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Eye of the Dragon
Divide the circle below in two pieces. Then, put the pieces together to get a circle with a dragon, such that the dragon’s eye is at the center of the new circle.
SOLUTION
The solution is shown below.
Tag: Geometry
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Cut the Pizza
Cut a circular pizza into 12 congruent slices, such that exactly half of them contain crust.
SOLUTION
First, cut the pizza into 6 congruent circular triangles, and then split each of them in half, as shown on the image below.
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Beautiful Geometry 1
Since 2018, Catriona Shearer, a UK teacher, has been posting on her Twitter various colorful geometry puzzles. In this mini-course, we cover some of her best problems and provide elegant solutions to them. Use the pagination below to navigate the puzzles, which are ordered by difficulty.
Example Problems
Rapunzel and the Prince
The evil witch has left Rapunzel and the prince in the center of a completely dark, large, square prison room. The room is guarded by four silent monsters in each of its corners. Rapunzel and the prince need to reach the only escape door located in the center of one of the walls, without getting near the foul beasts. How can they do this, considering they can not see anything and do not know in which direction to go?
SOLUTION
The prince must stay in the center of the room and hold Rapunzel’s hair, gradually releasing it. Then, Rapunzel must walk in circles around the prince, until she gets to the walls and finds the escape door.
Borromean Rings
Borromean rings are rings in the 3-dimensional space, linked in such a way that if you cut any of the three rings, all of them will be unlinked (see the image below). Show that rigid circular Borromean rings cannot exist.
SOLUTION
Assume the opposite. Imagine the rings have zero thickness and reposition them in such a way, that two of them, say ring 1 and ring 2, touch each other in two points. These two rings lie either on a sphere or a plane which ring 3 must intersect in four points. However, this is impossible.
Square Cake
There is a square cake at a birthday party attended by a dozen people. How can the cake be cut into twelve pieces, so that every person gets the same amount of cake, and also the same amount of frosting?
Remark: The decoration of the cake is put aside, nobody eats it.
SOLUTION
Divide the boundary of the cake into twelve equal parts, then simply make cuts passing through the separation points and the center. This way all tops and bottoms of the formed pieces will have equal areas, and also all their sides will have equal areas. Since all pieces have the same height, their volumes will be equal as well.
Four Points in the Plane
Find all configurations of four points in the plane, such that the pairwise distances between the points take at most two different values.
SOLUTION
All 5 configurations are shown below: a square, a rhombus with 60°-120°-60°-120°, an equilateral triangle with its center, an isosceles triangle with 75°-75°-30° and its center, and a quadrilateral with 75°-150°-75°-150°.
Bridge Over the River
Pinkbird is trying to get to Redbird across the river. Where should we place the bridge, so that the path between the two birds becomes as short as possible?
Remark: The bridge is exactly as long as the river is wide, and must be placed straight across it. Additionally, it has some positive width.
SOLUTION
Notice that no matter how the bridge gets placed over the river, the shortest path would be to go to its top left corner, then traverse it diagonally, then go from its bottom right corner to Redbird. The second part of the way has fixed length, so we must minimize the first part plus the third part. In order to do that, imagine we place the bridge, so that its top left corner is at the current position of Pinkbird – point A. If the bottom right corner ends up at point C, then we must connect C with the position of Redbird – point B, and wherever the line intersects the bottom shore – point D, that will be the best place for the bottom right corner of the bridge.
Every Acute Triangle
Consider an arbitrary acute triangle ABC. Let E be the intersection of the bisector at vertex C and the bisection of the side AB. Let F and G be the projections of E on AC and BC respectively.
Since E belongs to the bisection of AB, we must have AE = BE. Also, since E belongs to the bisector of C, we must have EF = EG. However, this would imply that triangles AEF and BGF are identical, and then AF = BF. We also have that CF = CG, which implies that AC = BC. The arbitrarily chosen triangle ABC is isosceles!
Can you find where the logic fails?
SOLUTION
The bisector of C and the bisection of AB always intersect outside the triangle, on the circumcircle. One of the points F and G always lies on the segment AC or BC, and the other one does not.
Obtuse Angle
Prove that among any 9 points in (3D) space, there are three which form an obtuse angle.
SOLUTION
Let the points be labeled A1, A2, … , A9, and P be their convex hull. If we assume that all angles among the points are not obtuse, then the interiors of the bodies P + A1, P + A2, … , P + A9 should be all disjoint. That is because, for every Ai and Aj, P must be bound between the planes passing through Ai and Aj which are orthogonal to the segment AiAj. However, all of these 9 bodies have the same volume and are contained in the body P + P, which has 8 times larger volume. This is a contradiction, and therefore our assumption is wrong.