414. Geometry 7.

by Virgil Nicula, Mar 21, 2015, 1:44 PM

PP1. Let a convex quadrilateral $ABCD$ with $AC\perp BD$ , $P\in AC\cap BD$ , the projections $(K,L,M,N)$ of $P$ on $(AB,BC,CD,DA)$ respectively

and $X\in KP\cap CD$ , $Y\in LP\cap DA$ , $Z\in MP\cap AB$ and $T\in NP\cap BC$ . Prove that there is a circle $w$ so that $\{K,L,M,N;X,Y,Z,T\}\subset w$ .

Proof 1. Let $\left\{\begin{array}{ccccccc} AB=a & ; & BC=b & ; & CD=c & ; & DA=d\\\\ PA=m & ; & PB=n & ; & PC=p & ; &PD=r\end{array}\right\|$ . Observe that $BKPL$ is cyclic $\implies$ $\widehat{BLK}\equiv\widehat{BPK}\equiv\widehat{KAC}$ $\implies$

$\widehat{BLK}\equiv\widehat{KAC}$ $\implies$ $ACLK$ is cyclic. Prove analogously that $BDML$ , $CANM$ and $DBKN$ are cyclically.

$\blacktriangleright$ Let $U\in KN\cap BD$ , $U'\in LM\cap BD$ . Apply the Menelaus' theorem to $:$

$\left\{\begin{array}{cccc} \overline{UKN}/\triangle ABD\ : & \frac {UB}{UD}\cdot\frac{ND}{NA}\cdot\frac {KA}{KB}=1 & \implies & \frac {UB}{UD}=\frac {NA}{ND}\cdot\frac {KB}{KA}=\left(\frac mr\right)^2\cdot\left(\frac nm\right)^2=\left(\frac nr\right)^2\\\\ \overline{U'LM}/\triangle BCD\ : & \frac {U'B}{U'D}\cdot\frac{MD}{MC}\cdot\frac {LC}{LB}=1 & \implies & \frac {U'B}{U'D}=\frac {MC}{MD}\cdot\frac {LB}{LC}=\left(\frac pr\right)^2\cdot\left(\frac np\right)^2=\left(\frac nr\right)^2\end{array}\right\|$ $\implies$ $U'\equiv U$ where $\frac {UB}{UD}=\left(\frac nr\right)^2\implies$ $\boxed{U\in KN\cap ML\cap BD}$ .

$\blacktriangleright$ Let $V\in KL\cap AC$ , $V'\in MN\cap AC$ . Apply the Menelaus' theorem to $:$

$\left\{\begin{array}{cccc} \overline{VKL}/\triangle ABC\ : & \frac {VA}{VC}\cdot\frac{LC}{LB}\cdot\frac {KB}{KA}=1 & \implies & \frac {VA}{VC}=\frac {LB}{LC}\cdot\frac {KA}{KB}=\left(\frac np\right)^2\cdot\left(\frac mn\right)^2=\left(\frac mp\right)^2\\\\ \overline{V'NM}/\triangle ACD\ : & \frac {V'A}{V'C}\cdot\frac{MC}{MD}\cdot\frac {ND}{NA}=1 & \implies & \frac {V'A}{V'C}=\frac {MD}{MC}\cdot\frac {NA}{ND}=\left(\frac rp\right)^2\cdot\left(\frac mr\right)^2=\left(\frac mp\right)^2\end{array}\right\|$ $\implies$ $V'\equiv V$ where $\frac {VA}{VC}=\left(\frac mp\right)^2\implies$ $\boxed{V\in KL\cap MN\cap AC}$ .

$\blacktriangleright$ Thus, $\left\{\begin{array}{ccc} BKND\ \mathrm{cyclic} & \implies & UK\cdot UN=UB\cdot UD\\\\ BTMD\ \mathrm{cyclic} & \implies & UL\cdot UM=UB\cdot UD\end{array}\right\|$ $\implies$ $UK\cdot UN=UL\cdot UM\implies$ $KLMN$ is cyclically. Denote $\left\{\begin{array}{c} m\left(\widehat{ABD}\right)=\alpha\\\\ m\left(\widehat{ACD}\right)=\beta\\\\ m\left(\widehat{KXD}\right)=\phi\end{array}\right\|$ where $\alpha +\beta =\phi$ and

$\left\{\begin{array}{cc} \sin\alpha =\frac ma\ ; & \cos\alpha =\frac na\\\\ \sin\beta =\frac rc\ ; & \cos\beta =\frac pc\end{array}\right\|$ $\implies$ $\sin\phi =\sin(\alpha +\beta )=$ $\sin\alpha \cos\beta +\sin\beta\cos\alpha =$ $\frac ma\cdot \frac pc+\frac rc\cdot\frac na\implies$ $\boxed{\sin\phi =\frac {mp+nr}{ac}}\ (*)$ . Apply the theorem of Sines in $\triangle PXD\ :$

$\frac {PX}{\sin\widehat{PDX}}=\frac {PD}{\sin\widehat{PXD}}\implies$ $\frac {PX}{\frac pc}=\frac {r}{\sin\phi}\implies$ $PX=\frac {pr}c\cdot \frac {ac}{mp+nr }\implies$ $PX=\frac {apr}{mp+nr }\implies$ $PK\cdot PX=\frac {mn}a\cdot \frac {apr}{mp+nr }\implies$ $\boxed{PK\cdot PX=\frac {mnpr}{mp+nr}}$ .

In conclusion, $PK\cdot PX=PL\cdot PY=PM\cdot PZ=PN\cdot PT=\frac {mnpr}{mp+nr}$ , i.e. there is a circle $w$ so that $\{K,L,M,N;X,Y,Z,T\}\subset w$ .

Remark. If the quadrilateral $ABCD$ is cyclically, then the points $X,Y,Z,T$ are the midpoints of the sides $[CD]\ ,\ [DA]\ ,\ [AB]\ ,\ [BC]$ respectively.

PP2. Let a scalene $\triangle{ABC}$ with incentre $I$ and circumcircle $w=\mathbb C(O)$ . Lines $AI, BI, CI$ intersect again $w$ at $D, E, F$ respectively.

The parallel lines from $I$ to $BC, AC, AB$ intersect $EF, DF, DE$ at $K, L, M$ respectively. Prove that $M\in KL$ (B.M.O. 2015 - Cyprus).

Proof. $\widehat{FEI}\equiv \widehat{ICB}\equiv \widehat{FIK}\implies KI$ is tangent to the circumcircle of $\triangle EFI$ at $I \Longrightarrow KI^2 =KE \cdot KF$ . Hence $K$ belongs

to the radical axis $d$ of $w$ and degenerate circle $I$ . Can prove similarly that $\{L,M\}\subset d$ , i.e. $M\in KL$ . Remark that $\overline{KLM} \perp OI$ .

PP3. Let the parallelogram $ABCD$ where denote $\left\{\begin{array}{ccc} B\in (AE) & ; & BE=BC\\\\ H\in (BD) & ; & CH\perp BD\\\\ F\in CH & ; &FE\perp AB\end{array}\right\|$ . Prove that the ray $[AF$ is the $A$ -bisector of $\triangle BAD$ (O.N.M. - Mexic).

Proof 1. Let $K\in AD\cap CE$ and $\left\{\begin{array}{c} AB=CD=a\\\\ AD=BC=b\end{array}\right\|$ . Thus, $\left\{\begin{array}{c} DK=DC=a\\\\ AK=AE=a+b\end{array}\right\|$ . Define the point $G$ so that $GK\perp AD$ and $GE\perp AB$ .

Therefore, $GK=GE$ , i.e. $E$ belongs to the bisector of $\widehat{BAD}$ . Therefore, $GD^2-GB^2=$ $\left(KG^2+KD^2\right)-\left(EG^2+EB^2\right)=$ $KD^2-EB^2=$

$CD^2-CB^2\iff$ $GD^2-GB^2=$ $CD^2-CB^2\iff$ $CG\perp BD\iff$ $G\in CH\iff G\equiv F\iff$ $[AF$ is the $A$ -bisector of $\triangle BAD$ .

Proof 2. Let $L\in AD$ so $FL\perp AD\ ,\ DL=x$ and $\left\{\begin{array}{ccc} AB=CD=a\\\\ and AD=BC=b\end{array}\right\|$ . So $CH\perp BD\iff HB^2-HD^2=b^2-a^2$ . Apply the Carnot's lemma to $\triangle ABD$ and

$\left\{\begin{array}{ccc} E\in AB & ; & FE\perp AB\\\\ H\in BD & ; & FH\perp BD\\\\ L\in DA & ; & FL\perp DA\end{array}\right\|\ :\ \left(EA^2-EB^2\right)+$ $\left(HB^2-HD^2\right)+\left(LD^2-LA^2\right)=0\iff$ $\left[(b+a)^2-b^2\right]+\left(b^2-a^2\right)+\left[x^2-(x+b)^2\right]=0\iff$

$2ba-2xb=0\iff \boxed{x=a}\ (*)$ . In conclusion, $AE=AL=a+b\implies$ the ray $[AF$ is the $A$ -bisector of $\triangle BAD$ .

PP4. A trapezoid $ABCD\ ,\ AD\parallel BC$ with $I\in AC\cap BD\ ;\ P\in (AB)\ ,\ R\in (CD)$ so that $I\in PR\ ;\ S\in AR\cap BD$ . Prove that $CS\parallel AB\iff \frac {PB}{PA}+\frac {BC}{AD}=1$ .

Proof. Denote $:\ K\in CS\cap AD\ ,\ L\in PR$ $\cap AD\ ;\ AD=a\ ,\ BC=b$ . Is well-known that the division $(A,K,D,L)$ is harmonically, i.e. $\frac {LD}{LA}=\frac {KD}{KA}$ .

Apply the Menelaus' theorem to the transversal $\overline{LIP}$ and $\triangle ABD\ :\ \frac {LD}{LA}\cdot\frac {PA}{PB}\cdot\frac {IB}{ID}=1\ \stackrel{(*)}{\iff}\ \frac {KD}{KA}$ $\cdot\frac {PA}{PB}\cdot\frac ba=1\iff$ $\frac {PB}{PA}=\frac ba\cdot\frac {KD}{KA}\ (1)$ .

Thus, $CS\parallel AB\iff$ $AK=b\iff$ $\frac {KD}{KA}=\frac {a-b}b\ \stackrel{(1)}{\iff}\ \frac {PB}{PA}=$ $\frac ba\cdot \frac {a-b}b\iff$ $\frac {PB}{PA}=\frac {a-b}a\iff$ $\frac {PB}{PA}+\frac ba=1\iff$ $\frac {PB}{PA}+\frac {BC}{AD}=1$ . .

PP5. Let $ABC$ be an $A$ -isosceles triangle with $AB=AC=b$ , $BC=2a$ and its incircle $w=\mathbb C(I,r)$ . Define the midpoint $D$ of $[BC]$ , its projection $T$ on $AB$ and the

circle $\alpha =\mathbb C(D,\rho )$ , where $DT=\rho$ . Consider the point $Q\in \alpha\cap w$ so that $AD$ separates $\{Q,C\}$ and $\{P,Q\}=AQ\cap w$ . Prove that $\boxed{AP=(b-a)\sqrt{\frac b{a+b}}}\ (*)$ .

Proof. Denote $F\in AB\cap w$ and $AD=h$ . Observe that $\boxed{b^2=h^2+a^2}\ (1)$ and $\left\{\begin{array}{cccc} ah=2\cdot [ABD]=b\rho & \implies & \boxed{\rho =\frac {ah}b} & (2)\\\\ \triangle AFI\sim \triangle ADB & \implies & \boxed{r=\frac {a(b-a)}h} & (3)\end{array}\right\|$ . Thus, $IQ=ID\implies$

$m\left(\widehat{IQD}\right)=m\left(\widehat{IDQ}\right)=\phi\implies$ $\rho =QD=2r\cos\phi\implies$ $\cos\phi =\frac {\rho}{2r}\ \stackrel{(2\wedge 3)}{=}\ \frac {ah}b\cdot$ $\frac h{2a(b-a)}\stackrel{(1)}{=}\ \frac {b^2-a^2}{2b(b-a)}$ $\implies$ $\boxed{\cos\phi =\frac {a+b}{2b}}\ (4)$ . Apply the generalized

Pythagoras's theorem to the side $AQ$ of $\triangle ADQ\ :\ AQ^2=QD^2+QA^2-2\cdot QD\cdot QA\cdot\cos\phi =$ $\rho^2+h^2-2\rho h\cos\phi =$ $\left(\frac {ah}b\right)^2+h^2-2\frac {ah}b\cdot h\cdot \frac {a+b}{2b}\implies$ $b^2\cdot AQ^2=h^2\left(a^2+b^2-a^2-ab\right)\implies$ $b\cdot AQ=h\sqrt{b(b-a)}\implies$ $\boxed{AQ=h\cdot \sqrt{\frac {b-a}b}}\ (5)$ . Since $AP\cdot AQ=AF^2=(b-a)^2$ obtain that

$AP=\frac {(b-a)^2}h\cdot\sqrt{\frac b{b-a}}=$ $(b-a)^2\cdot \sqrt{\frac b{\left(b^2-a^2\right)(b-a)}}\implies$ $AP=(b-a)\sqrt{\frac b{a+b}}$ .

PP6. Let $m,n$ be two lines (which can be parallels or concurrently) and a circle $w=C(I,r)$ which is tangent to the lines $m,n$ in the points $A\in m$ , $B\in n$ respectively.

Prove that for any points $M\in m$ , $N\in n$ the following sentencies are equivalently: $1^{\circ}.\ IM\perp AN\ ;\ \ \ \ 2^{\circ}.\ IN\perp BM\ ;\ \ \ \ 3^{\circ}.\ AM^2+BN^2=MN^2.$

A particular cases.

$a).\ m\parallel n$ . Let $AMNB$ be a trapezoid ( $AM\parallel BN$ ) and the middlepoint $I$ of the side $[AB]$ . Prove that $IM\perp AN\Longleftrightarrow IN\perp BM\Longleftrightarrow AM^2+BN^2=MN^2.$

$b).\ P\in m\cap n$ . Let $\triangle ABC$ with the incircle $w=C(I,r)$ which touches its sides in $D\in BC$ , $E\in CA$ , $F\in AB$ . Its exincircle $w_a=C(I_a,r_a)$ touches its sides in $D_1\in BC$ ,

$E_1\in CA$ and $F_1\in AB$ . Denote $L\in BC\cap EF$ and $L_1\in BC\cap E_1F_1$ . Prove that $:\ \left\{\begin{array}{ccc} IL\perp AD & \iff & LA^2=LD^2+AE^2\\\\ I_aL_1\perp AD_1 & \iff & L_1A^2=L_1D_1^2+AE_1^2\end{array}\right\|$ . A remarkable property !

Proof.

$\bullet\ \ (1^{\circ})\Longleftrightarrow IA^2-IN^2=$ $MA^2-MN^2\Longleftrightarrow$ $r^2-(r^2+BN^2)=MA^2-MN^2\Longleftrightarrow$ $AM^2+BN^2=MN^2\Longleftrightarrow (3^{\circ})$ . Thus, $(1^{\circ})\Longleftrightarrow (3^{\circ}).$

$\bullet\ \ (2^{\circ})\Longleftrightarrow IB^2-IM^2=$ $NB^2-NM^2\Longleftrightarrow$ $r^2-(r^2+AM^2)=NB^3-NM^2\Longleftrightarrow$ $AM^2+BN^2=MN^2\Longleftrightarrow (3^{\circ})$ . Thus, $(2^{\circ})\Longleftrightarrow (3^{\circ}).$

Therefore, $(1^{\circ})\Longleftrightarrow (2^{\circ})\Longleftrightarrow (3^{\circ}).$

Lemma. In a right trapezoid $ABCD$ with $AB\parallel CD\ ,\ AB\perp AD$ and $BA=BC$ there is the relation $AC^2=2\cdot BA\cdot CD$ .

Proof 1. Let $M$ be the midpoint of $[AC]$ . Therefore, $\triangle ABM\sim \triangle CAD\implies$ $\frac {AB}{CA}=\frac {AM}{CD}\implies$ $\frac {AB}{CA}=\frac {AC}{2\cdot CD}\implies$ $\boxed{AC^2=2\cdot AB\cdot CD}$ .

Proof 2. Apply the Euler's in $ABCD\ :\ BD^2+AC^2=AD^2+BC^2+2\cdot CD\cdot AB\iff$

$AD^2+AB^2+AC^2=$ $AD^2+BC^2+2\cdot AB\cdot CD$ $\iff$ $\boxed{AC^2=2\cdot AB\cdot CD}$ .

Proof 3. Let $E\in AB$ so that $CE\perp AB$ . Thus, $CB^2=CE^2+EB^2\iff$ $AB^2=AD^2+(AB-DC)^2\iff$ $AD^2+DC^2=2\cdot AB\cdot CD\iff$ $\boxed{AC^2=2\cdot AB\cdot CD}$ .

Proof 4. Let $E\in AB$ so that $CE\perp CA$ . Construct the rectangle $DAEF$ and $G\in (CF)$ so that $GF=CD$ . Thus, $DA^2=DC\cdot DG\iff$ $AC^2-DC^2=DC\cdot DG\iff$

$AC^2=DC(DC+DG)\iff$ $AC^2=DC(DC+CF)\iff$ $\AC^2=CD\cdot DF\iff$ $AC^2=CD\cdot AE\iff$ $\boxed{AC^2=2\cdot AB\cdot CD}$ .

PP7. Let the incircle $w=C(I,r)$ of $\triangle ABC$ and $\left|\begin{array}{c} D\in w\cap BC\\\\ E\in w\cap CA\\\\ F\in w\cap AB\end{array}\right|$ . For $P\in \overarc[]{EF}$ denote

$\left\{\begin{array}{ccc} X\in BC\ ,\ PX\perp BC & ; & PX=x\\\\ Y\in CA\ ,\ PY\perp CA & ; & PY=y\\\\ Z\in AB\ ,\ PZ\perp AB & ; & PZ=z\end{array}\right\|$ . Prove that $\boxed{\sqrt x\cdot\cos\frac A2=\sqrt y\cdot\cos\frac B2+\sqrt z\cdot\cos\frac C2}$ .

Proof (Miguel O. Sanchez). Apply the upper lemma to the trapezoids $PIDX$ , $PIEY$ and $PIFZ$ $\implies \boxed{\frac {PD}{\sqrt x}=\frac {PE}{\sqrt y}=\frac {PF}{\sqrt z}=\sqrt{2r}}\ (1)$ .

Observe that $\boxed{\frac {EF}{\cos\frac A2}=\frac {FD}{\cos \frac B2}=\frac {DE}{\cos\frac C2}=2r}\ (2)$ . Apply the Ptolemy's theorem to the cyclical quadrilateral $DEPF$ and

the proportions $(1)$ and $(2)\ :\ \underline{PD}\cdot EF=$ $\underline{PE}\cdot DF+\underline{PF}$ $\cdot DE\ \stackrel{(1\wedge 2)}{\iff}\ \sqrt x$ $\cdot\cos\frac A2=\sqrt y\cdot\cos\frac B2+\sqrt z\cdot\cos\frac C2$ .

PP8. Let the circles $\left\{\begin{array}{ccc} w_1=C\left(O_1,R\right) & ; & O_1(0,R)\\\\ w_2=C\left(O_2,R\right) & ; & O_2\left(R,0\right)\\\\ w_3=C\left(O_3,\rho\right) & ; & O_3\left(2R-\rho ,0\right)\end{array}\right\|$ , where $\rho <2R$ and $w_3$ is exterior tangent to $w_1$ .

Find the length $r$ of the radius for the circle $w=C(I,r)$ which is exterior tangent to $\left\{w_1,w_3\right\}$ and is interior tangent to $w_2$ .

Proof. Suppose w.l.o.g. $R=3$ , i.e. $\left\{\begin{array}{ccc} w_1=C\left(O_1,3\right) & ; & O_1(0,3)\\\\ w_2=C\left(O_2,3\right) & ; & O_2\left(3,0\right)\\\\ w_3=C\left(O_3,\rho\right) & ; & O_3\left(6-\rho ,0\right)\end{array}\right\|\ ,\ \rho <6$ . So $w_3$ is exterior tangent to $w_1\iff$ $O_1O_3=3+\rho$ $\iff$

$(3+\rho )^2=3^2+(6-\rho )^2\iff$ $\boxed{\rho =2}$ . Let $I(x,y)$ . $\left\{\begin{array}{cccccccc} IO_1=3+r & \implies & x^2+(y-3)^2=(3+r)^2 & \implies & x^2+y^2-6y & = & r^2+6r & (1)\\\\ IO_2=3-r & \implies & (x-3)^2+y^2=(3-r)^2 & \implies & x^2+y^2-6x & = & r^2-6r & (2)\\\\ IO_3=2+r & \implies & (x-4)^2+y^2=(2+r)^2 & \implies & x^2+y^2-8x & = & r^2+4r-12 & (3)\end{array}\right\|$ $\implies$

$\odot\begin{array}{cccccc} \nearrow & (1)-(2)\ : & 6(x-y)=12r & \implies & x-y=2r & \searrow\\\\ \searrow & (2)-(3)\ : & 2x=-10r+12 & \implies & \boxed{x=6-5r} & \nearrow\end{array}\odot$ $\implies$ $\boxed{y=6-7r}$ . Using the relation $(1)$ obtain that

$(6-5r)^2+(6-7r)^2-6(6-7r)=r^2+6r\iff$ $(6-5r)^2+(6-7r)(-7r)=r^2+6r\iff$ $25r^2-60r+36-42r+49r^2-r^2-6r=0\iff$

$73r^2-108r+36=0$ with $\Delta^{\prime}=$ $54^2-73\cdot 36=36(81-73)=$ $36\cdot 8=12^2\cdot 2$ . Thus, $r=\frac {54-12\sqrt 2}{73}$ , i.e. $\boxed{r=\frac {6\left(9-2\sqrt 2\right)}{73}}$ .

PP9. Let a semicircle $w=\mathbb S(O,r)$ with the diameter $[AB]$ and $\{F,E\}\subset w$ for which $F\in\overarc[]{AE}$ . Let

$\left\{\begin{array}{cc} P\in FF\cap AB\ ; & Q\in EE\cap AB\\\\ X\in PE\cap QF\ ; & Y\in AE\cap BF\end{array}\right\|$ so that $A\in (FB)$ and $B\in (AQ)$ . Prove that $XY\perp AB$ .

Proof. I denoted by $TT$ - the tangent to $w$ in the point $T\in w$ . Let $\left\{\begin{array}{ccc} C\in AF\cap BE & ; & Z\in PF\cap QE\\\\ \left|\begin{array}{c} D\in (AB)\\\\ CD\perp AB\end{array}\right| & ; & OA=OB=r\end{array}\right\|$ . Observe that $Y$ is the orthocenter of $\triangle ABC$ .

Denote $\left\{\begin{array}{c} U\in PF\cap AY\\\\ V\in QE\cap AY\end{array}\right\|$ . Thus, $\left\{\begin{array}{ccccc} \widehat{UCF}\equiv\widehat{FBA}\equiv\widehat{AFP}\equiv \widehat{UFC} & \implies & \widehat{UCF}\equiv \widehat{UFC} & \stackrel{(FC\perp FY)}{\implies} & UF=UC=UY\\\\ \widehat{VCE}\equiv\widehat{EAB}\equiv\widehat{BEQ}\equiv \widehat{VEC} & \implies & \widehat{VCE}\equiv \widehat{VEC} & \stackrel{(FC\perp FY)}{\implies} & VE=VC=VY\end{array}\right\|$ $\implies$

$U\equiv V\equiv Z$ - the midpoint of $CY$ . Therefore, $\left\{\begin{array}{cccc} \triangle ZDP\sim \triangle OFP & \iff & \frac {PD}{PF}=\frac {ZD}{OF} & (1)\\\\ \triangle ZDQ\sim \triangle OEQ & \iff & \frac {QD}{QE}=\frac {ZD}{OE} & (2)\end{array}\right\|$ . Apply the Ceva's theorem to $\triangle PZQ$ and the points

$\{D,E,F\}\ :\ \frac {DP}{DQ}$ $\cdot\frac {EQ}{EZ}\cdot\frac {FZ}{FP}\ \stackrel{(EZ=FZ)}{=}\ \frac {DP}{FP}\cdot$ $\frac {EQ}{DQ}\ \stackrel{(1\wedge 2)}{=}\ \frac {ZD}{OF}\cdot \frac {OE}{ZD}\ \stackrel{(OE=OF)}{=}\ 1\ \implies$ $X\in PE\cap QF\cap ZD$ , i.e. $XY\perp AB$ . An equivalent enunciation.

Let $\triangle ABC$ with the orthocenter $H$ . Denote the midpoint $M$ of $[AH]$ and $\left\{\begin{array}{ccc} E\in BH\cap CA & ; & F\in CH\cap AB\\\\ P\in MF\cap BC & ; & Q\in ME\cap BC\end{array}\right\|$ . Prove that $PE\cap QF\cap AH\ne\emptyset$ .

PP10. Let $\triangle ABC$ with circumcircle $w=\mathbb C(O,R)$ . Let the projections $E$ , $F$ of $B$ , $C$ on $AC$ , $AB$ . Prove that $\boxed{O\in EF\iff b^2+c^2=4R^2+a^2\iff [ABC]=R^2\tan A}$ .

Proof 1. Let $S\in EF\ ,\ AS\perp EF$ .Prove easily that $AO\perp EF$ . Thus, $m\left(\widehat{AFE}\right)=C\implies$ $AS=AF\sin C=b\cos A\sin C\implies$ $\boxed{AS\cdot 2R=bc\cos A}\ (*)$ .

In conclusion, $O\in EF\iff AS=R\ \stackrel{(*)}{\iff}\ 2R^2=$ $bc\cos A\iff (2R)^2=2bc\cos A\iff$ $4R^2=b^2+c^2-a^2\iff b^2+c^2=4R^2+a^2$ .

Proof 2. Denote $\{A,L\}=\{A,O\}\cap w$ . Since $AO\perp EF$ and $CA\perp CL$ then $OECL$ is a cyclical quadrilateral, i.e. $AC.AE=AL.AO\iff$ $bc\cos A=2R^2\iff$

$2bc\cos A=(2R)^2\iff$ $4R^2+a^2 = b^2 + c^2$ . Remark. In any $\triangle ABC$ there is the relation $AS=\frac {b^2+c^2-a^2}{4R}$ , where $S$ is the projection of $A$ on $EF$ .

Proof 3. Let $D\in AH\cap BC$ and $BD = h_a$ . Observe that $FAO\sim CAD$ , i.e. $\cos A=\frac{AF}{AC}=\frac {AO}{AD}\implies$ $R=h_a\cos A\implies$

$4R^2=2\cdot 2Rh_a\cos A=2bc\cos A= a^2+c^2-b^2\iff$ $4R^2+a^2 = b^2+c^2$ . I used the following identity $:\ 2Rh_a=bc$ .

Proof 4. The area $2\cdot [FAE] =EF\cdot AO = AH\sin A\cdot R =$ $2R^2\cos A\sin A$ . On other hand, $2[FAE]=AF\cdot AE\cdot \sin A=b\cos A\cdot c\cos A\cdot \sin A$ .

In conclusion, $2R^2\cos A\sin A=b\cos A \cdot c\cos A\cdot \sin A\iff$ $2R^2=bc\cos A\iff$ $4R^2=2bc\cos A=b^2+c^2-a^2\iff$ $4R^2+a^2=b^2+c^2$ .

PP11. Let $\triangle ABC$ and $E\in (AC)$ , $F\in (AB)$ , $D\in EF\cap BC$ so that $B\in (DC)$ and $[AEF]=[BDF]=[BCEF]$ . Find the ratio $\frac {FD}{FE}$ .

Proof (Ruben Dario). Suppose w.l.o.g. $[AEF]=[BDF]=[BCEF]=1$ and let $[ADF]=x$ . Thus, $[ADF][BEF]=[AFE][BDF]\iff$

$[BEF]=\frac 1x$ and $[BCE]=[BCEF]-[BEF]=1-\frac 1x$ and $[BDE]=[BAE]=1+\frac 1x\iff$ $\boxed{AD\parallel BE}\ (*)$ . Therefore, $\frac x1=\frac {[AFD]}{[AFE]}=$

$\frac {FD}{FE}=\frac {AD}{BE}\ \stackrel{(*)}{=}\ \frac {CA}{CE}=$ $\frac {[CAB]}{[CEB]}=\frac 2{1-\frac 1x}\implies$ $x=\frac 2{1-\frac 1x}\implies$ $x\left(1-\frac 1x\right)=2\implies$ $x-1=2\implies x=3$ , i.e. $\frac {FD}{FE}=3$ .

An easy extension. Let $\triangle ABC$ and $E\in (AC)$ , $F\in (AB)$ , $D\in EF\cap BC$ so that $B\in (DC)$ and $[AEF]=a$ , $[BDF]=b$ and $[BCEF]=c$ . Find the ratio $\frac {FD}{FE}$ .

Proof 1. Let $[ADF]=x$ , i.e. $\frac {FD}{FE}=\frac {[AFD]}{[AFE]}=\frac xa$ . Thus, $[ADF][BEF]=[AFE][BDF]\iff$ $[BEF]=\frac {ab}x$ , $[BCE]=[BCEF]-[BEF]=c-\frac {ab}x$ , $[ADB]=x+b$

and $[BAE]=a+\frac {ab}x$ .Apply Menelaus to $\overline{DEF}/\triangle ABC\ :\ \frac {DB}{DC}\cdot \frac {EC}{EA}\cdot\frac {FA}{FB}=1\iff$ $\frac {[ADB]}{[ADC]}\cdot \frac {[BEC]}{[BEA]}\cdot\frac {[DFA]}{[DFB]}=1\iff$ $\frac {x+b}{x+a+b+c}\cdot \frac {cx-ab}{a(x+b)}\cdot\frac {x}{b}=1\iff$

$x(cx-ab)=ab(x+a+b+c)\iff$ $\boxed{cx^2-2abx-ab(a+b+c)=0}$ with $\Delta^{\prime}=ab(a+c)(b+c)$ . In conclusion, $\boxed{\frac {FD}{FE}=\frac xa=\frac {ab+\sqrt{ab(a+c)(b+c)}}{ac}}\ (3)$ .

Proof 2. $[BEF]=u\ ,\ [BEC]=v$ $\implies$ $\left\{\begin{array}{cccc} u+v & = & c & (1)\\\\ xu & = & ab & (2)\end{array}\right\|$ and $\frac {b+u}{c-u}\ \stackrel{(1)}{=}\ \frac {b+u}v=\frac {[EBD]}{[EBC]}=\frac {BD}{BC}=$ $\frac {[ABD]}{[ABC]}=\frac {x+b}{a+c}\implies$ $\frac {b+u}{c-u}=\frac {x+b}{a+c}\implies$

$(b+u)(a+c)=(x+b)(c-u)\ \stackrel{(2)}{\implies}\ (xb+ab)(a+c)=(x+b)(cx-ab)\implies$ $cx^2-2abx-ab(a+b+c)=0$ with $\Delta^{\prime}=ab(a+c)(b+c)\implies (3)$ .

Particular case. If $a=b$ , then the ratio $\frac {FD}{FE}$ becomes $\frac {a^2+\sqrt{a^2(a+c)^2}}{ac}=\frac {a^2+a(a+c)}{ac}\implies$ $\frac {FD}{FE}=1+\frac {2a}c$ .

PP12. Let $w$ be circumcircle of $\triangle ABC$ and $w_x=\mathbb C(X,x)$ , $w_y=\mathbb C(Y,y)$ , $w_z=\mathbb C(Z,z)$ what are interior tangent

to $w$ and tangent to $BC$ , $CA$ , $AB$ respectively so that their centers are exterior for $\triangle ABC$ . Prove that $16xyz=Rr^2$ .

Proof. Is well known that the arrowlength of the arc $\overarc {BC}$ so that $A\not\in\overarc {BC}$ is equal to $\frac {r_a-r}2$ , where $r_a$ is the $A$ -exradius a.s.o. Therefore, $\left\{\begin{array}{ccc} 4x & = & r_a-r\\\ 4y & = & r_b-r\\\ 4z & = & r_c-r\end{array}\right\|$ . Since

$r_a(s-a)=rs\implies \boxed{r_a-r=\frac {ar_a}s}\ (*)$ a.s.o. Obtain that $64xyz=\prod\left(r_a-r\right)=\prod\frac {ar_a}s=$ $\frac {abc\cdot r_ar_br_c}{s^3}=\frac {4Rrs\cdot s^2r}{s^3}=4Rr^2\implies$ $16xyz=Rr^2$ .

PP13. Let $C$ -isosceles $\triangle ACB$ with the incenter $I$ and the circumcenter $O\ .$ Denote the lengths $l_b$ and $l_c$ of the the $B$ -bisector

and the $C$ -bisector respectively. Prove that $l_b=2l_c\iff MI=MO\ ,$ where $M$ is the midpoint of $[AB]\ .$

Method 1 (synthetic). Denote $A=2x$ where $x<45^{\circ}$ and $X$ on the $A$ -bisector $[AD]\ ,$ where $D\in (BC)\ .$ Thus, $XM\parallel BC$ and $AD=2\cdot CM\iff XD=CM\ .$

In conclusion, $CDMX$ is an isosceles trapezoid, where $m\left(\widehat {BCM}\right)=m\left(\widehat {ADC}\right)=3x\ .$ Therefore, $B+m\left(\widehat {BCM}\right)=2x+3x=90^{\circ}\iff$ $x=18^{\circ}\ ,$

what means that $\boxed{A=B=36^{\circ}\ ,\ C=108^{\circ}}\ .$ In conclusion, $m\left(\widehat{AOB}\right)=360^{\circ}-2C= 144^{\circ}\ ,$ what means that $AIBO$ is a rhombus, i.e. $MI=MO\ .$

Method 2 (trigonometric). Let $A=2x\ ,\ x<45^{\circ}$ and the $A$ -bisector $[AD]\ ,\ D\in (BC)\ .$ Thus, $2=\frac {AD}{CM}=$ $\frac {AD}{AC}\cdot \frac {AC}{CM}=$ $\frac {\sin C}{\sin \widehat{ADC}}\cdot \frac {1}{\sin A}=$ $\frac {\sin 4x}{\sin 3x}\cdot\frac {1}{\sin 2x}=$

$\frac {2\sin 2x\cos 2x}{\sin 3x\sin 2x}\implies$ $\sin 3x=\cos 2x\iff$ $x=18^{\circ}\iff$ $\boxed{A=B=36^{\circ}\ ,\ C=108^{\circ}}\ .$ So $m\left(\widehat{AOB}\right)=360^{\circ}-2C= 144^{\circ}\ ,$ i.e. $AIBO$ is a rhombus, i.e. $MI=MO\ .$

Method 3 (metric). Let $A$ -bisector $AD\ ,\ D\in (BC)$ and $CA=CB=a\ ,$ $AB=c\ .$ Thus, $AD=2\cdot CM$ $\iff$ $AD^2=4\cdot CM^2\iff$ $\frac {ac^2(2a+c)}{(a+c)^2}=$ $4a^2-c^2\iff$

$ac^2=(2a-c)\left(a^2+2ac+c^2\right)\iff$ $2a^3+3a^2c-ac^2-c^3=0$ $\iff$ $(2a+c)\left(a^2+ac-c^2\right)=0\iff$ $\frac ac=\frac {-1+\sqrt 5}{2}$ $\iff$ $\cos A=$ $\frac {AM}{AC}=$ $\frac {c}{2a}=\frac {1}{-1+\sqrt 5}=$

$\frac {1+\sqrt 5}{4}=$ $\cos 36^{\circ}\implies x=18^{\circ}\iff$ $\boxed{A=B=36^{\circ}\ ,\ C=108^{\circ}}\ .$ Thus, $m\left(\widehat{AOB}\right)=360^{\circ}-2C= 144^{\circ}\ ,$ i.e. $AIBO$ ia a rhombus, i.e. $MI=MO\ .$

PP14. Prove that in a triangle $ABC$ exists the chain of equivalencies $l_b=2h_a\iff \cos\frac {A-C}{2}=\frac a{2b}\iff$ $2\sin\frac B2=\frac a{a+c}\iff$

$a^3c+\left(a^2-c^2\right)^2=b^2(a+c)^2\ ,$ where $h_a$ is the length of the $A$ -altitude and $l_b$ is the length of the $B$ -bisector.

Particular case. If $b=c\ ,$ then there is the chain of equivalencies $l_b=2h_a\iff \frac ac=\frac {1+\sqrt 5}{2}\iff$ $\boxed{B=C=36^{\circ}\ ,\ C=108^{\circ}}\ .$

Proof. Let $P\in BC$ so that $AP\perp BC$ and $L\in BC$ so that $B\in (CL)$ and $BL=BA\ .$ Since $BD\parallel AL$ obtain that $\frac {BD}{AL}=\frac {CB}{CL}\ .$ Thus, $\boxed{l_b=2h_a}\iff$ $\frac {2h_a}{AL}=\frac {a}{a+c}$ $\iff$

$\sin\frac B2=\frac {AP}{AL}\iff$ $\boxed{2\sin\frac B2=\frac a{a+c}}\ (*)\ .$ Thus, $\sin\widehat{LAC}=\sin\left(\frac B2+C\right)\iff$ $\sin\widehat{LAC}=\cos\frac {A-C}{2}\ .$ Theorem of Sines in $\triangle LAC\ :\ \frac {CL}{\sin \widehat{LAC}}=\frac {AC}{\sin\frac B2}\iff$

$\frac {a+c}{\cos\frac {A-C}{2}}=\frac b{\sin \frac B2}\stackrel{(*)}{\iff}$ $\boxed{\cos\frac {A-C}{2}=\frac a{2b}}\ .$ Using $\sin^2\frac B2=\frac {(s-a)(s-c)}{ac}$ obtain that $2\sin\frac B2=\frac a{a+c}\iff$ $\frac {4(s-a)(s-c)}{ac}=\frac {a^2}{(a+c)^2}\iff$

$\left[b^2-(a-c)^2\right](a+c)^2=a^3c\iff$ $\boxed{a^3c+\left(a^2-c^2\right)^2=b^2(a+c)^2}\ .$

PP15. In $\triangle ABC$ denote the midpoint $M$ of $[BC]$ and the projection $P$ of $A$ on $BC\ .$ Prove that $l_b=2m_a$ $\iff$

$m\left(\widehat{MAP}\right)=\frac {|A-3C|}{2}\ ,$ where $m_a$ is the length of the $A$ -median and $l_b$ is the length of the $B$ -bisector.

Proof. Denote the midpoint $X$ of the $B$ -bisector $[BD]\ ,$ where $D\in (AC)\ .$ Thus, $XM\parallel AC$ and $BD=2\cdot AM\iff XD=AM\ .$ Therefore, $ADMX$ is an isosceles

trapezoid, i.e. $m\left(\widehat {CAM}\right)=m\left(\widehat {ADB}\right)=$ $\frac B2+C\ .$ In conclusion, $\left(\widehat {MAP}\right)=\left(\widehat {MAC}\right)-\left(\widehat {CAM}\right)=$ $\left|\left(90^{\circ}-C\right)-\left(\frac B2+C\right)\right|\implies$ $m\left(\widehat{MAP}\right)=\frac {|A-3C|}{2}\ .$

PP16. Let $\triangle ABC$ with the circumcircle $w\ .$ For an interior point $P$ denote $\{B,U\}=\{B,P\}\cap w\ ,$ $\{C,V\}=\{C,P\}\cap w\ .$

Consider a point $R\in w$ so that the sideline $BC$ separates $A\ ,$ $R$ and denote $X\in AB\cap RV\ ,$ $Y\in AC\cap RU\ .$ Prove that $P\in XY\ .$

Proof 1. Apply the Pascal's theorem to the cyclical $ABURVC\ :\ \left\{\begin{array}{cc} \nearrow & X\in AB\cap RV\\\\ \rightarrow & P\in BU\cap VC\\\\ \searrow & Y\in UR\cap CA\end{array}\right|$ $\implies P\in XY\ .$

Proof 2. Let $\left\{\begin{array}{c} L\in AP\cap BC\\\\ M\in BP\cap CA\\\\ N\in CP\cap AB\end{array}\right|$ and apply the Menelaus' theorem to the transversals $:\ \begin{array}{ccc} \overline{BPM}/\triangle ALC\ \implies & \frac {BL}{BC}\cdot\frac {MC}{MA}\cdot \frac {PA}{PL}=1\ \implies & \frac {MC}{MA}=\frac {BC}{BL}\cdot\frac {PL}{PA}\ (1)\\\\ \overline{CPN}/\triangle ALB\ \implies & \frac {CL}{CB}\cdot\frac {NB}{NA}\cdot \frac {PA}{PL}=1\ \implies & \frac {NB}{NA}=\frac {CB}{CL}\cdot\frac {PL}{PA}\ (2)\end{array}\ .$ Apply the

well-known property in an cyclic quadrilateral $XYZT\ :\ I\in XZ\cap YT\ \implies\ \frac {IX}{XT\cdot XY} =$ $\frac {IY}{YX\cdot YZ}=\frac {IZ}{ZY\cdot ZT}=\frac {IT}{TZ\cdot TX}\ .$ $\left\{\begin{array}{cc} AVBR\ : & \frac {XB}{XA}=\frac {BR\cdot BV}{AR\cdot AV}\\\\ AVBC\ : & \frac {NA}{NB}=\frac {AC\cdot AV}{BC\cdot BV}\end{array}\right|\bigodot\implies$

$\frac {XB}{XA}=\frac {BR}{AR}\cdot\frac {NB}{NA}\cdot\frac {AC}{BC}\stackrel{(2)}{\implies}\frac {XB}{XA}=\frac {BR}{AR}\cdot \frac {CA}{CL}\cdot \frac {PL}{PA}\ (3)\ .$ $\left\{\begin{array}{cc} AUCR\ : & \frac {YC}{XA}=\frac {CR\cdot CU}{AR\cdot AU}\\\\ AUCB\ : & \frac {MA}{MC}=\frac {AB\cdot AU}{CB\cdot CU}\end{array}\right|$ $\bigodot\implies$ $\frac {YC}{YA}=\frac {CR}{AR}\cdot\frac {MC}{MA}\cdot\frac {AB}{CB}\stackrel{(1)}{\implies}\frac {YC}{YA}=\frac {CR}{AR}\cdot\frac {BA}{BL}\cdot\frac {PL}{PA}\ (4)\ .$

In conclusion, $P\in XY\iff$ $\frac {XB}{XA}\cdot LC+\frac {YC}{YA}\cdot LB=$ $\frac {PL}{PA}\cdot BC\stackrel{(3\wedge 4)}{\iff}$ $\frac {BR}{AR}$ $\cdot \frac {CA}{CL}\cdot \frac {PL}{PA}\cdot LC+\frac {CR}{AR}\cdot\frac {BA}{BL}\cdot\frac {PL}{PA}\cdot LB=\frac {PL}{PA}\cdot BC\iff$

$\frac {BR}{AR}\cdot CA+\frac {CR}{AR}\cdot BA=BC\iff$ $BR\cdot CA+CR\cdot BA=AR\cdot BC\ ,$ what is truly - the Ptolemy's theorem in $ABRC\ .$

PP17. In $\triangle ABC$ se considera bisectoarele $BE$ si $CF\ ,$ unde $E\in (AC)$ si $F\in (AB)\ .$ Notam mijloacele

$M\ ,$ $N$ ale segmentelor $[BE]\ ,$ $[CF]$ respectiv. Sa se arate ca $EN\cap FM\cap BC\ne\emptyset\Longleftrightarrow A=60^\circ\ .$

Proof. Let the incenter $I\in BE\cap CF$ and $U\in FM\cap BC\ ,$ $V\in EV\cap BC\ .$ Observe that

$\blacktriangleright\ M\in (BI)\iff$ $BM<BI \iff$ $\frac 12\cdot BE<\frac {a+c}{2s}\cdot BE \iff$ $s<a+c \iff b<a+c\ .$

$\blacktriangleright\ N\in (CI) \iff$ $CN<CI \iff \frac 12\cdot CF<\frac {a+b}{2s}\cdot CF \iff s<a+b \iff c<a+b\ .$

$\frac {IB}{a+c}=\frac {IE}{b}=\frac {BE}{2s}\implies$ $\frac {MB}{MI}=\frac {MB}{IB-MB}=$ $\frac {\frac 12\cdot BE}{\frac {a+c}{2s}\cdot BE-\frac 12\cdot BE}=$ $\frac {\frac 12}{\frac {a+c}{2s}-\frac 12}=$ $\frac {s}{a+c-s}=\frac {s}{s-b}\implies$ $\boxed{\frac {MB}{MI}=\frac {s}{s-b}}\ (1)\ .$

$\frac {IC}{a+b}=\frac {IF}{b}=\frac {CF}{2s}\implies$ $\frac {NC}{NI}=\frac {NC}{IC-NC}=$ $\frac {\frac 12\cdot CF}{\frac {a+b}{2s}\cdot CF-\frac 12\cdot CF}=$ $\frac {\frac 12}{\frac {a+b}{2s}-\frac 12}=$ $\frac {s}{a+b-s}=\frac {s}{s-c}\implies$ $\boxed{\frac {NC}{NI}=\frac {s}{s-c}}\ (2)\ .$

Apply the Menelaus' theorem to the transversals $:$

$\blacktriangleright\ \overline{FMU}/\triangle BIC\ :\ \frac {FI}{FC}\cdot$ $\frac {UC}{UB}\cdot\frac {MB}{MI}=1 \stackrel{(1)}{\implies\ }$ $\frac {c}{2s}\cdot\frac {UC}{UB}\cdot\frac {s}{s-b}=1\implies$ $\frac {UB}{UC}=\frac {c}{2(s-b)}\ .$

$\blacktriangleright\ \overline{ENV}/\triangle BIC\ :\ \frac {EI}{EB}$ $\cdot\frac {VB}{VC}\cdot\frac {NC}{NI}=1 \stackrel{(2)}{\implies\ }$ $\frac {b}{2s}\cdot\frac {VB}{VC}\cdot\frac {s}{s-c}=1 \implies$ $\frac {VB}{VC}=\frac{2(s-c)}{b}\ .$

In conclusion, $EN\cap FM\cap BC\ne\emptyset\iff$ $U\equiv V\iff$ $\frac {UB}{UC}=\frac {VB}{VC}\iff$ $\frac {c}{2(s-b)}=\frac{2(s-c)}{b}\iff$ $bc=4(s-b)(s-c)\iff bc=a^2-(b-c)^2\iff$ $a^2=b^2-bc+c^2\iff A=60^{\circ}\ .$

Generalization. Let $\triangle ABC$ and $E\in (AC)\ ,\ F\in (AB)$ so that $\left\{\begin{array}{c} EA=m\cdot EC\\\\ FA=n\cdot FB\end{array}\right\|\ ,$ where $\{m,n\}\subset \mathbb R^*_+\ ,\ |m-n|<1\ .$

Denote the midpoints $M\ ,\ N$ of $[BE]\ ,\ [CF]$ respectively. Prove that $EN\cap FM\cap BC\ne\emptyset\Longleftrightarrow m^2+n^2=mn+1\ .$ See here

PP18. Let an acute $\triangle{ABC}\ ,$ the foot $D$ of the $C$ -bisector and the circumcenter $O$ of $\triangle ABC\ .$

Let $P\in OD$ for what $CP\perp AB\ .$ Prove that $P$ belongs to the circumcircle of $\triangle AOB\iff$ $C=\frac {\pi}3$

Proof. $OA=OB\implies \frac {PA}{PB}=\frac {DA}{DB}=$ $\frac {CA}{CB}\implies\frac {PA}{PB}=\frac {CA}{CB}\implies$ $P\ ,$ $C$ belong to the Appolonius' circle $w$ w.r.t. $[AB]$ and the ratio $\frac {XA}{XB}=\frac ba\ ,$ $(\forall ) X\in w\ .$

Since the center of $w$ belongs to $AB$ and $CP\perp AB$ obtain that $P$ is the symmetrical point of $C$ w.r.t. $AB\ .$ Hence the circumcircle $\beta$ of $\triangle AOB$ is symmetric w.r.t. $AB$

with the circumcircle of $\triangle ABC\ ,$ i.e. the orthocenter $H$ of $\triangle ABC$ belongs to $\beta\ .$ In conclusion, $180^{\circ}-C=2C\implies C=60^{\circ}\ .$

This post has been edited 167 times. Last edited by Virgil Nicula, Jun 18, 2016, 9:36 PM

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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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391345 views moment
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blog
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the visits tho
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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

414. Geometry 7.

by Virgil Nicula, Mar 21, 2015, 1:44 PM

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