385. Some nice geometry problems.

by Virgil Nicula, Aug 17, 2013, 2:27 AM

PP1. Let $P$ be an interior point of $\triangle ABC$ for which let $\left\{\begin{array}{c} M\in BP\cap AC\\\\ E\in AP\cap BC\\\\ D\in CP\cap BA\end{array}\right\|$ . For $F\in (AE)$ and $G\in (CD)$ construct

$\left\{\begin{array}{c} H\in BF\cap AC\\\\ I\in BG\cap AC\\\\ L\in HE\cap ID\end{array}\right\|$ . Prove that $\boxed{L\in BP\iff \frac {FA}{FE}\cdot [CME]=\frac {GC}{GD}\cdot [AMD]}$ , where $[XYZ]$ is the area of $\triangle XYZ$ .

Remarks.

$\blacktriangleright$ IF $P$ is the centroid of $\triangle ABC$ , THEN there is the equivalence $\boxed{L\in BP\iff FG\parallel AC}\ .$

$\blacktriangleright$ IF $P$ is the incenter of $\triangle ABC$ , THEN there is the equivalence $\boxed{L\in BP\iff \frac {FA}{FE}\cdot \frac {PE}{PA}=\frac {GC}{GD}\cdot\frac{PD}{PC}}\ .$

$\blacktriangleright$ IF $P$ is orthocenter of $\triangle ABC$ , THEN there is the equivalence $\boxed{L\in BP\iff \frac {FA}{FE}\cdot \cos^2C=\frac {GC}{GD}\cdot\cos^2A}$ . (Juan Felipe Crisostomo Ramos).

Proof. Let $\left\{\begin{array}{c} X\in ID\cap BP\\\ Y\in HE\cap BP\end{array}\right\|$ and $\left\{\begin{array}{ccc} \frac {DB}{m}=\frac {DA}{1}=\frac {BA}{m+1} & ; & \frac {EB}{n}=\frac {EC}1=\frac {BC}{n+1}\\\\ \frac {FA}p=\frac {FE}1=\frac {AE}{p+1} & ; & \frac {GC}{q}=\frac {GD}1=\frac {CD}{q+1}\end{array}\right\|$ . Apply the Ceva's theorem to $P/\triangle ABC\ :\ \frac {MA}{MC}\cdot\frac {EC}{EB}\cdot\frac {DB}{DA}=1\iff$

$\frac {MA}{n}=\frac {MC}{m}=\frac {AC}{m+n}$ . Apply the Menelaus' theorem to $:\ \left\{\begin{array}{cccc} \overline{BFH}/\triangle AEC\ : & \frac {BE}{BC}\cdot\frac {HC}{HA}\cdot\frac {FA}{FE}=1 & \iff & \frac {HC}{n+1}=\frac {HA}{np}=\frac {AC}{n+1+np}\\\\ \overline{BGI}/\triangle ADC\ : & \frac {BD}{BA}\cdot\frac {IA}{IC}\cdot\frac {GC}{GD}=1 & \iff & \frac {IA}{m+1}=\frac {IC}{mq}=\frac {AC}{m+1+mq}\\\\ \overline{IXD}/\triangle ABM\ : & \frac {IM}{IA}\cdot\frac {DA}{DB}\cdot\frac {XB}{XM}=1 & \iff & \frac {XB}{XM}=\frac {(m+n)(m+1)}{m+1-nq}\\\\ \overline{HYE}/\triangle CBM\ : & \frac {HM}{HC}\cdot\frac {EC}{EB}\cdot\frac {YB}{YM}=1 & \iff & \frac {YB}{YM}=\frac {(m+n)(n+1)}{n+1-mp}\end{array}\right\|$ .

In conclusion, $L\in PB\iff$ $X\equiv Y\iff$ $\frac {XB}{XM}=\frac {YB}{YM}\iff$ $\frac {m+1}{m+1-nq}=\frac {n+1}{n+1-mp}\iff$ $\boxed{mp(m+1)=nq(n+1)}\ (*)\iff$

$\frac {FA}{FE}\cdot\frac {CM\cdot CE}{a}=\frac {GC}{GD}\cdot\frac {AM\cdot AD}{c}\iff$ $\frac {FA}{FE}\cdot\frac {2[CME]}{a\cdot\sin C}=\frac {GC}{GD}\cdot\frac {2[AMD]}{c\cdot\sin A}\iff$ $\frac {FA}{FE}\cdot [CME]=\frac {GC}{GD}\cdot [AMD]$ because $\frac {\sin A}{a}=\frac {\sin C}{c}$ .

IF $P:=H$ , THEN $\left\{\begin{array}{ccc} 2[CME]=CM\cdot CE\cdot\sin C=a\cos C\cdot b\cos C\cdot\sin C=2S\cdot \cos^2C\\\\ 2[AMD]=AM\cdot AD\cdot\sin A=c\cos A\cdot b\cos A\cdot\sin A=2S\cdot \cos^2A\end{array}\right\|$ , where $S=[ABC]$ and $2S=ab\sin C=bc\sin A$ .

Remark. The relation $(*)$ is equivalently with $\boxed{\frac {FA}{FE}\cdot\frac {CE}{CB}\cdot MC=\frac {GC}{GD}\cdot\frac {AD}{AB}\cdot MA}$ .If $P$ belongs to the $B$ -bisector, then previous relation becomes $\boxed{CE\cdot\frac {FA}{FE}=AD\cdot\frac {GC}{GD}}$ .

PP2 (Julio Orihuela Bastidas). Let $\triangle ABC$ with the circumcircle $w=C(O,R)$ , the incircle $C(I,r)$ and the $A$ -excircle $C(I_a,r_a)$ .

Let the Euler's lines $d$ , $d_a$ of the triangles $BIC$ , $BI_aC$ respectively and $M\in IO\cap d$ , $N\in I_aO\cap d_a$ . Prove that $\frac {MI}{MO}=\frac {NI_a}{NO}$ .

Proof. Let the midpoint $D$ of $[BC]$ and $\{A,S\}=\{A,I\}\cap w$ . Is well-known that $S$ is the midpoint of $[II_a]$ and $BICI_a$ is inscribed in the circle with the diameter $[II_a]$ . Thus,

$S$ is the common circumcenter for $\triangle BIC$ and $BI_aC$ . Let the centroid $G$ of $\triangle BIC$ , i.e. $G\in (ID)$ and $GI=2\cdot GD$ . Let the centroid $G_a$ of $\triangle BI_aC$ , i.e. $G_a\in (I_aD)$ and

$G_aI_a=2\cdot G_aD$ . Therefore, $d\equiv GS$ and $d_a\equiv G_aS$ . Apply the Menelaus' theorem to the transversals: $\left\{\begin{array}{cccccc} \overline{SGM}/\triangle DIO\ : & \frac {SD}{SO}\cdot\frac {MO}{MI}\cdot\frac {GI}{GD}=1 & \implies & \frac {MI}{MO}=2\cdot \frac {SD}{SO}\\\\ \overline{S_aN}/\triangle DI_aO\ : & \frac {S_aD}{SO}\cdot\frac {NO}{NI_a}\cdot\frac {G_aI_a}{G_aD}=1 & \implies & \frac {NI_a}{NO}=2\cdot \frac {SD}{SO}\end{array}\right\|$

$\implies$ $\boxed{\frac {MI}{MO}=\frac {NI_a}{NO}=2\cdot \frac {SD}{SO}=\frac {r_a-r}{R}}$ .

PP3. Let $\triangle ABC$ with incenter $I$ , $M\in (AB)$ , $N\in (AC)$ so that $I\in MN$ . Know the value of $\angle BAC$ and $\left\{\begin{array}{c} AB=c\ ;\ AC=b\\\\ IM=m\ ;\ IN=n\end{array}\right\|$ . Find $BC=a=f(b,c,m,n)$ .

Proof. Let $AM=x$ and $AN=y$ . Thus, $MB=c-x$ and $NC=b-y$ . Since $\frac xm=\frac yn=\lambda$ and using an well-known property $b\cdot\frac {MB}{MA}+c\cdot\frac {NC}{NA}=a$ obtain $\frac {b(c-x)}{x}+$

$\frac {c(b-y)}{y}=a\iff$ $\frac 1x+\frac 1y=\frac {2s}{bc}$ , where $a+b+c=2s$ . Thus, $\frac 1{m\lambda}+\frac 1{n\lambda}=\frac {2s}{bc}\iff$ $\boxed{\frac{\lambda}{bc} =\frac {m+n}{2mns}}\ (*)$ .Apply the generalized Pythagoras' theorem in $\triangle MAN\ :$

$(m+n)^2=x^2+y^2-2xy\cos A\stackrel{(*)}{\iff}$ $(2mns)^2=(bc)^2\left(m^2+n^2-2mn\cos A\right)\iff$ $\left\{\begin{array}{c} a+b+c=\frac {bc}{mn}\sqrt {m^2+n^2-2mn\cos A}\\\\ b^2+c^2=a^2+2bc\cos A\end{array}\right\|\implies a=f(b,c,m,n)$ .

Particular case. IF $a=b=c$ , THEN $\boxed{a=\frac {3mn}{\sqrt{m^2+n^2-mn}}}$ (Miguel Ochoa Sanchez).

PP4. Let $\triangle ABC$ with $A$ -bisector $AD$ , where $D\in BC$ . For $P\in (AD$ (ray) let $\left\{\begin{array}{c} M\in AB\cap CP\\\\ N\in AC\cap BP\end{array}\right\|$ . Prove that $\frac {1}{AB^2}-\frac 1{AC^2}=$ $\frac {1}{AM^2}-\frac 1{AN^2}$ $\iff$ $AB=AC$ .

Proof. Denote $AM=m$ and $AN=n$ . Observe that $(m-c)(n-b)\ge 0$ . Apply the Ceva's theorem to $P/\triangle ABC\ :\ \frac {DB}{DC}$ $\cdot \frac {NC}{NA}\cdot\frac {MA}{MB}=1\iff$

$\frac cb\cdot\frac {n-b}{n}\cdot\frac {m}{m-c}=1\iff$ $mc(n-b)=nb(m-c)\iff$ $\boxed{\frac 1c-\frac 1m=\frac 1b-\frac 1n}\ (*)$ . Therefore, $\frac {1}{AB^2}-\frac 1{AC^2}=$ $\frac {1}{AM^2}-\frac 1{AN^2}$ $\iff$

$\frac 1{c^2}-\frac 1{b^2}=\frac 1{m^2}-\frac 1{n^2}\iff$ $\frac 1{c^2}-\frac 1{m^2}=\frac 1{b^2}-\frac 1{n^2}\stackrel{(*)}{\iff}$ $\frac 1c+\frac 1m=\frac 1b+\frac 1n\iff$ $\left\{\begin{array}{c} \frac 1c-\frac 1m=\frac 1b-\frac 1n\\\\ \frac 1c+\frac 1m=\frac 1b+\frac 1n\end{array}\right\|\bigoplus\iff$ $\frac 2c=\frac 2b\iff c=b$ .

PP5. Let $\triangle{ABC}$ with midpoint $D$ of $[BC]$ , midpoint $M$ of $[AD]$ and $N\in BM\cap AC$ . Show $AB$ is tangent to circuncircle of $\triangle{NBC}\iff$ $\frac{{BM}}{{MN}} =\frac{({BC})^2}{({BN})^2}$ .

Proof (without the Menelaus' theorem). Let the middlepoint $P$ of $[CN]$ . Thus, $\left\{\begin{array}{ccccc} \triangle BNC\ :\ DB=DC\ ,\ PN=PC & \implies & DP\parallel BN & \implies & \boxed {BN=2\cdot DP}\\\\ \triangle ADP\ :\ MA=MD\ ,\ MN\parallel DP & \implies & NA=NP & \implies & \boxed {DP=2\cdot MN}\end{array}\right\|$ and

$\left\{\begin{array}{ccccc} BN=2\cdot DP=4\cdot MN & \iff & BM+MN=4\cdot MN &\iff & \boxed {\frac{MB}{MN}=3}\\\\ NA=NP=PC & \iff & AC=3\cdot AN & \iff & \boxed {\frac{MB}{MN}=\frac{AC}{AN}=3}\end{array}\right\|$ . In conclusion, $AB$ is tangent to the

circumcircle of $\triangle BNC\iff$ $\triangle ABC\sim \triangle ANB$ $\iff$ $\frac{AB}{AN}=\frac{BC}{NB}=\frac{CA}{BA}$ $\iff$ $\left(\frac{BC}{NB}\right)^2=\frac{AB}{AN}\cdot \frac{CA}{BA}$ $\iff$ $\left(\frac{BC}{BN}\right)^2=\frac{AC}{AN}$ $\iff$ $\left(\frac{BC}{BN}\right)^2=\frac{MB}{MN}\ .$

PP6. Let $H$ be the orthocenter of an acute $\triangle ABC$ . The circle centered at the midpoint of $BC$ and passing through $H$ intersects

$BC$ at $A_{1}$ , $A_{2}$ . Similarly define the pairs $B_{1}$ , $B_{2}$ and $C_{1}$ , $C_{2}$ . Prove that $A_{1}$ , $A_{2}$ , $B_{1}$ , $B_{2}$ , $C_{1}$ , $C_{2}$ are concyclically.

Proof. Let the midpoints $D$ , $E$ , $F$ of $[BC]$ , $[CA]$ , $[AB]$ respectively. Prove easily $\boxed {\ HA^2 + a^2 = HB^2 + b^2 = HC^2 + c^2 = 4R^2\ }\ (*)$ .

Thus, $4\cdot\overline {BA_1}\cdot\overline {BA_2} = 4\cdot\left(BD^2 - HD^2\right) =$ $a^2 - 4\cdot HD^2 =$ $a^2 - 2\cdot\left(HB^2 + HC^2\right) + a^2$ $\stackrel {(*)}{\ \ \implies\ \ }$ $\overline {BA_1}\cdot\overline {BA_2} = \frac 12\cdot \left(a^2 + b^2 + c^2\right) - 4R^2$ $\implies$

$\boxed {\ \overline {BA_1}\cdot\overline {BA_2} = 4R^2\prod\cos A\ }$ because prove easily $\boxed {\ \left(a^2 + b^2 + c^2\right) - 8R^2 = 8 R^2\prod\cos A\ } > 0$ (remark that $\{A_1,A_2\}\subset (BC)$ a.s.o.). Thus,

$\overline {BA_1}\cdot \overline {BA_2} = - \overline {A_1B}\cdot\overline {A_1C} = R^2 - OA_1^2\ \implies\ 4R^2\prod\cos A = R^2 - OA_1^2$ $\implies$ $OA_1^2 = R^2\left(1 - 4\prod\cos A\right)$ (symmetrically in $a$ , $b$ , $c$ ).

Thus, $\boxed {\ \rho = OA_1 = OA_2 = OB_1 = OB_2 = OC_1 = OC_2 = R\sqrt {1 - 4\cos A\cos B\cos C}\ }$ .

PP7. Let $\triangle ABC$ with circumcircle $w$ and the symmedian point $S$ . For $P\in w$ let $\left\{\begin{array}{c} X\in BC\cap PS\\\ Y\in CA\cap PS\\\ Z\in AB\cap PS\end{array}\right\|$ . Prove that $\frac{3}{\overline{PS}}=\frac{1}{\overline{PX}}+\frac{1}{\overline {PY}}+\frac{1}{\overline{PZ}}$ (segments are orientated).

Proof. Let $\delta_d(X)$ - the distance of $X$ to $d$ . Thus, $\frac{3}{\overline{PS}}=\frac{1}{\overline{PX}}+\frac{1}{\overline {PY}}+\frac{1}{\overline{PZ}}\iff$ $3=\sum\frac {\overline{PS}}{\overline{PX}}\iff$ $3=\sum\frac {\overline{PX}-\overline{SX}}{\overline{PX}}\iff$ $\boxed{\sum\frac {\overline{SX}}{\overline{PX}}=0}\ (*)\ \iff$

$\sum\frac {\delta_{BC}(S)}{\delta_{BC}(P)}=0\iff$ $\sum\frac {a}{\delta_{BC}(P)}=0\ (1)$ because is well-known that $\frac { \delta_{BC}(S)}{a}=\frac { \delta_{CA}(S)}{b}=\frac { \delta_{AB}(S)}{c}$ . The relation $(1)$ is barycentric orientated. (see PP16 from here).

Generalization. Let $\triangle ABC$ with circumcircle $w$ and let $M$ with barycentric coordinates $(x,y,z)$ w.r.t. $\triangle ABC$ . Let isogonal conjugate $S$

of $M$ w.r.t. $\triangle ABC$ and $P\in w$ . The line $PS$ cuts $BC$ , $CA$ , $AB$ at $X$ , $Y$ , $Z$ respectively. Prove that $\dfrac{x+y+z}{\overline{PS}}=\dfrac{x}{\overline{PX}}+\dfrac{y}{\overline{PY}}+\dfrac{z}{\overline{PZ}}$ .

PP8. Let $\triangle ABC$ with the incenter $I$ . Let $\left\{\begin{array}{c} D\in BI\cap AC\\\ E\in CI\cap AB\end{array}\right\|$ and suppose that $\left\{\begin{array}{c} m\left(\widehat{CED}\right)=18^{\circ}\\\\ m\left(\widehat{BDE}\right)=24^{\circ}\end{array}\right\|$ . Prove that $B=12^{\circ}$ .

Proof. Let $B=2x$ and apply theorem of Sines in $:\ \left\{\begin{array}{cccc} \triangle DIE\ : & \frac {ID}{IE}=\frac {\sin\widehat{IED}}{\sin\widehat{IDE}} & \implies & \frac {ID}{IE}=\frac {\sin 18^{\circ} }{\sin 24^{\circ}}\\\\ \triangle IAD\ : & \frac {IA}{ID}=\frac {\sin\widehat{IDA}}{\sin\widehat{IAD}} & \implies & \frac {IA}{ID}=\frac {\sin (84^{\circ}-x)}{\sin \frac A2}\\\\ \triangle IAE\ : & \frac {IE}{IA}=\frac {\sin\widehat{IAE}}{\sin\widehat{IEA}} & \implies & \frac {IE}{IA}=\frac {\sin \frac A2}{\sin (42^{\circ}+x)}\end{array}\right\|\bigodot\implies$ $\sin 18^{\circ}\sin (84^{\circ}-x)=\sin 24^{\circ}\sin (42^{\circ}+x)\iff$

$\boxed{f(x)\equiv\sin 18^{\circ} \cos (6^{\circ}+x)-\sin 24^{\circ}\cos (48^{\circ}-x)=0}\ (*)$ . The function $f$ on $\left(0,48^{\circ}\right)$ is strict decreasing and $2f(6)=2\sin 18^{\circ}\cos 12^{\circ}-2\sin 24^{\circ}\cos 42^{\circ}=$

$\sin 30^{\circ}+\sin 6^{\circ}-\sin 66^{\circ}+\sin 18^{\circ}=$ $\frac 12+\sin 18^{\circ}-2\sin 30^{\circ}\cos 36^{\circ}\implies$ $2f(x)=\frac 12+\sin 18^{\circ}-\cos 36^{\circ}=$ $\frac 12+\frac {\sqrt 5-1}{4}-\frac {\sqrt 5+1}{4}=0\implies$ $f(6^{\circ})=0$ , i.e. $B=12^{\circ}$

Lemma. Let a quadrilateral $ABCD$ which is inscribed in the circle $w$ and $\left\{\begin{array}{c} E\in AB\cap CD\\\\ F\in AD\cap BC\end{array}\right\|$ . Prove that $p_w(E)+p_w(F)=EF^2$ , where $p_w(X)$ is the power of $X$ w.r.t. $w$ .

Proof 1. Let $\left\{\begin{array}{c} m\left(\widehat{CEF}\right)=u\\\\ m\left(\widehat{CFE}\right)=v\end{array}\right\|$ . Thus $u+v=A$ and $p_w(E)+p_w(F)=EF^2\iff$ $EC\cdot ED+FC\cdot FB=EF^2\iff$ $\frac {EC}{EF}\cdot\frac {ED}{EF}+\frac {FC}{FE}\cdot\frac {FB}{FE}=1\iff$

$\frac {\sin v}{\sin A}\cdot\frac {\sin (D-u)}{\sin B}+\frac {\sin u}{\sin A}\cdot\frac {\sin (B-v)}{\sin D}=1\iff$ $\sin v\sin (D-u)+\sin u\sin (B-v)=\sin A\sin B\iff$ $\cos (D-u-v)-\cos (D+v-u)+$

$\cos (B-u-v)-\cos (B+u-v)=$ $\cos (A-B)-\cos (A+B)\iff$ $\cos (D-A)-\cos (D+v-u)+$ $\cos (B-A)-\cos (B+u-v)=$ $\cos (A-B)-\cos (A+B)$ ,

what is truly because $\left\{\begin{array}{c} A+C=B+D=180^{\circ}\\\\ (D-A)+(A+B)=180^{\circ}\\\\ (D+v-u)+(B+u-v)=180^{\circ}\end{array}\right\|$ .

Proof 2. Let $\left\{\begin{array}{ccc} AB=a & BC=b & CD=c\\\\ DA=d & AC=e & BD=f\end{array}\right\|$ and $G\in AC\cap BD$ . I"ll use the relations $\left\{\begin{array}{cc} \frac {GA}{da}=\frac {GB}{ab}=\frac {GC}{bc}=\frac {GD}{cd}=\frac {e}{da+bc}=\frac f{ab+cd}=\sqrt{\frac {-p_w(G)}{abcd}}\\\\ \frac {EA}{de}=\frac {EB}{bf}=\frac {EC}{be}=\frac {ED}{df}=\frac {a}{de-bf}=\frac {c}{df-be}=\sqrt{\frac {p_w(E)}{bdef}}\\\\ \frac {FA}{ae}=\frac {FB}{af}=\frac {FC}{ce}=\frac {FD}{cf}=\frac {d}{ae-cf}=\frac {b}{af-ce}=\sqrt {\frac {p_w(F)}{acef}}\end{array}\right\|$ . Thus, $\left\{\begin{array}{c} p_w(E)=\frac {a^2bdef}{(de-bf)^2}\\\\ p_w(F)=\frac {acd^2ef}{(ae-cf)^2}\end{array}\right\|\bigodot\implies$ $\boxed{p_w(E)+p_w(F)=adef\cdot\left[\frac {ab}{(de-bf)^2}+\frac {cd}{(ae-cf)^2}\right]}\ (1)$ . Apply the generalized Pythagoras' theorem to $[EF]$ of $\triangle EAF\ :$

$EF^2=$ $AE^2+AF^2-2\cdot AE\cdot AF\cdot\cos A\iff$ $EF^2=\left(\frac {ade}{de-bf}\right)^2+\left(\frac {ade}{ae-cf}\right)^2-2\cdot \frac {ade}{de-bf}\cdot \frac {ade}{ae-cf}\cdot\frac {a^2+d^2-f^2}{2ad}\iff$

$\boxed{EF^2=ade\cdot\frac {ade\left[(ae-cf)^2+(de-bf)^2\right]-e(de-bf)(ae-cf)\left(a^2+d^2-f^2\right)}{(de-bf)^2(ae-cf)^2}}\ (2)$ . Thus, $p_w(E)+p_w(F)=EF^2\iff$ $adef\cdot\left[\frac {ab}{(de-bf)^2}+\frac {cd}{(ae-cf)^2}\right]=$

$ade\cdot\frac {ade\left[(ae-cf)^2+(de-bf)^2\right]-e(de-bf)(ae-cf)\left(a^2+d^2-f^2\right)}{(de-bf)^2(ae-cf)^2}\iff$ $f\left[ab(ae-cf)^2+cd(de-bf)^2\right]=$ $ade\left[(ae-cf)^2+(de-bf)^2\right]-$

$e(de-bf)(ae-cf)\left(a^2+d^2-f^2\right)\iff$ $f\left[ab(ae-cf)^2+cd(de-bf)^2\right]-$ $ade\left[(ae-cf)^2+(de-bf)^2\right]+$ $e(de-bf)(ae-cf)\left(a^2+d^2-f^2\right)=0\iff$

$a(bf-de)(ae-cf)^2+d(cf-ae)(de-bf)^2+$ $e(de-bf)(ae-cf)\left(a^2+d^2-f^2\right)=0\iff$ $a(ae-cf)+d(de-bf)-$ $e\left(a^2+d^2-f^2\right)=0\iff$

$a^2e-acf+d^2e-bdf-a^2e-d^2e+ef^2=0\iff$ $ef=ac+bd$ (Ptolemy's theorem), what is true.

PP9 (Edson Curahua). Let $\triangle ABC$ with the centroid $G$ . Prove that $m\left(\widehat{AGB}\right)=2C\ \iff\ c^4=a^4+b^4-a^2b^2$ .

Proof 1 (Josue Garcia Piscoya). Let $\left\{\begin{array}{c} E\in AL\cap BC\\\\ F\in BL\cap AC\end{array}\right\|$ , where $L$ is the Lemoine's point of $\triangle ABC$ . Prove easily that $CELF$ is cyclically (with the circumcircle $w$ ) and

$\left\{\begin{array}{c} p_w(B)=BE\cdot BC\\\\ p_w(A)=AF\cdot AC\end{array}\right\|$ . Thus, $\left\{\begin{array}{ccc} \frac {EB}{c^2}=\frac {EC}{b^2}=\frac a{b^2+c^2} & \implies & p_w(B)=\frac {a^2c^2}{b^2+c^2}\\\\ \frac {FA}{c^2}=\frac {FC}{a^2}=\frac b{a^2+c^2} & \implies & p_w(A)=\frac {b^2c^2}{a^2+c^2}\end{array}\right\|$ . Apply upper lemma to the cyclic $CELF$ and get $p_w(B)+p_w(A)=AB^2\iff$

$\frac {a^2c^2}{b^2+c^2}+\frac {b^2c^2}{a^2+c^2}=c^2\iff$ $\frac {b^2}{a^2+c^2}+\frac {a^2}{b^2+c^2}=1\iff$ $b^2\left(b^2+c^2\right)+a^2\left(a^2+c^2\right)=\left(a^2+c^2\right)\left(b^2+c^2\right)\iff$ $c^4=a^4+b^4-a^2b^2$ .

Proof 2. I"ll use the well-known relation in any $\triangle ABC\ :\ \boxed{4S=\left(a^2+b^2-c^2\right)\tan C}$ . Prove easily that $GA^2+GB^2-c^2=\frac 19\left(a^2+b^2-5c^2\right)$ . Thus,

$3[AGB]=[ACB]\iff$ $3\left(GA^2+GB^2-c^2\right)\tan 2C=$ $\left(a^2+b^2-c^2\right)\tan C\iff$ $\frac 13\left(a^2+b^2-5c^2\right)\cdot\frac {2\tan C}{1-\tan^2C}=$ $\left(a^2+b^2-c^2\right)\tan C\iff$

$2\left(a^2+b^2-5c^2\right)=3\left(1-\tan^2C\right)\left(a^2+b^2-c^2\right)\iff$ $3\left(a^2+b^2-c^2\right)\tan^2C=a^2+b^2+7c^2\iff$ $12S\tan C=a^2+b^2+7c^2\iff$

$48S^2=\left(a^2+b^2+7c^2\right)\left(a^2+b^2-c^2\right)\iff$ $3\left(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4\right)=$ $a^4+b^4+2a^2b^2-7c^4+6c^2\left(a^2+b^2\right)\iff$ $c^4=a^4+b^4-a^2b^2$ .

Proof 3. Let the midpoints $D$ , $E$ of $[CB]$ , $[CA]$ respectively and $\left\{\begin{array}{ccc} m\left(\widehat{GAB}\right)=x & \implies & m\left(\widehat{GAC}\right)=A-x\\\\ m\left(\widehat{GBA}\right)=y & \implies & m\left(\widehat{GBC}\right)=B-y\end{array}\right\|$ . Thus $x+y+2C=180^{\circ}\implies$

$\boxed{\tan 2C=\frac {\tan x+\tan y}{\tan x\tan y -1}}\ (*)$ . Thus, $\left\{\begin{array}{ccccccc} DB=DC & \implies & c\sin x=b\sin (A-x) & \implies & c\tan x=b(\sin A-\cos A\tan x) & \implies & \tan x=\frac {b\sin A}{c+b\cos A}\\\\ EA=EC & \implies & c\sin y=a\sin (B-y) & \implies & c\tan y=a(\sin B-\cos B\tan y) & \implies & \tan y=\frac {a\sin B}{c+a\cos B}\end{array}\right\|$ . Observe

that $\left\{\begin{array}{c} b\sin A=a\sin B\\\\ b\cos A+a\cos B=c\end{array}\right\|\implies$ $\left\{\begin{array}{c} (c+b\cos A)+(c+a\cos B)=3c\\\\ (c+b\cos A)\cdot(c+a\cos B)=2c^2+ab\cos A\cos B\end{array}\right\|$ . Thus, $\left\{\begin{array}{c} \tan x+\tan y=\frac {6S}{2c^2+ab\cos A\cos B}\\\\ \tan x\cdot\tan y=\frac {ab\sin A\sin B}{2c^2+ab\cos A\cos B}\end{array}\right\|$ .

From $(*)$ obtain that $\frac {2\tan C}{1-\tan^2C}=$ $\frac {6S}{ab\sin A\sin B-2c^2-ab\cos A\cos B}=$ $\frac {6S}{ab\cos C-2c^2}=\frac {12S}{a^2+b^2-5c^2}=$ $\frac {3\left(a^2+b^2-c^2\right)\tan C}{a^2+b^2-5c^2}\implies$

$2\left(a^2+b^2-5c^2\right)=3\left(1-\tan^2C\right)\left(a^2+b^2-c^2\right)\iff$ $3\left(a^2+b^2-c^2\right)\tan^2C=a^2+b^2+7c^2\iff$ $12S\tan C=a^2+b^2+7c^2\iff$

$48S^2=\left(a^2+b^2+7c^2\right)\left(a^2+b^2-c^2\right)\iff$ $3\left(2\sum b^2c^2-\sum a^4\right)=$ $a^4+b^4+2a^2b^2-7c^4+6c^2\left(a^2+b^2\right)\iff$ $c^4=a^4+b^4-a^2b^2$ .

PP10 (Josue Garcia Piscoya). Let $\triangle ABC$ with $\left\{\begin{array}{c} A'\in BC\\\\ B'\in CA\\\\ C'\in AB\end{array}\right|$ so that $A'B'C'\sim ABC$ and $\frac {BC}{B'C'}=k$ . Let $P\in BC\cap B'C'$ and $x=m\left(\widehat{B'PC}\right)$ . Prove that $\cos x=\frac k2$ .

Proof 1. Prove easily that $m\left(\widehat{BC'A'}\right)=A+x$ and $m\left(\widehat{CB'A'}\right)=A-x$ . The circumradii of $A'B'C'$ , $AB'C'$ , $A'BC'$ and $A'B'C$ have same length $r$ .

Thus, $B'C'=2r\sin A$ and $BC=A'B+A'C=2r[\sin (A+x)+\sin (A-x)]\iff$ $2kr\sin A=4r\sin A\cos x$ . Hence $\cos x=k/2$ .

Proof 2. Prove easily that $\left\{\begin{array}{ccc} m\left(\widehat{AC'B'}\right)=B-x & ; & m\left(\widehat{AB'C'}\right)=C+x\\\\ m\left(\widehat{BA'C'}\right)=C-x & ; & m\left(\widehat{BC'A'}\right)=A+x\\\\ m\left(\widehat{CB'A'}\right)=A-x & ; & m\left(\widehat{CA'B'}\right)=B+x\end{array}\right\|$ . The circumradii of $A'B'C'$ , $AB'C'$ , $A'BC'$ and $A'B'C$ have same length $r$ .

I"ll use the identities $\left\{\begin{array}{c} \sum\sin 2A=4\sin A\sin B\sin C\\\\ S=2R^2\sin A\sin B\sin C\ \ (1)\end{array}\right\|$ in any $\triangle ABC$ , where $R$ is the length of circumradius. Prove easily

$\boxed{\sum\sin A\sin (B-x)\sin (C+x)=(1+2\cos 2x)\sin A\sin B\sin C}\ (*)$ . Thus, $R=kr$ and $[ABC]=[A'B'C']+\sum[AB'C']\iff$ $k^2\prod\sin A=$

$\prod \sin A+\sum\sin A\sin (B-x)\sin (C+x)\stackrel{(*)}{\iff}$ $\left(k^2-1\right)\prod\sin A=(1+2\cos 2x)\prod\sin A\iff$ $k^2=2(1+\cos 2x)\iff$ $k=2\cos x\iff\cos x=\frac k2$ .

$\blacktriangleright$ Remark. $4\sum\sin A\sin (B-x)\sin (C+x)=$ $2\sum\sin A[\cos (B-C-2x)+\cos A]=$ $\sum\sin 2A+\sum 2\sin (B+C)\cos (B-C-2x)=$

$\sum\sin 2A+\sum 2\sin (B+C)[\cos (B-C)\cos 2x+\sin (B-C)\sin 2x]=$ $4\prod\sin A+\cos 2x\sum 2\sin (B+C)\cos (B-C)+$ $\sin 2x\sum 2\sin (B+C)\sin (B-C)=$

$4\prod\sin A+\cos 2x\sum (\sin 2B+\sin 2C)+\sin 2x\sum (\cos 2B-\cos 2C)=$ $4\prod\sin A+2\cos 2x\sum \sin 2A=$ $4\prod\sin A+8\cos 2x\prod\sin A=$

$4(1+2\cos 2x)\prod\sin A\implies\ (*)$ .

PP11 (Miguel Ochoa Sanchez). Let $\triangle ABC$ with the orthocenter $H$ . Let the midpoint $M$ of $[BC]$ and $Q\in AM$ so that $HQ\perp AM$ . Prove that $MQ\cdot MA=MB^2$ .

Proof 1. Let $AD$ , $BE$ , $CF$ be altitudes of $\triangle ABC$ , where $D\in BC$ , $E\in CA$ , $F\in AB$ . Let $S\in EF\cap BC$ . Is wellknown $SH\perp AM$ , i.e. $S\in HQ$ and $AQDS$ is cyclic ;

the division $(B,C;S,D)$ is harmonically and $MB=MC\implies$ $MB^2=MD.MS$ . Since $AQDS$ is cyclically obtain that $MD.MS=MQ.MA$ . Thus, $MQ.MA=MB^2$ .

Proof 2. I"ll use same notations from upper proof. Thus, $DHQM$ is cyclically $\implies AH.AD= AQ.AM = (AM-MQ).AM\implies$

$MQ.MA = AM^2 - AH.AD =$ $AD^2 + DM^2 - AH.AD =$ $DM^2 + DA.DH =$ $DM^2 + DB.DC = MB^2\implies$ $MQ.MA=MB^2$ .

PP12 (Miguel Ochoa Sanchez). Let a parallelogram $ABCD$ and let $M\in (AD)\ ,\ N\in (AB)$ so that $[CBN]=[CNM]=[CMD]=1$ . Prove that $[MAN]=\sqrt 5-2$ .

Proof. $\left\{\begin{array}{ccccc} MA=x & , & MD=y & ; & \frac xy=m\\\\ NA=u & , & NB=v & ; & \frac uv=n\end{array}\right\|$ and $S=[ABCD]=\sigma +\sum_{k=1}^3\sigma_k$ , where $\left\{\begin{array}{ccc} \sigma=[MAN] & ; & \sigma_1=[CBN]\\\\ \sigma_2=[CNM] & ; & \sigma _3=[CMD]\end{array}\right\|\implies$

$\frac {\sigma}{xu}=\frac {\sigma_1}{v(x+y)}=\frac {\sigma_3}{y(u+v)}=\frac {S}{2(x+y)(u+v)}=$ $\frac {S-\left(\sigma +\sigma_1+\sigma_3\right)}{2(x+y)(u+v)-\left[xu+v(x+y)+y(u+v)\right]}=$ $\frac {\sigma_2}{x(u+v)+yu}=p$ , where

$\boxed{2p=\sin A}\implies$ $\frac {\sigma}{mn}=\frac {\sigma_1}{m+1}=\frac {\sigma_2}{mn+m+n}=\frac {\sigma_3}{n+1}=\frac {S}{2(m+1)(n+1)}=2p$ . Thus, $\sigma_1=\sigma_2=\sigma_3\iff$ $m+1=mn+(m+n)=n+1\iff$

$m=n\ ,\ n^2=1-n\iff$ $m=n=\frac {\sqrt 5-1}{2}\implies$ $\sigma =\frac {n^2}{n+1}=\frac {1-n}{1+n}=\frac {1-\frac {\sqrt 5-1}{2}}{1+\frac {\sqrt 5-1}{2}}=\frac {3-\sqrt 5}{1+\sqrt 5}=\sqrt 5-2\implies$ $\boxed{\sigma=\sqrt 5-2}$ .

An easy extension (Virgil Nicula). Let $ABCD$ be a parallelogram with $M\in (AD)\ ,\ N\in (AB)$ .

Prove that $\ \ \boxed{[CBN]=x\ ,\ [CNM]=y\ ,\ [CMD]=z\ \implies\ [MAN]=-(x+z)+\sqrt {y^2+4xz}}$ .

PP13. Let a parallelogram $ABCD$ with $O\in AC\cap BD$ and $F\in (AO)$ so that $\widehat{FBA}\equiv\widehat{FBO}$ . Suppose that $\frac {AB}{AD}=\frac {BD}{AC}$ . Find the ratio $\frac {FC}{FA}$ .

Proof. Let $\left\{\begin{array}{c} AB=b\ ;\ AD=a\\\\ AC=e\ ;\ BD=f\end{array}\right\|$ . I"ll use Euler's relation $\boxed{e^2+f^2=2\left(a^2+b^2\right)}\ (*)$ . So $\frac {AB}{AD}=\frac {BD}{AC}\iff$ $\frac ba=\frac fe\iff$ $\frac ea=\frac fb=k\iff$ $\left\{\begin{array}{c} e=ak\\\ f=bk\end{array}\right\|\stackrel{(*)}{\implies}$

$k=\sqrt 2$ . Thus, $\widehat{FBA}\equiv\widehat{FBO}\iff$ $\frac {FA}{FO}=\frac {BA}{BO}\iff$ $\frac {FA}{2b}=\frac {FO}{f}=\frac e{2(2b+f)}\implies$ $\boxed{FA=\frac {eb}{2b+f}}\ (1)\implies$ $FC=AC-AF=e-\frac {eb}{2b+f}\implies$

$\boxed{FC=\frac {e(b+f)}{2b+f}}\ (2)\stackrel{(1\wedge 2)}{\implies}\ \frac {FC}{FA}=$ $1+\frac fb\implies$ $\boxed{\frac {FC}{FA}=1+\sqrt 2}$ .

Lemma. Prove $(\forall)\ x\in\mathbb R$ there are $g(x)\equiv\sin ^2x+\sin^2\left (60^{\circ}+x\right)+\sin^2\left (60^{\circ}-x\right)=\frac 32$ and $f(x)\equiv\sin ^4x+\sin^4\left (60^{\circ}+x\right)+\sin^4\left (60^{\circ}-x\right)=\frac 98$ .

Proof 1 (Ruben Dario). Let an equilateral $\triangle ABC$ with $BC=a$ and $P\in (BC)$ so that $\left\{\begin{array}{ccc} PB & = & m\\\ PC & = & n\\\ AP & = & p\end{array}\right\|\implies$ $\boxed{m+n=a}$ and Stewart's $\implies\boxed{a^2=p^2+mn}\ (*)$

$\blacktriangleright\ m^2+n^2+a^2 =$ $(m+n)^2-2mn+a^2 =$ $2a^2-2mn = 2(a^2-mn)\stackrel{(*)}{\implies}$ $\boxed{m^2+n^2+a^2=2p^2}\ (1)$

Otherwise. Apply the Pythagoras theorem $\left\{\begin{array}{ccc} \triangle ABP\ : & \implies & p^2=a^2+m^2-am\\\\ \triangle ACP\ : & \implies & p^2=a^2+n^2-an\end{array}\right\|\bigoplus\implies$ $2p^2=2a^2+m^2+n^2-a(m+n)\implies$ $m^2+n^2+a^2=2p^2$ .

$\blacktriangleright$ The circumcircle of $\triangle ABC$ meet again $AP$ in $R$ . Then $\frac {RB}{m}=\frac {RC}{n}=\frac {RA}{a}=\frac {p^2+mn}{pa}$ and $\left\{\begin{array}{c} a^2=RA^2+RB^2-RA\cdot RB\\\\ a^2=RA^2+RC^2-RA\cdot RC\end{array}\right\|\bigoplus\implies$ $\boxed{RA^2+RB^2+RC^2=2a^2}$ .

Otherwise. Let $a=R\sqrt 3$ . Let $m\left(\widehat {RAB}\right)=x$ . Then $\left\{\begin{array}{c} m\left(\widehat{RAC}\right)=60^{\circ}-x\\\\ m\left(\widehat{ACR}\right)=60^{\circ}+x\end{array}\right\|$ . Thus, $RA^2+RB^2+RC^2=2a^2\iff$

$4R^2\left[\sin^2 \left(60^{\circ}+x\right)+\sin^2 x+\sin ^2\left(60^{\circ}-x\right)\right]=6R^2\iff$ $\sin^2\left(60^{\circ}+x\right)+\sin^2 x+\sin ^2\left(60^{\circ}-x\right)=\frac 32$ .

$\blacktriangleright\ m^4+n^4+a^4 =$ $(m^2+n^2)^2-2m^2n^2+a^4 \stackrel{(1)}{=}$ $(2p^2-a^2)^2 - 2m^2n^2+a^4 =$ $4p^4-4a^2p^2+2a^4-2m^2n^2 =$

$4p^4+2a^2(a^2-2p^2)-2m^2n^2=$ $4p^4-2(p^2+mn)(p^2-mn)-2m^2n^2=2p^4\implies$ $\boxed{m^4+n^4+a^4=2p^4}\ (2)$ .

Let $m\left(\widehat{BAP}\right)=x$ and $\left\{\begin{array}{c} m\left(\widehat{CAP}\right)=60^{\circ}-x\\\\ m\left(\widehat{APC}\right)=60^{\circ}+x\end{array}\right\|$ . Apply the theorem ofSines in $:\ \left\{\begin{array}{ccccc} \triangle ABP\ : & \implies & k\cdot \frac mp & = & \sin x\\\\ \triangle ACP\ : & \implies & k\cdot\frac np & =& \sin\left(60^{\circ}-x\right)\\\\ \triangle ABP\ : & \implies & k\cdot\frac ap & = & \sin\left(60^{\circ}+x\right)\end{array}\right\|$ ,

where $k=\frac {\sqrt 3}{2}$ . In conclusion, $f(x)=\sin ^4x+\sin^4\left (60^{\circ}+x\right)+\sin^4\left (60^{\circ}-x\right)=$ $k^4\cdot\left(\frac {m^4+n^4+a^4}{p^4}\right)\stackrel{(2)}{\implies}$ $f(x)=2k^4=2\cdot \left(\frac {\sqrt 3}{2}\right)^4=\frac {2\cdot 9}{16}\implies$ $f(x)=\frac 98$ .
Proof 2. $2g(x)=\left(1-\cos 2x\right)+\left[1-\cos\left (120^{\circ}+2x\right)\right]+\left[1-\cos\left (120^{\circ}-2x\right)\right]\iff$ $2g(x)=3-\cos 2x-2\cos 120^{\circ}\cos 2x\iff$ $\boxed{g(x)=\frac 32}$ .

$f(x)=\sin ^4x+\cos^4\left (30^{\circ}-x\right)+\cos^4\left (30^{\circ}+x\right)\iff$ $4f(x)=(1-\cos 2x)^2+\left[1+\cos\left (60^{\circ}-2x\right)\right]^2+\left[1+\cos\left (60^{\circ}+2x\right)\right]^2\iff$ $4f(x)=(1-\cos 2x)^2+$

$\left[1+\sin\left (30^{\circ}+2x\right)\right]^2+$ $\left[1+\sin\left (30^{\circ}-2x\right)\right]^2\iff$ $4f(x)=3+2\left[\sin\left (30^{\circ}+2x\right)+\sin\left (30^{\circ}-2x \right)-\cos 2x\right]+$ $\cos^22x+\sin^2\left (30^{\circ}+2x\right)+\sin^2 \left (30^{\circ}-2x\right)\iff$

$4f(x)=3+2\left[2\sin 30^{\circ}\cos 2x-\cos 2x\right]+$ $\cos^22x+\sin^2\left (30^{\circ}+2x\right)+\sin^2 \left (30^{\circ}-2x\right)\iff$ $4f(x)=3+\cos^22x+\sin^2\left (30^{\circ}+2x\right)+\sin^2 \left (30^{\circ}-2x\right)\iff$

$8f(x)=6+3+\cos 4x-\cos \left(60^{\circ}+4x\right)-$ $\cos \left(60^{\circ}-4x\right)\implies$ $8f(x)=9+\cos 4x-2\cos 60^{\circ}\cos 4x\iff$ $\boxed{f(x)=\frac 98}$ .

PP14. Let an equilateral $\triangle ABC$ with $BC=a$ and centroid $G$ . For $d\ ,\ G\in d$ let $\left\{\begin{array}{c} M\in d\cap BC\\\\ N\in d\cap CA\\\\ P\in d\cap AB\end{array}\right\|$ . Find the sum $\frac 1{GM^4}+\frac 1{GN^4}+\frac 1{GP^4}$ .

Proof. Let the midpoints $D$ , $E$ and $F$ of the sides $[BC]$ , $[CA]$ and $[AB]$ respectively and $m\left(\widehat{GMC}\right)=x$ . Suppose w.l.o.g. $a=1$ and $B\in (MC)$ . Thus, $GD=\frac 1{2\sqrt 3}$ and

$\left\{\begin{array}{c} m\left(\widehat{MGF}\right)=30^{\circ}+x\\\\ m\left(\widehat{EGN}\right)=30^{\circ}-x\end{array}\right\|$ . Thus, $\frac 1k=GD=GM\cdot\sin x=GN\cdot \sin \left(30^{\circ}-x\right)=GP\cdot \sin \left(30^{\circ}+x\right)$ , where $\boxed{k=2\sqrt 3}$ . In conclusion, $\frac 1{GM^4}+\frac 1{GN^4}+\frac 1{GP^4}=$

$k^4\left[\sin ^4x+\cos^4\left (30^{\circ}-x\right)+\cos^4\left (30^{\circ}+x\right)\right]$ , Using the upper lemma obtain that $\frac 1{GM^4}+\frac 1{GN^4}+\frac 1{GP^4}=k^4f(x)=\frac {9k^4}{8}=144\cdot\frac 98\implies$ $\boxed{\frac 1{GM^4}+\frac 1{GN^4}+\frac 1{GP^4}=\frac {162}a}$

PP15. Let $\triangle ABC$ with the circumcircle $w=C(O,R)$ and the $A$ -excircle $w_a=C(I_a,r_a)$ a.s.o. Let the radical axes $\delta_a$ , $\delta_b$ , $\delta_c$ between $w$ and $A$ -excircle, $B$ -excircle and $C$ -excircle

respectively and $\left\{\begin{array}{ccc} B_1\in \delta_a\cap AB & ; & C_2\in \delta_a\cap AC\\\\ C_1\in \delta_b\cap BC & ; & A_2\in \delta_b\cap BA\\\\ A_1\in \delta_c\cap CA & ; & B_2\in \delta_c\cap CB\end{array}\right\|$ . Prove that $\left(\frac {1}{AB_1}+\frac 1{AC_2}\right) +\left(\frac {1}{BC_1}+\frac 1{BA_2}\right) +\left(\frac {1}{CA_1}+\frac 1{CB_2}\right)=\frac 8s$ , where $2s=a+b+c$ .

Proof. Denote the power $p_w(X)$ of the point $X$ w.r.t. the circle $w$ . Let $M$ be the midpoint of $[AC]$ and $T$ be the point where $w_a$ touches again $BC$ . Thus,

$C_2\in \delta_a\iff$ $p_w\left(C_2\right)=p_{w_a}\left(C_2\right)\iff$ $C_2O^2-R^2=C_2T^2\ (*)\iff$ $\left(OM^2+C_2M^2\right)-R^2=C_2T^2\iff$ $C_2M^2-C_2T^2=OC^2-OM^2\iff$

$(C_2M+C_2T)(C_2M-C_2T)=CM^2\iff$ $(AT-AM)(2\cdot C_2A-MA-s)=CM^2\iff$ $\left(s-\frac b2\right)\left(2\cdot AC_2-\frac b2-s\right)=\left(\frac b2\right)^2\iff$

$(a+c)\left[4\cdot AC_2-(a+2b+c)\right]=b^2\iff$ $\boxed{AC_2=\frac {s^2}{a+c}}$ . Prove analogously $\boxed{AB_1=\frac {s^2}{a+b}}$ . In conclusion, $\left(\frac {1}{AB_1}+\frac 1{AC_2}\right) +\left(\frac {1}{BC_1}+\frac 1{BA_2}\right) +$

$\left(\frac {1}{CA_1}+\frac 1{CB_2}\right)=$ $\sum\left(\frac {1}{AB_1}+\frac 1{AC_2}\right)=$ $\sum \left(\frac {a+b}{s^2}+\frac {a+c}{s^2}\right)=$ $\sum\frac {2a+b+c}{s^2}=\frac 8s$ .

PP16. Let $\triangle ABC$ and the circles $\left\{\begin{array}{cccc} w_a=C\left(A,h_a\right)\ : & \{A_1, A_2\}=BC\cap w_a & ; & A_2\in (BA_1)\\\\ w_b=C\left(B,h_b\right)\ : & \{B_1, B_2\}=CA\cap w_b & ; & B_2\in (CB_1)\\\\ w_c=C\left(C,h_c\right)\ : & \{C_1, C_2\}=AB\cap w_c & ; & C_2\in (AC_1)\end{array}\right\|$ . Let $\left\{\begin{array}{c} U\in BB_2\cap CC_1\\\\ V\in CC_2\cap AA_1\\\\ W\in AA_2\cap BB_1\end{array}\right\|$ . Prove that $AU\cap BV\cap CW\ne\emptyset$ .

Proof. Let $\left\{\begin{array}{c} X\in AU\cap BC\\\\ Y\in BV\cap CA\\\\ W\in CW\cap AB\end{array}\right\|$ and apply the Ceva's theorem to $U$ and $\triangle ABC\ :\ \frac {XB}{XC}\cdot\frac {B_2C}{B_2A}\cdot\frac {C_1A}{C_1B}=1\iff$ $\frac {XB}{XC}\cdot\frac {b-h_a}{h_a}\cdot \frac{h_a}{c-h_a}=1$ $\iff$

$\frac {XB}{XC}=$ $\frac {c-h_a}{b-h_a}=\frac {2Rc-2Rh_a}{2Rb-2Rh_a}=$ $\frac {2Rc-bc}{2Rb-bc}\implies$ $\boxed{\frac {XB}{XC}=\frac {c(2R-b)}{b(2R-c)}}$ and analogously $\frac {YC}{YA}=\frac {a(2R-c)}{c(2R-a)}$ and

$\frac {ZA}{ZB}=\frac {b(2R-a)}{a\b(2R-b)}$ . In conclusion, $\frac {XB}{XC}\cdot\frac {YC}{TA}\cdot\frac {WA}{WB}=1\iff$ $AU\cap BV\cap CW\ne\emptyset$ . I used the relation $\boxed{2Rh_a=bc}$ .

PP17. Let $\triangle ABC$ and $\left\{\begin{array}{c} M\in (BC)\\\\ N\in (CA)\\\\ P\in (AB)\end{array}\right\|$ for which denote $\left\{\begin{array}{c} X\in BX\cap CP\\\\ Y\in CP\cap AM\\\\ Z\in AM\cap BN\end{array}\right\|$ . Suppose that $\left\{\begin{array}{c} 4=[ANZ]=[BPX]=[CMY]\\\\ 20=[APXZ]=[BMYX]=[CNZY]\end{array}\right\|$ . Find the area $[XYZ]=x$ .

Proof 1. Let $[CNZ]=u$ and $[CYZ]=v$ . Thus, $u+v=20$ , $\frac {[ZAB]}{[ZAN]}=\frac {ZB}{ZN}=\frac {[ZCB]}{[ZCN]}\implies$ $\frac {24}{4}=\frac {x+v+24}{u}$ and $\frac {[ZBA]}{[ZBM]}=\frac {ZA}{ZM}=\frac {[ZCA]}{[ZCM]}\implies$

$\frac {24}{x+20}=\frac {u+4}{v+4}$ . Therefore, $\left\{\begin{array}{c} v=20-u\\\\ x+v+24=6u\\\\ 24(v+4)=(x+20)(u+4)\end{array}\right\|\iff$ $\left\{\begin{array}{c} v=20-u\\\\ x+44=7u\\\\ 24(24-u)=(x+20)(u+4)\end{array}\right\|\iff$ $\left\{\begin{array}{c} v=20-u\\\\ x=7u-44\\\\ 24(24-u)=(7u-24)(u+4)\end{array}\right\|\iff$

$\left\{\begin{array}{c} v=20-u\\\\ x=7u-44\\\\ u^2+4u-96=0\end{array}\right\|\iff$ $\left\{\begin{array}{c} u=8\\\\ v=12\\\\ x=12\\\\ \end{array}\right\|$ , i.e. $\boxed{[XYZ]=12}$ . Prove easily that if $[ANZ]=[BPX]=[CMY]=a$

and $[APXZ]=[BMYX]=[CNZY]=b$ , then there is the relation $\boxed{a\left(2x+4a+5b\right)^2=(2a+b)\left(10a^2+13ab+4b^2\right)}$ .

Proof 2. Observe that $\left\{\begin{array}{c} \frac {CM}{CB}=\frac {[ACM]}{[ACB]}=\frac {28}{x+72}\\\\ \frac {PB}{PA}=\frac {[CPB]}{[CPA]}=\frac {28}{x+44}\\\\ \frac {YA}{YM}=\frac {[CYA]}{[CYM]}=\frac {24}{4}\end{array}\right\|$ . Apply the Menelaus' theorem to $\overline{CYP}$ and the triangle $ABM\ :\ \frac {CM}{CB}$ $\cdot\frac {PB}{PA}\cdot\frac {YA}{YM}=1\iff$

$\frac {28}{x+72}\cdot\frac {28}{x+44}\cdot 6=1\iff$ $(x+72)(x+44)=6\cdot 28^2\iff$ $x^2+116x-1536=0\begin{array}{ccc} \nearrow & x_1=12 & \searrow\\\\ \searrow & x_2=-128 & \nearrow\end{array}\odot\implies$ $x=12$ , i.e. $\boxed{[XYZ]=12}$ .

Proof 3. Observe that $\left\{\begin{array}{c} \frac {BM}{BC}=\frac {[ABM]}{[ABC]}=\frac {x+44}{x+72}\\\\ \frac {NC}{NA}=\frac {[BNC]}{[BNA]}=\frac {x+44}{28}\\\\ \frac {ZA}{ZM}=\frac {[BZA]}{[BZM]}=\frac {24}{x+20}\end{array}\right\|$ . Apply the Menelaus' theorem to $\overline{BZN}$ and $\triangle ACM\ :\ \frac {CM}{CB}$ $\cdot\frac {PB}{PA}\cdot\frac {YA}{YM}=1\iff$

$\frac {x+44}{x+72}\cdot\frac {x+44}{28}\cdot \frac {24}{x+20}=1\iff$ $6(x+44)^2=7\cdot (x+20)(x+72)\iff$ $x^2+116x-1536=0\begin{array}{ccc} \nearrow & x_1=12 & \searrow\\\\ \searrow & x_2=-128 & \nearrow\end{array}\odot\implies$ $x=12$ , i.e. $\boxed{[XYZ]=12}$ .

Routh relation. $\left\{\begin{array}{c} \frac {MB}{m}=\frac {MC}{1}=\frac {BC}{m+1}\\\\ \frac{NC}{n}=\frac {NA}{1}=\frac {CA}{n+1}\\\\ \frac {PA}{p}=\frac {PB}{1}=\frac {AB}{p+1}\end{array}\right\|\implies$ $\boxed{\frac {[XYZ]}{[ABC]}=\frac {(1-mnp)^2}{(1+m+mn)(1+n+np)(1+p+pm)}}$ .

Proof. Apply the Menelaus' theorem to the transversals $:\ \left\{\begin{array}{cccc} \overline{BZN}/\triangle ACM\ : & \frac {ZA}{ZM}=\frac {BC}{BM}\cdot\frac {NA}{NC}=\frac {m+1}{mn} & \implies & \frac {ZA}{m+1}=\frac {ZM}{mn}=\frac {AM}{1+m+mn}\\\\ \overline{CXP}/\triangle BAN\ : & \frac {XB}{XN}=\frac {CA}{CN}\cdot\frac {PB}{PA}=\frac {n+1}{np} & \implies & \frac {XB}{n+1}=\frac {XN}{np}=\frac {BN}{1+n+np}\\\\ \overline{AYM}/\triangle CBP\ : & \frac {YC}{YP}=\frac {AB}{AP}\cdot\frac {MC}{MB}=\frac {p+1}{pm} & \implies & \frac {YC}{p+1}=\frac {YP}{pm}=\frac {CP}{1+p+pm}\end{array}\right\|$ .

From the relation $[XYZ]=[ABC]-([ABM]+[BCN]+[CAP])+([CXN]+[AYP]+[BZM])$ , where

$\left\{\begin{array}{ccc} \frac {[CXN]}{[ABC]}=\frac {[CXN]}{[CBN]}\cdot\frac {[CBN]}{[CBA]}=\frac {XN}{BN}\cdot\frac {CN}{CA} & \implies & \frac {[CXN]}{[ABC]}=\frac {np}{1+n+np}\cdot\frac {n}{n+1}\\\\ \frac {[AYP]}{[ABC]}=\frac {[AYP]}{[APC]}\cdot\frac {[APC]}{[ABC]}=\frac {PY}{PC}\cdot\frac {AP}{AB} & \implies & \frac {[AYP]}{[ABC]}=\frac {pm}{1+p+pm}\cdot\frac {p}{p+1}\\\\ \frac {[BZM]}{[ABC]}=\frac {[BZM]}{[BAM]}\cdot\frac {[BAM]}{[BAC]}=\frac {MZ}{MA}\cdot\frac {BM}{BC} & \implies & \frac {[BZM]}{[ABC]}=\frac {mn}{1+m+mn}\cdot\frac {m}{m+1}\end{array}\right\|$ obtain that $:$

$\frac{[XYZ]}{[ABC]}=1-\left(\frac{m}{m+1}+\frac {n}{n+1}+\frac {p}{p+1}\right)+$ $\left(\frac {n^2p}{(n+1)(1+n+np)}+\frac {p^2m}{(p+1)(1+p+pm)}+\frac {m^2n}{(m+1)(1+m+mn)}\right)=$

$1-\left[ \frac {m}{m+1}\left(1-\frac {mn}{1+m+mn}\right)+\frac {n}{n+1}\left(1-\frac {np}{1+n+np}+\frac {p}{p+1}\right)\left(1-\frac {pm}{1+p+pm}\right) \right]=$ $1-\left(\frac {m}{1+m+mn}+\frac n{1+n+np}+\frac p{1+p+pm}\right)=$

$\frac {\prod (1+m+mn)-\sum m(1+n+np)(1+p+pm)}{\prod (1+m+mn)}=$ $\frac {\left(1+s_1+2s_2+4s_3+\sum mn^2+2s_1s_3+s_2s_3+s_3^2\right)-\left(s_1+2s_2+\sum m^2p+6s_3+2s_1s_3+s_2s_3\right)}{\prod (1+m+mn)}=$

$\frac {1-2s_3+s_3^2}{\prod (1+m+mn)}=$ $\frac {\left(1-s_3\right)^2}{\prod (1+m+mn)}\implies$ $\frac {[XYZ]}{[ABC]}=\frac {(1-mnp)^2}{(1+m+mn)(1+n+np)(1+p+pm)}$ , where $\left\{\begin{array}{ccc} s_1 & = & m+n+p\\\\ s_2 & = & mn+np+pm\\\\ s_3 & = & mnp\end{array}\right\|$ .

This post has been edited 312 times. Last edited by Virgil Nicula, Feb 11, 2018, 3:14 PM

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August 2018

462. Diverse probleme de geometrie.

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456* Integrals.

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

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

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

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449. Working page I.

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

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307. Very nice proofs !

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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).

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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

385. Some nice geometry problems.

by Virgil Nicula, Aug 17, 2013, 2:27 AM

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