446. Mathematics "instant" for the middle school.

by Virgil Nicula, Jun 22, 2016, 12:05 AM

P0. Prove that in any $\triangle ABC$ there is the inequality $\boxed{\sin\frac A2\le \frac a{b+c}\le\sqrt{1-\sin B\sin C}}\ (*)\ .$

Proof 1. Let $w=\mathbb C(I,r)$ be the incircle of $\triangle ABC$ and $D\in BC\cap AI\ .$ Thus, $\sin\frac A2=\frac r{IA}\le\frac {ID}{IA}=\frac {BD}{BA}=\frac {\frac {ac}{b+c}}c=\frac a{b+c}\implies$ $\sin\frac A2\le \frac a{b+c}$ with equality iff $AB=AC\ .$

Otherwise. Let $A$ -excircle $w_a=\mathbb C\left(I_a,r_a\right)\ .$ Thus, $I_aA\ge h_a+r_a\implies$ $\sin\frac A2=$ $\frac {r_a}{I_aA}\le$ $\frac {r_a}{h_a+r_a}=$ $\frac {ar_a}{ah_a+a r_a}=$ $\frac {ar_a}{2(s-a)r_a+ar_a}=\frac a{b+c}\implies$ $\sin\frac A2\le\frac a{b+c}\ .$

Proof 2. $\sin\frac A2\le \frac a{b+c}\iff$ $\sin^2\frac A2\le \left(\frac a{b+c}\right)^2\iff$ $\frac {(s-b)(s-c)}{bc}\le \frac {a^2}{(b+c)^2}\iff$ $\frac {bc-s(s-a)}{bc}\le \frac {a^2}{(b+c)^2}\iff$ $1-\frac {a^2}{(b+c)^2}\le\frac {s(s-a)}{bc}\iff$

$\frac {4s(s-a)}{(b+c)^2}\le\frac {s(s-a)}{bc}\iff$ $4bc\le (b+c)^2\iff$ $(b-c)^2\ge 0\ ,$ what is true. I'll have equality iff $B=C\ .$

Proof 3. $\sin\frac A2\le \frac a{b+c}\iff$ $\sin\frac A2\le\frac {\sin A}{\sin B+\sin C}\iff$ $\sin\frac A2\le\frac {2\sin \frac A2\cos\frac A2}{2\cos\frac A2\cos\frac {B-C}2}\iff$ $\cos\frac {B-C}2\le 1\ ,$ what is true. I'll have equality iff $B=C\ .$

Remark 1. $\frac a{b+c}\le\sqrt{1-\sin B\sin C}\iff$ $\sin B\sin C\le 1-\frac {a^2}{(b+c)^2}\iff$ $\frac {2S}{ac}\cdot\frac {2S}{ab}\le \frac {4s(s-a)}{(b+c)^2}\iff$ $\frac {(s-b)(s-c)}{bc}\le \left(\frac a{b+c}\right)^2\iff$ $\sin\frac A2\le\frac a{b+c}\ .$

Remark 2. Prove easily the identities $\boxed{\begin{array}{ccccccc} 1-\sin B\sin C & = & \sin^2\frac A2 & + & \left(\frac {b-c}a\right)^2\cdot \cos^2\frac A2 & \ge & \sin^2\frac A2\\\\ \left(\frac a{b+c}\right)^2 & = & \sin^2\frac A2 & + & \left(\frac {b-c}{b+c}\right)^2\cdot \cos^2\frac A2 & \ge & \sin^2\frac A2\\\\ 1-\sin B\sin C & = & \left(\frac a{b+c}\right)^2 & + & \frac {4s(s-a)(b-c)^2}{a^2(b+c)^2}\cdot\cos^2\frac A2 & \ge & \left(\frac a{b+c}\right)^2\end{array}}\ .$

Extension. Let $L\in (BC)$ be a point for what denote $\left\{\begin{array}{ccc} m\left(\widehat{LAC}\right) & = & x\\\ m\left(\widehat{LAB}\right) & = & y\end{array}\right\|\ ,$ i.e. $x+y=A\ .$ Prove that there is the inequality $\boxed{b\sin x+c\sin y\le a}\ (*)\ .$

Proof 1. Let $AL=l\ ,$ $P\in BC$ so that $AP\perp BC$ and area $S$ of $\triangle ABC\ .$ Thus, $ah_a=2S=cl\sin y+bl\sin x=$ $l(c\sin y+b\sin x)$ $\implies$

$\frac a{b\sin x+c\sin y}=$ $\frac l{h_a}\ge 1$ $\implies$ $b\sin x+c\sin y\le a\ .$ In the particular case $x=y=\frac A2$ obtain that $\sin\frac A2\le\frac a{b+c}\ .$

Proof 2. $(\exists )\ |u|\le \frac A2$ so that $\left\{\begin{array}{ccc} x & = & \frac A2-u\\\\ y & = & \frac A2+u\end{array}\right\|\ .$ Therefore, $b\sin x+c\sin y\le a$ $\iff$ $\sin B\sin\left(\frac A2-u\right)+$ $\sin C\sin\left(\frac A2+u\right)\le$ $\sin A$ $\iff$

$\sin B\cos\left(\frac {B+C}2+u\right)+$ $\sin C\cos\left(\frac {B+C}2-u\right)\le$ $\sin A$ $\iff$ $\sin\left(\frac{3B+C}2+u\right)+$ $\cancel{\sin\left(\frac {B-C}2-u\right)}+$ $\sin\left(\frac {3C+B}2-u\right)-$

$\cancel{\sin\left(\frac {B-C}2-u\right)}\le$ $2\sin A\iff$ $2\sin (B+C)\cos\left(\frac {B-C}2+u\right)\le 2$ $\sin A$ $\iff$ $\cos\left(\frac {B-C}2+u\right)\le 1\ ,$ what is true.

P1 (Cristian Tello). Let $\triangle ABC$ with $BA\perp BC$ and the circle $w$ with the diameter $[AB]\ .$ The bisector of $\widehat{BAC}$ cut again

$w$ in $E$ and $BC$ at $F\ .$ Let $AE=a\ ,$ $EF=b$ and $m\left(\widehat{ACB}\right)=\theta\ .$ Prove that $\boxed{\sin\theta =\frac {a-b}{a+b}}\ (*)\ .$

Proof 1. Let $m\left(\widehat{FAB}\right)=m\left(\widehat{FAC}\right)=\alpha\implies$ $\left\{\begin{array}{c} AF\cos\widehat{FAB}=AB\implies AF\cos \alpha =AB\\\\ AB\cos \widehat{BAE}=AE\implies AB\cos\alpha =AE\end{array}\right\|$ $\bigodot\implies$ $\cos^2\alpha =\frac {AE}{AF}\iff$ $\cos^2\alpha =\frac a{a+b}\iff$ $2\cos^2\alpha =\frac {2a}{a+b}$

$\iff$ $1+\cos 2\alpha =\frac {2a}{a+b}\iff$ $\cos 2\alpha =\frac {2a}{a+b}-1\iff$ $\boxed{\cos 2\alpha =\frac {a-b}{a+b}}\ (1)\ .$ In conclusion, $\theta +2\alpha =\frac {\pi}2\implies$ $\sin\theta =\cos 2\alpha\ \stackrel{(1)}{\iff}\ \sin\theta =\frac {a-b}{a+b}\ ,$ i.e. relation $(*)\ .$

Proof 2. $AE^2=EB\cdot EF=ab\ ,$ i.e. $AE=\sqrt{ab}$ and $\tan\alpha =\frac {EF}{EA}=\frac b{\sqrt {ab}}\implies$ $\boxed{\tan\alpha =\sqrt {\frac ba}}\implies$ $\underline{\underline{\sin\theta}} =\cos 2\alpha=\frac {1-\tan^2\alpha }{1+\tan^2\alpha}=$ $\frac {1-\frac ba }{1+\frac ba}=\underline{\underline{\frac {a-b}{a+b}}}\ ,$ i.e. relation $(*)\ .$

Proof 3. Let $\{A,D\}=\{A,C\}\cap w\ ,$ $G\in BD\cap AF$ and symmetric point $K$ of $F$ w.r.t. $B\ ,$ i.e. $AK=AF=a+b\ .$ Prove easily that $m\left(\widehat{KAG}\right)=m\left(\widehat{GBF}\right)=2\alpha$

$\implies$ $AKBG$ is cyclic $\implies$ $KG\perp AF$ and $\widehat{AKG}\equiv\widehat{ABG}\equiv\widehat{ACB}\ ,$ i.e. $m\left(\widehat{AKG}\right)=\theta\ .$ In conclusion, $\sin\theta =\frac {AG}{AK}\implies$ $\sin\theta =\frac {a-b}{a+b}$ (David Rodriguez).

P2 (Miguel Ochoa Sanchez). Let the incircle $w=\mathbb C(I,1)$ of the square $ABCD$ and a point $X\in w$ for which denote its projections

$(U,V,W,Z)$ on $(AB,BC,CD,DA)$ respectively where $XU=b\ ,$ $XV=c\ ,$ $XW=d\ ,$ $XZ=a\ .$ Prove that $a^3+b^3+c^3+d^3=10\ .$

Proof. Let midpoints $(M,N,P,Q)$ of $([AB],[BC],[CD],[DA])$ respectively and the projections $(K,L)$ of $X$ on $(MP,NQ)$ respectively. Thus, $XK^2+XL^2=XI^2\iff$

$\boxed{(1-c)^2+(1-b)^2=1}\ (*)\ .$ Apply the theorem of cathetus in $:\ \left\{\begin{array}{cccc} \triangle MXP\ : & XK^2=KM\cdot KP & \implies & (1-c)^2=bd\\\\ \triangle NXQ\ : & XL^2=LN\cdot LQ & \implies & (1-b)^2=ac\end{array}\right\|$ $\bigoplus\ \stackrel{(*)}{\implies}\ ac+bd=1\ .$

Therefore, if denote $s\equiv a+c=b+d=2\ ,$ $u\equiv ac=(1-b)^2=(d-1)^2\ ,$ $v\equiv bd=(1-c)^2=(a-1)^2\ ,$ where $u+v=1\ ,$ then obtain

$a^3+b^3+c^3+d^3=\left(a^3+c^3\right)+\left(b^3+d^3\right)=$ $\left(s^3-3su\right)+\left(s^3-3sv\right)=$ $2s^3-3s(u+v)=16-6\cdot 1=10\implies a^3+b^3+c^3+d^3=10\ .$

P3 (Edson Curahua Ortega). In the first quadrant $\overline O\overarc{AB}$ of the circle $w=\mathbb C(O,R)\ ,$ where $OA\perp OB\ ,$ inscribe circles $w_1=\mathbb C(I,x)$ and $w_2=\mathbb C(J,y)$

so that $:\ w_1$ and $w_2$ are exterior tangent one to other and both are interior tangent to $w\ ;\ J\in (OB)$ and $2y<R\ .$ Prove that $\boxed{\ 2x\ +\ y\ =\ R\ }\ (*)\ .$

Proof. Let $C$ be the projection of $I$ on $OB\ .$ Observe that $OC=x\ ,OI=R-x\ ,\ IC^2=OI^2-OC^2=(R-x)^2-x^2$ $\implies$ $IC^2=R^2-2Rx\ ;$ $CJ=R-(x+y)\ ,$

$IJ=x+y\ .$ Thus, $IC\perp OB\iff$ $IO^2-IJ^2=CO^2-CJ^2\iff$ $(R-x)^2-(x+y)^2=x^2-[R-(x+y)]^2\iff$ $R^2-2Rx=-R^2+2R(x+y)\iff$

$2R^2=2Rx+2R(x+y)\iff$ $R=x+(x+y)\iff$ $2x+y=R\ .$

P4. Prove that $(\forall )$ $\triangle ABC$ there is the chain of the equivalencies $:\ \boxed{A=2B\ \iff\ l_a=\frac {bc}a\ \iff\ a^2=b(b+c)}\ ,$ where $l_a$ is the length of the interior $A$ -bisector.

Proof 1. Let $AD$ - the interior $A$ -bisector of $\triangle ABC\ ,$ where $D\in (BC)\ .$ Prove easily that $\frac {DC}b=\frac {DB}c=\frac a{b+c}\implies$ $\boxed{DC=\frac {ab}{b+c}}\ (1)\ .$ Thus, $A=2B\iff$

$\widehat{CAD}\equiv\widehat{CBA}\iff$ $\triangle CAD\sim\triangle CBA\iff$ $\frac{CA}{CB}=\frac {AD}{BA}=\frac {CD}{CA}\ \stackrel{(1)}{\iff}\ \frac ba=\frac {l_a}c=\frac {\frac {ab}{b+c}}b=\frac a{b+c}\iff$ $\odot\begin{array}{cccccc} \nearrow & l_a & = & \frac {bc}a & (2) & \searrow\\\\ \searrow & a^2 & = & b(b+c) & (3) & \nearrow\end{array}\odot$

Proof 2. Denote $X\in AB$ so that $A\in (BX)$ and $AX=AC=b\ .$ Thus, $CX=CB=a$ and $\triangle XCA\sim\triangle XBC\iff$ $\frac {XC}{XB}=\frac {XA}{XC}\iff$ $\frac a{b+c}=\frac ba\iff$ $a^2=b(b+c)\ .$

Proof 3. Theorem of SINES $\ : a^2=b^2+bc\iff$ $\sin^2A-\sin^2B=\sin B\sin C\iff$ $\sin (A+B)\sin (A-B)=\sin B\sin C\iff$ $\sin (A-B)=\sin B\iff$ $A=2B\ .$

Remark. $B=2C\iff AB=2\cdot DM\ ,$ where $M$ is the midpoint of $[BC]$ and $D$ is the projection of $A$ on $BC\ .$ Indeed, I"ll use only two relations $:$

$\boxed{B = 2C\iff b^2 = c^2+ ac}\ (1)$ and $\boxed{2a\cdot DM = b^2-c^2}\ (2)\ .$ Therefore, $c = 2\cdot HM\stackrel{(2)}{\iff}\ ac = b^2 - c^2\iff$ $b^2=c^2+ac\ \stackrel{(1)}{\iff}\ B = 2C .$

P5. Let $\triangle ABC$ with $a>\max\{b,c\}$ and $\{P,Q\}\subset [BC]$ so that $BP=AP$ and $CQ=AC\ .$ Prove that $AB\perp AC\iff PQ^2=2\cdot BQ\cdot CP\iff m\left(\widehat{PAQ}\right)=45^{\circ}\ .$

Proof. Prove easily that $Q\in (BP)\ ,$ $QP=b+c-a$ and $\left\{\begin{array}{ccc} BQ=a-b\\\\ PC=a-c\end{array}\right\|\ .$ In conclusion, $PQ^2=2\cdot BQ\cdot CP\iff$ $(b+c-a)^2=2\cdot (a-b)(a-c)\iff$

$(b+c)^2-\cancel{2a(b+c)}+\cancel{a^2}=\cancel 2a^2-\cancel{2a(b+c)}+2bc\iff$ $b^2+c^2=a^2\iff$ $AB\perp AC\ .$ Thus, $\left\{\begin{array}{cccccc} m\left(\widehat{QAB}\right)=A-m\left(\widehat{QAC}\right)=A-\left(90^{\circ}-\frac C2\right)=\frac {A-B}2\\\\ m\left(\widehat{PAC}\right)=A-m\left(\widehat{PAB}\right)=A-\left(90^{\circ}-\frac B2\right)=\frac {A-C}2\end{array}\right\|\implies$

$m\left(\widehat{PAQ}\right)=A-m\left(\widehat{QAB}\right)-m\left(\widehat{PAC}\right)=$ $A- \frac {A-B}2-\frac {A-C}2=\frac {B+C}2\implies$ $\boxed{m\left(\widehat{PAQ}\right)=\frac {B+C}2}\ (*)\ .$ Thus, $AB\perp AC\ \stackrel{(*)}{\iff}\ m\left(\widehat{PAQ}\right)=45^{\circ}\ .$

P6. Let $\triangle ABC$ with the excircles $w_b=\mathbb C\left(I_b,r_b\right),\ w_c=\left(I_c,r_c\right).$ Denote $\left\{\begin{array}{ccccc} D\in AC\cap w_b & ; & E\in AB\cap w_c\\\\ \{D,F\}=ED\cap w_b & ; & \{E,G\}=ED\cap w_c\end{array}\right\|$ . Prove that $FD=GE.$

Proof. Let $\left\{\begin{array}{ccc} X & \in & AC\cap w_c\\\\ Y & \in & AB\cap w_b\end{array}\right\|.$ Is well known that $\left\{\begin{array}{ccc} AX=AE=s-b\\\\ AY=AD=s-c\end{array}\right\|.$ Thus, $\left\{\begin{array}{ccc} DX=DA+AX=(s-c)+(s-b) & \implies & DX=a\\\\ EY=EA+AY=(s-b)+(s-c) & \implies & EY=a\end{array}\right\|\implies$

$\boxed{EY=DX=a}\ (*)$ $\implies$ $\left\{\begin{array}{ccc} DX^2 & = & DE\cdot DG\\\\ EY^2 & = & ED\cdot EF\end{array}\right\|\ \stackrel{(*)}{\implies}\ \cancel{DE}\cdot DG=$ $\cancel{ED}\cdot EF\implies$ $DG=EF\implies$ $\cancel{DE}+EG=\cancel{ED}+DF\implies$ $EG=FD.$

P7 (Miguel Ochoa Sanchez). Let $\triangle ABC$ and $D\in (AC)$ so that $m\left(\widehat{ADB}\right)=30^{\circ}$ and $m\left(\widehat{ABD}\right)=2\cdot m\left(\widehat{DBC}\right).$ Prove that $BD^2=b^2-c^2\iff C=18^{\circ}.$

Proof 1. Let $E\in AC$ so that $BE\perp AC$ and suppose w.l.o.g. $BE=1,$ i.e. $\boxed{\ BD\ =\ 2\ }.$ Observe that $\left\{\begin{array}{ccc} m\left(\widehat{DBC}\right) & = & 30^{\circ}-C\\\ m\left(\widehat{ABD}\right) & = & 60^{\circ}-2C\end{array}\right\|$ and $m\left(\widehat{EBA}\right)= 2C.$ Therefore,

$1=c\cdot \cos 2C=a\cdot \sin C$ and $\left\{\begin{array}{ccc} AE & = & c\cdot \sin 2C\\\ CE & = & a\cdot\cos C\end{array}\right\|\ .$ Observe that $\boxed{c=\frac 1{\cos 2C}}\ (1)$ and $b=CE-AE=a\cdot\cos C-c\cdot \sin 2C\implies$ $\boxed{b=\cot C-\tan 2C}\ (2)\ .$

Therefore, $BD^2=b^2-c^2\iff b^2-c^2=4\iff\ \stackrel{(1\wedge 2)}{\iff}\ \left(\cot C-\tan 2C\right)^2=4+\frac 1{\cos^22C}\iff$ $\frac {\cos^23C}{\sin^2C\cos^22C}=4+\frac 1{\cos^22C}\iff$

$\cos^23C=(2\sin C\cos 2C)^2+\sin^2C\iff$ $\cos^23C=(\sin 3C-\sin C)^2+\sin^2C\iff$ $\cos^23C=\sin^23C+2\sin^2C-2\sin C\sin 3C\iff$

$\cos^23C-\sin^23C=(1-\cos 2C)-(\cos 2C-\cos 4C)\iff$ $\cos 6C+2\cos 2C=1+\cos 4C\iff$ $\cancel{\cos 2C}\left(4\cos^22C-3\right)+2\cancel{\cos 2C}=2\cos\cancel{^2}2C\iff$

$4\cos^22C-3+2=2\cos 2C\iff$ $4t^2-2t-1=0$ where $\boxed{t=\cos 2C}\iff$ $t=\frac {1+\sqrt 5}4\iff$ $\cos 2C=\cos 36^{\circ}\iff$ $2C=36^{\circ}\iff \boxed{\ C\ =\ 18^{\circ}\ }\ .$

Otherwise. $BD^2=b^2-c^2\iff b^2-c^2=4\iff\ \stackrel{(1\wedge 2)}{\iff}\ \left(\cot C-\tan 2C\right)^2=4+\frac 1{\cos^22C}\iff$ $\left(\cot C-\tan 2C\right)^2=5+\tan^22C\iff$ $\frac 1{\tan^2C}-\frac {2\tan 2C}{\tan C}=5$

$\iff$ $1-2\tan C\tan 2C=5\tan^2C\iff$ $\frac {4\tan^2C}{1-\tan^2C}+5\tan^2C=1\iff$ $5\tan^4C-10\tan^2C+1=0\iff$ $5\cdot \left(\frac {1-t}{1+t}\right)^2-10\cdot \frac {1-t}{1+t}+1=0\ ,$ where $t=\cos 2C,$

$\tan^2C=\frac {1-t}{1+t}\iff$ $5(1-t)^2-10(1-t^2)+(1+t)^2=0\iff$ $16t^2-8t-4=0\iff$ $4t^2-2t-1=0\iff$ $t=\frac {1+\sqrt 5}4\iff$ $\cos 2C=\cos 36^{\circ}\iff$ $C=18^{\circ}.$

P8 (Miguel Ochoa Sanchez). Let $A$ -rightangled $\triangle ABC$ and $H\in (BC)$ so that $AH\perp BC.$ The bisectors of the

angles $\widehat{BAH},$ $\widehat{CAH}$ meet $BC$ at the points $D,$ $E$ respectively. Prove that $[ABD]+[ACE]\ge [ADE]\sqrt 2\ .$

Proof 1. Let $\left\{\begin{array}{c} m\left(\widehat{DAB}\right)=m\left(\widehat{DAH}\right)=\gamma\\\\ m\left(\widehat{EAC}\right)=m\left(\widehat{EAH}\right)=\beta\end{array}\right\|\ ,\ \beta +\gamma=45^{\circ}\ ,\ \left\{\begin{array}{c} B=m\left(\widehat{HAC}\right)=2\beta\\\\ C=m\left(\widehat{HAB}\right)=2\gamma\end{array}\right\|\ .$ Thus, $(b+c)^2\le 2\left(b^2+c^2\right)=2a^2\implies\boxed{b+c\le a\sqrt 2}\ (*)$

and $\left\{\begin{array}{c} m\left(\widehat{BAE}\right)=m\left(\widehat{BEA}\right)=\beta +2\gamma\implies BE=BA=c\ \mathrm{and}\ CE=a-c\\\\ m\left(\widehat{CAD}\right)=m\left(\widehat{CDA}\right)=\gamma +2\beta\implies CD=CA=b\ \mathrm{and}\ BD=a-b\end{array}\right\|\implies$ $DE=b+c-a$ and $\left\{\begin{array}{c} \frac {[ABD]}{[ADE]}=\frac {DB}{DE}=\frac {a-b}{b+c-a}\\\\ \frac {[ACE]}{[ADE]}=\frac {CE}{DE}=\frac {a-c}{b+c-a}\end{array}\right\|\bigoplus\implies$

$\frac {[ABD]+[ACE]}{[ADE]}=$ $\frac {2a-(b+c)}{(b+c)-a}=$ $\frac {a-(b+c-a)}{b+c-a}=$ $\frac a{(b+c)-a}-1\ \stackrel{(*)}{\ge}$ $\frac a{a\sqrt 2-a}-1=$ $\frac 1{\sqrt 2-1}-1=$ $\sqrt 2\implies$ $[ABD]+[ACE]\ge [ADE]\sqrt 2$ (equality iff b=c).

Proof 2. Suppose w.l.o.g. $AH=1$ and I'll use same notations from upper proof. Denote $:$

$\left\{\begin{array}{ccccc} \tan \gamma = x & \implies & HD=x\ ,\ HB=\tan 2\gamma =\frac {2x}{1-x^2} & \implies & DB=HB-HD=\frac {2x}{1-x^2}-x\\\\ \tan\beta =y & \implies & HE=y\ ,\ HC=\tan 2\beta =\frac {2y}{1-y^2} & \implies & CE=HC-HE=\frac {2y}{1-y^2}-y\end{array}\right\|$ where $\beta +\gamma =45^{\circ}$ $\implies$ $\tan(\beta +\gamma )=1$ $\implies$ $\frac {\tan \beta +\tan\gamma}{1-\tan\beta\tan\gamma}=$

$\frac {y+x}{1-xy}=1$ $\implies$ $x+y+xy=1\implies$ $\boxed{(1+x)(1+y)=2}\ (*)\ .$ Thus, $\frac {[ABD]+[ACE]}{[ADE]}=\frac {BD+CE}{DE}=$ $\frac {\frac {2x}{1-x^2}+\frac {2y}{1-y^2}-(x+y)}{x+y}=$ $\frac {\frac {2x}{1-x^2}+\frac {2y}{1-y^2}}{x+y}-1=$

$\frac {2\cancel{(x+y)}-2xy\cancel{(x+y)}}{\cancel{(x+y)}\left(1-x^2\right)\left(1-y^2\right)}-1=$ $\frac {\cancel 2(1-xy)}{\cancel{(1+x)(1+y)}(1-x)(1-y)}-1\ \stackrel{(*)}{=}\ \frac {1-xy}{(1-x)(1-y)}-1\ .$ From the relation $(*)$ obtain that $\boxed{y=\frac {1-x}{1+x}}\ .$ Thus, $\frac {[ABD]+[ACE]}{[ADE]}=$

$\frac {1-xy}{(1-x)(1-y)}-1=\frac {1-x\cdot \frac {1-x}{1+x}}{(1-x)(1-\frac {1-x}{1+x})}-1=$ $\frac {x^2+1}{2x(1-x)}-1\ .$ So $\frac {[ABD]+[ACE]}{[ADE]} \ge \sqrt 2\iff$ $\frac {x^2+1}{2x(1-x)}\ge 1+\sqrt2\iff$ $x^2+1\ge 2\left(1+\sqrt 2\right) \left(x-x^2\right)$ $\iff$

$\left(3+2\sqrt 2\right)x^2-2\left(1+\sqrt 2\right)x+1\ge 0\iff$ $\left[x\left(1+\sqrt 2\right)-1\right]^2\ge 0$ what is truly. Have equality iff $x=\frac 1{1+\sqrt 2},$ i.e. $x=y=\sqrt 2-1$ what means $\gamma=\beta =\frac {\pi}8\ .$ Very nice problem!

Proof 3. Observe that $\left\{\begin{array}{ccccccccccccccccccccc} \frac {DB}a & = & \frac {DH}b & = & \frac {BH}{a+b} & = & \frac {\frac {c^2}a}{a+b} & = & \frac {a^2-b^2}{a(a+b)} & = & \frac {a-b}{a} & \implies & \frac {DB}a & = & \frac {a-b}a & = & \boxed{\ DB=a-b\ }\ (1) & \implies & \boxed{DH=\frac {b(a-b)}a\ } & (2.1)\\\\ \frac {CE}a & = & \frac {EH}c & = & \frac {CH}{a+c} & = & \frac {\frac {b^2}a}{a+c} & = & \frac {a^2-c^2}{a(a+c)} & = & \frac {a-c}{a} & \implies & \frac {EC}a & = & \frac {a-c}a & = & \boxed{\ EC=a-c\ }\ (3) & \implies & \boxed{\ EH=\frac {c(a-c)}a\ } & (2.2)\end{array}\right\|$

$\implies$ $DE=HD+HE\ \stackrel{2.1 \wedge\ 2.2}{=}\ \frac {b(a-b)}a+\frac {c(a-c)}a=$ $\frac {a(b+c)-\left(b^2+c^2\right)}a=$ $\frac {a(b+c)-a^2}a=b+c-a\implies$ $\boxed{\ DE=b+c-a\ }\ (4)\ .$ Denote $A_1=[ABD]\ ,$ $A_2=[ACE]$

and $A_3=[ADE]\ .$ Thus, $\frac {A_1}{BD}=\frac {A_3}{DE}=\frac {A_2}{CE}\iff$ $\frac {A_1}{a-b}=\frac {A_3}{b+c-a}=\frac {A_2}{a-c}=\frac {A_1+A_2}{2a-b-c}\implies$ $\frac {A_1+A_2}{A_3}=$ $\frac {2a-b-c}{b+c-a}=\frac {2-t}{t-1}\ ,$ where $t=\frac {b+c}a\ .$ From the simple inequality

$(b+c)^2\le 2\left(b^2+c^2\right)=2a^2$ obtain that $t=\frac {b+c}a\le \sqrt 2\ ,$ i.e. $\boxed{\ t\le\sqrt 2\ }\ (5)\ .$ Hence, $\frac {A_1+A_2}{A_3}=$ $\frac {2-t}{t-1}=$ $-1+\frac 1{t-1}\ \stackrel{(5)}{\ge}\ -1+\frac 1{\sqrt 2-1}=$ $\frac {\left(1-\sqrt 2\right)+1}{\sqrt 2-1}=$ $\frac {2-\sqrt 2}{\sqrt 2-1}=$ $\sqrt 2\implies$

$\frac{A_1+A_2}{A_3}\ge \sqrt 2$ $\implies$ $A_1+A_2\ge A_3\sqrt 2\implies$ $[ABD]+[ACE]\ge [ADE]\sqrt 2\ .$ Very nice proof!

P9 (Kunihiko Chikaya). Ascertain the natural numbers $n=\overline{ab24}$ what are multiples of $17,$ where $\{a,b\}$ are digits in the base $10\ ,$ i.e. $\{a,b\}\subset \overline{0,9}\ .$

Proof. Define the equivalence $:\ (\forall\ )\ \{x,y\} \subset\mathbb Z\ , \boxed{x\ \sim\ ~y\iff 17\ |\left(x-y\right)}\ .$ Thus, $n=\overline{ab24}=$ $100(10a+b)+24=$ $\left(\cancel{17\cdot 6}-2\right)\left(\cancel{17a}-7a+b\right)+\left(\cancel{17}+7\right)\ \sim\ (14a-2b+7)\ \sim$

$17a-(14a-2b+7)\implies \boxed{n\ \sim (3a+2b-7)}\ .$ Observe that $\{a,b\}\subset\overline{0,9}\implies$ $17\ |\ (3a+2b-7)\subset [-7,38]\cap Z=\{0,17,34\}\ ,$ i.e. $\boxed{(3a+2b)\in \{7,24,41\}}\ .$

Therefore $:\ 3a+2b=7\iff (a,b)=(1,2)\ ;\ 3a+2b=24\iff (a,b)\in\left\{(2,9),(4,6),(6,3),(8,0)\right\}\ ;\ 3a+2b=41\iff (a,b)=(9,7)\ .$

In conclusion, $(a,b)\in \{(1,2),(2,9),(4,6),(6,3),(8,0),(9,7)\}\ .$

P10 (Luis A. S. Cruz, Peru). Let $\triangle ABC$ with $AB\perp AC$ and $AB=AC\ .$ Choose the points $\{M,N\}\subset (BC)$ so that $BM<BN$ and $m\left(\widehat{MAN}\right)=\frac {\pi}4\ .$ Prove that $MB^2+NC^2=MN^2\ .$

Proof 1. Let $O$ be the midpoint of the side $[BC]\ .$ Suppose w.l.o.g. $AB=AC=AO=1\ .$ Hence $BC=\sqrt 2$ and there are $\{\alpha ,\beta\}\subset \left(0,\frac {\pi}4\right)$ so that $\alpha +\beta =\frac {\pi}4$ and

$\left\{\begin{array}{ccccccccc} OM=x & \implies & MB=1-x & ; & m\left(\widehat{OAM}\right)=\alpha & \implies & m\left(\widehat{BAM}\right)=\beta & \implies & x=\tan\alpha\\\\ ON=y & \implies & NC=1-y & ; & m\left(\widehat{OAN}\right)=\beta & \implies & m\left(\widehat{CAN}\right)=\alpha & \implies & y=\tan\beta\end{array}\right\|\ .$ Thus, $MB^2+NC^2=MN^2\iff$ $(1-x)^2+(1-y)^2=(x+y)^2\iff$

$2-2(x+y)+\cancel{x^2}+\cancel{y^2}=$ $\cancel{x^2}+2xy+\cancel{y^2}\iff$ $(x+y)+xy=1\iff$ $(1+x)(1+y)=2\iff$ $(1+\tan \alpha )(1+\tan\beta )=2\iff$ $(\cos\alpha +\sin\alpha )(\cos\beta+\sin\beta )=2\cos\alpha\cos\beta\iff$

$\cos \alpha\sin\beta +\sin\alpha\cos\beta=\cos\alpha\cos\beta -\sin\alpha\sin\beta\iff$ $\sin\left(\alpha +\beta\right)=\cos\left(\alpha +\beta\right)\iff$ $\sin\frac {\pi}4=\cos\frac{\pi}4\ ,$ what is true. In conclusion $,\ \boxed{MB^2+NC^2=MN^2}\ .$

Proof 2.

Application (Nelson C. D. Velasquez, Peru). Let the circle $w=\mathbb C(O,R)$ with the diameter $[AB]$ and for a point $S\in w$ construct $:$ the bisectors $[OP\ ,\ [OQ$ of the angles $\widehat{AOS}\ ,\ \widehat {BOS}$

respectively, where $\{P,Q\}\subset w\ ;$ the circle $\alpha =\mathbb C(I,m)$ what is interior tangent to $w$ at $P$ and to the lines $OA\ ,\ OS$ at $M\ ,U$ respectively $;$ the circle $\beta =\mathbb C(J,n)$ what is interior

tangent to $w$ at $N$ and to the lines $OB\ ,\ OS$ at $N\ ,V$ respectively. Prove that $MA^2+NB^2=MN^2\ .$

Proof. Let the midpoint $C$ of the semicircle $\overarc {AB}$ so that $AB$ separates $S\ ,\ C\ .$ Prove easily or is well-known that $C\in PM\cap QN\ .$

Observe that $OP\perp OQ\implies m\left(\widehat{MCN}\right)=m\left(\widehat{PCQ}\right)=\frac {\pi}4\ .$ Apply to $\triangle ACB$ the previous problem P10.

P11 (Ceran Mathematic, Turkey). In the regular pentagon $ABCDE$ denote $:\ P\in AC\cap BE\ ;$ the areas $[PAB]=x\ ,$ $[PBC]=y\ ,$ $[PCD]=z\ .$ Prove that there is the relation $\boxed{\frac 1x=\frac 1y+\frac 1z}\ .$

Proof. Observe that $CE\parallel AB\implies ABCE$ is an isosceles trapezoid $\implies$ $[PAE]=[PBC]=y\implies \boxed{xz=y^2}\ (1)$ and $[ABC]=[CDE]=[PCD]=z\implies$

$\boxed{x+y=z}\ (2)\ .$ From the relations $(1)$ and $(2)$ obtain that $\left\{\begin{array}{ccc} z-x & = & y\\\\ xz & = & y^2\end{array}\right\|\implies$ $\frac {z-x}{xz}=\frac y{y^2}\implies$ $\frac 1x-\frac 1y=\frac 1y\implies \frac 1x=\frac 1y+\frac 1z\ .$

P12 (Tezcan SAHIN, Turkey). Let $\triangle\ ABC$ and $D\in (BC)$ so that $\left\{\begin{array}{ccccc} m\left(\widehat{DAB}\right) & = & 90^{\circ}-C & ; & S_b=[DAB]\\\ m\left(\widehat{DAC}\right) & = & 90^{\circ}-B & ; & S_c=[DAC]\end{array}\right\|\ .$ Prove that $\boxed{\frac {S_b}{S_c}=\frac {\sin 2C}{\sin 2B}}\ .$

Proof. Apply an well-known relation $:\ \frac {S_b}{S_c}=\frac {[DAB]}{[DAC]}=\frac {DB}{DC}=\frac {AB}{AC}\cdot\frac {\sin\widehat{DAB}}{\sin\widehat{DAC}}=$ $\frac {\sin C}{\sin B}\cdot\frac {\cos C}{\cos B}=$ $\frac {\sin 2C}{\sin 2B}\ .$ Remark that the circumcenter $O$ of $\triangle ABC$ belongs to $AD\ .$

Particular case. If $\left\{\begin{array}{ccc} B & = & 63^{\circ}\\\ C & = & 75^{\circ}\end{array}\right\|\ ,$ then $\frac {S_b}{S_c}=$ $\frac {\sin \left(2\cdot 75^{\circ}\right)}{\sin\left(2\cdot 63^{\circ}\right)}=$ $\frac {\sin 150^{\circ}}{\sin 126^{\circ}}=$ $\frac {\sin 30^{\circ}}{\sin 54^{\circ}}=\frac 1{2\cos 36^{\circ}}\ .$ Since $\cos 36^{\circ}=\frac {1+\sqrt 5}4\ ,$ obtain that $\frac {S_b}{S_c}=\frac 1{2\cdot\frac {1+\sqrt 5}4}=\frac{\sqrt 5-1}2\implies$ $\boxed{\frac {S_b}{S_c}=\frac{\sqrt 5-1}2}\ .$

P13 (Kunihiko Chikaya). Let $\triangle ABC$ with $a=5\ ,\ b=4$ and $c=3\ .$ Find the area $[IOG]\ ,$ where $G$ -centroid, $O$ -circumcenter $($ midpoint of $[BC])$ and $I$ -incenter (standard notations).

Proof 1. Denote $D\in AI\cap BC\ ,$ where $DO=CD-CO=\frac {ab}{b+c}-\frac a2=\frac {a(b-c)}{2(b+c)}\implies$ $\boxed{DO=\frac {a(b-c)}{2(b+c)}}\ (1)\ .$ From $\frac {IA}{b+c}=\frac {ID}a=\frac {AD}{a+b+c}$ obtain that $\boxed{\frac {AI}{AD}=\frac {b+c}{a+b+c}}\ (2)\ .$

Hence $:\ \left\{\begin{array}{ccccc} \frac {[IOG]}{[IOA]} & = & \frac {OG}{OA} & = & \frac 13\\\ \frac {[OAI]}{[OAD]} & = & \frac {AI}{AD} & \stackrel{(2)}{=} & \frac 7{12}\\\ \frac {[ADO]}{[ABC]} & = & \frac {DO}{BC} & \stackrel{(1)}{=} & \frac 1{14}\end{array}\right\|$ $\bigodot\implies$ $\frac {[IOG]}{\cancel{[IOA]}}\cdot\frac {\cancel{[OAI]}}{\cancel{[OAD]}}\cdot\frac {\cancel{[ADO]}}{[ABC]}=\frac 13\cdot\frac 7{12}\cdot \frac 1{14}=\frac 1{72}\implies$ $\frac {[IOG]}{[ABC]}=\frac 1{72}\implies$ $[IOG]=\frac {[ABC]}{72}=\frac 6{72}=\frac 1{12}\implies$ $\boxed{\ [IOG]=\frac 1{12}\ }\ .$

Proof 2. Observe that $[IOG]=\frac 13\cdot [AIO]$ and $m\left(\widehat{IAO}\right)=45^{\circ}-C\ .$ Hence $[AIO]=\frac 12\cdot AI\cdot AO\cdot\sin\left(45^{\circ}-C\right)=$

$\frac 12\cdot (s-a)\sqrt 2\cdot \frac {\cancel a}2\cdot \frac {\sqrt 2}2\cdot\left(\frac b{\cancel a}-\frac c{\cancel a}\right)=$ $\frac 18\cdot (b+c-a)(b-c)=\frac{2\cdot 1}8=\frac 14\implies [AIO]=\frac 14$ $\implies$ $[IOG]=\frac 13\cdot\frac 14\implies$ $[IOG]=\frac 1{12}\ .$

Proof 3. Denote the projection $P$ of $A$ on $BC$ and $R\in IO\cap AP\ .$ Prove easily that $AR=r$ and $OP=\frac {b^2-c^2}{2a}\ .$ Since $AI$ is the bisector of $\widehat{PAO}$ obtain that $\frac {IR}r=\frac {IO}R=\frac {RO}{R+r}$

and $\frac {[AIO]}{[ARO]}=\frac {OI}{OR}=\frac {R}{R+r}\implies$ $[AIO]=\frac R{R+r}\cdot [ARO]=\frac R{R+r}\cdot \frac {AR\cdot PO}2=$ $\frac R{R+r}\cdot\frac {r\left(b^2-c^2\right)}{4a}=$ $\frac {\frac a2}{\frac a2+(s-a)}\cdot \frac {(s-a)\left(b^2-c^2\right)}{4a}=$ $\frac {{\cancel a}(s-a)\left(b^2-c^2\right)}{4\cancel a(b+c)}=$

$\frac {(s-a)(b-c)}4=\frac {(b+c-a)(b-c)}8\implies$ $[AIO]=\frac 14\implies$ $[IOG]=\frac 13\cdot\frac 14\implies$ $[IOG]=\frac 1{12}\ .$

P13 (An easy extension). Let $\triangle ABC$ with the centroid $G\ ,$ the circumcircle $\Omega =\mathbb C(O,R)$ and the incircle $w=\mathbb C(I,r)\ .$ Prove that the area $\boxed{\ [IOG]=\frac {|(a-b)(b-c)(c-a)|}{24r}\ }\ .$

Proof.

P14 (Seventh class). https://scontent.fotp3-1.fna.fbcdn.net/v/t1.0-9/21616185_1538076272897682_2381212052937400875_n.jpg?oh=cec95381137b423dda44f11a6bdf2ffe&oe=5A5B06DA

Proof 1. The upper square $ABCD$ with $AB=x$ is placed to south-west. Consider a broken line $AXYC$ from $A$ to $C$ so that $XA\perp XY\perp YC\ ,$ where $AX=m ,$ $XY=n\ ,$ $YC=p\ .$

Construct the rectangle $AUCV$ where $U\in CY$ so that $UA\perp UC$ and $V\in AX$ so that $VA\perp VC\ .$ Observe that $UC=AX+YC=m+p\ ,$ $VC=n$ and $2x^2=AC^2=CU^2+CV^2=$
$(m+p)^2+n^2\implies$ $\boxed{\ x=\sqrt{\frac {(m+p)^2+n^2}2}\ }\ .$ In the particular case $m=14\ ,\ n=7\ ,\ p=9$ obtain $2x^2=23^2+7^2=$ $529+49=578=2\cdot 289=$ $2\cdot 17^2\implies$ $\boxed{\ x=17\ }\ .$ Very nice !

Remark. Construct only $U\in CY$ so that $UA\perp UC\ .$ Thus $UC=UY+YC=AX+YC=m+p$ and $UA=XY=n\ .$ Hence $2x^2=AC^2=UC^2+UA^2=$ $(m+p)^2+n^2$ a.s.o.

P15 (Marian Dinca). Prove that $(\forall )\ \triangle ABC$ there is the following implication $:\ \boxed{\ a=\max\{a,b,c\}\implies m_a\sqrt 3\le s\ }$ (standard notations). Nice inequality!

Proof 1. $a=\max\{a,b,c\}\implies\ \left\{\begin{array}{ccc} a\ge b & \iff & ab\ge b^2\\\\ a\ge c & \iff & ac\ge c^2\end{array}\right\|\stackrel{\bigoplus}{\implies}\boxed{\ \begin{array}{ccc} a(b+c) & \ge & b^2+c^2\\\\\ 2a & \ge & b+c\end{array}\ }\ (*)\ .$ So $m_a\sqrt 3\le s$ $\iff$ $3\cdot 4a^2\le 4s^2\iff$ $3\left[2\left(b^2+c^2\right)-a^2\right]\le (a+b+c)^2$ $\iff$

$6\left(b^2+c^2\right)-3a^2\le$ $a^2+2a(b+c)+2bc+b^2+c^2\iff$ $4a^2+2a(b+c)+2bc-5\left(b^2+c^2\right)\ge 0\iff$ $\left[2a-(b+c)\right]^2+6a(b+c)-(b+c)^2+2bc-5\left(b^2+c^2\right)\ge 0\iff$

$\left[2a-(b+c)\right]^2+6a(b+c)-\left(b^2+c^2\right)-\cancel{2bc}+\cancel{2bc}-5\left(b^2+c^2\right)\ge 0\ \stackrel{(*)}{\iff}\ \left[(2a-(b+c)\right]^2+6\left[a(b+c)-\left(b^2+c^2\right)\right]\ge 0\ ,$ what is true from the relation $(*)\ .$

Proof 2. $\underline{\underline{m_a\sqrt 3\le s}}\iff$ $3\cdot \left[2\left(b^2+c^2\right)-a^2\right]\le (a+b+c)^2\stackrel{:(b+c)^2}{\implies}\ 3\le 3\cdot \frac {2\left(b^2+c^2\right)}{(b+c)^2}\le$ $3\cdot \left(\frac a{b+c}\right)^2+\left(\frac a{b+c}+1\right)^2\ \stackrel{t(b+c)=a}{\implies}\ 3\le 3t^2+(t+1)^2\iff$

$3t^2+(t+1)^2-3\ge 0$ $\iff 4t^2+2t-2\ge 0\iff$ $2t^2+t-1\ge 0\iff$ $(t+1)(2t-1)\ge 0\iff t\ge \frac 12\iff \frac a{b+c}\ge \frac 12\ \stackrel{(*)}{\iff}\ \underline{\underline{2a\ge b+c}}\ ,$ what is true.

This post has been edited 404 times. Last edited by Virgil Nicula, Jan 13, 2019, 12:49 PM

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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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Bro this blog is ripped
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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

My (math)blog

446. Mathematics "instant" for the middle school.

by Virgil Nicula, Jun 22, 2016, 12:05 AM

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