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?

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.

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.

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.

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.

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?

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.

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.

NASA and the Meteor

NASA locates a meteor in outer space and concludes that it has either a cubical or spherical shape. In order to determine the exact shape, NASA lands a spacecraft on the meteor and lets a rover travel from the spacecraft to the opposite point on the planet. By measuring the relative position of the rover with respect to the spacecraft throughout its travel on the planet (in 3D coordinates), can NASA always determine the shape, no matter the route taken by the rover?

The answer is NO. 

The question is equivalent to analyzing the intersection of a cube and a sphere which share a common center. Thus the question gets reduced to figuring out whether such intersection, which is a curve, can connect two opposite points on the sphere/cube.

Let the edge of the cube has length 1. If you pick the radius of the sphere equal to √2/2, the intersection will consist of 6 circles inscribed in the sides of the cube. Then the rover can just move along these circles from one point to its opposite and NASA won’t be able to figure out the exact shape.

Remark: It is not hard to see that 2:√2 is the only edge-radius ratio, for which NASA can’t figure out the shape.

Worm in an Apple

There is a perfectly spherical apple with radius 50mm. A worm has entered the apple, made a tunnel of length 99mm through it and left. Prove that we can slice the apple in two pieces through the center, so that one of them is untouched by the worm.

Let the entering point is A, the leaving point is B and the center of the apple is C. Consider the plane P containing the points A, B and C and project the worm’s tunnel on it. Since 99 < 2×50, the convex hull of the tunnel’s projection will not contain the center C. Therefore we can find a line L through C, such that the tunnel’s projection is entirely in one of the semi-planes of P with respect to L. Now cut the apple with a slice orthogonal to P passing through the line L and you are done.