This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 12

2017 Romania National Olympiad, 1

Prove that the line joining the centroid and the incenter of a non-isosceles triangle is perpendicular to the base if and only if the sum of the other two sides is thrice the base.

2006 Mathematics for Its Sake, 2

The cevians $ AP,BQ,CR $ of the triangle $ ABC $ are concurrent at $ F. $ Prove that the following affirmations are equivalent. $ \text{(i)} \overrightarrow{AP} +\overrightarrow{BQ} +\overrightarrow{CR} =0 $ $ \text{(ii)} F$ is the centroid of $ ABC $ [i]Doru Isac[/i]

2012 Grigore Moisil Intercounty, 3

Let $ M,N,P $ on the sides $ AB,BC,CA, $ respectively, of a triangle $ ABC $ such that $ AM=BN=CP $ and such that $$ AB\cdot \overrightarrow{AT} +BC\cdot \overrightarrow{BT} +CA\cdot \overrightarrow{CT} =0, $$ where $ T $ is the centroid of $ MNP. $ Prove that $ ABC $ is equilateral.

2008 Alexandru Myller, 1

$ O $ is the circumcentre of $ ABC $ and $ A_1\neq A $ is the point on $ AO $ and the circumcircle of $ ABC. $ The centers of mass of $ ABC, A_1BC $ are $ G,G_1, $ respectively, and $ P $ is the intersection of $ AG_1 $ with $ OG. $ Show that $ \frac{PG}{PO}=\frac{2}{3} . $ [i]Gabriel Popa, Paul Georgescu[/i]

2014 Romania National Olympiad, 3

Let $ P,Q $ be the midpoints of the diagonals $ BD, $ respectively, $ AC, $ of the quadrilateral $ ABCD, $ and points $ M,N,R,S $ on the segments $ BC,CD,PQ, $ respectively $ AC, $ except their extremities, such that $$ \frac{BM}{MC}=\frac{DN}{NC}=\frac{PR}{RQ}=\frac{AS}{SC} . $$ Show that the center of mass of the triangle $ AMN $ is situated on the segment $ RS. $

2018 Ramnicean Hope, 2

Let be the points $ M,N,P, $ on the sides $ BC,AC,AB $ (not on their endpoints), respectively, of a triangle $ ABC, $ such that $ \frac{BM}{MC} =\frac{CN}{NA} =\frac{AP}{PB} . $ Denote $ G_1,G_2,G_3 $ the centroids of $ APN,BMP,CNM, $ respectively. Show that the $ MNP $ has the same centroid as $ G_1G_2G_3. $ [i]Ovidiu Țâțan[/i]

Brazil L2 Finals (OBM) - geometry, 1998.2

Let $ABC$ be a triangle. $D$ is the midpoint of $AB$, $E$ is a point on the side $BC$ such that $BE = 2 EC$ and $\angle ADC = \angle BAE$. Find $\angle BAC$.

1998 Brazil National Olympiad, 2

Let $ABC$ be a triangle. $D$ is the midpoint of $AB$, $E$ is a point on the side $BC$ such that $BE = 2 EC$ and $\angle ADC = \angle BAE$. Find $\angle BAC$.

2002 District Olympiad, 3

Let $ G $ be the center of mass of a triangle $ ABC, $ and the points $ M,N,P $ on the segments $ AB,BC, $ respectively, $ CA $ (excluding the extremities) such that $$ \frac{AM}{MB} =\frac{BN}{NC} =\frac{CP}{PA} . $$ $ G_1,G_2,G_3 $ are the centers of mass of the triangles $ AMP, BMN, $ respectively, $ CNP. $ Pove that: [b]a)[/b] The centers of mas of $ ABC $ and $ G_1G_2G_3 $ are the same. [b]b)[/b] For any planar point $ D, $ the inequality $$ 3\cdot DG< DG_1+DG_2+DG_3<DA+DB+DC $$ holds.

2014 Contests, 2

Let $\Delta A_1A_2A_3, \Delta B_1B_2B_3, \Delta C_1C_2C_3$ be three equilateral triangles. The vertices in each triangle are numbered [u]clockwise[/u]. It is given that $A_3=B_3=C_3$. Let $M$ be the center of mass of $\Delta A_1B_1C_1$, and let $N$ be the center of mass of $\Delta A_2B_2C_2$. Prove that $\Delta A_3MN$ is an equilateral triangle.

2010 Victor Vâlcovici, 3

$ A',B',C' $ are the feet of the heights of an acute-angled triangle $ ABC. $ Calculate $$ \frac{\text{area} (ABC)}{\text{area}\left( A'B'C'\right)} , $$ knowing that $ ABC $ and $ A'B'C' $ have the same center of mass. [i]Carmen[/i] and [i]Viorel Botea[/i]

2014 Israel National Olympiad, 2

Let $\Delta A_1A_2A_3, \Delta B_1B_2B_3, \Delta C_1C_2C_3$ be three equilateral triangles. The vertices in each triangle are numbered [u]clockwise[/u]. It is given that $A_3=B_3=C_3$. Let $M$ be the center of mass of $\Delta A_1B_1C_1$, and let $N$ be the center of mass of $\Delta A_2B_2C_2$. Prove that $\Delta A_3MN$ is an equilateral triangle.