7. Some interesting "slicing" problems.

by Virgil Nicula, Apr 19, 2010, 3:06 PM

PP1. Let $\triangle ABC$ and $D\in (AC)$ such that $m(\angle CBD) - m(\angle ABD) = 60^\circ$ , $m(\angle BDC) = 30^\circ$ and $BD^2=BA\cdot BC$ . Find $A$ .

Proof 1 (trigonometric). Denote $m(\angle ABD)=x$ , i.e. $m(\angle CBD)=60^{\circ}+x$ and $m(\angle BAD)=30^{\circ}-x$ . Apply the Sinus' theorem

in $\triangle BCD$ and $\triangle ABD$ . The given relation becomes $:\ \ \frac {BD}{BC}=\frac {AB}{BD}\iff$ $\frac {\sin\widehat {BCD}}{\sin\widehat{ADB}}=\frac {\sin\widehat {BAD}}{\sin\widehat{BDC}}\iff$ $\frac {\cos x}{\sin 30^{\circ}}=\frac {\sin 30^{\circ}}{\sin (30^{\circ}-x}$

$\iff$ $4\cos x\sin (30^{\circ}-x)=1\iff$ $2\left[\sin 30^{\circ}+\sin (30^{\circ}-2x)\right]=1\iff$ $\sin (30^{\circ}-2x)=0\iff$ $x=15^{\circ}\iff$ $A=15^{\circ}$ .

Proof 2 (synthetic). Let $B'$ be the reflection of $B$ in $AC$ . Thus, $BDB'$ is an equilateral triangle and $BB'$ is the isogonal line of $BD$ .

Since $BB'=BD$ the given relation becomes $\frac{AB}{BB'}=\frac{BD}{BC}$ , i.e. $\triangle ABD \sim \triangle BB'C$ , i.e. $BD=AD$ . Further, just angle chasing.

PP2. Let $ABC$ be a triangle with $A=70^{\circ}$ and the incenter $I$ for which $IA=a-b$ . Find $B$ .

Proof 1. Let $D\in AC$ so that $A\in (CD)$ and $AD=AI$ . So $\triangle DAI$ is $A$ -isosceles with $m(\widehat{ADI})=$

$m(\widehat{AID})=\frac {35^{\circ}}{2}$ . Since $CBD$ is $C$ -isosceles obtain that $m(\widehat{IBC})=m(\widehat{ADI})$ . Thus, $m(\widehat{ABC})=35^{\circ}$ .

Proof 2. Let $D\in [CB$ so that $CD=CA$ . Thus, $AI=DI=BD$ and $m(\angle CDI)=$

$m(\angle CAI)=35^\circ$ , i.e. $\triangle BDI$ is $D$ -isosceles $\implies$ $DI\parallel AB$ , i.e. $B=35^\circ$ .

PP3. Let $ABC$ be a triangle. Denote the midpoint $M$ of $[BC]$ and suppose that $\widehat{MAC}=B$ and $\widehat{BAM}=105^\circ$ . Find $m\left(\widehat{ABC}\right)$ .

Method 1 (trigonometric). $\triangle CAM\sim\triangle CBA\implies$ $\frac ba=\frac {a}{2b}\implies$ $a\stackrel{(*)}{=}b\sqrt 2$ . Apply the Sinus' theorem in $\triangle ABM\ :$

$\frac {b}{\sin B}=\frac {a}{\sin (105+B)}\stackrel{(*)}{\implies}$ $\sqrt 2\cdot\sin B=\cos (15+B)\iff$ . The function $f(x)=\sqrt 2\cdot\sin x-\cos (15+x)$ is strict

increasing on $\left[0,\frac{\pi}{2}\right]$ $\implies$ the equation $f(x)=0$ has at least one zero. Observe that $f(B)=f(30^{\circ})=0$ . In conclusion, $B=30^{\circ}$ .

Method 2 ("slicing" - Sunken Rock). The sideline $AC$ is tangent to the circle of $\triangle AMB\implies$ $a=b\sqrt 2$ and $c=m_a\sqrt 2$ . Construct equilateral

$\triangle ABD$ so that $AB$ doesn't separates $D\ ,\ C$ . Thus, $AM=AB\cdot\frac {\sqrt 2}{2}$ $\implies$ $AM=AD\cdot\cos 45^{\circ}$ . From $m(\angle MAD)=45^\circ$

obtain that $MD = MA$ and $MD\perp MA$ $\implies$ $\triangle BAM \equiv \triangle BDM$ , i.e. $AMDB$ is a kite with $AD\bot BC$ , i.e. $B=30^\circ$ .

Method 3 (trigonometric). $\triangle CAM\sim\triangle CBA\implies$ $\frac ba=\frac {m_a}{c}=\frac {a}{2b}$ $\implies$ $\boxed{\frac {c}{m_a}=\frac ab=\sqrt 2}$ . Apply the generalized Pytagoras' theorem to the side

$[BM]$ in $\triangle ABM\ :\ \frac {a^2}{4}=c^2+m_a^2-2cm_a\cdot\cos 105^{\circ}\iff$ $a^2=$ $4c^2+2\left(b^2+c^2\right)-a^2-8\cdot c\cdot \frac {c}{\sqrt 2}\cdot\frac {\sqrt 2}{2}\left(\frac 12-\frac {\sqrt 3}{2}\right)\iff$

$a^2=b^2+3c^2+c^2\left(\sqrt 3-1\right)\iff$ $a^2=\frac {a^2}2+\left(2+\sqrt 3\right)\cdot c^2\iff$ $a^2=2\left(2+\sqrt 3\right)\cdot c^2\implies$ $\boxed{a=b\cdot\sqrt 2=\left(1+\sqrt 3\right)\cdot c}$ .

In conclusion, the triangle with this property is similarly with the triangle $\left(\sqrt 2+\sqrt 6,1+\sqrt 3,\sqrt 2\right)$ . Apply the Cosinus' theorem to $\widehat{ABC}$

in $\triangle ABC\ :\ \cos B=\frac {2(1+\sqrt 3)^2+2-(1+\sqrt 3)^2}{2\cdot \sqrt 2(1+\sqrt 3)\cdot \sqrt 2}=$ $\frac {(1+\sqrt 3)^2+2}{4(1+\sqrt 3)}=$ $\frac {2\sqrt 3\cdot (1+\sqrt 3)}{4(1+\sqrt 3)}\implies$ $\cos B=\frac {\sqrt 3}{2}\implies$ $B=30^{\circ}$ . Vezi link - urile (1) , (2) , (3) , (4) .

PP4. Let $A$ -right $\triangle ABC$ for which denote $D\in (AC)$ so that $BA+AD=BC$ . Prove that $m(\angle CBD)=15^{\circ}\ \Longrightarrow\ C=30^{\circ}$ .

Method 1. Denote $C=x$ . Prove easily that $x<\frac {\pi}{3}$ and $BA+AD=BC\Longleftrightarrow$ $\tan\left(x+\frac {\pi}{12}\right)=\frac {c}{a-c}=$ $\frac {\frac ca}{1-\frac ca}\Longleftrightarrow$ $\boxed {\tan\left(x+\frac {\pi}{12}\right)=\frac {\sin x}{1-\sin x}\ }$ $\Longleftrightarrow$

$\frac {\sin\left(x+\frac {\pi}{12}\right)}{\sin x}\cdot\frac {1-\sin x}{\cos\left(x+\frac {\pi}{12}\right)}=1$ $\Longleftrightarrow$ $f(x)\cdot g(x)=1$ , where $f(x)=\frac {\sin\left(x+\frac {\pi}{12}\right)}{\sin x}$ and $g(x)=\frac {1-\sin x}{\cos\left(x+\frac {\pi}{12}\right)}$ . But $f(x)=\cos \frac {\pi}{12}+\sin \frac {\pi}{12}\cdot\cot x$ is

decreasing and $g'(x)\ .a.s.\ -\cos x\cos\left(x+\frac {\pi}{12}\right)+(1-\sin x)\sin \left(x+\frac {\pi}{12}\right)=$ $\sin \left(x+\frac {\pi}{12}\right)-\cos \frac {\pi}{12}=$ $\cos\left(\frac {5\pi}{12}-x\right)-\cos$ $\frac {\pi}{12}\ .a.s.\ \left(x-\frac {\pi}{3}\right)<0$ .

Observe that the functions $f$ , $g$ are positively and decreasing. Hence the function $f\cdot g$ is decreasing $\left(\searrow\right)$ , i.e. our equation has exactly one zero.

Extension. For any $\alpha\in\left[\frac {\pi}{12},\frac {\pi}{8}\right]$ the equation $\tan (x+\alpha )=\frac {\sin x}{1-\sin x}$ has exactly one zero in the interval $\left[\frac {\pi}{4}-\alpha\ ,\ \frac {\pi}{2}-2\alpha\right]$ .

Proof. Denote analogously $f(x)=\frac {\sin (x+\alpha)}{\sin x}$ , $g(x)=\frac {1-\sin x}{\cos (x+\alpha )}$ si $h(x)=f(x)(g(x)-1$ . The equation becomes $f(x)g(x)=1$ or $h(x)=0$ . Observe that

$f(x)=\cos \alpha +\sin\alpha\cdot\cot x$ is decreasing and $g'(x)\ .a.s.\ -\cos x\cos\ (x+\alpha )+(1-\sin x)\sin (x+\alpha )=$ $\sin (x+\alpha )-\cos\alpha =$ $\left[\cos\left(\frac {\pi}{2}-x-\alpha\right)-\cos\alpha\right]\ .a.s.$

$\left[x-\left(\frac {\pi}{2}-2\alpha\right)\right]<0$ . Thus, the functions $f$ , $g$ are positively and decreasing. Therefore, the function $f\cdot g$ is decreasing $\left(\searrow\right)$ , i.e. our equation has at least one zero. But

$h\left(\frac {\pi}4-\alpha\right)=$ $\frac {1-2\sin\left(\frac {\pi}{4}-\alpha\right)}{\sin\left(\frac {\pi}{4}-\alpha \right)}\ .a.s.$ $\left[\sin \frac {\pi}6-\sin\left(\frac {\pi}{4}-\alpha \right)\right]=$ $2\sin\frac {\alpha-\frac {\pi}{12}}{2}\cos\frac {\frac {5\pi}{12}-\alpha}{2}\ge 0$ and $h\left(\frac {\pi}{2}-2\alpha\right)=$ $\frac {\cos\alpha}{\cos 2\alpha}\cdot\frac {1-\cos 2\alpha }{\sin\alpha }-1=$ $\frac {2\sin \alpha \cos\alpha}{\cos 2\alpha }-1=$

$\tan 2\alpha -1\le 0$ .In concluzie $h\left(\frac {\pi}{4}-\alpha\right)\cdot h\left(\frac {\pi}{2}-2\alpha\right)\le 0$ , what means the equation $h(x)=0$ has exactly one zero in the interval $\left[\frac {\pi}{4}-\alpha\ ,\ \frac {\pi}{2}-2\alpha\right]$ .

Remark. For $\alpha =\frac {\pi}{12}\implies x=\frac {\pi}{6}$ and for $\alpha =\frac {\pi}{8}\implies x=\frac {\pi}{4}$ .

Method 2. Let circumcentre $O$ , the incentre $I$ for $\triangle ABC$ and $E\in AI$ , $OE\perp BC$ . Thus, $AD=a-c$ , $BD^2=2c^2+a(a-2c)$ , $m(\angle DBE)=60^{\circ}$ , $BE=\frac {a}{\sqrt 2}$ and

$AE=\frac {b+c}{\sqrt 2}$ . Apply generalized Pytagoras' theorem $:\ \left\|\ \begin{array}{cc} \triangle DBE\ : & DE^2=\frac {a^2}{2}+(a-c)^2+c^2-\frac {a}{\sqrt 2}\cdot BD\\\\ \triangle DAE\ : & DE^2=(a-c)^2+\frac {(b+c)^2}{2}-(b+c)(a-c)\end{array}\ \right\|$ $\Longrightarrow$ $\frac {a^2}{2}+(a-c)^2+c^2-\frac {a}{\sqrt 2}\cdot BD=$

$(a-c)^2+\frac {(b+c)^2}{2}-(b+c)(a-c)$ $\Longrightarrow$ $a\cdot BD=\sqrt 2\cdot \left[a(b+c)-2bc\right]$ $\Longrightarrow$ $a^2\cdot \left[2c^2+a(a-2c)\right]=$ $2\cdot \left[ac+b(a-2c)\right]^2$ $\Longrightarrow$ $a^3(a-2c)=$

$4abc(a-2c)+2b^2(a-2c)^2$ $\Longrightarrow$ $a=2c$ sau $a^3=4abc+$ $2b^2(a-2c)$ . Observe that $a^3=4abc+2b^2(a-2c)$ $\Longleftrightarrow$ $a(b^2+c^2)=$

$4abc+2b^2(a-2c)$ $\Longleftrightarrow$ $a\left(b^2-c^2\right)+4bc(a-b)=0$ ceea ce este absurd ( $c\ <\ b\ <\ a$ ). In conclusion, $a=2c$ ceea ce inseamna $C=30^{\circ}$ .

Remark. In this case $\triangle BDE$ is equilateral and $\triangle CED$ este $E$ -isosceles, i.e. $B$ , $I$ , $D$ , $C$ belong to the circle with the centre $E$ .

PP5. Let $A$ -right $\triangle ABC$ for which exists $D\in (AC)$ so that $m(\angle DBC)=\frac C2$ . Prove that $BA+AD=BC\ \Longleftrightarrow\ C\ \in \left\{30^{\circ} , 45^{\circ}\right\}$ .

Method 1. Suppose that $BA+AD=BC$ . Denote $E$ for which $DE\perp BC$ si $x=m(\angle DBC)$ . Observe that $DC=b+c-a$ si $\triangle CDE\sim\triangle CBA$ , i.e.

$\frac {b+c-a}{a}=\frac {DE}{c}=\frac {EC}{b}$ . Obtain $DE=\frac {c(b+c-a)}{a}$ , $EC=\frac {b(b+c-a)}{a}\ \Longrightarrow$ $BE=BC-EC=a-\frac {b(b+c-a)}{a}$ , i.e. $BE=\frac {c^2-bc+ab}{a}$ . Thus,

$\tan x=\frac {DE}{BE}$ , i.e. $\boxed {\ \tan x=\frac {c(b+c-a)}{c^2+b(a-c)}\ }$ . For $C=2x$ obtain $: \tan x=\tan\frac C2=\frac {\sin C}{1+\cos C}=$ $\frac {\frac ca}{1+\frac ba}=\frac {c}{a+b}$ $\Longrightarrow$ $\frac {c}{a+b}=$ $\frac {c(b+c-a)}{c^2+b(a-c)}$ $\Longleftrightarrow$ $c^2+b(a-c)=$

$(a+b)(b+c-a)\ \stackrel{(a^2=b^2+c^2)}{\Longleftrightarrow}$ $2c^2-2bc=ac-ab$ $\Longleftrightarrow$ $(a-2c)(b-c)=0$ . In conclusion, $\boxed {\ BA+AD=BC\ \Longleftrightarrow\ \tan \frac C2=\frac {c(b+c-a)}{c^2+b(a-c)}\ \Longleftrightarrow\ C\ \in \left\{30^{\circ} , 45^{\circ}\right\}\ }$ .

Method 2. Suppose well-known or prove easily that in $\triangle ABC$ exists the equivalence $B=2C$ $\Longleftrightarrow$ $b^2=c(c+a)\ \vee\ a=c$ . Thus, $C=2(\angle DBC)$ $\stackrel{(\triangle DBC)}{\Longleftrightarrow}$

$BD^2=DC\cdot (DC+BC)\ (1)$ . Thus, $BA+AD=BC$ $\Longleftrightarrow$ $AD=a-c$ , $DC=$ $b+c-a$ $\stackrel{(1)}{\Longleftrightarrow}$ $c^2+(a-c)^2=$ $(b+c-a)(b+c)$ $\Longleftrightarrow$ $c^2+(a-c)^2=$

$(b+c-a)(b+c)$ $\Longleftrightarrow$ $\ 2c^2+a^2-$ $2ac=b^2+c^2+2bc-ab-$ $ac\ \stackrel{(a^2=b^2+c^2)}{\Longleftrightarrow}$ $2c^2-ac=2bc-ab$ $\Longleftrightarrow$ $2c(b-c)=$ $a(b-c)$ $\Longleftrightarrow$ $b=$ $c\ \vee\ a=2c$ $\Longleftrightarrow$ $C\ \in \left\{30^{\circ} , 45^{\circ}\right\}$ .

Method 3. Let $m(\angle DBC)=x$ , i.e. $C=2x$ . Thus, $D\in (AC)$ $\Longleftrightarrow$ $x<30^{\circ}$ , i.e. $\underline{\overline{\left\|\ C\ <\ 60^{\circ}\ \right\|}}$ . Thus, $BA+AD=BC$ $\Longleftrightarrow$ $BD\cdot (\sin 3x+$ $\cos 3x)=BC$ $\Longleftrightarrow$

$\sin 2x(\sin 3x+$ $\cos 3x)=\sin 3x$ $\Longleftrightarrow$ $2\cos x(\sin 3x+\cos 3x)=$ $1+2\cos 2x$ $\Longleftrightarrow$ $\sin 4x+\sin 2x+\cos 4x=$ $1+\cos 2x\ \stackrel{(2x=C)}{\Longleftrightarrow}\ \sin 2C+\sin C=$ $1-\cos 2C+\cos C$

$\Longleftrightarrow$ $\sin 2C-\cos C=$ $2\sin^2C-\sin C$ $\Longleftrightarrow$ $\cos C(2\sin C-1)=$ $\sin C(2\sin C-1)$ $\Longleftrightarrow$ $\tan C=1\ \ \vee\ \sin C=$ $\frac 12$ $\Longleftrightarrow$ $C\in\left\{30^{\circ},45^{\circ}\right\}$ .

Method 4. Prove easily that $C\ \in \left\{30^{\circ}\ ,\ 45^{\circ}\right\}$ $\Longrightarrow$ $BA+AD=BC$ . Suppose that $BA+AD=BC$ , i.e. $AD=a-c$ . Let circumcentre $O$ , incentre $I$ for $\triangle ABC$ and $E\in AI$

so that $OE\perp BC$ . So $BE=\frac {a}{\sqrt 2}$ , $AE=\frac {b+c}{\sqrt 2}$ , $BD^2=c^2+(a-c)^2=2c^2+a(a-2c)$ si $\left\|\ \begin{array}{ccc} m(\angle DBE) & = & 45^{\circ}+\frac C2\\\\ m(\angle DAE) & = & 45^{\circ}\end{array}\ \right\|$ . Apply the generalized Pytagoras' theorem :

$\left\|\ \begin{array}{cc} \triangle DBE\ : & DE^2=BD^2+BE^2-2\cdot BD\cdot BE\cdot \cos (\angle DBE)\\\\ \triangle DAE\ : & DE^2=AD^2+AE^2-2\cdot AD\cdot AE\cdot \cos (\angle DAE)\end{array}\ \right\|$ $\Longrightarrow$ $BD^2+BE^2-2\cdot BD\cdot BE\cdot \cos (\angle DBE)=$ $AD^2+AE^2-2\cdot AD\cdot AE\cdot \cos (\angle DAE)$

$\Longrightarrow$ $c^2+\frac {a^2}{2}-2\cdot BD\cdot\frac {a}{\sqrt 2}\cdot \cos \left(45^{\circ}+\frac C2\right)=$ $\frac {(b+c)^2}{2}-2\cdot (a-c)\cdot \frac {b+c}{\sqrt 2}\cdot \cos 45^{\circ}$ $\Longrightarrow$ $a\sqrt 2\cdot BD\cdot \cos \left(45^{\circ}+\frac C2\right)=$ $a(b+c)-2bc$ $\Longrightarrow$

$a^2\cdot BD^2\left[1+\cos \left( 90^{\circ}+C\right)\right]=$ $[ac+b(a-2c)]^2$ $\Longrightarrow$ $a^2\left[2c^2+a(a-2c)\right]\left(1-\frac ca\right)=$ $[ac+b(a-2c)]^2$ $\Longrightarrow$ $a(a-c)\left[2c^2+a(a-2c)\right]=$ $[ac+b(a-2c)]^2$ $\Longrightarrow$

$2ac^2(a-c)+a^2(a-c)(a-2c)=$ $a^2c^2+2abc(a-2c)+b^2(a-2c)^2$ $\Longrightarrow$ $ac^2(a-2c)+a^2(a-c)(a-2c)=$ $2abc(a-2c)+b^2(a-2c)^2$ $\Longrightarrow$ $\boxed {\ a=2c\ }\ \ \vee$

$ac^2+\left(b^2+c^2\right)(a-c)=$ $2abc+b^2(a-2c)$ . Suppose $a\ne 2c$ . In this case obtain $2ac^2+ab^2-b^2c-c^3=$ $ab^2-2b^2c+2abc\ \Longrightarrow$ $2ac^2-c^3=-b^2c+2abc\ \Longrightarrow$

$2a(b-c)=b^2-c^2$ $\Longrightarrow$ $\boxed {\ b=c\ }\ \ \vee\ \ 2a=b+c$ . Se observa ca $2a>b+c$ . In conclusion, $c\in\left\{b\ ,\ \frac a2\right\}$ , i.e. $C\ \in \left\{30^{\circ} , 45^{\circ}\right\}$ .

Method 5. Denote $x=m(\angle DBC)$ . Thus, $\tan B=\frac bc$ and $\tan (B-x)=\frac {a-c}{c}$ . Hence $\frac bc=\tan [x+(B-x)]=$ $\frac {\tan x+\tan (B-x)}{1-\tan x\cdot\tan (B-x)}=$ $\frac {\tan x+\frac {a-c}{c}}{1-\tan x\cdot\frac {a-c}{c}}=$

$\frac {c\cdot\tan x+(a-c)}{c-(a-c)\cdot\tan x}$ $\Longrightarrow$ $b\cdot\left[c-(a-c)\cdot\tan x\right]=$ $c\left[c\cdot \tan x+(a-c)\right]$ $\Longleftrightarrow$ $\tan x=\frac {c(b+c-a)}{c^2+b(a-c)}$ . For $C=2x$ obtain : $\tan x=\tan\frac C2=$ $\frac {\sin C}{1+\cos C}=$

$\frac {\frac ca}{1+\frac ba}=$ $\frac {c}{a+b}$ $\Longrightarrow$ $\frac {c}{a+b}=$ $\frac {c(b+c-a)}{c^2+b(a-c)}$ $\Longleftrightarrow$ $c^2+$ $b(a-c)=(a+b)(b+c-a)$ $\stackrel{(a^2=b^2+c^2)}{\Longleftrightarrow}$ $2c^2-2bc=ac-ab$ $\Longleftrightarrow$ $(a-2c)(b-c)=0$ .

In conclusion, $BA+AD=BC$ $\Longleftrightarrow$ $\tan \frac C2=$ $\frac {c(b+c-a)}{c^2+b(a-c)}$ $\Longleftrightarrow$ $C\ \in \left\{30^{\circ} , 45^{\circ}\right\}$ .

PP6 (British math Olimpiad 2013). The point $P$ lies inside of $\triangle ABC$ so that $\widehat{ABP} \equiv\widehat{ACP}$ . The point $Q$ is such that $PBQC$ is a parallelogram. Prove that $\widehat{BAQ} \equiv\widehat{CAP}$ .

Proof 1. Awesome problem ! In my diagram, $P$ is to the left of $Q$ . Complete the parallelogram $BASP$ . Then note that $BPQC$ is also a parallelogram, leading us to the fact that $ASCQ$

is also a parallelogram. Therefore, $\angle ABP=\angle ASP=\angle ACP$ so that $ASCP$ is a cyclic quadrilateral. Now, $\angle QAC=\angle ACS=\angle APS=\angle BAP$ . So we are done.

Proof 2. Construct the parallelogram $PARC$ . Observe that $BARQ$ is also a parallelogram. Thus, $\triangle ABP\equiv \triangle RQC$ and $ARCQ$ is cyclically

because $\widehat{RAC}\equiv\widehat{ACP}\equiv$ $\widehat{ABP}\equiv\widehat{RQC}\implies$ $\widehat{RAC}\equiv \widehat{RQC}$ . Hence $\widehat{CAQ}\equiv\widehat{CRQ}\equiv$ $\widehat{PAB}\implies$ $\widehat{CAQ}\equiv\widehat{PAB}\implies$ $\widehat{CAP} \equiv\widehat{BAQ}$

Proof 3. Let $\left\{\begin{array}{c} m\left(\widehat{BAQ}\right)=x\\\\ m\left(\widehat{CAP}\right)=y\end{array}\right|\ .$ Thus, $\left\{\begin{array}{cc} QB=PC\ ; & \widehat{ABP}\equiv\widehat{ACP}\\\\ QC=PB\ ; & \widehat{ABQ}\equiv\widehat{ACQ}\end{array}\right|\ .$ Apply the theorem of Sines $:\ \left\{\begin{array}{ccc} \triangle ABQ\ : & \frac {QA}{QB}=\frac {\sin\widehat {ABQ}}{\sin x}\\\\ \triangle ACQ\ : & \frac {QC}{QA}=\frac {\sin (A-x)}{\sin \widehat{ACQ}}\\\\ \triangle APB\ : & \frac {PA}{PB}=\frac {\sin\widehat {ABP}}{\sin (A-y)}\\\\ \triangle APC\ : & \frac {PC}{PA}=\frac {\sin y}{\sin \widehat{ACP}}\end{array}\right|\ \bigodot\ \implies$

$\frac {\sin (A-x)}{\sin x}=$ $\frac {\sin (A-y)}{\sin y}\iff$ $\sin A\cdot\cot x-\cos A=\sin A\cot y-\cos A\iff$ $\cot x=\cot y\iff$ $x=y\iff$ $\widehat{BAQ} \equiv\widehat{CAP}$ .

PP7. Let an isosceles $\triangle ABC$ with $AB=AC$ . The $B$ -angled (interior) bisector meet $AC$ at $D$ so that $BD + DA = BC$ . Find $A$ .

Proof 1. Let $B=2x$ , the circle $w=C(D,r)$ with $DA=r$ and $\{M,N\}\subset BC\cap w$ , so that $M\in (BN)$ . Prove easily that $\triangle MDN$ is $D$ -isosceles with $m\left(\widehat{DMN}\right)=$

$m\left(\widehat{DNM}\right)=4x$ $\implies$ $N\in (MC)$ and $ND=NC\implies$ $NC=DA\implies$ $BN=BD\implies$ $m\left(\widehat{BDN}\right)=4x\implies$ $9x=180^{\circ}\implies$ $x=20^{\circ}\implies$ $A=100^{\circ}$ .

Proof 2 (trigonometric). Let $A=4x$ . Thus, $m\left(\widehat{ABD}\right)=45^{\circ}-x$ , $m\left(\widehat{BDC}\right)=45^{\circ}+3x$ and $C=90^{\circ}-2x$ . Apply theorem of Sines in the triangles $ABD$

and $ABC\ :\ BD+DA=BC\implies$ $\frac {BD+DA}{AB}=\frac {BC}{AB}\implies$ $\frac {\sin 4x+\sin\left(45^{\circ}-x\right)}{\sin \left(45^{\circ}+3x\right)}=$ $\frac {\sin 4x}{\cos 2x}\implies$ $\sin 4x+\sin\left(45^{\circ}-x\right)=2\sin 2x\sin\left(45^{\circ}+3x\right)\implies$

$\sin 4x+\cos\left(45^{\circ}+x\right)=\cos\left(45^{\circ}+x\right)-\cos\left(45^{\circ}+5x\right)\implies$ $\sin 4x=\sin \left(5x-45^{\circ}\right)\implies$ $4x+\left(5x-45^{\circ}\right)=180^{\circ}\implies$ $9x=225^{\circ}\implies$ $x=25^{\circ}\implies$ $\boxed{A=100^{\circ}}$ .

Proof 3. Let $A=2x$ and $AB=AC=b$ , $BC=b$ . Thus, $\boxed{2\sin x=\frac ab}\ (*)$ , $DA=\frac {b^2}{a+b}$ and $BD=a-DA\implies$ $BD=\frac {a^2+ab-b^2}{a+b}$ . Apply the theorem of Cosines

to $\widehat{BAC}$ in $\triangle ABD\ :$ $\cos 2x=\frac {b^2(a+b)^2+b^2-\left(a^2+ab-b^2\right)}{2b^3(a+b)}=$ $\frac {b^4+4ab^3+2a^2b^2-2a^3b-a^4}{2b^3(a+b)}\implies$ $\cos 2x=\frac {b^3+3ab^2-a^2b-a^3}{2b^3}\implies$

$2\cos 2x=1+3\left(\frac ab\right)-\left(\frac ab\right)^2-\left(\frac ab\right)^3\stackrel{(*)}{\implies}$ $2\cos 2x=1+6\sin x-4\sin^2x-8\sin^3x\implies$ $2\cos 2x=1+6\sin x-2(1-\cos 2x)-4\sin x(1-\cos 2x)\implies$

$-1+2\sin x+4\sin x\cos 2x=0\implies$ $-1+2\sin x+2(\sin 3x-\sin x)=0\implies$ $\sin 3x=\frac 12\implies$ $3x=150^{\circ}\implies$ $x=50^{\circ}\implies A=100$ .

PP8. Let $\triangle ABC$ with $A=20^{\circ}$ . Its $B$ -bisector cut $AC$ at $D$ . Prove that $\boxed{AD=DB+BC\iff C\in \left\{80^{\circ},40^{\circ}\right\}}$ .

Proof. Let $B=2x$ . Thus, $AD=DB+BC\iff$ $\frac {AD}{DB}=1+\frac {BC}{BD}\iff$ $\frac {\sin x}{\sin 20}=1+\frac {\sin(x+20)}{\sin(2x+20)}$ $\iff$ $\sin x\sin (2x+20)=\sin 20[\sin (2x+20)+\sin (x+20)]\iff$

$2\sin x\sin (2x+20)=$ $\cos 2x-\cos (2x+40)+$ $\cos x-\cos (x+40)\iff$ $2\sin x\sin (2x+20)=[\cos 2x-\cos (x+40)]+$ $[\cos x-\cos (2x+40)]\iff$

$\sin x\sin (2x+20)=$ $\sin\left(\frac {3x}{2}+ 20\right)\sin\left(20-\frac x2\right)+$ $\sin\left(\frac {3x}{2}+20\right)\sin\left(\frac x2+20\right)$ $\iff$ $\sin x\sin (2x+20)=$ $\sin\left(\frac {3x}{2}+ 20\right)\left[\sin\left(20-\frac x2\right)+\sin\left(\frac x2+20\right)\right]\iff$

$\sin \frac x2\cos\frac x2\sin (2x+20)=$ $\sin\left(\frac {3x}{2}+ 20\right)\sin 20\cos\frac x2\iff$ $\sin\frac x2\sin (2x+20)=$ $\sin\left(\frac {3x}{2}+ 20\right)\sin 20\iff$ $\cos\left(\frac {3x}{2}+20\right)-\cos\left(\frac {5x}{2}+20\right)=$

$\cos\frac {3x}{2}-\cos\left(\frac {3x}{2}+40\right)\iff$ $\cos\left(\frac {3x}2+20\right)+\cos\left(\frac {3x}2+40\right)=$ $\cos\left(\frac {5x}2+20\right)+\cos \frac {3x}{2}$ $\iff$ $\cos\left[3\left(\frac {x}{2}+10\right)\right]\cos 10=$ $\cos (2x+10)\cos\left(\frac x2+10\right)$ $\iff$

$\left[4\cos^2\left(\frac x2+10\right)-3\right]\cos 10=\cos (2x+10)\iff$ $[2\cos (x+20)-1]\cos 10=$ $\cos (2x+10)\iff$ $\cos (x+30)+\cos (x+10)=$ $\cos 10+\cos (2x+10)\iff$

$x\in\{40^{\circ},60^{\circ}\}$ $\iff$ $C\in \left\{80^{\circ},40^{\circ}\right\}$ .

This post has been edited 114 times. Last edited by Virgil Nicula, Nov 22, 2015, 9:43 PM

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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).

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440. Piece of Poetry.

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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.

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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

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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).

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El_Ectric:
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You're welcome.
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Nice blog!
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Dear Virgil !

Congratulations for this blog !
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7. Some interesting "slicing" problems.

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