Clock-Tower School Juniors 2007-14 (Romania) 14p

geometry problems from Junior Round of Clock-Tower School (Romanian) - Școala cu ceas

with aops links

Junior Round lasted only years 2007-14

collected inside aops here

2007 - 2014

2007 Clock-Tower School Junior p2

Consider a triangle $ABC$ with $AB=AC,$ and a point $M$ on the segment $BC.$ Let $N,P$ be the projections of $M$ on $AB,$ respectively, $AC,$ let $Q$ be the intersection of $BP,CN,$ and $R$ be the intersection of $AQ$ with $NP.$ Show that $MR$ bisects $\angle NMP .$

Francisc Bozgan

2007 Clock-Tower School Junior p6

Consider $1210$ points inside a circle of radius $1.$ Show that there are at least $11$ of these points that are inside a circle of radius $1/10.$

Vasile Pop

2008 Clock-Tower School Junior p1

Consider a circle of center $O$ and a chord $AB$ of it (not a diameter). Take a point $T$ on the ray $OB$. The perpendicular at $T$ onto $OB$ meets the chord $AB$ at $C$ and the circle at $D$ and $E$. Denote by $S$ the orthogonal projection of $T$ onto the chord $AB$. Show that $AS \cdot BC = TE \cdot TD$

2008 Clock-Tower School Junior p4

Consider any $25$ points, three by three non-collinear, in the interior of a square of side length $3$. Show that there exist, four among them that form a quadrilateral perimeter less than $5$.

2008 Clock-Tower School Junior p6

Let $P$ be the set of all points of the plane, and $O \in P$  fixed.  The function $f: P - \{O\} \to R$  has the property:

For any four distinct points $A,B,C,D  \in P - \{O\}$ with $\vartriangle AOB \sim \vartriangle COD$, $f(A)-f(B)+f(C) - f(D) =0$ occurs. 

Prove the function $f$ is constant.

2009 Clock-Tower School Junior p1

Consider a triangle $ABC$ with the property that there is a point $D$ inside it so that  $\angle DAC  =  \angle DCA  = 30^o$ and  $\angle DBA = 60^o$. Let $E$ be the midpoint of $[BC]$ and $F \in [AC]$ such that $CF = \frac13 AC$ . Show that $DE \perp EF$.

2010 Clock-Tower School Junior p1

Let be a regular polygon $A_1A_2...A_{2010}$ having the center at the point $O$. On each of the segments $OA_k$, with $k = 1, 2, ..., 2010$, the point $B_k$ is considered such that $\frac{OB_k}{OA_k}=\frac{1}{k}$. Determine the ratio between the area of the polygon $B_1B_2... B_{2010}$ and that of the polygon $A_1A_2...A_{2010}$ .

2011 Clock-Tower School Junior p4

Consider a triangle $ABC$ and the points $M \in (BC), N  \in (AC), P \in (AB)$ such that $\angle BMP = \angle CNM =\angle APN$ and $BM = CN = AP$. Prove that the triangle $ABC$ is equilateral.

2012 Clock-Tower School Junior p2

Let $ABC$ be a right-angled triangle, where$A$ is the right angle. On the catheti $(AB)$ and $(AC)$ we consider points $D$ and $E$, respectively, such us the perpendicular lines through $D$ and $A$ to $(BE)$ intersect $(BC)$ in $F$ and $G$, respectively and  $[FG]=[GC]$.

 a) Prove that $BD\cdot GC = AD \cdot  BF$ .

 b) If $EG  \perp BC$  , prove that $AE < EC$ .

Ştefan SMĂRĂNDOIU – Rm. Vâlcea

2012 Clock-Tower School Junior p6

Let $ABC$ be a triangle with $\angle B=\angle C$  . On $(AB)$ we consider a point $D$ and on $(AC - [AC]$ we consider a point $E$ , such as $D$ is symmetric of $E$ with respect to $F$ , where $\{F\}= DE \cap BC$  . Prove that $[BD]=[CE$].

Ştefan SMĂRĂNDOIU – Rm. Vâlcea

2013 Clock-Tower School Junior p4

In the acute triangle $ABC$, let $O$ and $I$ be the centers of the circumscribed circles and inscribed respectively , and $D$ and $E$ the feet of the heights of $A$ and $B$. respectively. Line $CI$ cuts the circle around the triangle $ABC$ in $K$, and $\{P\}=OK\cap AB$ . If $OK =IK$  , show that the triangle $DEP$ is equilateral.

Dorian CROITORU, Chisinau

2013 Clock-Tower School Junior p8

Let $\angle XOY$ be a right angle and $[OP$ its bisector. On the right half $(OX$ take a point $A$ so that $\angle OAP$ is an obtuse angle, and on the half-ray $(OY$ a point$B$ is taken so that $\angle OBP$ is a acute angle and $[PA]=[PB]$. Find the measure of the angle $\angle APB$.

Ştefan SMĂRĂNDOIU, Rm. Vâlcea

2014 Clock-Tower School Junior p3

Let $ABCD$ be a convex quadrilateral with $AB = AD$. Let $E$ be the midpoint of the arc $BC$ of the circle $\odot (ABC)$ which does not contain point $A$, and $F$ be the midpoint of the arc $CD$ of the circle $\odot (ACD)$ which does not contain point $A$. Prove that $EF \perp AC$.

L. Ploscaru

2014 Clock-Tower School Junior p6

The triangle $ABC$ is right with $D$ in the midpoint  of the hypotenuse $[AC]$. $E$ is a point such that so that $D\in (BE)$ and $[AC]= [DE]$ . Knowing that $[AB]=[AE]$ , prove that $\angle BAE  = 120^o$.

Ștefan SMĂRĂNDOIU, Râmnicu Vâlcea

source: https://pregatirematematicaolimpiadejuniori.wordpress.com/scoala-cu-ceas/