435. Geometry problems in space.

by Virgil Nicula, Dec 4, 2015, 12:13 PM

P1 (Relatiile lui Euler). Intr-un tetraedru $ABCD$ suma intre patratele a doua muchii opuse si de patru ori patratul bimedianei corespunzatoare este egal cu suma patratelor celorlalte muchii.

Proof. Notam $\left\{\begin{array}{ccc} AB=a\ ; & BC=b\ ; & CD=c\\\\ DA=d\ ; & AC=e\ ; & BD=f\end{array}\right\|$ si alegem muchiile opuse $AC$ si $BD$ ale caror mijloace le notam $M\ ,$ $N$ respectiv. Aplicam teorema medianei in triunghiurile

$\left\{\begin{array}{ccccc} \triangle ABC\ : & 4\cdot MB^2 =2\cdot\left(BA^2+BC^2\right)-AC^2 & \iff & 4\cdot MB^2=2\left(a^2+b^2\right)-e^2 & (1)\\\\ \triangle ADC\ : & 4\cdot MD^2 =2\cdot\left(DA^2+DC^2\right)-AC^2 & \iff & 4\cdot MD^2=2\left(d^2+c^2\right)-e^2 & (2)\end{array}\right\|$ si in triunghiul $BMD\ :\ \boxed{4\cdot MN^2 =2\cdot\left(MB^2+MD^2\right)-f^2}\ (3)\ .$

Adunam relatiile $(1)$ si $(2)$ la relatia $(3)$ inmultita cu doi $:\ 8\cdot MN^2=2\left(a^2+b^2\right)+2\left(c^2+d^2\right)-2\left(e^2+f^2\right)\ ,$ adica $\boxed{e^2+f^2+4\cdot MN^2=a^2+b^2+c^2+d^2}\ (*)\ .$

Observatie. In particular, daca un varf al tetraedrului este "tras" in planul fetei opuse se obtin trei identitati pentru un patrulater oarecare din plan. Astfel, daca

$ABCD$ este un patrulater convex cu mijloacele $(E,F,X,Y,Z,T)$ ale diagonalelor $[AC]\ ,$ $[BD]$ si ale laturilor $[AB]\ ,$ $[BC]\ ,$ $[CD]\ ,$ $[DA]$ respectiv,

atunci exita relatiile $:\ \left\{\begin{array}{cccc} AC^2+BD^2+4\cdot EF^2 & = & AB^2+BC^2+CD^2+DA^2 & (1)\\\\ AB^2+CD^2+4\cdot XZ^2 & = & BC^2+AD^2+AC^2+BD^2 & (2)\\\\ AD^2+BC^2+4\cdot YT^2 & = & AB^2+CD^2+AC^2+BD^2 & (3)\end{array}\right\|\ .$ Daca $ABCD$ este un trapez cu $AB\parallel CD\ ,$

atunci exista relatia $\boxed{e^2+f^2=2ac+b^2+d^2}\ (*)\ .$ Daca $ABCD$ este un paralelogram, atunci exista relatia $\boxed{e^2+f^2=2\left(a^2+b^2\right)}\ (**)\ .$

P2. I"ll use the notations from the previous problem. Prove that in any tetrahedron $ABCD$ there are the equivalencies $:\ \left\{\begin{array}{cccccc} a^2+c^2 & = & b^2+d^2 & \iff & AC\perp BD & (1)\\\\ a^2+c^2 & = & e^2+f^2 & \iff & AD\perp BC & (2)\\\\ b^2+d^2 & = & e^2+f^2 & \iff & AB\perp CD & (3)\end{array}\right\|\ .$

Proof. Equivalence $(1)\ .$ Let the midpoints $(M,N,P,R)$ of $([AB]\ ,\ [BC]\ ,\ [CD]\ ,\ [DA])\ .$ Apply the Euler's relations $\left\{\begin{array}{cccc} a^2+c^2+4\cdot MP^2 & = & b^2+d^2+e^2+f^2\\\\ b^2+d^2+4\cdot NR^2 & = & a^2+c^2+e^2+f^2\end{array}\right\|\ .$

Thus, $a^2+c^2=b^2+d^2\iff$ $\left\{\begin{array}{cccc} 4\cdot MP^2 & = & e^2+f^2\\\\ 4\cdot NR^2 & = & e^2+f^2\end{array}\right\|\iff$ $MP=NR\iff$ the parallelogram $MNPR$ is a rectangle $\iff$ $MN\perp MR\iff$ $AC\perp BD\ .$

Application. Ascertain the area of the trapezoid $ABCD\ ,$ where $AB\parallel CD$ and $AB=52\ ,$ $BC=12\ ,$ $CD=39\ ,$ $DA=5\ .$

Proof 1. Notam $E\in AB\ ,\ CE\parallel AD\ .$ Thus, $AE=39\ ,\ EB=13\ ,\ EB=13\ ,\ CE=5\ , BC=12\ ,$ i.e. $BCE$ is $C$ -right triangle and

$AE=3\cdot BE\implies [AECD]=6\cdot [BCE]\ .$ In conclusion, $[BCE]=30$ and $[ABCD]=[BCE]+[AECD]=7\cdot [BCE]=210\ .$

Proof 2. Denote $X\in D\cap BC$ and suppose w.l.o.g. $CD<AB\ .$ I"ll use the standard notations for a convex quadrilateral. Denote $XD=x\ ,$ $XC=y\ .$ Observe that

$\frac {XD}{XA}=\frac {XC}{XB}=$ $\frac {DC}{AB}\iff$ $\frac {x}{x+5}=\frac {y}{y+12}=$ $\frac {39}{52}=\frac 34$ $\odot\begin{array}{ccccc} \nearrow & x=15 & \iff & XA=20 & \searrow\\\\ \searrow & y=36 & \iff & XB=48 & \nearrow\end{array}\odot$ Observe that $XD^2+XC^2=CD^2\ ,$ i.e. $XA\perp XB$ a.s.o.

Proof 3. Calculate $\left\{\begin{array}{ccccccc} e^2+f^2 & = & 2ac+b^2+d^2 & = & (24+1)\cdot \cdot 13^2 & = & 5^2\cdot 13^2\\\\ a^2+c^2 & = & 39^2+52^2 & = & \left(3^2+4^2\right)\cdot 13^2 & = & 5^2\cdot 13^2\end{array}\right\|\implies$ $e^2+f^2=a^2+c^2\iff$ $AD\perp BC$ a.s.o.

P3. Let a plane $\pi\ ,$ $\{B,C\}\subset\pi$ and $A\not\in \pi$ for which denote $P\in\pi$ so that $AP\perp \pi$ and $P\not\in BC\ .$ Let $\left\{\begin{array}{ccccccc} m\left(\widehat{BAC}\right) & = & \alpha & ; & m\left(\widehat{BPC}\right) & = & \phi\\\\ m\left(\widehat{ABP}\right) & = & \beta & ; & m\left(\widehat{ACP}\right) & = & \gamma\end{array}\right\|\ .$

Prove that the relation $\boxed{\cos\alpha =\cos\beta\cos\gamma\cos\phi +\sin\beta\sin\gamma}\ (*)\ .$ Particular case. If $AB\parallel\pi\ ,$ i.e. $\beta =0\ ,$ then the relation $(*)$ becomes $\boxed{\cos\alpha =\cos\gamma\cos\phi}\ .$

Proof. Apply generalized Pythagoras's theorem to $[BC]$ in $:\ \left\{\begin{array}{cccc} \triangle BAC\ : & AB^2+AC^2-2\cdot AB\cdot AC\cdot\cos\alpha & = & BC^2\\\\ \triangle BPC\ : & PB^2+PC^2-2\cdot PB\cdot PC\cdot\cos\phi & = & BC^2\end{array}\right\|$ $\implies$ $AB^2+AC^2-2\cdot AB\cdot AC\cdot\cos\alpha =$

$PB^2+PC^2-2\cdot PB\cdot PC\cdot\cos\phi$ $\iff$ $\left(AB^2-PB^2\right)+\left(AC^2-PC^2\right)-2\cdot AB\cdot AC\cdot\cos\alpha =$ $-2\cdot PB\cdot PC\cdot\cos\phi$ $\iff$ $2\cdot AP^2-2\cdot AB\cdot AC\cdot\cos\alpha =$

$-2\cdot PB\cdot PC\cdot\cos\phi$ $\iff$ $AB\cdot AC\cdot\cos\alpha=$ $AP^2+PB\cdot PC\cdot\cos\phi$ $\iff$ $\cos\alpha =$ $\frac {AP}{AB}\cdot\frac {AP}{AC}+\frac {PB}{AB}\cdot \frac {PC}{AC}\cdot\cos\phi$ $\iff$ $\cos\alpha=\sin\beta\sin\gamma+\cos\beta\cos\gamma\cos\phi\ .$

P4. Let a cube $ABCDA'B'C'D'$ with $AB=1$ , where the base is $[ABCD]$ and the vertical edges are $[AA']\ ,$

$[BB']\ ,\ [CC']$ and $[DD']\ .$ Prove that the distance between the diagonals $AC$ and $D'B$ is equally to $\boxed{\frac {\sqrt 6}6}$ .

Proof 1. Observe that $AC=\sqrt 2$ and $D'B=\sqrt 3$ . Let $I\in AC\cap BD$ and $J\in D'B$ so that $IJ\perp D'B$ . Prove easily that $AC\perp (BD'D)\ ,$

$IJ$ is common perpendicular line of the choice diagonals and $\triangle BIJ\sim\triangle BD'D\ .$ Hence $\frac {IJ}{D'D}=\frac {BI}{BD'}$ $\implies$ $\frac {IJ}1=\frac {\frac {\sqrt 2}2}{\sqrt 3}=\frac 1{\sqrt 6}\implies$ $\boxed{IJ=\frac {\sqrt 6}6}$ .

Remark. Can obtain $IJ$ with the "power" of $B$ w.r.t. the circumcircle of $IDD'J$ which has the diameter $[D'I]\ :\ BI\cdot BD=BJ\cdot BD'\implies$ $\frac {\sqrt 2}2\cdot \sqrt 2=$

$BJ\cdot \sqrt 3\implies$ $BJ=\frac 1{\sqrt 3}\ .$ Using the Pitagora' theorem in $J$ -right $\triangle BIJ$ obtain that $IJ^2=IB^2-BJ^2=\frac 12-\frac 13=\frac 16\implies$ $IJ=\frac {\sqrt 6}6$ .

Proof 2. Let $O\in AC\cap BD$ and the midpoint $M$ of the edge $[DD']$ . Thus, $BD'\parallel MO$ and $MO\subset (AMC)\implies$ $BD'\parallel (AMC)$ . Denote the distance

$d$ between the diagonals $AC\ ,\ BD'$ and now preserve from figure only the tetrahedron $\tau\equiv ABCM$ . Prove easily that $3\cdot V(\tau )=\frac 14$ and $MA=MC=\frac {\sqrt 5}2\ ,$

$AC=\sqrt 2\ ,$ $\delta_{AC}(M)=\frac {\sqrt 3}2\implies$ the area of the face $AMC$ is $S=[AMC]=\frac {\sqrt 6}4$ . The distance $d$ between the choice diagonals is in fact the distance from

the vertex $B$ to the plane $(AMC)$ . In conclusion, $3\cdot V(\tau )=d\cdot S\implies\frac 14=d\cdot \frac {\sqrt 6}4\implies d=\frac 1{\sqrt 6} \implies\boxed{d=\frac {\sqrt 6}6}\ .$

Extension. Let a rectangular parallelipipedon $ABCDA'B'C'D'$ with the base $[ABCD]$ and the vertical edges $[AA']$ , $[BB']$ , $[CC']$ , $[DD']$ so that

$\left\{\begin{array}{ccc} AB & = & a\\BC & = & b\\\ AA' & = & c\end{array}\right\|$ . Prove that the distance $d(c)$ between the diagonals $AC$ si $D'B$ is given by the relation $\boxed{d(c)=\frac {abc}{\sqrt{4a^2b^2+c^2\cdot \left(a^2+b^2\right)}}}\ .$

Proof. Let $O\in AC\cap BD$ and the midpoint $M$ of the edge $[DD']$ . Thus, $BD'\parallel MO$ and $MO\subset (AMC)\implies$ $BD'\parallel (AMC)$ . Now preserve from the figure only

the tetrahedron $\tau\equiv ABCM$ . Observe that $\left\{\begin{array}{ccc} MA^2 & = & b^2+\frac {c^2}4\\\\ MC^2 & = & a^2+\frac {c^2}4\\\\ AC^2 & = & a^2+b^2\end{array}\right\|\ .$ The distance of $M$ to the plane $(ABC)$ is $\delta_{ABC}(M)=\frac c2$ and $\boxed{V(\tau )=\frac {abc}{12}}$ . Let $d=d(c)$

be the distance between choice diagonals $AC$ , $BD'$ is in fact the distance from the vertex $B$ to the face $(AMC)$ . Prove easily that the area $S$ of the face $(AMC)$ is

$\boxed{S=\frac 14\cdot \sqrt{4a^2b^2+c^2\left(a^2+b^2\right)}}\ (*)\ .$ In conclusion, $\boxed{3\cdot V(\tau )=d\cdot S}\implies\frac {abc}4=$ $\frac {d\cdot \sqrt{4a^2b^2+c^2\left(a^2+b^2\right)}}4\implies$ $\boxed{d(c)=\frac {abc}{\sqrt{4a^2b^2+c^2\cdot \left(a^2+b^2\right)}}}\ .$

Remark 1. I"ll prove the relation $(*)$ with the well known identity $16S^2=2\cdot \sum b^2c^2-\sum a^4$ for any $\triangle ABC$ . Indeed, in our case

$16S^2=2\left(b^2+\frac {c^2}4\right)\left(a^2+\frac {c^2}4\right)+2\left(a^2+b^2\right)\left(a^2+b^2+\frac {c^2}2\right)-$ $\left(a^2+b^2\right)^2-\left(a^2+\frac {c^2}4\right)^2-\left(b^2+\frac {c^2}4\right)^2=$

$2a^2b^2+\left[\frac {c^2}2\cdot\left(a^2+b^2\right)+\frac {c^4}8\right]+2\left(a^2+b^2\right)^2+c^2\left(a^2+b^2\right)-\left(a^2+b^2\right)^2-\left(a^4+b^4\right)-\left[\frac {c^2}2\cdot\left(a^2+b^2\right)+\frac {c^4}8\right]=$

$2a^2b^2+\left(a^2+b^2\right)^2+c^2\left(a^2+b^2\right)-\left(a^4+b^4\right)=$ $4a^2b^2+c^2\left(a^2+b^2\right)\implies$ $4S=\sqrt{4a^2b^2+c^2\left(a^2+b^2\right)}$ .

Remark 2. I"ll prove the relation $(*)$ with the remarkable identity (prove easily!) $16S^2=\sum a^2\left(b^2+c^2-a^2\right)$ for any $\triangle ABC$ . Indeed, in our case

$16S^2=2b^2\left(a^2+\frac {c^2}4\right)+2a^2\left(b^2+\frac {c^2}4\right)+\frac {c^2}2\cdot\left(a^2+b^2\right)\implies$ $16S^2=4a^2b^2+c^2\left(a^2+b^2\right)\implies$ $4S=\sqrt{4a^2b^2+c^2\left(a^2+b^2\right)}\ .$

P5 (29th IMO 1988 shortlist). $ABCD$ is a tetrahedron. Show that any plane through the midpoints of $[AB]$ and $[CD]$ divides the tetrahedron into two parts of equal volume.

Proof. Denote the given plane $\pi$ and suppose w.l.o.g. $\{M,N,P,Q,R,S\}\subset\pi$ , where $M$ , $N$ are the midpoints of $[AB]$ , $[CD]$ respectively , $P\in (BC)$ so that $\boxed{\frac {PB}{PC}=p>1}$ ,

i.e. $\boxed{\frac {PB}p=\frac {PC}1=\frac {BC}{p+1}}\ (*)$ , $R\in (AD)$ , $Q\in AC$ so that $C\in (AQ)$ and $S\in BD$ so that $D\in (BS)$ . Observe that $\left\{\begin{array}{ccc} Q & \in & MP\cap NR\cap AC\\\\ S & \in & MR\cap NP\cap BD\end{array}\right\|$ . Apply the Menelaus'

theorem to $:\ \left\{\begin{array}{ccccccc} \overline{MPQ}/\triangle ABC\ : & \frac {QC}{QA}\cdot\frac {MA}{MB}\cdot\frac {PB}{PC}=1 & \implies & \frac {QA}{QC}=p & \implies & \frac {QA}p=\frac {QC}1=\frac {AC}{p-1} & (1)\\\\ \overline{RNQ}/\triangle ACD\ : & \frac {QA}{QC}\cdot\frac {NC}{ND}\cdot\frac {RD}{RA}=1 & \implies & \frac {RA}{RD}=p & \implies & \frac {RA}p=\frac {RD}1=\frac {AD}{p+1} & (2)\end{array}\right\|$ . Let $\delta_{XYZ}(P)$ be the distance of $P$ to the plane $XYZ\ ;$

let $v_1$ , $v_2$ be the volumes of the pyramids $QAMR$ , $QCPN$ . Therefore, $\left\{\begin{array}{ccccc} \frac {v_1}{[ABCD]}=\frac {[AMR]\cdot \delta_{ABD}(Q)}{[ABD]\cdot\delta_{ABD}(C)}=\frac {AM}{AB}\cdot\frac {AR}{AD}\cdot\frac {QA}{CA} & \stackrel{1\wedge 2}{=} & \frac 12\cdot\frac p{p+1}\cdot\frac p{p-1} & \implies & v_1=\frac {p^2}{2\left(p^2-1\right)}\cdot [ABCD]\\\\ \frac {v_2}{[ABCD]}=\frac {[CPN]\cdot \delta_{BCD}(Q)}{[CBD]\cdot\delta_{BCD}(A)}=\frac {CP}{CB}\cdot\frac {CN}{CD}\cdot\frac {QC}{AC} & \stackrel{1\wedge 2}{=} & \frac 1{p+1}\cdot\frac 12\cdot \frac 1{p-1} & \implies & v_2=\frac {1}{2\left(p^2-1\right)}\cdot [ABCD]\end{array}\right\|$

In conclusion, the required volume is $v=v_1-v_2$ , where $\frac {v}{[ABCD]}=\frac {p^2}{2\left(p^2-1\right)}- \frac {1}{2\left(p^2-1\right)}=\frac 12$ , i.e. $[ABCD]=2\cdot v$ . The case $p<1$ is similarly with upper case. Suppose

now $p=1$ , i.e. $MPNR$ is a parallelogram. Let the midpoint $X$ of $[AC]$ . In this case $\frac v{[ABCD]}=\frac {v(PCNMXR)+v(MXRA)}{[ABCD]}=\frac 38+\frac 18=\frac 12\implies$ $[ABCD]=2\cdot v\ .$

This post has been edited 120 times. Last edited by Virgil Nicula, Jun 18, 2016, 6:09 PM

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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
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by OlympusHero, Sep 14, 2022, 4:44 AM
blog
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the visits tho
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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

435. Geometry problems in space.

by Virgil Nicula, Dec 4, 2015, 12:13 PM

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