Ten Lanterns

You have ten lanterns, five of which are working, and five of which are broken. You are allowed to choose any two lanterns and make a test that tells you whether there is a broken lantern among them or not. How many tests do you need until you find a lantern you know for sure is working?

Remark: If the test detects that there are broken lanterns, it does not tell you which ones and how many (one or two) they are.

You need 6 tests:

(1, 2) → (3, 4) → (5, 6) → (7, 8) → (7, 9) → (8, 9)

If at least one of these tests is positive, then you have found two working lanterns.

It all of these tests are negative, then lantern #10 must be working. Indeed, since at least one lantern in each of the pairs (1, 2), (3, 4), (5, 6) is not working. Therefore, there are at least 2 working lanterns among #7, #8, #9, #10. If #10 is not working, then at least one of the pairs (7, 8), (7, 9), or (8, 9) must yield a positive test, which is a contradiction.

Buffalo Buffalo Buffalo

The sentence below is grammatically correct. Can you explain it?

Buffalo buffalo Buffalo buffalo buffalo buffalo Buffalo buffalo.

The sentence says that buffalo (animals) from Buffalo (city, US), which are buffaloed (intimidated) by Buffalo (city, US) buffalo (animals), themselves buffalo (intimidate) buffalo (animals) from Buffalo (city, US).

Diagonal in a Rectangle

A 1000 × 1004 rectangle is split into 1 × 1 squares. How many of these squares does the main diagonal of the large rectangle pass through?

Notice that the number of small squares the main diagonal passes through is equal to the number of horizontal and vertical lines it intersects. Indeed, every time the diagonal goes through the interior of one square to the interior of another, it must intersect one of these lines.

There are 1000 + 1004 = 2004 lines which are intersected by the main diagonal. However, on four occasions (which is the greatest common divisor of 1000 and 1004), the main diagonal intersects one horizontal and one vertical line at the same time, which results in double-counting., so we must subtract 4 from the answer.

Therefore, the answer is 1000 + 1004 – 4 = 2000.

Rhombuses

A regular hexagon is split into small equilateral triangles and then the triangles are paired arbitrarily into rhombuses. Show that this results into three types of rhombuses based on orientation, with equal number of rhombuses from each type.

Color the rhombuses based on their type and imagine the diagram represents a structure of small cubes arranged in a larger cube. If you look at the large cube from three different angles, you will see exactly the three types of rhombuses on the diagram.

Alternatively, the problem can be proven more rigorously by considering the three sets of non-intersecting broken lines connecting the pairs of opposite sides of the hexagon, as shown on the image below. The type of each rhombus is determined by the types of the broken lines passing through it. Therefore, there are n² rhombuses of each type, where n is the length of the hexagon’s sides.

Unravel the Rope

Is it true that for every closed curve in the plane, you can use a rope to recreate the layout, so that the rope can be untangled?
Said otherwise, you have to determine at each intersection point of the closed curve, which of the two parts goes over and which one goes under, so that there aren’t any knots in the resulting rope.

Start from any point of the curve and keep moving along it, so that at each non-visited intersection you go over, until you get back to where you started from.

Blue and Red Points

You have 100 blue and 100 red points in the plane, no three of which lie on one line. Prove that you can connect all points in pairs of different colors so that no two segments intersect each other.

Connect the points in pairs of different colors so that the total length of all segments is minimal. If any two segments intersect, you can swap the two pairs among these four points and get a smaller total length.