188. Some problems from vectorial geometry.

by Virgil Nicula, Dec 10, 2010, 8:02 PM

PP0. Let $I$, $O$ be the incenter and the circumcenter of $\triangle ABC$ . Express $\overrightarrow{OI}$ with $\overrightarrow{OA}\ ,\ \overrightarrow{OB}\ ,\ \overrightarrow{OC}$ and $a\ ,\ b\ ,\ c$ .

Proof. $a\cdot \overrightarrow{XA}+b\cdot \overrightarrow{XB}+c\cdot \overrightarrow{XC}=(a+b+c)\cdot\overrightarrow{XI}$ for any $X\in (ABC)$ - plane.

In the particular case $X:=O$ obtain $\boxed {a\cdot \overrightarrow{OA}+b\cdot\overrightarrow{OB}+c\cdot \overrightarrow{OC}=(a+b+c)\cdot \overrightarrow{OI}}$ .


PP1 (own). Let $ABC$ be a triangle with the centroid $G$ and the Lemoine's point $S$ . Prove that there is the relation

$\left(b^2-a^2\right)\cdot\overrightarrow{AB}+\left(c^2-b^2\right)\cdot\overrightarrow{BC}+\left(a^2-c^2\right)\cdot\overrightarrow{CA}=$ $3\cdot\left(a^2+b^2+c^2\right)\cdot \overrightarrow {GS}$ .

Proof. I"ll use the barycentrical coordinates w.r.t. the triangle $ABC$ . Therefore, $G(1,1,1)$ and $S\left(a^2,b^2,c^2\right)$ . From the remarkable identity

$\sum a^2\cdot \overrightarrow {XA}=\left(\sum a^2\right)\cdot\overrightarrow{XS}$ for $X:=G$ obtain that $\boxed{\sum a^2\cdot \overrightarrow {GA}=\left(\sum a^2\right)\cdot\overrightarrow{GS}}\ (1)$ and from the remarkable identity

$\sum \overrightarrow {XA}=$ $3\cdot\overrightarrow{XG}$ for $X:=S$ obtain that $\boxed {\sum\overrightarrow {SA}=3\cdot\overrightarrow{SG}}\ (2)$ . Observe that $3\cdot \overrightarrow{GA}=$ $\overrightarrow {BA}+\overrightarrow{CA}$ a.s.o. Using the relations

$(1)$ and $(2)$ obtain that $\sum a^2\cdot \left(\overrightarrow {BA}+\overrightarrow{CA}\right)=3\cdot\left(\sum a^2\right)\cdot \overrightarrow {GS}$ , i.e. $\sum \left(b^2-a^2\right)\cdot\overrightarrow {AB}=3\cdot\left(\sum a^2\right)\cdot\overrightarrow {GS}$ .


PP2 (own). Let $ABC$ be a triangle with the centroid $G$ , the incenter $I$ and the Lemoine's point $S$ .

Prove that exists the relation $(a-b)(a-c)\cdot\overrightarrow{IA}+\left(a^2+b^2+c^2\right)\cdot \overrightarrow{SI}+3bc\cdot\overrightarrow{GI}=\overrightarrow 0$ .

Proof. From the remarkable identities obtain that $\left\|\begin{array}{c} \sum \overrightarrow{IA}=3\cdot\overrightarrow {IG}\\\\ \sum a^2\cdot\overrightarrow {IA}=\left(\sum a^2\right)\cdot {IS}\end{array}\right\|$ . Thus, $(a-b)(a-c)\cdot\overrightarrow{IA}+$ $\left(\sum a^2\right)\cdot \overrightarrow{SI}+3bc\cdot\overrightarrow{GI}=$

$\left(a^2-ac-ab+bc\right)\cdot\overrightarrow {IA}-$ $\left(\sum a^2\cdot\overrightarrow {IA}\right)+$ $bc\cdot\left(\sum\overrightarrow{AI}\right)=$ $\sum a(b+c)\cdot\overrightarrow{AI}=$ $-(b+c)\cdot\left(\sum a\cdot\overrightarrow{IA}\right)=\overrightarrow 0$ .


PP3 (own). Let $ABC$ be a triangle with the centroid $G$ and the incenter $I$ . Define the vectors $v_1=a\cdot \overrightarrow{GA}+b\cdot \overrightarrow{GB}+c\cdot \overrightarrow{GC}$

and $v_2=\overrightarrow {IA}+\overrightarrow {IB}+\overrightarrow {IC}$ . Prove that $v_1$ and $v_2$ are colinearly and $3\cdot v_1+2s\cdot v_2=\overrightarrow 0$ , where $2s=a+b+c$ .

Proof. Since $v_1\equiv \sum a\cdot \overrightarrow {GA}=2s\cdot\overrightarrow{GI}$ and $v_2\equiv \sum \overrightarrow{IA}=3\cdot\overrightarrow{IG}$ obtain $3\cdot v_1+2s\cdot v_2=\overrightarrow 0$ , i.e. $v_1$ and $v_2$ are colinearly and $3\cdot|v_1|=2s\cdot |v_2|$ .


PP4 (C. Cosnita). Let $w$ be a circle which pass through the centroid $G$ of the triangle $ABC$ . Denote the

power $p_w(X)$ of the point $X$ w.r.t. the circle $w$ . Prove that $p_w(A)+p_w(B)+p_w(C)=3\cdot p_w(G)$ (constant).

Proof. Let $w=C(P,\rho )$ be a circle for which $G\in w$ , i.e. $PG=\rho$ . Observe that $PA^2+PB^2+PC^2=3\cdot PG^2+3\cdot p_w(G)$ .

Therefore, $\sum p_w(A)=\sum\left(PA^2-\rho^2\right)=$ $\sum PA^2-3\rho^2=$ $3\cdot PG^2+3\cdot p_w(G)-3\rho^2=$ $3\cdot p_w(G)=\frac 13\cdot\sum a^2$ .


PP5. Let $ABC$ be a triangle. For a point $M$ define the points $\left\|\begin{array}{ccc} \{D_1,D_2\}\subset BC & , & D_1D_2\parallel BC\\\\ \{E_1,E_2\}\subset CA & , & E_1E_2\parallel CA\\\\ \{F_1,F_2\}\subset AB & , & F_1F_2\parallel AB\end{array}\right\|$ and $\{M\}=D_1D_2\cap E_1E_2\cap F_1F_2$ .

Prove that the power of $M$ w.r.t. the circumcircle $w$ of $\triangle ABC$ is given by the relation $p_w(M)=\overrightarrow {MD_1}\cdot \overrightarrow{MD_2}+\overrightarrow {ME_1}\cdot \overrightarrow{ME_2}+\overrightarrow {MF_1}\cdot \overrightarrow{MF_2}$ .

Proof. Let $M(x,y,z)$ and denote $A_1\in AM\cap BC$ . Suppose w.l.o.g. $D_1\in AB$ and $D_2\in AC$ . From the relations $\frac {D_1M}{BA_1}=\frac {AM}{AA_1}=\frac {D_2M}{CA_1}$ obtain

that $\overrightarrow{D_1M}\cdot\overrightarrow{D_2M}=\left(\frac {AM}{AA_1}\right)^2\cdot\overrightarrow{A_1B}\cdot\overrightarrow{A_1C}$ . Since $x\cdot \overrightarrow{MA}+(y+z)\cdot\overrightarrow{AA_1}=\overrightarrow 0$ obtain that $\left(\frac {MA}{AA_1}\right)^2=(y+z)^2$ and from the relation

$y\cdot\overrightarrow{A_1B}+z\cdot\overrightarrow{A_1C}=\overrightarrow 0$ obtain that $\overrightarrow {A_1B}=-\frac {z}{y+z}\cdot\overrightarrow{BC}$ and $\overrightarrow{A_1C}=-\frac {y}{y+z}\cdot\overrightarrow{BC}$ . Therefore, $\overrightarrow{MD_1}\cdot\overrightarrow{MD_2}=-yza^2$ . Show

analogously that $\overrightarrow{ME_1}\cdot\overrightarrow{ME_2}=-zxb^2$ and $\overrightarrow{MF_1}\cdot\overrightarrow{MF_2}=-xyc^2$ . In conclusion, $\sum \overrightarrow{MD_1}\cdot\overrightarrow{MD_2}=-\sum yza^2=p_w(M)$ .


PP6 (own). Let $ABC$ be a triangle. Consider three given real numbers $\{\alpha , \beta ,\gamma\}$ so that $\alpha +\beta +\gamma \ne 0$ . Ascertain the

locus of a mobile point $M$ for which exists the relation $\alpha\cdot \overrightarrow {MB}\cdot\overrightarrow{MC}+\beta\cdot \overrightarrow {MC}\cdot\overrightarrow{MA}+\gamma\cdot \overrightarrow {MA}\cdot\overrightarrow{MB}=k$ (constant).

Proof. Suppose w.l.o.g. that $\alpha +\beta +\gamma=1$ . From the generalized Pytagoras' theorem obtain that $MA^2+MB^2-c^2=2\cdot\overrightarrow{MA}\cdot\overrightarrow{MB}$ a.s.o.

and $\sum\alpha\cdot \overrightarrow{MB}\cdot\overrightarrow{MC}=$ $\frac 12\cdot\sum \alpha\cdot\left(MB^2+MC^2-a^2\right)=$ $\frac 12\sum (\beta+\gamma )\cdot MA^2-\frac 12\cdot\left(\sum\alpha a^2\right)$ . Consider $\boxed {\ F\left(\frac {\beta +\gamma}{2}, \frac {\gamma +\alpha}{2},\frac {\alpha +\beta}{2}\right)\ }$ .

For any point $M$ exist the relations $\frac 12\cdot\sum (\beta +\alpha )\cdot \overrightarrow{MA}=\overrightarrow {MF}$ and $\frac 12\cdot\sum (\beta +\gamma )\cdot MA^2=MF^2-p_w(F)$ , where $p_w(F)$ is the power of $F$

w.r.t. the circumcircle $w$ of $\triangle ABC$ . In conclusion, $MF$ is constant , i.e. the required locus is a circle with the center in the fixed point $F$ which was defined upper.


PP7. Let $\triangle ABC$ and three lines $d_k\ ,\ k\in\overline{1,3}$ for what denote $\left\{\begin{array}{cc} d_1\ : & \left|\begin{array}{cc} M\in AB\cap d_1\ ; & \overline{MB}=m\cdot\overline{MA}\\\ N\in AC\cap d_1\ ; &\overline{NC}=n\cdot \overline{NA}\end{array}\right|\\\\ d_2\ : & \left|\begin{array}{cc} P\in BC\cap d_2\ ; & \overline{PC}=p\cdot\overline{PB}\\\ Q\in BA\cap d_2\ ; &\overline{QA}=q\cdot \overline{QB}\end{array}\right|\\\\ d_3\ : & \left|\begin{array}{cc} R\in CA\cap d_3\ ; & \overline{RA}=r\cdot\overline{RC}\\\ S\in CB\cap d_3\ ; &\overline{SB}=s\cdot \overline{SC}\end{array}\right|\end{array}\right|$ .

Prove that $\boxed{d_1\cap d_2\cap d_3\ne\emptyset\ \iff\ 1+mpr+nqs=mq+ps+rn}$ .

Proof. Denote $\overline {OX}=X$ , where $O$ is the origin of the vectorial system and $\overline{XY}=Y-X$ . Thus, $\left\{\begin{array}{cc} (1-m)M=B-mA\ ; & (1-n)N=C-nA\\\ (1-p)P=C-pB\ ; & (1-q)Q=A-qB\\\ (1-r)R=A-rC\ ; & (1-s)S=B-sC\end{array}\right\|$ . Denote $U\in AC\cap d_2$

and $V\in AB\cap d_3$ , $\left\{\begin{array}{c} Y\in d_1\cap d_3=MN\cap SR\\\ Z\in d_1\cap d_2=MN\cap PQ\end{array}\right\|$ and I"ll show $Y=Z$ , i.e. $\frac {\overline{YM}}{\overline{YN}}=\frac {\overline{ZM}}{\overline{ZN}}$ . Thus $\underline{V\in RS}\iff$ $\frac {VB}{VA}\cdot\frac {RA}{RC}\cdot\frac {SC}{SB}=1\implies$ $\boxed{\frac {VB}{VA}=\frac sr}\ (1)$ . Since

$R=\frac {A-rC}{1-r}$ and $N=\frac {C-nA}{1-n}$ obtain $\frac {RA}{RN}=\frac {A-R}{N-R}=$ $\frac {A-\frac {A-rC}{1-r}}{\frac {C-nA}{1-n}-\frac {A-rC}{1-r}}=\frac {r(1-n)}{1-nr}\implies$ $\boxed{\frac {RA}{RN}=\frac {r(1-n)}{1-nr}}\ (2)$ . From $(1)$ obtain $V=\frac {sA-rB}{s-r}$ and

$M=\frac {B-mA}{1-m}\implies$ $\frac {VM}{VA}=\frac {M-V}{A-V}=$ $\frac {\frac {B-mA}{1-m}-\frac {sA-rB}{s-r}}{A-\frac {sA-rB}{s-r}}=\frac {s-mr}{r(1-m)}\implies$ $\boxed{\frac {VM}{VA}=\frac {s-mr}{r(1-m)}}\ (3)$ . Hence $V\in YR\iff$ $\frac {YM}{YN}\cdot\frac {RN}{RA}\cdot\frac {VA}{VM}=1\implies$

$\frac {YM}{YN}=\frac {RA}{RN}\cdot\frac{VM}{VA}$ . From $(2)$ and $(3)$ obtain that $\boxed{\frac {YM}{YN}=\frac {(1-n)(s-mr)}{(1-m)(1-nr)}}\ (4)$ . I"ll proceed analogously to assess the ratio $\frac {ZM}{ZN}$ . Thus, $\underline{U\in PQ}\iff$

$\frac {UA}{UC}\cdot\frac {PC}{PB}\cdot\frac {QB}{QA}=1\implies$ $\boxed{\frac {UA}{UC}=\frac qp}\ (5)$ and $U=\frac {pA-qC}{p-q}$ . Since $M=\frac {B-mA}{1-m}$ and $Q=\frac {A-qB}{1-q}$ obtain that $\frac {QM}{QA}=$ $\frac {M-Q}{A-Q}=$ $\frac {\frac {B-mA}{1-m}-\frac {A-qB}{1-q}}{A-\frac {A-qB}{1-q}}=$ $\frac {1-mq}{q(1-m)}\implies$

$\boxed{\frac {QM}{QA}=\frac {1-mq}{q(1-m)}}\ (6)$ . Since $N=\frac {C-nA}{1-n}$ and $U=\frac {pA-qC}{p-q}$ obtain that $\frac {UA}{UN}=\frac {A-U}{N-U}=$ $\frac {A-\frac {pA-qC}{p-q}}{\frac {C-nA}{1-n}-\frac {pA-qC}{p-q}}=\frac {q(1-n)}{p-nq}\implies$ $\boxed{\frac {UA}{UN}=\frac {q(1-n)}{p-nq}}\ (7)$ . Thus,

$\underline{U\in QZ}\iff$ $\frac {ZM}{ZN}\cdot\frac {UN}{UA}\cdot\frac {QA}{QM}=1\implies$ $\frac {ZM}{ZN}=\frac {QM}{QA}\cdot\frac {UA}{UN}$ . From $(6)$ and $(7)$ obtain $\boxed{\frac {ZM}{ZN}=\frac {(1-n)(1-mq)}{(1-m)(p-nq)}}\ (8)$ . In conclusion, $(4)$ and $(8)$ express that

$\frac {YM}{YN}=\frac {ZM}{ZN}\iff$ $\frac {YM}{YN}=\frac {(1-n)(s-mr)}{(1-m)(1-nr)}=\frac {(1-n)(1-mq)}{(1-m)(p-nq)}\iff$ $(1-mq)(1-nr)=(p-nq)(s-mr)\iff$ $1+mpr+nqs=mq+ps+rn$ .
This post has been edited 53 times. Last edited by Virgil Nicula, Nov 26, 2015, 6:35 PM

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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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