407. Cyclical quadrilaterals.

by Virgil Nicula, Feb 11, 2015, 10:52 AM

PP1. Let a cyclical quadrilateral $ABCD$ with the area $[ABCD]=S$ and its circumcircle $w=\mathbb C(O,R)$ .

Denote $\left\{\begin{array}{ccccc} AB=a & ; & BC=b & ; & AC=e\\\\ CD=c & ; & DA=d & ; & BD=f\end{array}\right\|$ and $\left\{\begin{array}{c} G\in AC\cap BD\\\\ E\in AB\cap CD\\\\ F\in AD\cap BC\end{array}\right\|$ . Prove that:

$1.\ \blacktriangleright\ \frac {GA}{da}=\frac {GB}{ab}=$ $\frac {GC}{bc}=\frac {GD}{cd}=$ $\frac e{ad+bc}=\frac f{ab+cd}=$ $\sqrt{\frac {ac+bd}{(ad+bc)(ab+cd)}}=$ $\sqrt {-\frac {p_w(G)}{abcd}}\equiv k$ , where $p_w(X)$ - power of $X$ w.r.t. $w$ and $\boxed{k^2=\frac {ac+bd}{(ad+bc)(ab+cd)}}\ .$

$2.\ \blacktriangleright\ \frac {EA}{de}=\frac {EB}{bf}=$ $\frac {EC}{be}=\frac {ED}{df}=$ $\frac a{bf-de}=\frac c{be-df}=\frac 1{k\left(b^2-d^2\right)}$ because $(1)\implies\frac c{be-df}=$ $\frac c{b\cdot k(ad+bc)-d\cdot k(ab+cd)}=$ $\frac c{k\left(b^2c-cd^2\right)}=$ $\frac 1{k\left(b^2-d^2\right)}\ .$

$3.\ \blacktriangleright\ \frac {FA}{ae}=\frac {FB}{af}=$ $\frac {FC}{ce}=\frac {FD}{cf}=$ $\frac d{ae-cf}=\frac b{af-ce}=\frac 1{k\left(a^2-c^2\right)}$ because $(1)\implies\frac b{af-ce}=$ $\frac b{a\cdot k(ab+cd)-c\cdot k(ad+bc)}=$ $\frac b{k\left(a^2b-bc^2\right)}=$ $\frac 1{k\left(a^2-c^2\right)}\ .$

$4.\ \blacktriangleright\ m\left(\widehat{AGB}\right)=\phi\implies 2kR\sin\phi =1$ and $[EGF]=\frac {2abcdS}{\left(a^2-c^2\right)\left(b^2-d^2\right)}$ (Miguel Ochoa Sanchez).

Proof. Can prove easily $(1)\ ,\ (2)\ ,\ (3)$ . I" ll show only $(4)$ . Thus, $S=[ABCD]=[ABC]+[ADC]\implies$ $S=\frac 1{4R}\cdot (bcf+adf)\implies$ $4RS=f(ad+bc)=$ $\frac {ef}k\stackrel{(1)}{\implies}$

$\boxed{4RS=\frac {ef}k}\ (*)$ . But $2S=ef\sin\phi\implies$ $\boxed{2kR\sin\phi =1}\ .$ Thus, $\left\{\begin{array}{ccc} S^2 & = & (s-a)(s-b)(s-c)(s-d)\\\\ 4RS & = & \sqrt{(ac+bd)(ab+cd)(ad+bc)}\end{array}\right\|$ , where $2s=a+b+c+d$ .

$\blacktriangleright\ 4R\cdot [EBF]=2R\cdot BE\cdot BF\cdot\sin B\ \stackrel{(2\wedge 3)}{=}\ e\cdot BE\cdot BF=$ $e\cdot \frac {bf}{k\left(b^2-d^2\right)}\cdot\frac {af}{k\left(a^2-c^2\right)}\implies$ $4R\cdot [EBF]=\frac {abef^2}{k^2\left(a^2-c^2\right)\left(b^2-d^2\right)}\iff$

$\boxed{4R\cdot [EBF]=\frac {abf(ab+cd)(ad+bc)}{\left(a^2-c^2\right)\left(b^2-d^2\right)}}\ (5)\ .$

$\blacktriangleright\ 4R\cdot [GBE]=2R\cdot BG\cdot BE\cdot\sin \widehat{GBE}=$ $d\cdot BG\cdot$ $BE\ \stackrel{(1\wedge 2)}{=}\ d$ $\cdot abk\cdot \frac {bf}{k\left(b^2-d^2\right)}$ $\implies$ $\boxed{4R\cdot [GBE]=\frac {ab^2df}{b^2-d^2}}\ (6)\ .$

$\blacktriangleright\ 4R\cdot [GBF]=2R\cdot BG\cdot BF\cdot\sin \widehat{GBF}=$ $c\cdot BG\cdot BF\ \stackrel{(1\wedge 3)}{=}\ c$ $\cdot abk\cdot \frac {af}{k\left(a^2-c^2\right)}$ $\implies$ $\boxed{4R\cdot [GBF]=\frac {a^2bcf}{a^2-c^2}}\ (7)\ .$

$\blacktriangleright\ 4R\cdot [EGF]=4R\cdot\left([EBF]-[GBE]-[GBF]\right)=$ $\frac {abf(ab+cd)(ad+bc)}{\left(a^2-c^2\right)\left(b^2-d^2\right)}-$ $\frac {ab^2df}{b^2-d^2}-$ $\frac {a^2bcf}{a^2-c^2}=$ $\frac {abf}{\left(a^2-c^2\right)\left(b^2-d^2\right)}$

$\cdot \left[(ab+cd)(ad+bc)-bd\left(a^2-c^2\right)-ac\left(b^2-d^2\right)\right]=$ $\frac {2abcdf(ad+bc)}{\left(a^2-c^2\right)\left(b^2-d^2\right)}\implies$ $\boxed{4R\cdot [EGF]=\frac {2abcdf(ad+bc)}{\left(a^2-c^2\right)\left(b^2-d^2\right)}}\ (8)\ \implies$ $\frac {[EGF]}{[ABCD]}=$ $\frac {4R\cdot [EGF]}{4RS}=$

$\frac {2abcdf(ad+bc)}{\left(a^2-c^2\right)\left(b^2-d^2\right)}\cdot \frac k{ef}=$ $\frac {2abcd}{\left(a^2-c^2\right)\left(b^2-d^2\right)}\implies$ $[EGF]=\frac {2abcdS}{\left(a^2-c^2\right)\left(b^2-d^2\right)}\ .$

Remark. If $\triangle ABC$ is equilateral, then there is the relation $\frac {EA}{FC}=\left(\frac {EC}{FA}\right)^2$ . Indeed, denote $\left\{\begin{array}{c} \left(\widehat{ACE}\right)=\alpha\\\\ \left(\widehat{CAE}\right)=\beta\end{array}\right\|$ . Observe that $\alpha +\beta =60^{\circ}$ and

$\left\{\begin{array}{ccc} m\left(\widehat{ACE}\right)=\alpha=60^{\circ}-\beta=m\left(\widehat{CFA}\right) & \implies & \widehat{ACE}\equiv \widehat{CFA}\\\\ m\left(\widehat{CAF}\right)=\beta=60^{\circ}-\alpha=m\left(\widehat{AEC}\right) & \implies & \widehat{AEC}\equiv \widehat{CAF}\end{array}\right\|$ . Therefore, $\triangle ACE\sim\triangle CFA\implies$ $\frac {CA}{FC}=\frac {AE}{CA}=\frac {CE}{FA}\implies$

$\odot\begin{array}{ccc} \nearrow & AE=CA\cdot\frac {EC}{FA} & \searrow\\\\ \searrow & FC=CA\cdot\frac {FA}{EC} & \nearrow\end{array}\odot\implies$ $\frac {EA}{FC}=\left(\frac {EC}{FA}\right)^2$ .

PP2. Let $ABCD$ be a circumscribed quadrilateral with $A=D$ . Denote its incircle $w=\mathbb C(I,r)$ . Prove that $\frac 1{AB}+\frac 1{DC}=\frac {\sin A}{r}$ .

Proof 1. Denote $[ABCD]=\mathbb A$ , the midpoint $S$ of $[AD]$ , the projections $(M,N)$ on the line $AB$ of the points $(B,C)$ respectively and

$\left\{\begin{array}{ccccc} AB=a & , & DC=b & ; & SA=SD=s\\\\ BM=m & , & CN=n & ; & AM=p\ ,\ DN=q\end{array}\right\|$ . Thus, $AB+CD=BC+AD=p$ and $\mathbb A=pr$ , i.e. $\boxed{\mathbb A=r(a+b)}\ (*)$ . Therefore,

$\mathbb A=[ABM]+[DCN]+[BCNM]\iff$ $2\mathbb A=MA\cdot MB+ND\cdot NC+MN\cdot (MB+NC)=$ $MB\cdot AN+NC\cdot MD=$ $m(2s-q)+n(2s-p)\implies$

$\boxed{2\mathbb A=2s(m+n)-(mq+np)}$ . Observe that $s=r\cot\frac A2$ and $\left\{\begin{array}{ccccc} \frac ma & = & \frac nb & = & \sin A\\\\ \frac pa & = & \frac qb & = & \cos A\end{array}\right\|$ . Thus, $2\mathbb A=2r(a+b)\sin A\cot\frac A2-$ $2ab\sin A\cos A\ \stackrel{(*)}{\iff}$

$2r(a+b)=2r(a+b)(1+\cos A)-2ab\sin A\cos A\iff$ $r(a+b)=ab\sin A\iff \frac 1a+\frac 1b=\frac {\sin A}r$ , i.e. $\frac 1{AB}+\frac 1{DC}=\frac {\sin A}{r}$ .

Proof 2 (Miguel Ochoa Sanchez). Denote $E\in BI\cap AD$ . Observe that $A+\frac B2+\frac C2=180^{\circ}\implies$ $\left\{\begin{array}{c} m\left(\widehat{IEA}\right)=\frac C2\\\\ \delta_{AE}(I)=\delta_{DC}(I)=r\end{array}\right\|$ $\implies$ $\triangle AIE\equiv \triangle DIC\implies$

$AE=DC\implies AE=b$ . Since the ray $[AI$ is $A$ -bisector in $\triangle ABE$ , then $AI=\frac {2\cdot AB\cdot AE\cdot \cos\frac A2}{AB+AE}\implies$ $AI=\frac {2ab\cos\frac A2}{a+b}\implies$ $\frac r{\sin\frac A2}=\frac {2ab\cos\frac A2}{a+b}\implies$

$r(a+b)=ab\sin A\implies$ $\frac 1a+\frac 1b=\frac {\sin A}r$ , i.e. $\frac 1{AB}+\frac 1{DC}=\frac {\sin A}{r}$ . Very nice!

PP3. Let $\triangle ABC$ and the points $\left\{\begin{array}{ccc} M\in (AB) & ; & N\in (AC)\\\\ D\in (BC) & ; & P\in AD\cap MN\end{array}\right\|$ . Prove that $\frac {MB}{MA}\cdot DC+\frac {NC}{NA}\cdot DB=\frac {PD}{PA}\cdot BC$ .

Proof 1. Denote $\{U,V\}\subset MN$ so that $BU\parallel AD\parallel CV$ . Thus, $\left\{\begin{array}{ccc} \frac {MB}{MA} & = & \frac {BU}{AP}\\\\ \frac {NC}{NA} & = & \frac {VC}{AP}\end{array}\right\|$ $\implies$ $\frac {MB}{MA}\cdot DC+\frac {NC}{NA}\cdot DB=$ $\frac {BU}{AP}\cdot DC+\frac {VC}{AP}\cdot DB=$

$\frac {BU\cdot DC+VC\cdot DB}{AP}\ \stackrel{(*)}{=}\ \frac {PD\cdot BC}{PA}\implies$ $\frac {MB}{MA}\cdot DC+\frac {NC}{NA}\cdot DB=\frac {PD}{PA}\cdot BC$ .

Lemma. Let $AXCY$ be a quadrilateral what is inscribed in the circle with the diameter $[AC]$ . Denote $\left\{\begin{array}{ccc} E\in AY & ; & CE\parallel AX\\\\ F\in AX & ; & CF\parallel AY\end{array}\right\|$ . Prove that $AE\cdot AY+AF\cdot AX=AC^2$ .

Proof 2. Observe that $\left\{\begin{array}{c} EC=AF\\\\ FC=AE\end{array}\right\|$ and generalized Pythagoras $:\ \left\{\begin{array}{ccc} EC/\triangle AEC\ : & EC^2=AE^2+AC^2-2\cdot AE\cdot AY\\\\ FC/\triangle AFC\ : & FC^2=AF^2+AC^2-2\cdot AF\cdot AX\end{array}\right\|$ $\bigoplus\implies$ $AE\cdot AY+AF\cdot AX=AC^2$ .

PP4 (an extension of the previous lemma). Let $ABCD$ be a cyclical quadrilateral with $A<90^{\circ}$ . Denote $\left\{\begin{array}{ccc} E\in AD & ; & CE\parallel AB\\\\ F\in AB & ; & CF\parallel AD\end{array}\right\|$ . Prove that $AE\cdot AD+AF\cdot AB=AC^2$ .

Proof 1. Denote the projections $X$ , $Y$ of $C$ on $AB$ , $AD$ respectively. Observe that $\left\{\begin{array}{ccc} \triangle CEY\sim\triangle CFX & \implies & \frac {CE}{CF}=\frac {CY}{CX}\\\\ \triangle CYD\sim\triangle CXB & \implies & \frac {CY}{CX}=\frac {DY}{BX}\end{array}\right\|$ $\implies$ $\frac {CE}{CF}=\frac {DY}{BX}\implies$ $\frac {AF}{AE}=\frac {DY}{BX}\implies$

$\boxed{AF\cdot BX=AE\cdot DY}\ (*)$ . Apply the previous lemma to the quadrilateral $AXCY$ what is inscribed in the circle with the diameter $[AC]\ :\ AC^2=AF\cdot AX+AE\cdot AY=$

$AF\cdot (AB+BX)+AE\cdot (AD-DY)=$ $AF\cdot AB+AE\cdot AD+$ $(AF\cdot BX-AE\cdot DY)\ \stackrel{(*)}{\implies}\ AC^2=AE\cdot AD+AF\cdot AB$ .

Proof 2 (Ruben Dario). Denote the second intersection $L$ of the circumcircle $w$ of $\triangle BCF$ with $AC$ . Thus, from the power $p_w(A)$ of the point $A$ w.r.t. $w$ get $\boxed{AF\cdot AB=AL\cdot AC}\ (1)$ .

Observe that $\triangle CLF\sim\triangle ADC$ , i.e. $\frac {CL}{AD}=\frac {FC}{CA}$ $\implies$ $\boxed{AE\cdot AD=CL\cdot CA}\ (2)$ . In conclusion, the sum of $(1)$ and $(2)$ becomes the required relation. Very nice proof !

Proof 3 (trigonometric). Denote $\left\{\begin{array}{ccc} m\left(\widehat{ADC}\right) & = & \phi\\\ m\left(\widehat{CAB}\right) & = & x\\\ m\left(\widehat{CAD}\right) & = & y\end{array}\right\|$ $\implies$ $\left\{\begin{array}{ccc} m\left(\widehat{CFB}\right) & = & x+y\\\ m\left(\widehat{ACB}\right) & = & \phi -x\\\ m\left(\widehat{ACD}\right) & = & 180^{\circ}-(\phi +y)\end{array}\right\|$ . Apply the theorem of Sines in the triangles $:$

$\left\{\begin{array}{cccccc} \triangle AEC\ \wedge\ \triangle AFC\ : & \frac {AE}{\sin x} & = & \frac {AF}{\sin y} & = & \frac {AC}{\sin (x+y)}\\\\ \triangle ABC\ \wedge\ \triangle ADC\ : & \frac {AB}{\sin (\phi -x)} & = & \frac {AD}{\sin(\phi +y)} & = & \frac {AC}{\sin \phi}\end{array}\right\|\ .$ Therefore, the required relation becomes $\frac {AE}{AC}\cdot \frac {AD}{AC}+\frac {AF}{AC}\cdot\frac {AB}{AC}=1\iff$

$\frac {\sin x}{\sin (x+y)}\cdot\frac {\sin (\phi +y)}{\sin \phi}+\frac {\sin y}{\sin (x+y)}\cdot\frac {\sin (\phi -x)}{\sin \phi}=1\iff$ $\boxed{\sin x\sin (\phi +y)+\sin y\sin (\phi -x)=\sin\phi\sin(x+y)}$ . Divide this relation

by the expression $\cos x\cos y\cos\phi$ and obtain that $\tan x(\tan\phi+\tan y)+\tan y(\tan\phi -\tan x)=\tan\phi (\tan x+\tan y)$ , what is truly.

Proof 4 (synthetic - own). Apply Ptolemy's theorem $:\ \boxed{CD}\cdot AB+\boxed{BC}\cdot AD=\boxed{BD}\cdot AC\ (*)$ . But $\triangle BCD\sim\triangle AEC\iff$

$\frac {\boxed{CD}}{EC}=\frac {\boxed{BC}}{AE}=$ $\frac {\boxed{BD}}{AC}$ , i.e. the relation $(*)$ is equivalently with $\boxed{AF}\cdot AB+\boxed{AE}\cdot AD=\boxed{AC}\cdot AC$ . I used $EC=AF$ in the first ratio.

PP5 (Miguel Ochoa Sanchez). Let a cyclic quadrilateral $ABDC$ for which denote $M\in AB\cap CD$ , $N\in AC\cap BD$ and $P\in AD\cap BC$ . Prove that $\frac {\cos B}{BN}+\frac {\cos C}{CM}=\frac {\cos A}{AP}$ .

Proof. $\left\{\begin{array}{ccccc} m\left(\widehat{DAB}\right)=m\left(\widehat{DCB}\right)=u\\\ m\left(\widehat{DAC}\right)=m\left(\widehat{DBC}\right)=v\end{array}\right\|\implies$ $\left\{\begin{array}{ccccc} m\left(\widehat{APB}\right)=C+v & ; & m\left(\widehat{APC}\right)=B+u\\\ m\left(\widehat{MBN}\right)=C+u & ; & m\left(\widehat{MCN}\right)=B+v\\\ m\left(\widehat{AMC}\right)=B-u & ; & m\left(\widehat{ANB}\right)=C+v\end{array}\right\|$ and $\boxed{(B+u)+(C+v)=(B+v)+(C+u)=\pi}\ (*)$ . Thus,

$\frac {\cos B}{BN}+\frac {\cos C}{CM}=\frac {\cos A}{AP}\iff$ $\frac {AP}{AB}\cdot\frac {BA}{BN}\cdot \cos B+\frac {AP}{AC}\cdot\frac {CA}{CM}\cdot \cos C=\cos A\iff$ $\frac {\sin\widehat{ABP}}{\sin\widehat{APB}}\cdot\frac {\sin\widehat{BNA}}{\sin\widehat{BAN}}\cdot \cos B+\frac {\sin\widehat{ACP}}{\sin\widehat{APC}}\cdot\frac {\sin\widehat{CMA}}{\sin\widehat{CAM}}\cdot \cos C=\cos A\iff$

$\frac {\sin B}{\sin (C+v)}\cdot\frac {\sin (C-v)}{\sin A}\cdot \cos B+\frac {\sin C}{\sin (B+u)}\cdot \frac {\sin (B-u)}{\sin A}\cdot\cos C=$ $\cos A\ \stackrel{(*)}{\iff}\ \boxed{\sin 2B\sin [(C-A)+u]+\sin 2C\sin (B-u)=\sin 2A\sin (B+u)}\iff$

$\sin 2B\left[\sin (C-A)+\cos (C-A)\tan u\right]+\sin 2C\left(\sin B-\cos B\tan u\right)=\sin 2A(\sin B+\cos B\tan u)\iff$

$\tan u\left[(\sin 2A+\sin 2C)\cos B-\sin 2B\cos (C-A)\right]=$ $\sin 2B\sin (C-A)+\sin B(\sin 2C-\sin 2A)\iff$

$\tan u\left[2\sin B\cos (A-C)\cos B-\sin 2B\cos (C-A)\right]=$ $\sin 2B\sin (C-A)-2\sin B\cos B\sin (C-A)$ , what is truly.

Remark. If $\triangle ABC$ is equilateral, then $\frac 1{BN}+\frac 1{CM}=\frac 1{AP}$ .

Lermma. The properties of the cyclic and orthodiagonal quadrilateral.

Let a cyclic, orthodiagonal and convex $ABCD$ with $(AC\perp BD)$ and the circumcircle circle $w=C(O,R)\ .$ I"ll use the standard notations : $AB=a$ , $BC=b$ , $CD=c$ ,

$DA=d$ , $AC=e$ , $BD=f\ .$ Denote $E\in AC\cap BD$ and the middlepoints $M$ , $N$ , $P$ , $Q$ of $[AB]$ , $[BC]$ , $[CD]$ , $[DA]$ respectively. Then there are the following properties :

$1.\blacktriangleright ME\perp CD\ ;\ \delta_{AB}(O)=\frac 12\cdot CD\ ,$ where $\delta_{AB}(O)$ is the distance of the point $O$ to the line $AB\ .$

$2.\blacktriangleright \delta_{AB}(O)+\delta_{BC}(O)+\delta_{CD}(O)+\delta_{DA}(O)=p\ .$

$3.\blacktriangleright \sigma [ABCD]=\frac 12(ac+bd)=\frac 12ef\ .$

$4.\blacktriangleright a^2+c^2=b^2+d^2=4R^2\ ;\ MP=NQ=\frac 12\sqrt{e^2+f^2}\ .$

$5\blacktriangleright e^2+f^2=4\left[R^2+p_w (E)\right]\ ,$ where $p_w(E)=R^2-OE^2$ is the power of the point $E$ w.r.t. the circle $w\ .$

$6.\blacktriangleright$ $M,N,P,Q$ and the projections of $P$ on the sides of the quadrilateral $ABCD$ belong to the circle with the centre in the

middlepoint of $[OE]$ and the length of the diameter equal to $R^2+p_w(E)=2R^2-OE^2$ (the circle of the eight points) .

This post has been edited 116 times. Last edited by Virgil Nicula, Aug 27, 2016, 1:46 PM

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292. Length of the angled bisector and some means.

291. Some integrals.

290. Characterize "OP perpendicular on OA" in ABC a.s.o.

289. Nice identities in an algebra/ geometry/trigonometry.

288. An inequality with two A-cevians in a triangle ABC.

287. Some metrical problems with or without trigonometry.

286. ABC is right triangle iff s=2R+r and other similar pp.

285. Nice and easy inequalities. For example, IMAC - 2009.

284. Ellipse (definition).

283. Some nice and easy problems.

282. Factorize of polynomials.

May 2011

281. Identity with R in an acute triangle (ILL - 1974).

280. Common roots of two polynomial equations.

279. O.M. Holland , O.M. Ireland 1988 and TST USA 2000, etc.

278. In memoria profesorului Alexandru Lupas.

277. A nice and difficult relation with Lemoine's axis.

276. Chinese TST 2009 and Second Round 2006.

275. Some interesting algebraic equations and systems.

274. An inequality in a triangle for its interior point.

273. The area of the triangle IOH. When I belongs to OH ?

272. Outside of triangle. Extension of Fermat's point.

271. Relations concerning Fermat points.

April 2011

270. Morrocan M.O. 2010 and Croatia TST 2009.

269. Some nice and easy properties in a trapezoid.

268. Saraghin contest, 2011.

267. Germany 2002, Russia 2003, IMO 2003, Aus.-Polish 1992.

266. Rectangles erected on sides of a triangle. Concurrence.

265. Really nice problem !

264. Identities with complex numbers.

263. Newton's sums. Applications.

262. IMO Shortlist 2005, Geometry 6th.

261. IMO 2009, problem 2 and JBMO 2007, problems 1 & 2.

260. Some nice, simple identities and their applications.

259. ABC equilateral triangle ==> DEF equilateral triangle.

258. Some difficult and nice problems with complex numbers.

257. Seeking roots of 2th equation with complex coef.

256. An extremum problems with complex numbers.

March 2011

255. A line which is parallelly with OH.

254. An interesting C. Mateescu's geometrical inequality.

253. An old algebraic inequality.

252. OI=f(A,R) in certain conditions.

251. Problems of Analytical Geometry.

250. An application of the Euler's relation.

249. An usual and nice metrical relations in quadrilateral.

248. Pagina instructiva.

247. An remarkable inequality with median.

246. R1-R2=OI (nice).

245. An inequality in an acute triangle.

244. Some properties of cyclic orthodiagonal quadrilaterals.

243. Nice ! (a+b) is constant.

242. A general geometrical problems.

241. Areas in a triangle.

240. Some interesting geometrical inequalities.

239. Algebraic marathon.

238. Some metrical problems.

237. Some problems with ... problems.

236. A locus with the metrical relation.

235. Proofs of some geometry problems with complex numbers.

234. Some simple and nice problems from contests.

233. An interesting identity and an easy & nice inequality.

February 2011

232. Some geometrical inequalities (own).

Pagina libera

231. A conditioned minimum value.

230. Some usual synthetical properties in a triangle.

229. Concurs online MATEFBC, ed. 5 -subiect 2.

228. OJM - Romania, 2010 problem 3.

227. IMO ShortList 2003 (problem 5).

226. Some inequalities from Calculus (Real Analysis).

225. An interesting problems from Kunny.

224. Some nice and easy inequalities in a triangle.

223. An inequality with complex numbers-RMO shortlist 2010.

222. An extension of probl. 2 from IMO 2009.

221. Some nice and simple "slicing" problems.

220. Some nice geometry problems from contest.

January 2011

219. A very nice geometrical inequality.

218. Some problems from the Suiss IMO Selection Team 2006.

217. A nice and interesting Murray S. Klamkin's problem.

216. Nice ! Find the length of hypotenuse (Belarus - 2010)

215. Limit of the unique root for polynominal equations.

214. A range of some functions.

213. Another classical "slicing" problems.

212. A simple applications of the Casey's theorem.

211. Pretty hard problem !

210. Tangential circle of the triangle ABC.

209. Nice problem, nice proof !

208. An interesting geometrical inequality.

207. Some interesting problems of Toshio Seimiya, Japan.

206. Four problems with extremum in a quadrilateral.

205. A "slicing" problem with 11 methods.

204. Saraghin 2010 (three geometry problems).

203. An application of the Pascal's permutation theorem.

December 2010

202. Concurrent lines in a right triangle.

201. Applications of Desarques' and Pappus' theorems.

200. Some nice applications of the inversion.

199. Some properties of symmedian. Two extensions.

198. Generalized Carnot's lemma.

197. Triangles outside of a triangle and concurrence.

195. Semicircle. Nice application of harmonical division.

194. The Lalescu's sequence. Properties.

193. An inequality with n! .

192. An easy and nice problem of Valentin Vornicu.

191. Six nice problems with the incircle of a triangle.

190. Very nice ! Coculescu's Contest seniors 2010.

189. Pascal's theorem and some nice and easy applications.

188. Some problems from vectorial geometry.

187. Distancies.

186. Without the l'Hospital's rules.

185. The Durrande's relation.

184. An interesting geometry problem from RMO 2010.

183. A remarkable characterization of equilateral triangle.

182. Another two nice problems with complex numbers (own).

181. Two special inequalities from CRUX (own).

180. Own proposed problem from CRUX (with complex numbers).

November 2010

179. The range of |z| , |z^2+a|=|bz+c|. Particular cases.

178. Some nice and simple geometry problems.

177. A nice proof of one identity in a triangle.

176. Another nice problems for middle school.

175. A Geometric Proof (without trig.) of Heron's Formula.

174. Some interesting properties of modulus (middle school).

173. An interesting trigonometric equations.

172. The Zelinsky's problem.

171. Another "slicing" proposed problems (middle school).

October 2010

170. A nice "slicing" proposed problem for middle school.

169. Some metrical problems in a circle (III).

168. Some properties of the Lemoine's point.

167. Nice problems - OIM, 1995 and ... Toshio Seimyia.

166. Some constant geometrical expressions.

165. Some metrical problems (5) in a circle (II).

164. Some metrical problems (5) in a circle.

163. Applications of Ptolemy's inequality. Ex. - China,1998.

162. Generalization of proposed problem from Mongolia, 2000.

161. A difficult inequality (barycentrical coordinates)

160. A very nice "slicing" problem.

159. Multiple derivatives in a point.

158. Very nice and difficult limits.

157. Steinbart's theorem. Applications.

156. Some problems with perpendicularities.

155. A constant ratio in a triangle ABC.

154. Difficult inequality (without derivatives eventually).

153. Geometrical minimum/maximum.

152. Concurrence, cyclically, metrics, symmedian a.s.o.

151. Miscellaneous problems for the admission at math.

150. Some (5) problems with fixed points.

149. Some (8) problems with collinearity or concurrence.

148. A special class of systems.

147. A problem with two polynominals.

146. Synthetic, metric, trigonometric and analitycal proofs.

145. Spain MO P5 - Point on median.

144. A conditioned concurrency in a triangle.

143. Assess of a ratio (nice and difficult problem).

142. Inscribed/circumscribed quadrilateral w.r.t. ABC.

141. N is interior <==> 1/3<a/b<3 (CRUX Mathematicorum).

140. A synthetical property <==> linear/angled relation.

139. A caracterization of a right angle in a triangle.

138. Generalized "slicing" problems (own).

137. Stewart's relation.

136. A geometrical interpretation of a system (own).

135. Three strong geometrical inequalities.

September 2010

134. Some problems with projections/perpendiculars (2).

133. Some problems with projections/perpendiculars (1).

132. Asupra unei inegalitati intr-un triunghi.

131. A stronger Form of the Steiner-Lehmus Theorem (own).

130. A triangle and a fixed point.

129. BX=CY and three properties.

128. Value of IG and two inequalities in neighbourhood of 0.

127. Pagina educativa.

126. Routh's theorem.

125. I like very much this Darij Grinberg's message.

124. A nice and easy problem with a right-angled triangle.

123. My extension and its proof.

122. An identity for an equilateral triangle. Extension.

121. Viete's relations with cos A , cos B , cos C.

120. Two equal neighbouring angles/sides in a quadrilateral.

119. Inspired by a Miculita's problem.

118. An usual remarkable geometrical locus.

117. Some applications of harmonic division/quadrilateral.

116. The length of AE (nine methods !).

115. An difficult equation with one radical.

114. A property of tangential triangle for ABC.

113. Range for a function.

112. The equivalence relation ".s.s."

111. Simple, but interesting ...

110. German TST 2004 - Exam V , Problem 1

109. Two equal areas ==> find A.

108. Old, short and nice proof of Euler's relation.

107. XY parallel BC and X,Y are isogonal/izotomic points.

106. OLM - Bucuresti (geometrie).

105. A problem with three circles.

104. Concursul Revistei de Matematica din Galati (RMG).

103. Metrical usual relations in triangle. Applications.

102. Characterization of a metrical relation in a triangle.

101. O problema cu numere complexe.

100. Bobiller's theorem.

99. Some simple and nice geometry problems.

98. Symmedian & median (metrical relations).

97. Problem of butterfly.

August 2010

96. Integrals.

95. Three circles in an angle.

94. Equations of angle bisectors (interior and exterior).

93. The Appolonius' construct problems.

92. CA , CB tangent to parabola - OIMU 2008 Problem 4.

91. JBMO 2010, Problem 3.

90. Nice and difficult inequality in ABC (own).

89. Similarity (parallel/antiparallel).

88. A nice geometric locus conditioned metrically.

87. An interesting vectorial identity.

86. A very nice maxmin.

85. Problem "slicing" : 60-24-12.

84. IRAN national math olympiad - 2010

83. A metrical characterization of concurrence.

82. AA' , BN , CM concur in Gergonne point.

81. Sawayama and Thebault’s theorem.

80. Three concurrent perpendiculars (the Carnot's lemma).

79. A quadrilateral with many perpendicularities.

78. Jakobi's theorem.

77. Two equivalent perpendicularities.

76. A parallelogram and a circle (nice !).

75. Extended Steinbart Theorem.

74. Inegalitatea Erdos-Mordell.

73. A nice problems with a square and a perpendicularity.

July 2010

72. Colinearity in a triangle - H , P , M .

70. Colinearity with pole/polar - H\in PQ .

71. An angle questions (synthetic/trigonometric proofs).

69*1. Slicing problem 3.

69. Position of x1, x2 w.r.t. a given real number k.

68. Some nice applications of "pole/polar".

67. Remarkable algebraic/geometrical inequalities (1).

66. Remarkable perpendicularities in a triangle.

65. Parallel to BC through I .

64. Problems of the type "instant" (at most one minute).

63. A nice problem with radical center (livetolove212).

62. Nice and easy problem with a right triangle.

61. IMO Shortlist 2009 (very nice problem and its proof).

60. Mediteranean M.C. 2010 Problem 3.

59. IMO 2010, problem 4.

58. A cevian and two incircles.

57. Opinii si comentarii relativ la inv. rom. II

56. A extremum problem with areas.

55. Two important circles - mixtilinear incircle/excircle.

54. Cinci grade de libertate.

53. An remarkable concurrency.

52. Harmonical division.

51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.

51. A nice problem with perpendicularity from Jaime.

50. IMAC 2010 seniors, the first day.

49. The midpoint of a cevian in an acute triangle.

June 2010

48. An inequality in a triangle : h_a+max{b,c} ...

47. Interesting and difficult identities in a triangle.

46. Outside of a quadrilateral.

45. \cos B+\cos C=1 <==> nice property.

44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.

43. A "slicing" problem with a parameter.

42. IMAC 2010 Problema 2.

41. ONM 2010 Calarasi Problema 3.

May 2010

40. Lema lui Haruki.

39. A nice and remarkable problem with Brocard's point.

38. Minimum area of triangle touching with semicircle.

37. Own extension of the Steiner-Lehmus' problem.

36. ONM (juniori) - 2010 Iasi, problema 4.

35. Nice problem from Arqady (Eduard_Tur).

34. IMO Shortlist 2002 geometry problem 7.

April 2010

32. Linkuri diverse.

31.Some nice metrical problems.

29. Inegalitate integrala de la Kunny.

28. A fost odată ...

27.Problem of Darij Grinberg (I - centroid of triangle DEF).

26. Outside of a triangle.

25. Geometric equations.

24. A fine and cool inequalities.

23. A very nice exponential inequality (own - from CRUX).

22. Some interesting limits (sequence/function).

21 bis. Some problems with complex numbers.

21. Probleme de geometrie (Yetty & I).

20. O ineg.cu suma de AG/AI .

19. Own proposed problems in CRUX Mathematicorum - Canada.

18. O problema a lui Toshio Seimiya.

17. O inegalitate "tare" intr-un triunghi.

16. Length of the (interior) angle bisector (10 proofs).

15. Inequalities stronger than Panaitopol's.

14. Opinii si comentarii relativ la inv. rom.

13. Inegalitatea de la Sorin Borodi.

12. Inegalitatea I. Maftei & S. Radulescu (2).

11. Inegalitatea I. Maftei & S. Radulescu (1).

10. Walker's inequality and its extension.

9. Inegalitatea HO+HA+HB+HC >= 3R .

8. The first problem Shortlist ONM 2010 Romania.

7. Some interesting "slicing" problems.

6. Interesting problems from JBMO.

5. Theoretical preliminary for inequalities.

4. Remarkable geometrical inequalities (2).

Submit

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391345 views moment
by ryanbear, May 9, 2023, 6:10 AM
We need virgil nicula to return to aops, this blog is top 10 all time.
by OlympusHero, Sep 14, 2022, 4:44 AM
blog
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Amazing blog.
by OlympusHero, May 13, 2021, 10:23 PM
the visits tho
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Holy- Darn this is good. shame it's inactive now
by the_mathmagician, Jan 17, 2021, 7:43 PM
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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

407. Cyclical quadrilaterals.

by Virgil Nicula, Feb 11, 2015, 10:52 AM

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