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RMM

Romanian Mathematical Magazine


Here are gonna be collected all the Problem Collections and the Marathons from the online magazine ''Romanian Mathematical Magazine''.

collected inside aops here



Geometry Problems till year 2023 
(selected from problem column - not 2D geometric inequalities)


Juniors

JP 057 Let ABC be an arbitrary triangle and I_a, I_b, I_c are excenters of ABC. I_aBC, I_bCA, I_cAB are the extriangles of ABC. Let h_i (i = 1, 2,3,...,9) the altitudes of extriangles. Prove that \prod _{i=1}^9 h_i =\left(\prod _{a,b,c}r_a \right)^3
by Mehmet Sahin - Ankara - Turkey

JP 150 Let be z_1,z_2, z_3 \in C^* different in pairs such that |z_1| = |z_2| = |z_3|. If (z_1 + z_2)(z_2 + z_3)(z_3 + z_1) + z_1z_2z_3 = 0, then z_1,z_2,z_3 are the affixes of an equilateral triangle.

by Marian Ursarescu - Romania

JP 156 Let ABC be a triangle having the area S. Let be A' \in  (BC) such that the incircles of \vartriangle AA'B, \vartriangle AA'C have the same radius. Analogous, we obtain the points B' \in (AC), C' \in (AB). Prove that: S =\frac{AA' \cdot BB' \cdot CC' }{s} where s is the semiperimeter of \vartriangle ABC.
by Marian Ursarescu - Romania

JP 167 Let OABC be a tetrahedron with \angle AOB = \angle BOC = \angle COA = 90^o and let P be any point inside the triangle ABC. Denote respectively by d_a, d_b, d_c the distances from P to faces (OBC), (OCA), (OAB). Prove that:

(a) d^2_a + d^2_b+ d^2_c= OP^2.

(b) d_ad_bd_c \le \frac{OA \cdot OB\cdot OC}{27} .

(c) OA \cdot  d^3_a + OB \cdot d^3_b+ OC \cdot d^3_c \ge OP^4.

by Nguyen Viet Hung - Hanoi - Vietnama

JP 194 In \vartriangle ABC, BE, CF are internal bisectors, E \in (AC), F \in (AB),O is circumcentre. Prove that: E,O, F collinear \Leftrightarrow  \cos A =\ cos B +\ cos C

by Marian Ursarescu - Romania

JP 199 Let SABCD be a pyramid with the base ABCD parallelogram and E any point which belongs to the side SC such that \frac{SE}{SC} = k. Through the vertex A and the point E we consider a variable plane which intersects the segment SB in M and the segment SD in N. Prove that \frac{V_{SAEMN}}{V_{SABCD}}\ge \frac{2k^2}{k + 1}

by Marian Ursarescu - Romania

JP 260 In \vartriangle ABC, N - Nagel’s point, BQ, CP - symedians. Prove that 
P, N, Q - collinear \Leftrightarrow \frac{1}{b^2r_b}+\frac{1}{c^2r_c}=\frac{1}{a^2r_a}

by Marian Ursarescu - Romania

JP 309 If m \in N, h_A, h_B, h_C, h_D are the lengths of heights of a tetrahedron [ABCD] having the radius of the inscribed sphere r, then
m+ \frac14 \left( \left( \frac{h_A - 3r}{h_A + 3r}\right)^{m+1}+\left(\frac{h_B - 3r}{h_B + 3r}\right)^{m+1} +\left( \frac{h_C - 3r}{h_C + 3r}\right)^{m+1}+\left( \frac{h_D - 3r}{h_D + 3r}\right)^{m+1} \right) \ge \frac{m+1}{7}

by D.M. Batinetu - Giurgiu, Daniel Sitaru - Romania


JP 338 In \vartriangle ABC, P,Q \in Int \,\, (\vartriangle ABC), \alpha, \beta, \gamma \in R,\alpha, \gamma \ne 1 such that \beta \overline{AB}+\gamma \overline{BP}+ \overline{PC} = \overline{0} and
\overline{AQ}+\alpha \overline{QB}+\overline{BC} = \overline{0}. Prove that A,P,Q are collinear if and only if \alpha+ \gamma = \beta +1

by Florica Anastase - Romania


JP 342 Let ABCDA'B'C'D' be a cube with length side 1 and M \in BC,N  \in DD', P  \in A'B'. Find minimum perimeter of \vartriangle MNP.

by Florentin Visescu - Romania

JP 352 If a, b, c \in C, |a| = |b| = |c| = 1 then 3|a + b + c| + 2(|a - b| + |b - c| + |c - a|) \ge 9 

by Daniel Sitaru - Romania

JP 353 In \vartriangle ABC, P \in  Int (\vartriangle ABC), \angle ABP  = 20^o, \angle PBC =\angle  PCB  = 10^o, \angle PCA  = 40^o. Prove that |AP|+ |BC| =\sqrt3 |AB|.

by Mehmet Sahin - Turkey
JP 354 In acute \vartriangle ABC,O - circumcenter, F,K \in (AB), M, L \in (BC), E,N \in  (CA),FOE, MON, LOK - are the antiparallels. Let \rho_a , \rho_b , \rho_c - inradii of \vartriangle AFE, \vartriangle BLK, \vartriangle CMN. Prove that \rho_a+\rho_b+\rho_c = R.

by Mehmet Sahin - Turkey

JP 357 In \vartriangle ABC, prove that inscribed circle of \vartriangle ABC passes through Nagel's point N_a if and only if s^2 +4r^2 = 16Rr.
by Marian Ursarescu - Romania

JP 367 Let a, b, c \in C^* be different in pairs, A(a),B(b),C(c), |a| = |b| = |c| = 1. If (ab)^3 + (bc)^3 + (ca)^3 = 3(abc)^2 then \vartriangle ABC is equilateral.

by Marian Ursarescu - Romania

JP 379 If ABCD tetrahedron AB = a, AD = b , AC = c ,BD = d , DC = e , CB = f , F - total area, then a^4 + b^4 + c^4 + d^4 + e^4 + f^4 \ge 2F^2

by D.M. Batinetu-Giurgiu, Daniel Sitaru - Romania

JP 389 A right parallelepiped ABCDA'B'C'D' has the basis ABCD rhombus, and areas of the two diagonals sections of the parallelepiped are F_1 and F_2 respectively. Let R_1 be the circumradius of \vartriangle ABC, R_2 circumradius of \vartriangle ABD and V volume of the right parallelepiped . Prove that R_1R_2F_1F_2 \ge V^2.
by Radu Diaconu - Romania

JP 391 In \vartriangle ABC, P - inner point, M, L \in [AB], D,E \in [BC], F,K \in [CA], AM = AF, BL = BE , CK = CD, |DE| = a_1, |FK| = b_1, |LM| = c_1, (M,P,F), (C, P, L), (D, P, K) - are collinear. Prove that F =\frac12 (a_1r_a + b_1r_b + c_1r_c)

by Mehmet Sahin - Turkey

JP 392 Let z_1, z_2, z_3 \in C^* be different in pairs such that |z_1| = |z_2| = |z_3|=1, A(z_1),B(z_2),C(z_3). Prove that, if \sum_{cyc}\frac{z_2z_3}{14z_2z_3 - z_2^2 - z_3^2}=\frac15 then AB = BC = CA.

by Marian Ursarescu - Romania


Seniors


SP 122 If z_1, z_2, z_3 \in C are different in pairs and |z_1| = |z_2| = |z_3| = 1 then |z_1 - z_3| + |z_2 - z_3| \le  3 + |z_1 + z_2|

by Marian Ursarescu - Romania

SP 235 Let be A(z_1), B(z_1), C(z_3), z_1, z_2, z_3 \in C-\{0\}, |z_1| = |z_2| = |z_3|, AB = c, BC = a, CA = b. If (b+c)z_Bz_C+(c+a)z_Cz_A+(a+b)z_Az_B = 0 then AB = BC = CA.

by Marian Ursarescu - Romania

SP 246 If ABCD bicentric quadrilateral, I incenter then:
(IA^2 + IC^2)(IB^2 + ID^2) \ge  AB \cdot BC \cdot CD \cdot  DA
by Daniel Sitaru - Romania


SP 250/ 323 (250) Let z_1, z_2, z_3 \in C-\{0\} be different in pairs, |z_1| = |z_2| = |z_3|=1, A(z_1), B(z_1), C(z_3). If |z_1 - z_2 - z_3| +|z_2 - z_1 - z_3| +|z_3 - z_2 - z_1| = 6 then AB = BC = CA.

reposted and rephrased as
(323) Let z_A, z_B, z_V \in C^* be different in pairs, |z_A| = |z_B| = |z_B|=1. If |z_A - z_B - z_C| +|z_B - z_C - z_A| +|z_C - z_A - z_B| = 6 then \vartriangle ABC is an equilateral triangle.


by Marian Ursarescu - Romania

SP 259 In \vartriangle ABC, \Gamma - Gergonne’s point and BN, CM symedians, M \in (AB), N \in (AC). Prove that B, \Gamma, N - collinear \Leftrightarrow \frac{r_b}{b^2}+\frac{r_c}{c^2} =\frac{r_a}{a^2}

by Marian Ursarescu - Romania

SP 310 In \vartriangle ABC, B' \in (AC) the contact point of the external circumscription circle of side AC and C' the contact point of the external circumscription circle of side AB. Prove that B'C' is tangent of the inscribed circle in ABC if and only if (s - b)^2 + (s - c)^2 = (s - a)^2

by Marian Ursarescu - Romania

My note: B' is touchpoint of B-excircle with AC, C' is touchpoint of C-excircle with AB, s is the semiperimeter

SP 348 In \vartriangle ABC prove that inscribed circle of \vartriangle ABC passes through the centroid G if and only if s^2 = 16Rr + 4r^2.

by Marian Ursarescu - Romania

SP 360 Let z_1, z_2, z_3 \in C^* be different in pairs such that |z_1| = |z_2| = |z_3|. If
\sum_{cyc}\frac{2z_1 - z_2 - z_3}{(z_1 - z_2)|z_1 - z_3| + (z_1 - z_3)|z_1 - z_2|}=\frac{1}{|z_1 - z_2|}+\frac{1}{|z_2 - z_3|}+\frac{1}{|z_3 - z_1|}, then z_1, z_2,z_3 are affixes on equilateral triangle.
by Marian Ursarescu - Romania

SP 371 Let ABCD be a tetrahedron, and let M be a point in space, M \not\in  \{A,B,C\}. Prove that \frac{MA}{MB +MC+MD}+\frac{MB}{MC +MD +MA}+\frac{MC}{MD +MA +MB}+
+\frac{MD}{MA +MB +MC} \ge \frac{R + r}{R} \ge \frac{4r}{R}

by D.M. Batinetu - Giurgiu, Neculai Stanciu - Romania

SP 382 Let z_1, z_2, z_3 \in C^* be different in pairs such that |z_1| = |z_2| = |z_3|=1, A(z_1),B(z_2),C(z_3). Prove that, if \sum_{cyc}\frac{z_2z_3}{(z_2 - z_3)^2[z_2(z_1 - z_3)^2 - z_3(z_1 + z_2)^2]}=\frac{1}{4z_1z_2z_3} then AB = BC = CA.

by Marian Ursarescu - Romania

SP 410 Let z_1, z_2, z_3 \in C^* be different in pairs such that |z_1| = |z_2| = |z_3|=1, A(z_1),B(z_2),C(z_3). Prove that:
\sum_{cyc}|2z_1-z_2-z_3|^4=243 \Rightarrow AB = BC = CA.

by Marian Ursarescu - Romania

SP 411 Let z_1, z_2, z_3 \in C^* be different in pairs such that |z_1| = |z_2| = |z_3|=1, A(z_1),B(z_2),C(z_3). Prove that:
\sum_{cyc}\frac{1}{8z_1z_2z_3 - (z_1^2+z_2z_3)(z_2+z_3)}=\frac{3}{10z_1z_2z_3} \Rightarrow AB = BC = CA.

by Marian Ursarescu - Romania

SP 423 Let z_1, z_2, z_3 \in C^* be different in pairs such that |z_1| = |z_2| = |z_3|=1, A(z_1),B(z_2),C(z_3). Prove that:
\sum_{cyc}\frac{z_2z_3}{3z_2z_3 - z_2^2 - z_3^2}=\frac34 \Leftrightarrow AB = BC = CA.

by Marian Ursarescu - Romania


SP 451 If ABCD is a convex quadrilateral such that AC \cap BD =\{O\}, AE = EC, BF = FD with order A - O - E - C respectively , B -F -O - D, EF \cap AB = \{J\}, EF \cap CD = \{K\}, CJ \cap BK = \{L\} and M the midpoint of KJ, then prove that O, M and L are collinear.

by Marius Dragan, Neculai Stanciu - Romania

SP 462 Let ABC be an triangle, D be a point on side BC and M be the symmetrical of A with respect to D. If \frac{BM^2}{AB} +\frac{CM^2}{AC} = AB+AC, then prove that AD is the bisector of the angle \angle A, or is the altitude from the vertex A.

by Neculai Stanciu - Romania


Undergraduate 

UP 056 Let ABC be a triangle and \Omega the first Brocard point of ABC. Let D,E, F are on the sides BC, CA, AB of ABC respectively. If \angle B \Omega D  =\angle  C \Omega E=\angle A \Omega F  = 90^o then prove that \frac{BD}{BC} + \frac{CE}{CA} + \frac{AF}{AB} = 2

by Mehmet Sahin - Ankara - Turkey

UP 063 Let SABC be a tetrahedron and let M be any point  inside the triangle ABC. The lines through M parallel with the planes SBC, SCA,SAB intersect SA,SB,SC at X,Y,Z, respectively.  Prove that:  Vol (MXY Z) \le \frac{2}{27} Vol (SABC).
Determine position of the point M such that the equality holds.

by Nguyen Viet Hung - Hanoi - Vietnam
UP 162 If ABCD is tetrahedron AB = a_1, AC = a_2, AD = a_3, BC = a_4, BD = a_5 ,CD = a_6 then \sum_{1\le i <j \le 6}(a_i + a_j)^2 \ge 4\sqrt3 S[ABCD] where S[ABCD] is total area of tetrahedron ABCD.

by Daniel Sitaru - Romania

UP 203 Given a triangle ABC with incenter I. The lines AI,BI,CI meet the sides BC,CA,AB at A',B',C' and meet the circumcircle at the second points A_1,B_1,C_1 respectively. Prove that:
(a) \frac{AI}{AA'} + \frac{BI}{BB'} + \frac{CI}{CC'} = 2,

(b) \frac{A_1I}{AI} + \frac{B_1I}{BI }+\frac{C_1I}{CI} = \frac{2R}{r} - 1.

by Nguyen Viet Hung - Hanoi - Vietnam

UP 294 In \vartriangle ABC, AD,BE,CF -medians, G centroid, AM = MG , M \in  (AG), and 2 \cot A = \cot B + \cot C. Prove that DEMF is a cyclic quadrilateral.

by Marian Ursarescu - Romania

UP 321 Let A_0A_1...A_n be an Euclidean n-simplex. We will use the following notations:
- O, V,R, r the centre if its circumscribed hypersphere, its volume, its circumradius and its inradius, respectively.
- O_i,R_i the centre and the radius of the hypersphere tangent to the circumscribed sphere of A_0,A_1,...,A_n in the vertex A_i and to the hyperplane A_0A_1...A_{i-1}A_{i+1}...A_n simultaneously.
With the above notations, the following identity holds: \sum_{i=0}^{n} \frac{1}{R_i}= \frac{n}{R} +\frac{1}{r}

by Vasile Jiglau - Romania
UP415 Let ABC denote a triangle and H its orthocenter. Let point M be the midpoint of the segment AH. Prove that:
(a) angle \angle BMC is acute.
(b) area \vartriangle BMC = 1/8 \cdot  AH^2 \cdot \tan \angle BMC.

by George Apostolopoulos - Greece

Geometry articles a selection
(to be updated)

Gakopoulos articles  (soon more)
Collected books
                             (to be added)


Marathons (google drive folder here)

Abstract Algebra Marathon Google Drive
Calculus Marathon Google Drive
Cyclic Inequalities Marathon Google Drive
Geometry Marathon Google Drive
Inequalities Marathon Google Drive
Math Adventures On CutTheKnot Google Drive
Triangle Marathon  Google Drive

Problem Column 2016-2023

2016: Problems & Solutions  problems 001-045
2017: Problems & Solutions  problems 046-105                                         
2018: Problems & Solutions  problems 106-165
2019: Problems & Solutions  problems 166-225
2020: Problems & Solutions  problems 226-285
2021: Problems & Solutions  problems 286-345
2022: Problems & Solutions  problems 346-405
2023: Problems & Solutions  problems 406-465



sources:
www.cut-the-knot.org  (Alexander Bogomolny)
www.ssmrmh.ro   (RMM = )



https://www.ssmrmh.ro/

4 comments:

  1. Τάκη,
    εξαιρετική ανάρτηση, σε ευχαριστούμε πολύ.

    ReplyDelete
  2. Υπέροχη προσπάθεια και εξαιρετικό υλικό.Τάκη σε ευχαριστούμε πολύ!

    ReplyDelete
  3. They released more of the calculus and triangle marathons

    ReplyDelete