343. Problems with collinear points and concurrent lines.

by Virgil Nicula, Jun 18, 2012, 1:36 AM

PP1. Let rectangle $ABCD$ with circumcircle $w=C(O,R)$ . For $P\in w$ , $Q\in (CD)$ let $E=\mathrm{pr}_{PA}(Q)$ , $F=\mathrm{pr}_{PB}(Q)$ , $G\in QE\cap AD$ , $H\in QF\cap BC$ . Prove $P\in GH$ .

Proof 1. $PEQF$ is inscribed in circle $\gamma$ with diameter $[PQ]$ , $ADQE$ is inscribed circle $\alpha$ with diameter $[AQ]$ and $BCQF$ is inscribed in circle $\beta$ with diameter $[BQ]$ .

Let $\{X_1,P\}=GP\cap w$ , $\{X_2,P\}=HP\cap w$ . Thus, $\left\{\begin{array}{ccccc} GX_1\cdot GP\stackrel{(w )}{=}GA\cdot GD\stackrel{(\alpha )}{=}GE\cdot GQ & \implies & X_1\in\odot (PEQ)\equiv\gamma & \implies & X_1\in \gamma\cap w\\\\ HX_2\cdot HP\stackrel{(w )}{=}HB\cdot HC\stackrel{(\beta )}{=}HF\cdot HQ & \implies & X_2\in\odot (PFQ)\equiv\gamma & \implies & X_2\in \gamma\cap w\end{array}\right|$

$\implies$ $\left\{X_1,X_2\right\}\subset w\cap \gamma\implies X_1\equiv X_2=X\in PG\cap PH\implies P\in GH$ .

Proof 2. Let $D'\in PD\cap BC$ and $C'\in PC\cap AD$ . Note $ QF\parallel PD $ , $ QE\parallel PC $ . Thus, $ \frac {D'H}{HC}=\frac {DQ}{QC}=\frac {DG}{GC'} $ . In conclusion, $P\in GH$ .


PP2. Let $\triangle ABC$ for which there is interior $P$ such that $(\forall )$ line which pass through $P$ and intersects

$AB$ , $AC$ at $E$ , $F$ respectively have $\frac{1}{AE} + \frac{1}{AF} = \frac{a+b+c}{bc}$ . Prove that $P$ is incenter $I$ of $\triangle ABC$ .

Proof. $\left\{\begin{array}{ccccccc} E:=B & \implies & AE=c\ ;\ \frac 1c+\frac {1}{AF}=\frac {2s}{bc} & \implies & AF=\frac {bc}{c+a} & \implies & P\in BI\\\\ F:=C & \implies & AF=b\ ;\ \frac 1{AE}+\frac 1b=\frac {2s}{bc} & \implies & AE=\frac {bc}{b+a} & \implies & P\in CI\end{array}\right|$ $\implies P\equiv I$ . Prove easily ($*$) that for any $E\in (AB)$

and $F\in (AC)$ so that $I\in EF$ exists the relation $b\cdot \frac {EB}{EA}+c\cdot \frac {FC}{FA}=a\iff$ $ b\cdot \left(\frac {EB}{EA}+1\right)+c\cdot \left(\frac {FC}{FA}+1\right)=2s\iff $ $ \frac{1}{AE} + \frac{1}{AF} = \frac{2s}{bc}$ . Denote $D\in BC\cap AI$ ,

$X\in BC\cap EF$ and $Y\in EF$ so that $AY\parallel BC$ . Suppose w.l.o.g. that $B\in (XC)$ . Thus, $b\cdot \frac {EB}{EA}+c\cdot \frac {FC}{FA}=$ $b\cdot \frac {XB}{AY}+c\cdot\frac {XC}{AY}=$ $\frac {b\cdot XB+c\cdot XC}{AY}=$

$\frac {b\cdot (XD-BD)+c\cdot (XD+DC)}{AY}=$ $\frac {(b+c)\cdot XD-b\cdot\frac {ac}{b+c}+c\cdot \frac {ab}{b+c}}{AY}=$ $(b+c)\cdot\frac {XD}{AY}=$ $(b+c)\cdot\frac {ID}{IA}=$ $(b+c)\cdot\frac {a}{b+c}=a\implies$ $\boxed{\ b\cdot \frac {EB}{EA}+c\cdot \frac {FC}{FA}=a\ }$ .


PP3. In a parallelogram $ABCD$ denote $E\in AB$ , $F\in AD$ so that $CE\perp AB$ , $CF\perp AD$ and $G\in EF\cap BD$ . Prove that $CA\perp CG$ .

Proof 1.Observe that the convex quadrilateral $AECF$ is inscribed in the circle $w$ with the diameter $[AC]$ . Denote $\left\{\begin{array}{c} AB=CD=a\\\\ AD=BC=b\end{array}\right\|$ and prove easily that

$\triangle CDF\sim\triangle CBE\iff \boxed{\frac ab=\frac {CD}{CB}=\frac {DF}{BE}=\frac {FC}{EC}}\ (*)$ . Define $G'\in EF\cap CC$ , where $CC$ is the tangent line to the circle $w$ in the point $C\in w$ ,

i.e. $CG'\perp CA$ . From an well-known property $\frac {G'F}{G'E}=\left(\frac {CF}{CE}\right)^2\stackrel{(*)}{\iff}$ $\boxed{\frac {G'F}{G'E}=\left(\frac ab\right)^2}\ (1)$ . Apply the Menelaus' theorem to the transversal $\overline {GBD}$ and $\triangle AEF\ :$

$\frac {GE}{GF}\cdot\frac {DF}{DA}\cdot\frac {BA}{BE}=1\iff$ $\frac {GF}{GE}=\frac {DF}{DA}\cdot\frac {BA}{BE}=\frac {DF}{BE}\cdot\frac {BA}{DA}\stackrel{(*)}{\iff}$ $\boxed{\frac {GF}{GE}=\left(\frac ab\right)^2}\ (2)$ . From the relations $(1)$ and $(2)$ obtain that

$\frac {G'F}{G'E}=\frac {GF}{GE}=\left(\frac ab\right)^2$ what means $G'\equiv G$ . In conclusion, $G\in BD\cap EF\cap CC$ , i.e. $CA\perp CG$ .

Remark. Apply the generalized Phytagoras' theorem to the sides $[BC]$ and $[CD]$ which belong to the triangles $ABC$ and $ACD$

respectively: $\left\{\begin{array}{c} BC^2=AB^2+AC^2-2\cdot AB\cdot AE\\\\ CD^2=AD^2+AC^2-2\cdot AD\cdot AF\end{array}\right\|\bigoplus\implies$ $\boxed{\ AB\cdot AE+AD\cdot AF=AC^2\ }$ .

Proof 2. Denote $O\in AC\cap BD$ . Observe that $w=C(O)$ is the circumcircle with the diameter $[AC]$ of $\triangle AEF$ and $O$ is the midpoint of $[BD]$ , where $B\in AE$

and $D\in AF$ . From an well-known property $G\in BD\cap EF\cap CC$ , where $CC$ is the tangent line to the circle $w=C(O)$ in the point $C\in w$ , i.e. $CA\perp CG$ .

An equivalent enunciation. Let $ABC$ be a triangle with the circumcircle $w=C(O,R)$ . A line $d$ for which $O\in d$ cut the sidelines $BC$ , $CA$ , $AB$ in the

points $D$ , $E$ , $F$ respectively. Denote $\{A,S\}=\{A,O\}\cap w$ and the tangent $t_a$ to $w$ in the point $S\in w$ . Prove that $OE=OF\iff D\in t_a$ .

PP4. Let an acute $\triangle ABC$ with the circumcircle $w=C(O)$ and $\left\{\begin{array}{c} E\in BO\cap AC\\\ F\in CO\cap AB\end{array}\right\|$ . Prove that $AE=AF\iff AB=AC$ .

Proof 1 (synthetic - Sorin Borodi). Denote the incenter $I$ of the triangle $ABC$ , $\{A,K\}=AI\cap w$ , the midpoints $M$ , $H$ and $S$ of $[BC]$ , $[OA]$ and $[EF]$ respectively.

$AE=AF\iff S\in AK\implies$ $\overline {HSM}$ is the Gauss' line of the complete quadrilateral $AFOE$ . Therefore, it is possibly iff $O\in AK$ , i.e. $ABC$ is $A$-isosceles.

Proof 2 (Tsikaloudakis). Suppose $OM<ON$ . Extend the ray $[OM$ at $OD=ON$ and let $E\in AD\cap BC$ . Denote $\left\{\begin{array}{c} m\left(\widehat{NDE}\right)=w\\\\ m\left(\widehat{NIC}\right)=\phi\\\\ m\left(\widehat{NMC}\right)=\theta\end{array}\right\|$ . Then triangles

$AEB$ , $EDN$ and $CMN$ are isosceles and $\omega = m\left(\widehat{INE}\right)=m\left(\widehat{IDE}\right) >\phi >\theta =m\left(\widehat{CMN}\right)= m\left(\widehat{MNC}\right)$ , that is inappropriate.

Proof 3 (trigonometric). Denote $D\in AO\cap BC$ and $\{A,L\}=AO\cap w$ . Thus, $\frac {DB}{DC}=\frac {BA\cdot BL}{CA\cdot CL}=\frac {c\cdot 2R\cos C}{b\cdot 2R\cos B}\implies$ $\boxed{\frac {DB}{DC}=\frac {c\cdot\cos C}{b\cdot\cos B}}$ . Obtain analogously that

$\left\{\begin{array}{ccc} \frac {EA}{EC}=\frac {c\cdot \cos C}{a\cdot\cos A} & \iff & \frac {EA}{c\cdot\cos C}=\frac {EC}{a\cdot\cos A}=\frac {b}{a\cdot \cos A+c\cdot\cos C}\\\\ \frac {FA}{FB}=\frac {b\cdot \cos B}{a\cdot\cos A} & \iff & \frac {FA}{b\cdot\cos B}=\frac {FB}{a\cdot\cos A}=\frac {c}{a\cdot \cos A+b\cdot\cos B}\end{array}\right\|$ . Therefore, $AE=AF\iff$ $\frac {\cos C}{a\cdot \cos A+c\cdot\cos C}=$ $\frac {\cos B}{a\cdot \cos A+b\cdot\cos B}\iff$

$\boxed{a\cdot\cos A(\cos B-\cos C)=(b-c)\cdot \cos B\cos C}\ (*)$ . Observe that $b\ge c\iff B\ge C\iff \cos B\le \cos C$ . In conclusion, if denote

$X=a\cdot\cos A(\cos B-\cos C)$ and $Y=(b-c)\cdot \cos B\cos C$ , then $X=Y$ and $XY\le 0$ , what means $X=Y=0$ , i.e. $b=c$ .



PP5. Let $\triangle ABC$ with the incenter $I$ and $P$ and $Q$ the intersections of the parallel to $BC$ that passes through $A$ with the

external bisectors of $\widehat{ABC}$ and $\widehat{ACB}$ respectively. Let $R$ such that $RP\perp BP$ and $RQ\perp CQ$ . Show that $RA = IA$ .

Proof. $\left\{\begin{array}{c} I_a\in PB\cap QC\\\\ D\in II_a\cap BC\end{array}\right\|$ . Prove easily $IB\parallel RP$ , $IC\parallel RQ$ and $\boxed{\triangle PRQ\sim \triangle BIC}\ (*)$ . Thus, $\{R,A,I,D,I_a\}$ are collinearly (using the Desarques' theorem).

Since $\frac {AP}{AQ}=\frac cb=\frac {DB}{DC}$ obtain $A\in PQ$ and $D\in BC$ are homologously in the similarity $(*)$ . Thus, $\frac {RA}{ID}=\frac {PQ}{BC}=$ $\frac {b+c}{a}=\frac {IA}{ID}$ (using the Van Aubel's relation) $\implies RA=IA$ .


Lemma (well-known "atomic bomb"). Let $ABCDEF$ be a cyclic hexagon. Then $AD\cap BE\cap CF\ne\emptyset\Longleftrightarrow AB\cdot CD\cdot EF=BC\cdot DE\cdot FA$ .

PP6. Let $\triangle ABC$ with incircle $w=C(I)$ which touches it in $D\in BC$ , $E\in CA$ and $F\in AB$ . Circle $w_1$ is tangent to $AB$ , $AC$ and is exterior tangent to $w$ in $K$ .

Circle $w_2$ is tangent to $BC$, $BA$ and is exterior tangent to $w$ in $M$ . Circle $w_3$ is tangent to $CA$ , $CB$ and is exterior tangent to $w$ in $N$ . Prove that $KD\cap ME\cap NF\ne\emptyset$ .

Proof. Apply above lemma to cyclic $KENDMF$, where $\left\{\begin{array}{c} KE=KF\\\ MF=MD\\\ ND=NE\end{array}\right\|$ . Relation $KE\cdot ND\cdot MF=EN\cdot DM\cdot FK$ is true. Hence $KF\cap EN\cap MD\ne \emptyset$ .

Remark. Since $KE=KF$ the ray $[DK$ is the bisector of $\widehat {EDF}$ a.s.o. Is evidently that the upper intersection is circumcenter $I$ of $\triangle DEF$ !


PP7. Let $ABCD$ be a parallelogram and let $M\in (AB)\ ,\ N\in (AD)$ be two points. Denote $\left\{\begin{array}{ccc} P\in CD & ; & MP\parallel AD\\\\ Q\in BC & ; & NQ\parallel AB\end{array}\right|$ and $\left\{\begin{array}{c} R\in MP\cap NQ\\\\ S\in AP\cap CN\end{array}\right|$ . Prove that $B\in RS$ .

Proof 1. Denote $AD=a\ ,\ AB=b$ and $\left\{\begin{array}{c} AN=x\in (0,a)\\\ AM=y\in (0,b)\end{array}\right|$ . Apply the Menelaus' theorem to the transversal $\overline {CSN}$ and $\triangle ADP\ :\ \frac {CP}{CD}$ $\cdot\frac {ND}{NA}\cdot \frac {SA}{SP}=1\iff$

$\frac {b-y}{1}\cdot\frac {a-x}{x}\cdot\frac {SA}{SP}=1\iff$ $\frac {SA}{SP}=\frac {x}{(b-y)(a-x)}\ (*)$ . Apply again the Menelaus' theorem (reciprocally) to the points $\{B,R,S\}$ and $\triangle AMP\ :$

$\frac {BM}{BA}\cdot \frac {SA}{SP}\cdot\frac {RP}{RM}=1\stackrel{(*)}{\iff}$ $\frac {b-y}{1}\cdot \frac {x}{(b-y)(a-x)}\cdot \frac {a-x}{x}=1$ , what is truly. In conclusion, the point $B$ belongs to the line $RS$ .

Proof 2. Denote $X\in AB\cap CN\ ,\ AD=a\ ,\ AB=b$ and $\left\{\begin{array}{c} AN=x\in (0,a)\\\ AM=y\in (0,b)\end{array}\right|$ . Apply the Menelaus' theorem (reciprocally)

to the points $\{B,R,S\}$ and $\triangle AMP\ :\ \frac {BM}{BA}\cdot \frac {SA}{SP}\cdot\frac {RP}{RM}=$ $\frac {b-y}{b}\cdot \frac {AX}{CP}\cdot \frac {a-x}{x}=$ $\frac {(a-x)(b-y)}{bx}\cdot\frac {AX}{CD}\cdot\frac {CD}{CP}=$

$\frac {(a-x)(b-y)}{bx}\cdot \frac {NA}{ND}\cdot\frac {b}{b-y}=$ $\frac {(a-x)(b-y)}{bx}\cdot\frac {x}{a-x}\cdot\frac {b}{b-y}=1\implies$ $\frac {BM}{BA}\cdot \frac {SA}{SP}\cdot\frac {RP}{RM}=1\implies$ $B\in RS$ .


PP8. Let $ABC$ be an acute triangle. Also, let $K$ and $L$ to be the two intersections of the perpendicular from $B$ with respect to side $AC$

with the circle of diameter $AC$, with $K$ closer to $B$ than $L$. Analogously, $X$ and $Y$ are the two intersections of the perpendicular from $C$

with respect to side $AB$ with the circle of diamter $AB$, with $X$ closer to $C$ than $Y$. Prove that the intersection of $XL$ and $KY$ lies on $BC$.

Proof. Let $H$ be the orthocenter of $ABC$, and let $D,E,F$ be its projections onto $BC$ , $AC$ , $AB$ respectively. Note that the circle with diameter $[AB]$ goes

through $D$ , $E$ and the circle with diameter $[AC]$ goes through $D$ , $F$. Now by power of point $HY\cdot HX=HD\cdot HA=HK\cdot HL$ . So $KLXY$ is cyclic.

Since $AF$ , $AE$ are perpendicular bisectors of $XY$ , $KL$ and $A$ is the center of this circle. Clearly $BX^2 = BF \cdot BA = BK \cdot BL$ i.e. $BK$ is tangent to the

circumcircle of $KLXY$ . Similarly, since $BY = BX$ , $BY^2 = BK \cdot BL$ , i.e. $BY$ also tangent. Similarly $CL$ and $CK$ are tangent to the circumcircle of $KLXY$ .

So using Pascal's theorem on $XXLLKY$ gives the intersections of $B\in XX\cap LK$ , $C\in LL\cap XY$ and $S\in XL\cap KY$ are collinear, which is the same as

$S\in BC$ , which is equivalent to the problem statement.


PP9. Let $ABCD$ be an isosceles trapezoid with $AB\parallel CD$ and $I\in AC\cap BD$ and which is inscrisbed in the circle $w$ . Denote $T\in AA\cap DD$ . Prove that $IT\parallel AB$ .

Proof 1. Prove easily that $\triangle ATD\sim\triangle CID$ . Thus, $\widehat{CID}\equiv\widehat{ATD}\iff$ $AIDT$ is cyclically and $TI$ is bisector of $\widehat{AID}$ , i.e. $TI\parallel AB$ .

Proof 2. Denote: $I\in AC\cap BD$ , $T\in AA\cap DD$ and $E\in AD\ ,\ F\in BC$ so that $I\in EF$ and $EF\parallel AB$ ; the circumcircles $w$ , $w_a$ , $w_d$

of the triangles $ABD$ , $AEI$ , $DEI$ respectively. The radical axis of $w$ , $w_a$ is $AT$ ; the radical axis of $w$ , $w_d$ is $DT$ ; the radical axis of $w_a$ ,

$w_d$ is $EF$ . In conclusion, the lines $AT$ , $DT$ and $EF$ are concurrently $\implies T\in EF$ , i.e. the point $T$ is the radical center of $w$ , $w_a$ , $w_d$ .


PP10. Let $\triangle ABC$ and $P\in (BC)\ ,\ R\in (AB)$ so that $Q\in PR\cap AC$ and $C\in (AQ)$ . For a $\triangle XYZ$ let its area is $S(XYZ)$ ,

length $r$ of its circumradius and ratio $f(XYZ)=\frac {S(XYZ)}{R^2}$ . Prove that $f(ABC)+f(PQC)=f(AQR)+f(BPR)\ .$

Proof. I"ll use identities $\boxed{\ S=2R^2\sin A\sin B\sin C}$ , i.e. $f(ABC)=2\prod\sin A$ and $\boxed{\sum \sin 2A=4\prod\sin A}\ (*)$ . Suppose w.l.o.g. that $C\in (AQ)$ . Denote

$\left\{\begin{array}{ccc} m\left(\widehat{AQR}\right)=y & ; & m\left(\widehat{ARQ}\right)=z\\\\ m\left(\widehat{BPR}\right)=x & ; & m\left(\widehat{BRP}\right)=\pi -z\\\\ m\left(\widehat{CPQ}\right)=x & ; & m\left(\widehat{CQP}\right)=y\end{array}\right\|$ where $\left\{\begin{array}{c} y=C-x\\\\ z=B+x\end{array}\right\|\ .$ Thus, $f(ABC)+f(PQC)=$ $f(AQR)+f(BPR)\iff$ $\sin A\sin B\sin C+\sin C\sin x\sin y=$

$\sin A\sin y\sin z+\sin B\sin x\sin z\iff$ $\sin A\sin B\sin C+\sin C\sin x\sin (C-x)=$ $\sin A\sin (C-x)\sin (B+x)+\sin B\sin x\sin (B+x)\iff$

$2\prod \sin A+\sin C[\cos (C-2x)-\cos C]=$ $\sin A[\cos (B-C+2x)+\cos A]+\sin B[\cos B-\cos (B+2x)]\iff$ $\underline{4\prod \sin A}+2\sin C\cos (C-2x)-\underline{\sin 2 C}=$

$2\sin A\cos (B-C+2x)+\underline{\sin 2A}+$ $\underline{\sin 2B}-2\sin B\cos (B+2x)\ \stackrel{(*)}{\iff}$ $\sin C\cos (C-2x)=$ $\sin A\cos (B-C+2x)-\sin B\cos (B+2x)\iff$

$\sin (2C-2x)+\sin 2x+\sin (2B+2x)-\sin 2x=\sin (2C-2x)+\sin (2B+2x)$ , what is truly for $(\forall )\ x\in\mathbb R$ .
This post has been edited 95 times. Last edited by Virgil Nicula, Nov 18, 2015, 9:31 AM

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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.
196. Similar rectangles outside of a triangle.
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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