## Puzzles in Geometry and Combinatorial Geometry Problem 1 Divide the given region in four congruent pieces.
The figure is obtained by joining three squares. Pieces are congruent if one can be obtained from the other using translations, rotations, and reflections (in other words, they have the same shape and same area). The solution is shown in the picture below.
Problem 2 Divide the given region in four congruent pieces.
The given figure is a trapezoid obtained by gluing together a square and an isosceles right triangle. The solution is shown in the picture below.
Problem 3 In the figure below, \( ABCD \) is a square, \( CE=BC \), and the arc \( DE \) is centered at \( B \).
Divide the given region in two congruent pieces using a curve that has only two common points with the given figure. The point \( P \) is the midpoint of the arc \( DE \) and \( Q \) is such that \( \angle BQP=90^{\circ} \) and \( QP=BP \). Problem 4 Divide the given region in five congruent pieces.
In the figure above \( ABKL \), \( CDIJ \), and \( EFGH \) are congruent squares and \( KBCJ \), \( DEHI \) are parallelograms with \( KB=BC=DE=EH \). The solution is shown in the picture below.
Problem 4 Given three squares of dimensions \( 2\times 2 \), \( 3\times 3 \), and \( 6\times 6 \), choose two of them and cut into two pieces each such that from the obtained \( 5 \) figures it is possible to assemble another square. The solution is shown in the picture below.
Problem 5 Does there exist a convex polygon that can be partitioned into non-convex quadrilaterals? The answer is no. Assume that, on the contrary it is possible to partition a polygon \( P \) into non-convex quadrilaterals. Let \( n \) be the number of quadrilaterals. Denote by \( S \) the total sum of all internal angles of all the quadrilaterals. Since the sum of internal angles of each quadrilateral is \( 360^{\circ} \) we have \( S=360^{\circ} \). However, each of the nonconvex angles has to be in the interior of \( P \), hence the sum of angles around the vertex of that angle has to be \( 360^{\circ} \). This immediately gives \( 360^{\circ}n \) as the sum of angles around such vertices. Since those are not the only vertices (at least the vertices of \( P \) will contribute to the sum \( S \)), we have that \( S> 360^{\circ} \) and this is a contradiction. Problem 6 Is it possible to divide a square into \( 10 \) convex pentagons?
Yes, see the following example: Problem 7 Let \( \triangle ABC \) be a triangle such that \( \angle A=90^{\circ} \). Determine whether it is possible to partition \( \triangle ABC \) into 2012 smaller triangles in such a way that the following two conditions are satisfied **(i)**Each triangle in the partition is similar to \( \triangle ABC \);**(ii)**No two triangles in the partition have the same area.
The required partition is always possible. We consider 2 cases: - \( 1^{\circ} \)
**The triangle is not isosceles:**First we construct the perpendicular from the vertex of the right angle. The triangle is divided into two similar, but non-congruent triangle. Now we divide smaller triangle, and keep going until the total number of triangle becomes \( 2012 \). - \( 2^{\circ} \)
**The triangle is isosceles:**Repeat the procedure from the previous problem until we get \( 2007 \) triangles. Then we divide one of the smallest triangles into \( 6 \) smaller and noncongruent triangles as shown in the second picture below.
Problem 8 A finite set of circles in the plane is called **(i)**No two circles intersect in more than one point;**(ii)**For every point \( A \) of the plane there are at most two circles passing through \( A \);**(iii)**Each circle from the set is tangent to exactly \( 5 \) other circles form the set.
Does there exist a nice set consisting of exactly **(a)**2011 circles?**(b)**2012 circles?
**(a)**We will prove that nice set can not contain an odd number of circles. Let \( n \) be a total number of circles. We will count the number of pairs \( (k,P) \) where \( k \) is a circle in the set and \( P \) a point at which \( k \) touches another circle. For each circle \( k \) there are exactly \( 5 \) such pairs, and hence the total number of pairs is \( 5n \). For each point \( P \) there are exactly \( 2 \) pairs corresponding to it. Hence the total number of pairs has to be even, but \( 5n \) can’t be even if \( n \) is odd.**(b)**The answer is yes. The following picture shows that there is a nice set consisting of exactly 12 circles.It is also possible to construct a nice set with \( 22 \) circles (the picture can be found below). Since \( 2012=164\cdot 12 + 2\cdot 22 \) we can make a nice set of \( 2012 \) circles by making a union of \( 164 \) discjoint nice sets of \( 12 \) circles each and 2 discjoint nice sets each of which contains \( 22 \) circles.
Problem 9 Six points are placed inside \( 4\times 3 \) rectangle. Prove that there are 2 among these points that are at a distance smaller than or equal to \( \sqrt 5 \). We can partition the \( 4\times 3 \) squares in 5 figures as shown in the picture below. Notice that among \( 6 \) chosen points at least two must be in the same figure in partition. It is not difficult to verify that the diameter of each of the figures is \( \sqrt 5 \). Problem 10 A finite set of unit circles is given in a plane such that the area of their union \( U \) is \( S \). Prove that there exists a subset of mutually disjoint circles such that the area of their union is greater than \( \frac{2S}9 \). (Vladimir Jankovic, IMO Shortlist 1981) Consider the partition of plane \( \pi \) into regular hexagons, each having inradius 2. Fix one of these hexagons, denoted by \( \gamma \). For any other hexagon \( x \) in the partition, there exists a unique translation \( \tau_x \) taking it onto \( \gamma \). Define the mapping \( \varphi:\pi\rightarrow\gamma \) as follows: If \( A \) belongs to the interior of a hexagon \( x \), then \( \varphi(A)=\tau_x(A) \) (if \( A \) is on the border of some hexagon, it does not actually matter where its image is). The total area of the images of the union of the given circles equals \( S \), while the area of the hexagon \( \gamma \) is \( 8\sqrt3 \). Thus there exists a point \( B \) of \( \gamma \) that is covered at least \( \frac{S}{8\sqrt3} \) times, i.e., such that \( \varphi^{-1}(B) \) consists of at least \( \frac{S}{8\sqrt3} \) distinct points of the plane that belong to some of the circles. For any of these points, take a circle that contains it. All these circles are disjoint, with total area not less than \( \frac{\pi}{8\sqrt3}S\geq \frac{2S}{9} \). |

2005-2017 IMOmath.com | imomath"at"gmail.com | Math rendered by MathJax Home | Olympiads | Book | Training | IMO Results | Forum | Links | About | Contact us |