416. Cateva solutii ale unei probleme de tip slicing.

by Virgil Nicula, Mar 23, 2015, 4:40 PM

P1. In triunghiul ABC punctul M este mijlocul laturii $[BC]$ si $C = 15^{\circ}$ . Stiind ca $m( \angle{AMB}) = 45^{\circ}$ , determinati masura unghiului $\angle BAC$ .

Desi problema este propusa la nivelul clasei a VI - a voi oferi mai multe solutii pentru aceasta frumoasa problema de tip slicing. Voi incepe

cu metodele metrice si/sau trigonometrice si sfarsind cu cele sintetice, de fapt singurele de interes estetic intr-o problema de tip slicing.

Metoda 1 (metrica). Aplicam teorema sinusurilor in $\triangle AMC\ :\ \frac {b}{\sin 45}=\frac {a}{2\sin 30}\ \Longrightarrow\ a=b\sqrt 2\ \ (1)$ . Aplicam teorema medianei in $\triangle ABC\ :\ 4m_a^2=$ $2\left(b^2+c^2\right)-a^2\stackrel{(1)}{\ =\ }$

$2\left(b^2+c^2\right)-2b^2\ \Longrightarrow$ $\ c=m_a\sqrt 2\ \ (2)$ . Teorema sinusurilor in $\triangle ABM\ :\ \frac {m_a}{\sin B}=$ $\frac {c}{\sin 45}\ \stackrel{(2)}{\Longrightarrow}\ \frac {m_a}{\sin B}=$ $\frac {m_a\sqrt 2}{\frac {\sqrt 2}{2}}\ \Longrightarrow\ \sin B=\frac 12\Longrightarrow B=30^{\circ}$ .

Metoda 2 (trig.). Aplicam in $\triangle ABC\ :\ \frac {MB}{MC}=$ $\frac {AB}{AC}\cdot\frac {\sin \widehat {MAB}}{\sin\widehat {MAC}}\Longrightarrow$ $1=\frac {\sin 15^{\circ}}{\sin B}\cdot \frac {\sin (B+45^{\circ})}{\sin 30^{\circ}}\Longrightarrow$ $\sin (B+45^{\circ})=2\sin B\cos 15^{\circ}\Longrightarrow$ $\sin\left(B+45^{\circ}\right)=$

$\sin \left(B+15^{\circ}\right)+\sin\left(B-15^{\circ}\right) \Longleftrightarrow$ $\sin (B+45^{\circ})-\sin\left(B-15^{\circ}\right)=$ $\sin (B+15^{\circ})\ \Longrightarrow$ $2\sin 30^{\circ}\cos \left(B+15^{\circ}\right)=$ $\sin \left(B+15^{\circ}\right)$ $\Longleftrightarrow$ $\tan\left(B+15^{\circ}\right)=1$ $\Longleftrightarrow$ $B=30^{\circ}$ .

Metoda 3 ("slicing"). Construim triunghiul echilateral $ACD$ astfel ca dreapta $BC$ sa separe punctele $A$ si $D$ . Asadar $DA=DC=AC\Longrightarrow$ $m(\angle MAD)=30^{\circ}$ , $m(\angle BCD)=45^{\circ}$

si semidreapta $[AM$ este bisectoarea $\angle CAD$ $\Longrightarrow$ $AM\perp CD$ si punctul $N\in AM\cap CD$ este mijlocul laturii $[CD]$ $\Longrightarrow$ segmentul $[MN]$ este linie mijlocie in $\triangle BCD$ $\Longrightarrow$

$MN\parallel BD$ $\Longrightarrow$ $m(\angle CBD)=45^{\circ}$ si $m(\angle ADB)=30^{\circ}$ $\Longrightarrow$ triunghiul $BDC$ este $A$ -dreptunghic isoscel $\Longrightarrow$ $DB=DC=DA$ $\Longrightarrow$ $DB=DA$ $\Longrightarrow$

triunghiul $ADB$ este $A$ -isoscel cu $m(\angle ADB)=30^{\circ}$ $\Longrightarrow$ $m(\angle BAD)=75^{\circ}$ $\Longrightarrow$ $m(\angle BAC)=135^{\circ}$ .

Metoda 4 ("slicing"). Construim $\triangle BCD$ echilateral astfel ca $BC$ sa nu separe $A$ , $D$ . Se observa ca $DM\perp BC$ si $m(\angle MDC)=$ $m(\angle MAC)=30^{\circ}$ ceea ce

inseamna ca $ADCM$ este inscriptibil. Rezulta $AD\perp AC$ , $m(\angle MDA)=$ $m(\angle MCA)=$ $m(\angle ADB)=15^{\circ}$ . Asadar semidreptele $[DA$ si $[MA$ sunt bisectoare

la $\angle BMD$ si \angle BDM respectiv ceea ce inseamna ca $A$ este centrul cercului inscris in $\triangle BDM$ si semidreapta $[BA$ este bisectoare la $\angle DBM$ , adica $B=30^{\circ}$ .

Si iata o metoda "killer", nu numai eficienta, dar si foarte scurta, fara puncte intermediare sau constructii auxiliare ...

Metoda 5 (metrica). Teorema sinusurilor in $\triangle AMC$ $\Longleftrightarrow$ $\frac {CM}{CA}=\frac {\sin \widehat {CAM}}{\sin\widehat {CMA}}$ $\Longrightarrow$ $\frac {\frac a2}{b}=\frac {\sin 30^{\circ}}{\sin 45^{\circ}}$ $\Longleftrightarrow$ $a=b\sqrt 2$ $\Longleftrightarrow$ $CA^2=CM\cdot CB$ $\Longleftrightarrow$

$\triangle CAM\sim \triangle CBA$ , adica dreapta $CA$ este tangenta in $A$ la cercul circumscris triunghiului $ABM$ $\Longleftrightarrow$ $\widehat {CBA}\equiv\widehat {CAM}$ $\Longleftrightarrow B=30^{\circ}$ .

O alta metoda ar fi utilizarea problemei/lemei foarte simple de "slicing" de aici

Metoda 6 ("slicing"). Fie $\triangle ADC$ cu $DA=DC\ ,\ DA\perp DC$ astfel ca $BC$ sa separe $A$ , $D$ . Din lema mentionata rezulta $MA=MD$ si $MC=CD$ . Fie mijlocul $N$ al lui

$[AD]$ . Insa $MA=MD$ $\Longrightarrow$ $MN\perp AD$ $\Longrightarrow$ $MN\parallel CD$ si $M$ , $N$ sunt mijloacele la $[BC]$ , $[AD]$ respectiv $\implies AB\parallel CD\implies$ $m(\widehat {ABC})=m(\widehat {DCB})$ $\Longrightarrow$ $B=30^{\circ}$ .

My (math)blog

Este binecunoscuta urmatoarea problema "slicing" intr-un triunghi : "In $\triangle ABC$ cu $B=30^{\circ}$ si $C=15^{\circ}\implies$ $m(\widehat {MAC})=30^{\circ}$ , unde

$M$ este mijlocul laturii $[BC]$ ". Cei care nu au reusit sa o rezolve in cel mult jumatate de ora pot gasi demonstratia pe my (math)blog.

Metoda 7 ("slicing"). Notam mijlocul $N$ al laturii $[AC]$ . Aplicam lema in $\triangle AMC$ si rezulta $m(\widehat{CMN})=30^{\circ}$ . Din $MN\parallel AB$ rezulta $B=30^{\circ}$ .

Metoda 8 ("slicing"). Construim triunghiul $BCD$ astfel ca dreapta $BC$ separa punctele $A$ , $D$ si $m(\angle CBD)=30^{\circ}$ , $m(\angle BCD)=15^{\circ}$ . Aplicam

lema in $\triangle BDC$ si rezulta $m(\widehat{MDC})=30^{\circ}$ . Deoarece triunghiurile $AMC$ si $DMC$ sunt simetrice fata de dreapta $BC$ rezulta ca $B=30^{\circ}$ .

P2. Let $\triangle ABC$ with $A=48^{\circ}\ ,\ B=12^{\circ}\ ,\ AB=\sqrt 3$ and $D\in (AB)$ so that $CD=1$ . Prove that $x\equiv\left(\widehat{BCD}\right)=6^{\circ}$ .

Proof 1. Apply the theorem of Sines in $:\ \left\{\begin{array}{cccc} \triangle ABC\ : & \frac {AC}{AB}=\frac {\sin\widehat{ABC}}{\sin\widehat{ACB}} & \iff & b=2\sin 12^{\circ}\\\\ \triangle ACD\ : & \frac {AC}{CD}=\frac {\sin\widehat{ADC}}{\sin\widehat{CAD}} & \iff & b=\frac {\sin (12^{\circ}+x)}{\sin 48^{\circ} }\end{array}\right\|$ $\implies$ $2\sin 12^{\circ}=\frac {\sin \left(12^{\circ}+x\right)}{\sin 48^{\circ}}\iff$ $2\sin 12^{\circ}\sin 48^{\circ}=\sin (12^{\circ}+x)\iff$

$\cos 36^{\circ}-\cos 60^{\circ}=\sin (12^{\circ}+x)\iff$ $\frac {1+\sqrt5}4-\frac 12=$ $\sin (12^{\circ}+x)\iff$ $\frac {-1+\sqrt 5}4=\sin (12^{\circ}+x)\iff$ $\sin 18^{\circ}=\sin (12^{\circ}+x)\iff$ $12^{\circ}+x=18^{\circ}\iff$ $x=6^{\circ}$ .

P3. Let a circle $w=\mathbb C(O,r)$ with the diameter $[AB]$ and from $C\in AB$ so that $A\in (BC)$ construct the tangents $CE$ ,

$CF$ to $w$ , where $\{E,F\}\subset w$ and let $D\in EF\cap AB$ . Prove that $(\forall )\ M\in w$ the ray $[MA$ is the bisector of $\widehat{CMD}$ .

Proof. $\left\{\begin{array}{ccccccc} \triangle CEA\sim\triangle CBE & \implies & \frac {CE}{CB}=\frac {EA}{BE}=\frac {AC}{EC} & \implies & \odot\begin{array}{ccc} \nearrow & CA=CE\cdot\frac {EA}{EB} & \searrow\\\\ \searrow & CB=CE\cdot \frac {EB}{EA} & \nearrow\end{array}\odot & \implies & \frac {CA}{CB}=\left(\frac {EA}{EB}\right)^2\\\\ \triangle ADE\sim\triangle EDB & \implies & \frac {AD}{ED}=\frac {DE}{DB}=\frac {AE}{EB} & \implies & \odot\begin{array}{ccc} \nearrow & DA=ED\cdot\frac {EA}{EB} & \searrow\\\\ \searrow & DB=ED\cdot \frac {EB}{EA} & \nearrow\end{array}\odot & \implies & \frac {DA}{DB}=\left(\frac {EA}{EB}\right)^2\end{array}\right\|$ $\implies$ $\boxed{\frac {DA}{DB}=\left(\frac {EA}{EB}\right)^2=\frac {CA}{CB}}$ .

Hence division $(A,B;C,D)$ is harmonic. Apply known property $: (\forall ) M\not\in AB$ we have $\boxed{MA\perp MB\iff \widehat{AMC}\equiv\widehat{AMD}}$ , i.e. the ray $[MA$ is the bisector of $\widehat{CMD}$ .

P4. Let a convex cyclical quadrilateral $ABCD$ with $T\in BB\cap DD$ so that $BD$ separates $A$ , $T$ , where $XX$ is the tangent line to the circle $w$ at the point $X\in w$ .

Then $\boxed{T\in AC\iff AB\cdot CD=AD\cdot BC}$ , i.e. $ABCD$ is a harmonic quadrilateral and in this case the division $(A,C;S,T)$ is harmonically, where $S\in AC\cap BD$ .

Proof. Suppose $T\in AC$ .Thus, $\left\{\begin{array}{ccccccc} \triangle TBC\sim\triangle TAB & \implies & \frac {TB}{TA}=\frac {BC}{AB}=\frac {CT}{BT} & \implies & \odot\begin{array}{ccc} \nearrow & TA=TB\cdot\frac {AB}{BC} & \searrow\\\\ \searrow & TC=TB\cdot \frac {BC}{AB} & \nearrow\end{array}\odot & \implies & \frac {TA}{TC}=\left(\frac {AB}{BC}\right)^2\\\\ \triangle TDC\sim\triangle TAD & \implies & \frac {TD}{TA}=\frac {DC}{AD}=\frac {CT}{DT} & \implies & \odot\begin{array}{ccc} \nearrow & TA=TD\cdot\frac {AD}{DC} & \searrow\\\\ \searrow & TC=TD\cdot \frac {DC}{AD} & \nearrow\end{array}\odot & \implies & \frac {TA}{TC}=\left(\frac {DA}{DC}\right)^2\end{array}\right\|$ $\implies$

$\left(\frac {AB}{BC}\right)^2=\frac {TA}{TC}=\left(\frac {DA}{DC}\right)^2\iff$ $\frac {AB}{BC}=\frac {DA}{DC}\iff$ $AB\cdot CD=AD\cdot BC$ . Thus, $\frac {SA}{SC}= \frac {AB\cdot AD}{CB\cdot CD}=\left(\frac {AB}{BC}\right)^2\implies$ $\frac {SA}{SC}=\frac {TA}{TC}$ , i.e. $(A,C;S,T)$ is harmonic.

Remark. Observe easily that the rays $[AS$ , $[BS$ , $[CS$ and $[DS$ are symmedians in the triangles $ABD$ , $BCA$ , $CDB$ and $DAC$ respectively. See here and here.

P5. Let $\triangle ABC$ with $A = 60^{\circ}$ . Let $AP$ bisect $\widehat { BAC}$ and let $BQ$ bisect $\widehat {ABC}$ , where $P\in BC$ and $Q\in AC$ . If $AB + BP = AQ + QB$ , then what are the angles of $\triangle ABC$ ?

Proof 1 (metric). Denote $l_b = BQ$ . Thus, $l_b = \frac {2ac}{a + c}\cdot\cos\frac B2$ . Therefore, $AB + BP = AQ + QB$ $\Longleftrightarrow$ $c + \frac {ac}{b + c} = \frac {bc}{a + c} + l_b$ $\Longleftrightarrow$ $c(c + a)(c + b) + ac(a + c) =$

$bc(b + c) + 2ac(b + c)\cdot\cos\frac B2$ $\Longleftrightarrow$ $ab + (a + c)^2 - b^2 = 2a(b + c)\cdot\cos\frac B2$ $\Longleftrightarrow$ $ab + 4p(p - b) = 2a(b + c)\cdot\cos\frac B2$ $\Longleftrightarrow$ $ab + 4ac\cdot\cos^2\frac B2 = 2a(b + c)\cdot \cos\frac B2$ $\Longleftrightarrow$

$b\left(1 - 2\cdot\cos\frac B2\right) = 2c\cdot\cos\frac B2\cdot\left(1 - 2\cdot\cos\frac B2\right)$ $\Longleftrightarrow$ $b = 2c\cdot\cos\frac B2$ $\Longleftrightarrow$ $\sin B = 2\cdot\sin C\cdot\cos\frac B2$ $\Longleftrightarrow$ $2\sin\frac B2\cos\frac B2 = 2\sin C\cos\frac B2$ $\Longleftrightarrow$ $\sin\frac B2 = \sin C$ $\Longleftrightarrow$

$B = 2C$ $\Longleftrightarrow$ $C = 40^{\circ}$ , $B = 80^{\circ}$ because $A = 60^{\circ}$ .

Proof 2 (trigonometric). Denote ${ x=m(\widehat ABQ})$ . I"ll apply the Sinus' theorem in the triangles $ABP$ , $ABQ$ : $AB+BP=AQ+QB$ $\Longleftrightarrow$ $1+\frac {BP}{BA}=\frac {AQ+QB}{AB}$ $\Longleftrightarrow$

$1+\frac {\sin 30}{\sin (30+2x)}=\frac {\sin x+\sin 60}{\sin (60+x)}$ $\Longleftrightarrow$ $\sin (60+x)\left[\sin (30+2x)+\frac 12\right]=\sin (30+2x)(\sin x+\sin 60)$ $\Longleftrightarrow$ $\cos (x-30)-\cos (90+3x)+\cos (30-x)=$

$\cos (x+30)-\cos (30+3x)+\cos (2x-30)-\cos (90+2x)$ $\Longleftrightarrow$ $\underline {\cos (x-30)}+\underline {\underline {\sin 3x}}+\underline {\underline {\underline {\cos (x-30)}}}=$ $\underline {\underline {\underline {\cos (x+30)}}}-\underline {\underline {\cos (30+3x)}}+\cos (2x-30)+\underline {\sin 2x}$ $\Longleftrightarrow$

$[\cos (x-30)-\cos (90-2x)]+[\sin 3x+\sin (60-3x)]=$ $[\cos (x+30)-\cos (x-30)]+\cos (2x-30)$ $\Longleftrightarrow$ $2\sin \left(30-\frac x2\right)\sin\left(60-\frac {3x}{2}\right)+$ $2\sin 30\cos (3x-30)=$

$\cos (2x-30)-2\sin 30\sin x$ $\Longleftrightarrow$ $\cos (3x-30)+2\sin\left(30-\frac x2\right)\sin \left(60-\frac {3x}{2}\right)=$ $\cos (2x-30)-\cos (90-x)$ $\Longleftrightarrow$ $\cos (3x-30)+2\sin\left(30-\frac x2\right)\sin \left(60-\frac {3x}{2}\right)$

$=2\sin \left(30+\frac x2\right)\sin\left(60-\frac {3x}{2}\right)$ $\Longleftrightarrow$ $\sin (120-3x)=2\sin\left(60-\frac {3x}{2}\right)\left[\sin\left(30+\frac x2\right)-\sin\left(30-\frac x2\right)\right]$ $\Longleftrightarrow$ $2\sin\left(60-\frac {3x}{2}\right)\cos\left(60-\frac {3x}{2}\right)=$

$2\sin\left(60-\frac {3x}{2}\right)\left[\sin\left(30+\frac x2\right)-\sin\left(30-\frac x2\right)\right]$ $\Longleftrightarrow$ $\sin\left(60-\frac {3x}{2}\right)=0\ \ \vee\ \ \cos\left(60-\frac {3x}{2}\right)=\sin\left(30+\frac x2\right)-\sin\left(30-\frac x2\right)$ . Thus, $\boxed {\ x=40\ }$

or $\sin\left(30+\frac {3x}{2}\right)+\sin\left(30-\frac x2\right)=\sin\left(30+\frac x2\right)$ $\Longleftrightarrow$ $2\sin\left(30+\frac x2\right)\cos x=\sin\left(30+\frac x2\right)$ $\Longleftrightarrow$ $x\in\emptyset$ .

$\bf\color{red}\mathrm{P6}\ (\Longrightarrow ).$ Let an equilateral $\triangle ABC$ with $D\in (AC)$ and the point $E$ so that the ray $[BE$ is the $B$ -bisector of $\triangle ABD$ and $AE\parallel BC\ .$ Prove that $AE+CD=BD\ .$

Proof 1 (trig). Let $m\left(\widehat{EBD}\right)=m\left(\widehat{EBA}\right)=x\ .$ So $AE+CD=BD\iff$ $\frac {AE}{AB}\cdot \frac {BC}{BD}+\frac {CD}{BD}=1$ $\iff$ $\frac {\sin x}{\sin (60-x)}\cdot \frac {\sin(60+2x)}{\sin 60}+$ $\frac {\sin (60-2x)}{\sin 60}=1\iff$

$\sin x\sin (60+2x)+\sin (60-2x)\sin (60-x)=$ $\sin (60-x)\sin 60$ $\iff$ $\underline{\cos (60+x)}-\underline{\underline{\cos (60+3x)}}+\cos x-\underline{\underline{\cos (120-3x)}}=$ $\cos x-\underline{\cos (120-x)}\ ,$ what is true.

Proof 2 (metric). Let $:\ F\in BE\cap AC\ ;\ AB=a\ ,\ BD=l\ ,\ AD=m\ ,\ ,$ i.e. $\boxed{CD=a-m}\ .$ Observe that $A=60^{\circ}\iff \boxed{l^2=a^2+m^2-am}\ (*)\ .$ Apply the theorem

of bisector in $\triangle ABD\ :\ \frac {FA}a=\frac {FD}l=\frac m{a+l}\implies \boxed{FA=\frac {am}{a+l}}\ (1)\ \implies FC=AC-AF=a-\frac {am}{a+l}\implies$ $\boxed{FC=\frac {a(a+l-m)}{a+l}}\ (2)\ .$ Apply again the

theorem of bisector in $\triangle ABE\ :\ \frac {AE}{AB}=\frac {FE}{FB}=$ $\frac {FA}{FC}\ \stackrel{1\wedge 2}{=}\ \frac m{a+l-m}\implies$ $\boxed{AE=\frac {am}{a+l-m}}\ (3)\ .$ In conclusion, $AE+CD=BD\iff$ $AE=BD-CD\ \stackrel{3}{\iff}$

$\frac {am}{a+l-m}=l+m-a\iff$ $[l+(m-a)]\cdot [(l-(m-a)]=am\iff$ $l^2-(a-m)^2=am\iff$ $l^2=(a-m)^2+am\iff$ $l^2=a^2+m^2-am\ ,$ what is true $(*).$

Proof 3 (synthetic). If denote $P\in AB$ so that $EP\perp AB$ and $G\in BD\cap AE\ ,$ then the point $F$ is the incenter of $\triangle ABG$ and $E$ is $B$ -excenter of $\triangle ABD\ .$

In conclusion, $AE=2\cdot AP=BD+AD-AB=BD-(AC-AD)=BD-CD\implies$ $AE=BD-CD\implies$ $\boxed{BD=AE+CD}\ .$

Remark. $M\in (AB)\ ,\ DM\parallel BC\implies AD=DM$ and from the trapezoid $ABCG$ obtain the relation $\frac 1{AG}+\frac 1{AB}=\frac 1{AM}\ ,$ i.e. $\boxed {\frac 1{AG}+\frac 1{AB}=\frac 1{AD}}\ .$

$\bf\color{red}\mathrm{P6}\ (\Longleftarrow ).$ Let an $A$ -isosceles $\triangle ABC\ ,\ AB\ge BC\ ,$ for what there are $D\in (AC)$ and the point $E$ so that the ray $[BE$ is

the $B$ -bisector of $\triangle ABD$ and $AE\parallel BC\ .$ Prove that $AE+CD=BD\implies AB=BC\ ,$ i.e. $\triangle ABC$ is equilateral $\ .$

Proof. Let $:\ F\in BE\cap AC\ ;\ AB=AC=b\ ,\ BC=a\ ,\ BD=l\ ,\ AD=m\ ,\ ,$ i.e. $\boxed{CD=b-m}\ .$ Observe that $l^2=b^2+m^2-2bm\cos A$ and $\cos A=\frac {2b^2-a^2}{2b^2}$

$\implies l^2=b^2+m^2-2bm\cdot \frac {2b^2-a^2}{2b^2}=b^2+m^2-m\cdot \frac {2b^2-a^2}{b}=b^2+m^2-2bm+\frac {a^2m}b\implies$ $\boxed{l^2=(b-m)^2+\frac {a^2m}{b}}\ (*)\ .$ Apply the theorem of bisector in $\triangle ABD\ :$

$\frac {FA}b=\frac {FD}l=\frac m{b+l}\implies \boxed{FA=\frac {bm}{b+l}}\ (1)\ \implies FC=AC-AF=b-\frac {bm}{b+l}\implies$ $\boxed{FC=\frac {b(b+l-m)}{b+l}}\ (2)\ .$ From $AE\parallel BC$ get $\frac {AE}{BC}=\frac {FA}{FC}\ \stackrel{1\wedge 2}{=}\ \frac m{b+l-m}$

$\implies$ $\boxed{AE=\frac {am}{b+l-m}}\ (3)\ .$ Thus, $BD=AE+CD\iff$ $BD^2=(AE+CD)^2\ \stackrel{3}{\iff}\ l^2=$ $\left[\frac {am}{b+l-m}+(b-m)\right]^2\ \stackrel{*}{\iff}\ (b-m)^2+\frac {ma^2}b=$ $(b-m)^2+$

$\frac {a^2m^2}{(l+b-m)^2}+\frac {2am(b-m)}{l+b-m}\iff$ $\frac ab=\frac {am}{(l+b-m)^2}+\frac {2(b-m)}{l+b-m}\iff$ $\frac ab=\frac {am+2(b-m)(l+b-m)}{(l+b-m)^2}$ $\iff$ $a(l+b-m)^2=abm+2b(b-m)(l+b-m)\iff$

$al^2+a(b-m)^2+2al(b-m)=abm+2b(b-m)^2+2bl(b-m)\iff$ $a\left[(b-m)^2+\frac {a^2m}{b}\right]+a(b-m)^2+2al(b-m)=abm+2b(b-m)^2+2bl(b-m)\iff$

E}{BC}
$(b-a)\cdot \left[\frac {am(a+b)}b+2(b-m)^2+2l(b-m)\right]=0\implies$ $a=b$ because $\frac {am(a+b)}b+2(b-m)^2+2l(b-m)>0\ .$

P7 (Murat Öz, Turkey). Let an $A$ -isosceles $\triangle ABC$ with $B=C=40^{\circ}\ .$ The $C$ -angled bisector meet $AB$ at $D$ and there is $E\in (BC)$ so that $BE=CD\ .$ Prove that $m\left(\widehat{CAE}\right)=20^{\circ}\ .$

Proof 1. Denote $m\left(\widehat{CAE}\right)=x$ (degrees) and apply theorem of Sines $:\ \frac {\sin\widehat{DBC}}{\sin\widehat{BDC}}=$ $\frac {DC}{BC}=$ $\frac {BE}{BC}=$ $\frac {AE}{AC}\cdot \frac {\sin\widehat{BAE}}{\sin\widehat{BAC}}=$ $\frac {\sin\widehat{ACE}}{\sin\widehat{AEC}}\cdot\frac {\sin\widehat{BAE}}{\sin\widehat{BAC}}\implies$ $\frac {\sin\widehat{DBC}}{\sin\widehat{BDC}}=$

$\frac {\sin\widehat{ACE}}{\sin\widehat{AEC}}\cdot\frac {\sin\widehat{BAE}}{\sin\widehat{BAC}}\implies$ $\frac {\cancel{\sin 40^{\circ}}}{\sin 60^{\circ}}=\frac {\cancel{\sin 40^{\circ}}}{\sin\left(40^{\circ}+x\right)}\cdot\frac {\sin\left(100^{\circ}-x\right)}{\sin 100^{\circ}}$ $\implies$ $\sin 60^{\circ}\cos \left(x-10^{\circ}\right)=\sin\left(40^{\circ}+x\right)\cos 10^{\circ}\iff$ $\cancel{\sin\left(50^{\circ}+x\right)}+\sin\left(70^{\circ}-x\right)=$

$\cancel{\sin\left(50^{\circ}+x\right)}+\sin\left(30^{\circ}+x\right)\iff$ $\cos\left(x+20^{\circ}\right)=\cos \left(60^{\circ}-x\right)\iff$ $20^{\circ}+x=60^{\circ}-x\iff$ $2x=40^{\circ}\iff$ $x=20^{\circ}\ .$

Proof 2 (Apostolis Manoloudis). Construct the bisector $BF$ of $\widehat{ABC}\ .$ Observe that $BE=DC=BF\implies BE=BF\ ,$ i.e. $\triangle EBF$ is isosceles

and $m\left(\widehat{BEF}\right)=80^{\circ}$ $\implies$ $m\left(\widehat{BAF}\right)+ m\left(\widehat{BEF}\right)= 180^{\circ}\ ,$ i.e. $ABEF$ is cyclic. In conclusion, $\left(\widehat{FAE}\right)=\left(\widehat{FBE}\right)=20^{\circ}\implies x=20^{\circ}\ .$

This post has been edited 190 times. Last edited by Virgil Nicula, Sep 23, 2017, 7:47 PM

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See my proof for PP.6 at http://forum.gil.ro/viewtopic.php?f=25&t=6884

Best regards,
sunken rock

by sunken rock, Aug 30, 2016, 11:04 AM

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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

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by mathMagicOPS, Jan 9, 2025, 3:40 AM
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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
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Amazing blog.
by OlympusHero, May 13, 2021, 10:23 PM
the visits tho
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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