin{array}{c} m\left(\widehat{BAP}\right)=18^{\circ}\\\\ m\left(\widehat{CAP}\right)=102^{\circ}\\\\ m\left(\widehat{ACP}\right)=24^{\circ}\end{array}\right\|$ . Prove that $PC=AB\ \iff\ x=12^{\circ}$

Proof. Denote $m\left(\widehat{BAP}\right)=x$ . Thus, $PC=AB\iff$ $\frac {PC}{AP}=\frac {AB}{AP}\iff$ $\frac {\sin 102^{\circ}}{\sin 24^{\circ}}=\frac {\sin (18^{\circ}+x)}{\sin x}\iff$ $\sin x\cos 12^{\circ}=\sin 24^{\circ}\sin (18^{\circ}+x)\iff$

$\sin x=2\sin 12^{\circ}\sin (18^{\circ}+x)\iff$ $\tan x=2\sin 12^{\circ}(\sin 18^{\circ}+\cos 18^{\circ}\tan x)\iff$ $\tan x=\frac {2\sin 18^{\circ}\sin 12^{\circ}}{1-2\sin 12^{\circ}\cos 18^{\circ}}=$ $\frac {2\sin 18^{\circ}\sin 12^{\circ}}{1-(\sin 30^{\circ}-\sin 6^{\circ})}=$

$\frac {2\sin 18^{\circ}\sin 12^{\circ}}{\sin 30^{\circ}+\sin 6^{\circ}}=$ $\frac {2\sin 18^{\circ}\sin 12^{\circ}}{2\sin 18^{\circ}\cos 12^{\circ}}\implies$ $\tan x=\tan 12^{\circ}\implies x=12^{\circ}$ .

PP10. Let $ABC$ be a triangle with the incircle $w=C(I,r)$ . Denote $\left\{\begin{array}{ccc} D\in BC\cap w & ; & M\in (DA\cap w\\\\ E\in CA\cap w & ; & N\in (EB\cap w\\\\ F\in AB\cap w & ; & P\in (FC\cap w\end{array}\right\|$ . Prove that $\boxed{\ \begin{array}{c} \frac {MA}{MD}=\frac {a(s-a)}{4(s-b)(s-c)}\\\\ \frac {MA}{MD}+\frac {NB}{NE}+\frac {PC}{PF}\ge \frac 32\end{array}\ }$ .

Proof. Apply the Stewart's relation to the cevian $AD$ in $\triangle ABC\ :\ \boxed{a\cdot AD^2+a(s-b)(s-c)=b^2(s-b)+c^2(s-c)}$ . Since $AM\cdot AD=AE^2=(s-a)^2$

obtain that $\frac {AM}{AD}=\frac {a\cdot AM\cdot AD}{a\cdot AD^2}=$ $\frac{a(s-a)^2}{b^2(s-b)+c^2(s-c)-a(s-b)(s-c)}\implies$ $\frac {MA}{MD}=\frac {a(s-a)^2}{b^2(s-b)+c^2(s-c)-a(s-b)(s-c)-a(s-a)^2}$ . Observe that

$b^2(s-b)+c^2(s-c)-a(s-b)(s-c)-a(s-a)^2=$ $b^2(s-b)+c^2(s-c)-a[bc-s(s-a)]-a(s-a)^2=$ $a^2(s-a)+b^2(s-b)+c^2(s-c)-abc$ .

Now I"ll use an well-known identity $\boxed{\sum a^2(s-a)=4sr(R+r)}$ . Hence $\frac {MA}{MD}=\frac {a(s-a)^2}{4sr(R+r)-4Rsr}\implies$ $\frac {MA}{MD}=\frac {a(s-a)^2}{4sr^2}\implies$ $\boxed{\frac {MA}{MD}=\frac {a(s-a)}{4(s-b)(s-c)}}$ a.s.o.

$\sum\frac {MA}{MD}=\frac 1{4(s-a)(s-b)(s-c)}\cdot\sum a(s-a)^2$ and $\sum a(s-a)^2=s^2\sum a-2s\sum a^2+\sum a^3=$ $2s^3-4s(s^2-r^2-4Rr)+8s^3-6s(s^2+r^2+4Rr)+$

$12Rsr=$ $2s(2r^2+8Rr-3r^2-12Rr+6Rr)=$ $2s(2Rr-r^2)\implies$ $\sum a(s-a)^2=2sr(2R-r)$ . So $\sum\frac {MA}{MD}=\frac {2sr(2R-r)}{4sr^2}=\frac {2R-r}{2r}\ge \frac 32$ because $R\ge 2r$ .

PP11. Let $\triangle ABC$ and its interior point $X$ for which denote $f(X)=(a\cdot XA)^2+(b\cdot XB)^2+(c\cdot XC)^2$ . Prove that $f(X)$ is minimum iff $X$ is the Lemoine's point of $\triangle ABC$ .

Proof. The power of Lemoine's point $L$ w.r.t. circumcircle $w$ is $p_w(L)=-3\cdot\left(\frac {abc}{a^2+b^2+c^2}\right)^2$ and for any $X$ there is the relation $f(X)=\left(a^2+b^2+c^2\right)\cdot \left[XL^2-p_w(L)\right]$ .

In conclusion, $f(X)$ is minimum if and only if $X:=L$ and in this case for any interior point $X$ we have $f(X)\ge \frac {3a^2b^2c^2}{a^2+b^2+c^2}=f(L)$ .

PP12. Prove that in any cyclical $ABCD$ exists the relation $MA^2\cdot [BCD]+MC^2\cdot [BAD]=MB^2\cdot [ADC]+MD^2\cdot [ABC]\ ,\ (\forall )\ M$ . Denote $[XYZ]$ - the area of $\triangle XYZ$ .

Proof. Denote $I\in AC\cap BD$ . Observe that $\left\{\begin{array}{c} \frac {IA}{[BAD]}=\frac {IC}{[BCD]}=\frac {AC}{S}\\\\ \frac {IB}{[ABC]}=\frac {ID}{[ADC]}=\frac {BD}{S}\end{array}\right\|\ (*)$ , where $S=[ABCD]$ . Apply the Stewart's relation to the common cevian $MI$ in the triangles:

$\blacktriangleright\ \triangle AMC \implies$ $MI^2\cdot AC+IA\cdot IC\cdot AC=MA^2\cdot IC+MC^2\cdot IA\stackrel{(*)}{\implies}$ $S\cdot\left(MI^2+IA\cdot IC\right)=MA^2\cdot [BCD]+MC^2\cdot [BAD]\ \ \ (1).$

$\blacktriangleright\ \triangle BMD \implies$ $MI^2\cdot BD+IB\cdot ID\cdot BD=MB^2\cdot ID+MD^2\cdot IB\stackrel{(*)}{\implies}$ $S\cdot\left(MI^2+IB\cdot ID\right)=MB^2\cdot [ADC]+MD^2\cdot [ABC]\ \ \ (2).$

Since $ABCD$ is cyclically get that $IA\cdot IC=IB\cdot ID=-p_w(I)$ , where $p_w(I)$ is the power of $I$ w.r.t. the circumcircle $w$ . From the relations

$(1)$ and $(2)$ obtain that $MA^2\cdot [BCD]+MC^2\cdot [BAD]=MB^2\cdot [ADC]+MD^2\cdot [ABC]=S\cdot\left[MI^2-p_w(I)\right]$ .

PP13. Let $\triangle ABC$ with $A=60^{\circ}$ , $b=16$ , $AD=20$ , where $E$ is midpoint of $[I_aI_c]$ and $D$ is projection of $E$ on $AB$ . Calculate $a$ and $c$ .

Proof 1. Let $\{U,V\}\subset AB$ so that $I_cU\perp AB$ and $I_aV\perp AB$ . Thus, $D$ is midpoint of $[UD]$ and $2\cdot UD=UV=$ $UB+BV=$ $(s-a)+(s-c)=b\implies$ $UD=\frac b2\implies$

$20=AD=AU+UD=(s-b)+\frac b2\implies$ $\boxed{a+c=40}$ . Remain to solve triangle with $A=60^{\circ}\ \wedge\ b=16\ \wedge\ a+c=40$ . Thus, $A=60^{\circ}\iff$ $a^2=b^2+c^2-bc\iff$

$(40-c)^2=16^2+c^2-16c\iff$ $1600-80c=256-16c\iff$ $c=\frac {1344}{64}\iff$ $\boxed{c=21\ \wedge\ a=19}$ .

Proof 2. Prove easily $E$ belongs to ircumcircle of $\triangle ABC$ and the bisector of $[AC]$ . If $F$ is projection of $E$ on $BC$ , then $DE=FE=x$ and $\triangle ADE\equiv \triangle CFE$ ,

i.e. $AD=CF=20\implies$ $c=AB=20+x$ and $a=BC=20-x$ . Thus, $a^2=b^2+c^2-bc\implies$ $(20-x)^2=16^2+(20+x)^2-16(20+x)\implies$ $(20+x)^2-(20-x)^2-16(4+x)=0\implies$

$80x-16(x+4)=0\implies$ $5x=x+4\implies$ $x=1\implies$ $a=19$ and $c=21$ .

Remark. Let midpoint $M$ of $[AC]$ and $\{X,Y\}\subset AC$ so that $I_cX\perp AC$ and $I_aY\perp AC$ . Since $XC=YA=s$ obtain $M$ is the midpoint and of $[XY]$ . Let $E'\in OM\cap w$ ,

where $w$ is circumcircle of $\triangle ABC$ . Thus, $E'\in I_aI_c$ and $E'\equiv E$ with $EM=\frac {r_a+r_c}{2}=\frac {bS}{2(s-a)(s-c)}=$ $\frac {b(s-b)}{2r}=\frac {2Rs(s-b)}{ac}=$ $2R\cos^2\frac B2=R(1+\cos B)$ .

PP14. The circle $w_1=C\left(I_1,r_1\right)$ is interior tangent to the circle $w_2=C\left(I_2,r_2\right)$ in the point $A$ . Find the maximum area of the triangle $XAY$ , where $X\in w_1$ and $Y\in w_2$ .

Proof 1. Denote $\{A,Y'\}=AY\cap w_1$ . Prove easily that $\frac {AY'}{AY}=\frac {r_1}{r_2}$ (constant). Thus, $[XAY]$ is maximum $\iff$ $[XAY']$ is maximum $\iff AX=XY'=Y'A$ .

Proof 2 (own). Consider the common tangent $t=AA$ to the circles $w_1$ and $w_2$ and a fixed point $T\in t$ so that the ray $[AX\subset \mathrm{int}\left(\widehat {TAY}\right)$ . Denote $\left\{\begin{array}{c} m\left(\widehat{TAX}\right)=x\\\\ m\left(\widehat{TAY}\right)=y\end{array}\right\|$ .

Suppose w.l.o.g. that $x<y$ . Prove easily that $\left\{\begin{array}{c} AX=2r_1\sin x\\\\ AY=2r_2\sin y\\\\ m\left(\widehat{XAY}\right)=y-x\end{array}\right\|$ and $[XAY]=\frac 12\cdot AX\cdot AY\cdot\sin \widehat{XAY}\implies$ $\boxed{[XAY]=2r_1r_2\sin x\sin y\sin (y-x)}$ .

Thus, $\sin x\sin y\sin (y-x)=$ $\sin x\sin (\pi -y)\sin (y-x)\le \frac {3\sqrt 3}{8}$ because $x+(\pi -y)+(y-x)=\pi$ with equality iff $x=\pi -y=y-x$ , i.e. $x=60^{\circ}$ and $y=120^{\circ}$ .

Remark. In any triangle $ABC$ we have $\sin A\sin B\sin C=\frac {abc}{8R^3}=$ $\frac {4RS}{8R^3}=\frac {S}{2R^2}=$ $\frac {sr}{2R^2}=$ $\frac 14\cdot \frac sR\cdot \frac {2r}{R}\le \frac 14\cdot\frac {3\sqrt 3}{2}\cdot 1=$ $\frac {3\sqrt 3}{8}$ .

PP15 (Ruben Auqui). Let $\triangle ABC$ with the excircles $\left\{\begin{array}{c} w_b=C(I_b,r_b)\\\\ w_c=C(I_c,r_c)\end{array}\right\|$ . Denote $\left\{\begin{array}{ccc} T\in AB\cap w_c & ; & \{P,T\}=TL\cap w_c\\\\ L\in AC\cap w_b & ; & \{Q,L\}=TL\cap w_b\end{array}\right\|$ . Prove that $TP=LQ$ .

Proof 1. $\left\{\begin{array}{ccc} E\in AC\cap w_c & \implies & LT\cdot LP=LE^2=a\\\\ F\in AB\cap w_b & \implies & TL\cdot TQ=TF^2=a\end{array}\right\|\implies$ $LT\cdot LP=TL\cdot TQ\Longrightarrow TQ=LP \Longrightarrow TP=LQ$ . Remark that $EF=TL$ .

Proof 2. Let $\left\{\begin{array}{ccc} U\in BC\cap w_c & ; & Y\in BC\ ,\ UY\perp BC\\\\ V\in BC\cap w_b & ; & X\in BC\ ,\ TX\perp BC\end{array}\right\|$ . Thus, $\left\{\begin{array}{c} LT\cdot LP=LI_c^2-r_c^2\\\\ TL\cdot TQ=TI_b^2-r_b^2\end{array}\right\|$ and $TP=LQ\iff$ $LP=TQ\iff$ $LT\cdot LP=TL\cdot TQ\iff$

$LI_c^2-r_c^2=TI_b^2-r_b^2\iff$ $\boxed{LI_c^2-TI_b^2=r_c^2-r_b^2}\ (*)$ . Prove easily that $\boxed{\cos C-\cos B=\frac {(s-a)(b-c)}{2Rr}}\ (1)$ and $\boxed{r_b\sin B-r_c\sin C=\frac {s(s-a)(b-c)}{2Rr}}\ (2)$ .

$\blacktriangleright\ LI_c^2=(UC-CY)^2+\left(r_c-LY\right)^2=$ $\left[s-(s-a)\cos C\right]^2+\left[r_c-(s-a)\sin C\right]^2\implies$ $\boxed{LI_c^2=s^2+r_c^2+(s-a)^2-2s(s-a)\cos C-2r_c(s-a)\sin C}\ (3)$ .

$\blacktriangleright\ TI_b^2=(VB-BX)^2+\left(r_b-TX\right)^2=$ $\left[s-(s-a)\cos B\right]^2+\left[r_b-(s-a)\sin B\right]^2\implies$ $\boxed{TI_b^2=s^2+r_b^2+(s-a)^2-2s(s-a)\cos B-2r_b(s-a)\sin B}\ (4)$ .

From the relations $(3)$ and $(4)$ obtain that $LI_c^2-TI_b^2=$ $r_c^2-r_b^2-2s(s-a)(\cos C-\cos B)-2(s-a)\left(r_c\sin C-r_b\sin B\right)$ .

Using the relations $(1)$ and $(2)$ obtain that $LI_c^2-TI_b^2=r_c^2-r_b^2$ . From the relation $(*)$ obtain the conclusion of the proposed problem, i.e. $TP=LQ$ .

Remark. $\frac {\cos C-\cos B}{\sin B-\sin C}=\frac {2\sin\frac {B+C}{2}\sin\frac {B-C}{2}}{2\sin\frac {B-C}{2}\cos\frac {B+C}{2}}=$ $\tan\frac {B+C}{2}=\cot\frac A2=$ $\frac {s-a}{r}\implies$ $\cos C-\cos B=\frac {s-a}{r}\cdot\frac {b-c}{2R}\implies \boxed{\cos C-\cos B=\frac {(s-a)(b-a)}{2Rr}}\ (1)$ .

$r_b\sin B-r_c\sin C=s\left(\tan\frac B2\sin B-\tan\frac C2\sin C\right)=$ $s\left(\frac {\sin\frac B2}{\cos\frac B2}\cdot 2\sin\frac B2\cos\frac B2-\frac {\sin\frac C2}{\cos\frac C2}\cdot2\sin\frac C2\cos\frac C2\right)=$ $2s\left(\sin^2\frac B2-\sin^2\frac C2\right)=$

$2s\left[\frac {(s-a)(s-c)}{ac}-\frac {(s-a)(s-b)}{ab}\right]=$ $\frac {2s(s-a)}{a}\cdot\left(\frac {s-c}c-\frac {s-b}b\right)\implies$ $\frac {2s(s-a)}{abc}\cdot [b(s-c)-c(s-b)]\implies$ $\boxed{r_b\sin B-r_c\sin C=\frac {s(s-a)(b-c)}{2Rr}}\ (2)$ .

Otherwise. $s(\cos C-\cos B)=r_b\sin B-r_c\sin C\iff$ $\cos C-\cos B=\tan\frac B2\sin B-\tan\frac C2\sin C\iff$

$\cos C-\cos B=2\sin^2\frac B2-2\sin^2\frac C2\iff$ $\cos C-\cos B=(1-\cos B)-(1-\cos C)$ , what is true.

PP16. Let $B$ -right-angled $\triangle ABC$ with incircle $w=C(I,r)$ . Denote $\left\|\begin{array}{c} D\in BC\cap w\\\ E\in AC\cap w\\\ F\in AB\cap w\end{array}\right\|$ and $\left\|\begin{array}{c} \{P,D\}=AD\cap w\\\ \{P,X\}=PC\cap w\\\ \{P,Y\}=PB\cap w\end{array}\right\|$ . Prove that $PB\perp PC\ \iff\ PD=3\cdot PA$ .

Proof. Prove easily $B=90^{\circ}\iff$ $a^2+c^2=b^2\iff$ $r=s-b\iff S=\frac {ac}{2}\iff$ $S=s(s-b)=(s-a)(s-c)\iff$ $\frac {s-a}{a}=\frac {s-b}{b-c}=\frac {c}{2(s-c)}\ (*)$ . Let

$T\in BC\cap EF$ and $L\in AC\cap PT$ . I"ll use properties: the division $\{B,C;D,T\}$ is harmonically, i.e. $\frac {DB}{DC}=\frac {TB}{TC}=$ $\frac {s-b}{s-c}$ ; $T$ belongs to the tangent $PP$ to $w$ at $P\in w$ . Suppose

w.l.o.g. that $b>c$ . Thus, $\frac {TB}{s-b}=\frac {TC}{s-c}=\frac a{b-c}\implies$ $TB=\frac {a(s-b)}{b-c}\stackrel{(*)}{\implies}$ $\boxed{TB=s-a}\implies DT=DB+BT=(s-b)+(s-a)\implies$ $\boxed{DT=PT=c}$ . Since

$\{B,C;D,T\}$ is harmonic, then $PB\perp PC\iff$ $\left\{\begin{array}{c} \widehat{BPD}\equiv\widehat{BPT}\\\\ \widehat{CPL}\equiv\widehat{CPD}\end{array}\right\|\iff$ $\frac {PD}{PT}=\frac {BD}{BT}\iff$ $\frac {PD}{c}=\frac {s-b}{s-a}\iff$ $\boxed{PD=\frac{c(s-b)}{s-a}}$ .Therefore, $I\in XY$ , the midpoint $H$

of $[PD]$ belongs to $\overline {XYT}$ and $AHBT$ is cyclic, i.e. $DH\cdot DA=DB\cdot DT\iff$ $\frac 12\cdot DP\cdot DA=c(s-b)\iff$ $DA=\frac {2c(s-b)}{\frac{c(s-b)}{s-a}}\iff$ $\boxed{DA=2(s-a)}$ . Thus,

$AP\cdot AD=AF^2\iff$ $AP=\frac {(s-a)^2}{2(s-a)}\iff$ $\boxed{PA=\frac {s-a}{2}}$ . In conclusion, $\frac {AD}{AP}=\frac {2(s-a)}{\frac {s-a}{2}}=4\iff$ $1+\frac {PD}{PA}=4\iff$ $\boxed{PD=3\cdot PA}$ .

PP17. Let $\triangle ABC$ with incircle $w=C(I,r)$ . Let $\left\|\begin{array}{c} D\in BC\cap w\\\ E\in AC\cap w\\\ P\in AD\cap w\end{array}\right\|$ . Prove that $PB\perp PC\ \iff\ AP+AE=PD$ .

Proof. Suppose w.l.o.g. that $b>c$ . Denote $\left\{\begin{array}{c}K\in BC\ ,\ AK\perp BC\\\\ L\in EF\cap BC\\\\ S\in LI\cap AD\\\\ \phi=m(\widehat{DLI})\end{array}\right\|$ . From the well-known property $LI\perp AD$ . obtain $L\in PP$ and $\sin\phi=\frac{SD}{LD}=\frac{KD}{AD}$ .

Prove easily $\left\{\begin{array}{c} KD=\frac{(b-c)(p-a)}{a}\\\\ LB=\frac{a(p-b)}{b-c}\\\\ LD=\frac{2(p-b)(p-c)}{b-c}\end{array}\right\|$ . Therefore, $\left\{\begin{array}{c}\boxed{PA+AE=PD}\\\\ \left[AP\cdot AD=(p-a)^{2}\right]\end{array}\right\|$ $\Longleftrightarrow$ $\left\{\begin{array}{c}PD-PA=p-a\\\\ PA\cdot (PA+PD)=(p-a)^{2}\end{array}\right\|$ $\Longleftrightarrow$ $PA\cdot (PA+PD)=$

$(PD-PA)^{2}$ $\Longleftrightarrow$ $PD=3\cdot PA$ $\Longleftrightarrow$ $\left|\begin{array}{c}AD=2(p-a)\\\\ \sin\phi =\frac{KD}{AD}\end{array}\right|$ $\Longleftrightarrow$ $\boxed{\sin\phi=\frac{b-c}{2a}}$ Because the division $(B,C,D,L)$ is harmonically obtain $:$

$\boxed{PB\perp PC}$ $\Longleftrightarrow$ $\widehat{BPD}\equiv\widehat{BPL}$ $\Longleftrightarrow$ $\frac{PD}{PL}=\frac{BD}{BL}$ $\Longleftrightarrow$ $2\cdot \frac{DS}{LD}=\frac{BD}{BL}$ $\Longleftrightarrow$ $2\sin\phi=\frac{(p-b)(b-c)}{a(p-b)}$ $\Longleftrightarrow$ $\boxed{\sin\phi =\frac{b-c}{2a}}$ .

Remark. The incircle $C(I,r)$ of an acute triangle $ABC$ is tangent (internally) to the circle with the diameter $[BC]$ if and only if $\boxed{a+h_{a}=b+c}$

and in this case the common tangent point of these circles belongs to the $A$ - Gergonne's line if and only if $a=4r$ .[/b]

PP18 (Miguel Ochoa Sanchez). Let a trapezoid $ABCD$ with $AD\parallel BC$ , $BC<AD$ and the points $\left\{\begin{array}{cc} M\in (AB)\ ; & Q\in (DC)\\\ N\in (AC)\ ; & P\in (DB)\end{array}\right\|$ so that

$MN=NP=PQ$ and the point $X\in AD\cap\overline{MN}$ so that $A\in (XD)$ . Prove that $MN\parallel AD$ or $AD=2\cdot BC$ .

Proof. Let $\left\{\begin{array}{c} AD=a\ ;\ BC=b\ ;\ a=xb\ ;\ x>1\\\\ MB=m\cdot MA\ ;\ QC=q\cdot QD\end{array}\right\|$ and $\left\{\begin{array}{c} Y\in AB\cap CD\ ;\ YA=x\cdot YB\\\\ I\in AC\cap BD\ ;\begin{array}{c} \nearrow IA=x\cdot IC\\\\ \searrow ID=x\cdot IB\end{array}\end{array}\right\|$ .Therefore,

$\blacktriangleright\ \frac {QY}{CD}=\frac {QC+CY}{CD}=\frac {q}{q+1}+\frac 1{x-1}=$ $\frac {1+xq}{(x-1)(q+1)}\implies$ $\frac {QD}{QY}=\frac {\frac {QD}{CD}}{\frac {QY}{CD}}=$ $\frac {\frac 1{q+1}}{\frac {1+xq}{(x-1)(q+1)}}\implies$ $\boxed{\frac {QD}{QY}=\frac {x-1}{1+xq}}\ \ (1)$ .

$\blacktriangleright\ \frac {MY}{AB}=\frac {MB+BY}{AB}=\frac {m}{m+1}+\frac 1{x-1}=$ $\frac {1+xm}{(x-1)(m+1)}\implies$ $\frac {MA}{MY}=\frac {\frac {MA}{AB}}{\frac {MY}{AB}}=$ $\frac {\frac 1{m+1}}{\frac {1+xm}{(x-1)(m+1)}}\implies$ $\boxed{\frac {MA}{MY}=\frac {x-1}{1+xm}}\ \ (2)$ .

Apply the Menelaus' theorem to the following transversals in the mentioned triangles :

$\blacktriangleright\ \overline{XMQ}/\triangle AYD\ :\ \frac {XA}{XD}\cdot\frac {QD}{QY}\cdot\frac {MY}{MA}=$ $1\ \stackrel{(1\wedge 2)}{\implies}\ \frac {XA}{XD}$ $\cdot\frac {x-1}{1+xq}\cdot\frac {1+xm}{x-1}=1$ $\implies$ $\boxed{\frac {XA}{XD}=\frac {1+xq}{1+xm}}\ (3)$ .

$\blacktriangleright\ \overline{ANI}/\triangle BMP\ :\ \frac {AM}{AB}\cdot\frac {IB}{IP}\cdot\frac {NP}{NM}=$ $\implies\frac {1}{m+1}\cdot\frac {IB}{IP}\cdot 1=1$ $\implies$ $IB=(m+1)\cdot IP$ . Since $ID=\frac ab\cdot IB$

obtain that $PD=ID-IP=\left(\frac ab-\frac 1{m+1}\right)\cdot IB\implies$ $\frac {PI}{PD}=\frac b{am+a-b}\implies$ $\boxed{\frac {PI}{PD}=\frac 1{x(m+1)-1}}\ (4)$ .

$\blacktriangleright\ \overline{DPI}/\triangle CQN\ :\ \frac {DQ}{DC}\cdot\frac {IC}{IN}\cdot\frac {PN}{PQ}=$ $\implies\frac {1}{q+1}\cdot\frac {IC}{IN}\cdot 1=1$ $\implies$ $IC=(q+1)\cdot IN$ . Since $IA=\frac ab\cdot IC$

obtain that $NA=IA-IN=\left(\frac ab-\frac 1{q+1}\right)\cdot IC\implies$ $\frac {NI}{NA}=\frac b{aq+a-b}\implies$ $\boxed{\frac {NI}{NA}=\frac 1{x(q+1)-1}}\ (5)$ .

$\blacktriangleright\ \overline{XNP}/\triangle AID\ :\ \frac {XA}{XD}\cdot\frac {PD}{PI}\cdot\frac {NI}{NA}$ $=1\ \stackrel{(3\wedge 4\wedge 5)}{\implies}\ \frac {1+xq}{1+xm}$ $\cdot\frac {x(m+1)-1}{x(q+1)-1}=1\implies$ $(1+xq)[x(m+1)-1]=(1+xm)[x(q+1)-1]\implies$

$x(m+1)+x^2q(m+1)-1-xq=x(q+1)+x^2m(q+1)-1-xm\implies$ $x^2(q-m)-2x(q-m)=0\implies$ $q=m\ \vee\ x=2$ , i.e. $MN\parallel AD\ \vee\ AD=2\cdot BC$ .

Remarks. We can estabilish $:\ \left\{\begin{array}{cccccccc} \blacktriangleright & \overline{XNQ}/\triangle ACD\ : & \frac {XA}{XD}\cdot\frac {QD}{QC}\cdot\frac {NC}{NA}=1 & \implies & \frac {NC}{NA}=\frac {XD}{XA}\cdot\frac {QC}{QD} & \implies & \boxed{\frac {NC}{NA}=\frac {q(1+xm)}{1+xq}} & (6)\ .\\\\ \blacktriangleright & \overline{XMP}/\triangle ABD\ : & \frac {XA}{XD}\cdot\frac {PD}{PB}\cdot\frac {MB}{MA}=1 & \implies & \frac {PB}{PD}=\frac {XA}{XD}\cdot\frac {MB}{MA} & \implies & \boxed{\frac {PB}{PD}=\frac {m(1+xq)}{1+xm}} & (7)\ .\end{array}\right\|$ . If $MN\parallel AD$ , then

$MN=PQ$ and prove easily that $MN=NP\iff$ the point $N$ belongs to the $B$ -median of $\triangle ABD$ . In conclusion, we get an easy construct of $\overline{MNPQ}$ . If $MN\not \parallel AD$ ,

then $AD=2\cdot BC$ when are given $\{b,m,q\}$ , prove easily that $:\ \frac 1q\cdot \frac {NC}{NA}=$ $\frac {1+2m}{1+2q}=m\cdot\frac {PD}{PB}\ ;\ XA=$ $\frac {b(2q+1)}{m-q}\ ;\ CZ=$ $\frac {bq(2m+1)}{m-q}$ , where $Z\in MN\cap BC$ .

PP19. Let an isosceles trapezoid $ABCD$ with $AB\parallel CD$ so that $[CD]$ is diameter of its circumcircle $w=C(O,R)$ .

For $E\in w$ so that $AB$ separates $E$ , $O$ let $\left\{\begin{array}{c} F\in ED\cap AB\\\\ G\in EC\cap AB\end{array}\right\|$ . Prove that $\frac {AF\cdot BG}{FG}$ is constant.

Proof. Let $\left\{\begin{array}{c} 45^{\circ}<m\left(\widehat{EDC}\right)=x<90^{\circ}\\\\ \delta_{CD}(A)=\delta_{CD}(B)=h<R\\\\ \frac 12\cdot (CD-AB)=d<R\end{array}\right\|$ . Thus, $h^2=BC^2-d^2\implies$ $h^2=d(2R-d)\implies$ $\boxed{h^2+d^2=2Rd}\ (*)$

and $\frac {h}{AF+d}=\tan x=\frac {GB+d}{h}\odot\begin{array}{ccc} \nearrow & AF=h\cot x-d & \searrow\\\\ \searrow & BG=h\tan x-d & \nearrow\end{array}\odot\implies$ $FG=AB-(AF+BG)=$ $2(R-d)-[h(\tan x+\cot x)-2d]\implies$

$FG=2R-h(\tan x+\cot x)$ . In conclusion, $\frac {AF\cdot BG}{FG}=\frac {h^2+d^2-hd(\tan x+\cot x)}{2R-h(\tan x+\cot x)}\stackrel{(*)}{\implies}$ $\boxed{\frac {AF\cdot BG}{FG}=d}$ (constant).

IF $h=24\ ,\ R=25\ \stackrel{(*)}{\implies}\ d^2-50d+24^2=0$ with $\Delta '=25^2-24^2=7^2\implies$ $\begin{array}{ccccc} \nearrow & d_1=25-7 & \implies & d_1=18<R & \searrow\\\\ \searrow & d_2=25+7 & \implies & d_2=32\not <R & \nearrow\end{array}\odot$ $\implies d=18\implies$ $\frac {AF\cdot BG}{FG}=18$ .

This post has been edited 409 times. Last edited by Virgil Nicula, Nov 26, 2015, 11:51 AM

Comment

0 Comments

August 2018

462. Diverse probleme de geometrie.

January 2018

461. Problems for the relaxation in ... weekend (II).

October 2017

460. Working page VI.

September 2017

459.Working page V.

August 2017

458. Working page IV.

457. Relatii metrice si inegalitati geometrice.

456. Probleme de Algebra.

July 2017

456* Integrals.

455. Problems for the relaxation in ... weekend!

May 2017

454. Mathematical note: "trig. ineg."- GMB 6/1992, page 201.

March 2017

453. Working page III.

December 2016

452. Working page II.

November 2016

451. ABC with C(O,R), C(I,r) and N-Nagel point ==> ON=R-2r.

450. Nota matematica: "Asupra unei inegalitati remarcabile".

October 2016

449. Working page I.

September 2016

448. Extensions or generalizations.

July 2016

447. Locuri geometrice remarcabile.

June 2016

446. Mathematics "instant" for the middle school.

445. Mathematics "instant" for the high school.

May 2016

444. Remarkable points in a triangle.

443. Problems of the type "instant" (continuare).

April 2016

442. Problems from and for baccalaureate II.

441. LaTeX (common simbols).

February 2016

440. Piece of Poetry.

January 2016

439. Properties of medians in triangle ABC and applications.

December 2015

438. Tangential quadrilateral. Zaslavsky's problem.

437. An equivalence relation over R.

436. Two characterizations of A-right triangle ABC.

435. Geometry problems in space.

November 2015

434. Problems with areas II.

433. Some properties of remarkable lines in a triangle II.

432. Geometry problems with perpendicularities II.

431. Trigonometry in geometry II.

430. Trigonometrical identities.

October 2015

429. Bretschneider's_formula

September 2015

428. Some metrical problems from the contests II.

July 2015

427. Selected nice or well-known geometry problems.

June 2015

426. Problems from and for baccalaureate.

May 2015

425. Geometrical extremity II.

424. Some nice problems of E. E. Q. Rodriguez and others.

March 2015

423. Geometry problems.

422. Geometrical problems.

420. Some geometry problems for high school II.

419. Some geometry problems for hight school I

418. Some nice problems with polynomials/equations III

417 <bis> 418 (educatie: "14.27.28.32.57"/\"127.248").

417. The incircle of ABC and other two incircles.

416. Cateva solutii ale unei probleme de tip slicing.

415. Geometry 8.

414. Geometry 7.

413. Geometry 6.

412. Geometry 5.

411. Geometry 4.

410. Geometry 3.

409. Geometry 2.

408. Geometry 1.

February 2015

407. Cyclical quadrilaterals.

November 2014

406. Trigonometry in geometry (middle or high school) II.

405. Art of problem solving (algebra).

403. Some nice geometry problems for the high school. II

402. Some interesting inequalities I.

401. Some nice problems with polynomials/equations II.

October 2014

400. Some nice geometry problems for the high school I.

September 2014

399. Algebra (I).

398. Geometry problems (II).

August 2014

July 2014

396. Own inegalities stronger than the Panaitopol's.

April 2014

395. Metrical relations in a quadrilateral.

394. A cevian in a triangle and two incircles.

393. Some problems with circles.

March 2014

392. Art of Problem Solving (geometry).

December 2013

391. CRUX Mathematicorum.

November 2013

390. Probleme de analiza matematica.

389. Some geometric properties in a triangle I.

October 2013

388. Some interesting "slicing" problems.

September 2013

387. Algebra (II).

August 2013

386. Barycentrical coordinates.

385. Some nice geometry problems.

384. Geometry problems (I).

383. Trigonometry in geometry (middle or high school) I.

July 2013

382. Peruan "jewels" (Israel Diaz, M. O. Sanchez a.s.o.)

381. Probleme date la diverse concursuri.

380. Functions. Range. Bijection and inverse.

June 2013

379. Geometrical extremity I.

378. Some metrical relations.

377. Some remarkable properties of the symmedian.

May 2013

376. Trigonometrical identities.

374. Some proofs of the Law of Cosines.

373. Extension of Chebyshev's inequality.

April 2013

372. The distancies : OI , OI_a , ON a.s.o.

371. Some problems from NMO (final stage) - 2013, Romania.

February 2013

370. Some metrical relations in a triangle.

January 2013

369. Mixtilinear circles of ABC.

364. Some interesting inequalities I

December 2012

363. Geometry problems with perpendicularities.

November 2012

362. Fermat's problem and other metrical problems.

October 2012

361. Arhimedes theorem of the broken chord in a circle.

360. Some problems for the middle school.

359. Ptolemy's theorem and generalized Ptolemy's relation.

358. Some properties of the remarkable lines in a triangle.

September 2012

357. Morley theorem.

356. Order relation between s & (2R+r) in an acute triangle.

August 2012

355. Some problems with polynomials or equations I

354. Gradul II - Univ. B.B. Cluj, aug. 1998 Prof I sau II.

353. Some problems of the analytical geometry.

July 2012

352. Some definite integrals.

351. Problems with areas.

350. Another (easy) integrals.

349. Another interesting limits.

348. Subiectele de la BAC 2012 /Mate-Info.

June 2012

347. Two isogonal lines in a triangle.

346. Some easy integrals.

345. Characteristic function of a set.

344. Some trigonometry problems.

343. Problems with collinear points and concurrent lines.

May 2012

342. Exponential or logarithmic equations/inequations.

April 2012

341. Some nice and interesting "slicing" problems.

340. Geometry problems for middle school.

339. Derivatives. Rolle's function.

338. ROU National Math Olympiad 2012.

February 2012

337. Nice metrical problems.

January 2012

336. Some problems of the analytical geometry.

335. Some difficult geometric/trigonometric inequalities.

334. 2011 Japan Mathematical Olympiad Finals Problem 1.

333. Problems with the common tangent for two circles.

332. Some metrical problems from the contests I.

December 2011

331. Some nice problems with polynomials/equations I.

November 2011

330. A triangle and an its transversal through a point.

329. Some problems with complex numbers.

328. A general identities with areas in a triangle ABC.

October 2011

327. Some geometrical/algebraic extremum problems.

326. Some problems from trigonometry.

325. Without the l'Hospital's rules.

324. Some determinants.

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

322. Incenter, Gergonne, Nagel a.s.o. of ABC.

321. Harmonical quadrilateral.

320. Some problems with metrical relations.

September 2011

319. Some problems with induction.

318. IRAN - 2011 (geometrie).

317. Some interesting geometry problems for middle school.

316. Some properties of isogonal rays in an angle of ABC.

August 2011

315. Indian Team Selection Test 2010 ST1 P1 and extension.

314. USAMTS 2009 Round 2 #5.

313. For middle school - some nice problems.

312. Very nice (famous trigonometrical problems) !

311. Some problems from USAMO and other contests.

310. Some properties of the tangential quadrilateral.

309. Beautiful problem ! IRAN Selection Test for IMO, 2004.

308. Some nice and easy properties in a triangle.

July 2011

307. Very nice proofs !

306. Some easy and nice "slicing" problems.

305. Position of the roots w.r.t. a given real number.

304. Integrals ("brothers") III.

303. Integrals II.

302. Some problems with sequencies of the "roots" .

301. Some trigonometrico-geometrical inequalities.

300. Lagrange's multiplier. Examples and problems.

299. A class of the recurrent sequencies.

298. Some nice problems from Crux Mathematicorum.

297. Nice inequalities with concrete numbers.

296. Some nice Vittasko's problems.

295. M1 3th subject, 2c - School Graduate, ROMANIA.

294. The Octav Halunga's inequality in a triangle.

June 2011

293. A difficult trigonometric equation (third point !).

292. Length of the angled bisector and some means.

291. Some integrals.

290. Characterize "OP perpendicular on OA" in ABC a.s.o.

289. Nice identities in an algebra/ geometry/trigonometry.

288. An inequality with two A-cevians in a triangle ABC.

287. Some metrical problems with or without trigonometry.

286. ABC is right triangle iff s=2R+r and other similar pp.

285. Nice and easy inequalities. For example, IMAC - 2009.

284. Ellipse (definition).

283. Some nice and easy problems.

282. Factorize of polynomials.

May 2011

281. Identity with R in an acute triangle (ILL - 1974).

280. Common roots of two polynomial equations.

279. O.M. Holland , O.M. Ireland 1988 and TST USA 2000, etc.

278. In memoria profesorului Alexandru Lupas.

277. A nice and difficult relation with Lemoine's axis.

276. Chinese TST 2009 and Second Round 2006.

275. Some interesting algebraic equations and systems.

274. An inequality in a triangle for its interior point.

273. The area of the triangle IOH. When I belongs to OH ?

272. Outside of triangle. Extension of Fermat's point.

271. Relations concerning Fermat points.

April 2011

270. Morrocan M.O. 2010 and Croatia TST 2009.

269. Some nice and easy properties in a trapezoid.

268. Saraghin contest, 2011.

267. Germany 2002, Russia 2003, IMO 2003, Aus.-Polish 1992.

266. Rectangles erected on sides of a triangle. Concurrence.

265. Really nice problem !

264. Identities with complex numbers.

263. Newton's sums. Applications.

262. IMO Shortlist 2005, Geometry 6th.

261. IMO 2009, problem 2 and JBMO 2007, problems 1 & 2.

260. Some nice, simple identities and their applications.

259. ABC equilateral triangle ==> DEF equilateral triangle.

258. Some difficult and nice problems with complex numbers.

257. Seeking roots of 2th equation with complex coef.

256. An extremum problems with complex numbers.

March 2011

255. A line which is parallelly with OH.

254. An interesting C. Mateescu's geometrical inequality.

253. An old algebraic inequality.

252. OI=f(A,R) in certain conditions.

251. Problems of Analytical Geometry.

250. An application of the Euler's relation.

249. An usual and nice metrical relations in quadrilateral.

248. Pagina instructiva.

247. An remarkable inequality with median.

246. R1-R2=OI (nice).

245. An inequality in an acute triangle.

244. Some properties of cyclic orthodiagonal quadrilaterals.

243. Nice ! (a+b) is constant.

242. A general geometrical problems.

241. Areas in a triangle.

240. Some interesting geometrical inequalities.

239. Algebraic marathon.

238. Some metrical problems.

237. Some problems with ... problems.

236. A locus with the metrical relation.

235. Proofs of some geometry problems with complex numbers.

234. Some simple and nice problems from contests.

233. An interesting identity and an easy & nice inequality.

February 2011

232. Some geometrical inequalities (own).

Pagina libera

231. A conditioned minimum value.

230. Some usual synthetical properties in a triangle.

229. Concurs online MATEFBC, ed. 5 -subiect 2.

228. OJM - Romania, 2010 problem 3.

227. IMO ShortList 2003 (problem 5).

226. Some inequalities from Calculus (Real Analysis).

225. An interesting problems from Kunny.

224. Some nice and easy inequalities in a triangle.

223. An inequality with complex numbers-RMO shortlist 2010.

222. An extension of probl. 2 from IMO 2009.

221. Some nice and simple "slicing" problems.

220. Some nice geometry problems from contest.

January 2011

219. A very nice geometrical inequality.

218. Some problems from the Suiss IMO Selection Team 2006.

217. A nice and interesting Murray S. Klamkin's problem.

216. Nice ! Find the length of hypotenuse (Belarus - 2010)

215. Limit of the unique root for polynominal equations.

214. A range of some functions.

213. Another classical "slicing" problems.

212. A simple applications of the Casey's theorem.

211. Pretty hard problem !

210. Tangential circle of the triangle ABC.

209. Nice problem, nice proof !

208. An interesting geometrical inequality.

207. Some interesting problems of Toshio Seimiya, Japan.

206. Four problems with extremum in a quadrilateral.

205. A "slicing" problem with 11 methods.

204. Saraghin 2010 (three geometry problems).

203. An application of the Pascal's permutation theorem.

December 2010

202. Concurrent lines in a right triangle.

201. Applications of Desarques' and Pappus' theorems.

200. Some nice applications of the inversion.

199. Some properties of symmedian. Two extensions.

198. Generalized Carnot's lemma.

197. Triangles outside of a triangle and concurrence.

195. Semicircle. Nice application of harmonical division.

194. The Lalescu's sequence. Properties.

193. An inequality with n! .

192. An easy and nice problem of Valentin Vornicu.

191. Six nice problems with the incircle of a triangle.

190. Very nice ! Coculescu's Contest seniors 2010.

189. Pascal's theorem and some nice and easy applications.

188. Some problems from vectorial geometry.

187. Distancies.

186. Without the l'Hospital's rules.

185. The Durrande's relation.

184. An interesting geometry problem from RMO 2010.

183. A remarkable characterization of equilateral triangle.

182. Another two nice problems with complex numbers (own).

181. Two special inequalities from CRUX (own).

180. Own proposed problem from CRUX (with complex numbers).

November 2010

179. The range of |z| , |z^2+a|=|bz+c|. Particular cases.

178. Some nice and simple geometry problems.

177. A nice proof of one identity in a triangle.

176. Another nice problems for middle school.

175. A Geometric Proof (without trig.) of Heron's Formula.

174. Some interesting properties of modulus (middle school).

173. An interesting trigonometric equations.

172. The Zelinsky's problem.

171. Another "slicing" proposed problems (middle school).

October 2010

170. A nice "slicing" proposed problem for middle school.

169. Some metrical problems in a circle (III).

168. Some properties of the Lemoine's point.

167. Nice problems - OIM, 1995 and ... Toshio Seimyia.

166. Some constant geometrical expressions.

165. Some metrical problems (5) in a circle (II).

164. Some metrical problems (5) in a circle.

163. Applications of Ptolemy's inequality. Ex. - China,1998.

162. Generalization of proposed problem from Mongolia, 2000.

161. A difficult inequality (barycentrical coordinates)

160. A very nice "slicing" problem.

159. Multiple derivatives in a point.

158. Very nice and difficult limits.

157. Steinbart's theorem. Applications.

156. Some problems with perpendicularities.

155. A constant ratio in a triangle ABC.

154. Difficult inequality (without derivatives eventually).

153. Geometrical minimum/maximum.

152. Concurrence, cyclically, metrics, symmedian a.s.o.

151. Miscellaneous problems for the admission at math.

150. Some (5) problems with fixed points.

149. Some (8) problems with collinearity or concurrence.

148. A special class of systems.

147. A problem with two polynominals.

146. Synthetic, metric, trigonometric and analitycal proofs.

145. Spain MO P5 - Point on median.

144. A conditioned concurrency in a triangle.

143. Assess of a ratio (nice and difficult problem).

142. Inscribed/circumscribed quadrilateral w.r.t. ABC.

141. N is interior <==> 1/3<a/b<3 (CRUX Mathematicorum).

140. A synthetical property <==> linear/angled relation.

139. A caracterization of a right angle in a triangle.

138. Generalized "slicing" problems (own).

137. Stewart's relation.

136. A geometrical interpretation of a system (own).

135. Three strong geometrical inequalities.

September 2010

134. Some problems with projections/perpendiculars (2).

133. Some problems with projections/perpendiculars (1).

132. Asupra unei inegalitati intr-un triunghi.

131. A stronger Form of the Steiner-Lehmus Theorem (own).

130. A triangle and a fixed point.

129. BX=CY and three properties.

128. Value of IG and two inequalities in neighbourhood of 0.

127. Pagina educativa.

126. Routh's theorem.

125. I like very much this Darij Grinberg's message.

124. A nice and easy problem with a right-angled triangle.

123. My extension and its proof.

122. An identity for an equilateral triangle. Extension.

121. Viete's relations with cos A , cos B , cos C.

120. Two equal neighbouring angles/sides in a quadrilateral.

119. Inspired by a Miculita's problem.

118. An usual remarkable geometrical locus.

117. Some applications of harmonic division/quadrilateral.

116. The length of AE (nine methods !).

115. An difficult equation with one radical.

114. A property of tangential triangle for ABC.

113. Range for a function.

112. The equivalence relation ".s.s."

111. Simple, but interesting ...

110. German TST 2004 - Exam V , Problem 1

109. Two equal areas ==> find A.

108. Old, short and nice proof of Euler's relation.

107. XY parallel BC and X,Y are isogonal/izotomic points.

106. OLM - Bucuresti (geometrie).

105. A problem with three circles.

104. Concursul Revistei de Matematica din Galati (RMG).

103. Metrical usual relations in triangle. Applications.

102. Characterization of a metrical relation in a triangle.

101. O problema cu numere complexe.

100. Bobiller's theorem.

99. Some simple and nice geometry problems.

98. Symmedian & median (metrical relations).

97. Problem of butterfly.

August 2010

96. Integrals.

95. Three circles in an angle.

94. Equations of angle bisectors (interior and exterior).

93. The Appolonius' construct problems.

92. CA , CB tangent to parabola - OIMU 2008 Problem 4.

91. JBMO 2010, Problem 3.

90. Nice and difficult inequality in ABC (own).

89. Similarity (parallel/antiparallel).

88. A nice geometric locus conditioned metrically.

87. An interesting vectorial identity.

86. A very nice maxmin.

85. Problem "slicing" : 60-24-12.

84. IRAN national math olympiad - 2010

83. A metrical characterization of concurrence.

82. AA' , BN , CM concur in Gergonne point.

81. Sawayama and Thebault’s theorem.

80. Three concurrent perpendiculars (the Carnot's lemma).

79. A quadrilateral with many perpendicularities.

78. Jakobi's theorem.

77. Two equivalent perpendicularities.

76. A parallelogram and a circle (nice !).

75. Extended Steinbart Theorem.

74. Inegalitatea Erdos-Mordell.

73. A nice problems with a square and a perpendicularity.

July 2010

72. Colinearity in a triangle - H , P , M .

70. Colinearity with pole/polar - H\in PQ .

71. An angle questions (synthetic/trigonometric proofs).

69*1. Slicing problem 3.

69. Position of x1, x2 w.r.t. a given real number k.

68. Some nice applications of "pole/polar".

67. Remarkable algebraic/geometrical inequalities (1).

66. Remarkable perpendicularities in a triangle.

65. Parallel to BC through I .

64. Problems of the type "instant" (at most one minute).

63. A nice problem with radical center (livetolove212).

62. Nice and easy problem with a right triangle.

61. IMO Shortlist 2009 (very nice problem and its proof).

60. Mediteranean M.C. 2010 Problem 3.

59. IMO 2010, problem 4.

58. A cevian and two incircles.

57. Opinii si comentarii relativ la inv. rom. II

56. A extremum problem with areas.

55. Two important circles - mixtilinear incircle/excircle.

54. Cinci grade de libertate.

53. An remarkable concurrency.

52. Harmonical division.

51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.

51. A nice problem with perpendicularity from Jaime.

50. IMAC 2010 seniors, the first day.

49. The midpoint of a cevian in an acute triangle.

June 2010

48. An inequality in a triangle : h_a+max{b,c} ...

47. Interesting and difficult identities in a triangle.

46. Outside of a quadrilateral.

45. \cos B+\cos C=1 <==> nice property.

44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.

43. A "slicing" problem with a parameter.

42. IMAC 2010 Problema 2.

41. ONM 2010 Calarasi Problema 3.

May 2010

40. Lema lui Haruki.

39. A nice and remarkable problem with Brocard's point.

38. Minimum area of triangle touching with semicircle.

37. Own extension of the Steiner-Lehmus' problem.

36. ONM (juniori) - 2010 Iasi, problema 4.

35. Nice problem from Arqady (Eduard_Tur).

34. IMO Shortlist 2002 geometry problem 7.

April 2010

32. Linkuri diverse.

31.Some nice metrical problems.

29. Inegalitate integrala de la Kunny.

28. A fost odată ...

27.Problem of Darij Grinberg (I - centroid of triangle DEF).

26. Outside of a triangle.

25. Geometric equations.

24. A fine and cool inequalities.

23. A very nice exponential inequality (own - from CRUX).

22. Some interesting limits (sequence/function).

21 bis. Some problems with complex numbers.

21. Probleme de geometrie (Yetty & I).

20. O ineg.cu suma de AG/AI .

19. Own proposed problems in CRUX Mathematicorum - Canada.

18. O problema a lui Toshio Seimiya.

17. O inegalitate "tare" intr-un triunghi.

16. Length of the (interior) angle bisector (10 proofs).

15. Inequalities stronger than Panaitopol's.

14. Opinii si comentarii relativ la inv. rom.

13. Inegalitatea de la Sorin Borodi.

12. Inegalitatea I. Maftei & S. Radulescu (2).

11. Inegalitatea I. Maftei & S. Radulescu (1).

10. Walker's inequality and its extension.

9. Inegalitatea HO+HA+HB+HC >= 3R .

8. The first problem Shortlist ONM 2010 Romania.

7. Some interesting "slicing" problems.

6. Interesting problems from JBMO.

5. Theoretical preliminary for inequalities.

4. Remarkable geometrical inequalities (2).

Submit

orzzzzzzzzz
by mathMagicOPS, Jan 9, 2025, 3:40 AM
this css is sus
by ihatemath123, Aug 14, 2024, 1:53 AM
391345 views moment
by ryanbear, May 9, 2023, 6:10 AM
We need virgil nicula to return to aops, this blog is top 10 all time.
by OlympusHero, Sep 14, 2022, 4:44 AM
blog
by tigerzhang, Aug 1, 2021, 12:02 AM
Amazing blog.
by OlympusHero, May 13, 2021, 10:23 PM
the visits tho
by GoogleNebula, Apr 14, 2021, 5:25 AM
Bro this blog is ripped
by samrocksnature, Apr 14, 2021, 5:16 AM
Holy- Darn this is good. shame it's inactive now
by the_mathmagician, Jan 17, 2021, 7:43 PM
godly blog. opopop
by OlympusHero, Dec 30, 2020, 6:08 PM
long blog
by MrMustache, Nov 11, 2020, 4:52 PM
372554 views!
by mrmath0720, Sep 28, 2020, 1:11 AM
wow... i am lost.

369302 views!

-piphi
by piphi, Jun 10, 2020, 11:44 PM
That was a lot! But, really good solutions and format! Nice blog!!!!
by CSPAL, May 27, 2020, 4:17 PM
impressive
awesome. 358,000 visits?????
by OlympusHero, May 14, 2020, 8:43 PM
Nice blog.
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

382. Peruan "jewels" (Israel Diaz, M. O. Sanchez a.s.o.)

by Virgil Nicula, Jul 30, 2013, 2:00 PM

0 Comments