Category: Mathematics

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Lost In the Forest

Section: Brain Teasers Category: Mathematics
Difficulty: Tags: ,

You are lost in the middle of a forest, and you know there is a straight road exactly 1 km away from you, but not in which direction. Can you find a path of distance less than 640 m which will guarantee you to find the road?

Imagine there is a circle with a radius of 100 m around you, and you are at its center O. Let the tangent to the circle directly ahead of you be t. Then, follow the path:

  1. Turn left 30 degrees and keep walking until you reach the tangent t at point A for a total of 100×2√3/3 meters, which is less than 115.5 meters.
  2. Turn left 120 degrees and keep walking along the tangent to the circle until you reach the circle at point B for a total of 100×√3/3 which is less than 58 meters.
  3. Keep walking around the circle along an arc of 210 degrees until you reach point C for a total of 100×7π/6 which is less than 366.5 meters.
  4. Keep walking straight for 100 meters until you reach point D on the tangent t.
unknown-author
Unknown Author September 22, 2023
3 Comments

An Ant’s Path

Section: Brain Teasers Category: Mathematics
Difficulty:

An ant is positioned at one of the vertices of a cube and wants to get to the opposite vertex. If the edges of the die have length 1, what is the shortest distance the ant needs to travel?

We unfold a cube to get a cross-shaped figure. Then, the problem is to find the shortest path between two points separated by a horizontal distance of 2 units and a vertical distance of 1 unit.

It is easy to see that the path in question is the one passing through the middle of the edge between the start and end points, and which has a distance of √5.

unknown-author
Unknown Author September 19, 2023
5 Comments

A Square and an Invisible Point

Section: Brain Teasers Category: Mathematics
Difficulty:

There is a square drawn on a piece of paper and also a point marked with invisible ink. You are allowed to draw 3 lines on the paper and for each of them you will be told whether the point is on its left, on its right, or lies on the line. Your task is to find out whether the point is inside the square, outside the square, or on its boundary. How do you do it?

Draw one of the diagonals of the square. Then, draw the 2 lines containing the sides of the square that are on the same side as the invisible point.

unknown-author
Unknown Author September 10, 2023
0 Comments

Cut the Pizza

Section: Brain Teasers Category: Mathematics
Difficulty: Tags:

Cut a circular pizza into 12 congruent slices, such that exactly half of them contain crust.

Remark: We say that a slice contains crust if it shares an arc with the boundary of the pizza (with non-zero measure).

First, cut the pizza into 6 congruent circular triangles, and then split each of them in half, as shown on the image below.

unknown-author
Unknown Author July 13, 2023
4 Comments

Spot the Robber

Section: Brain Teasers Category: Mathematics
Difficulty:

The streets of the city are a square grid that extends infinitely in all directions. One of the streets has a police officer stationed every 100 blocks and there is a robber is somewhere in the city.

Can you devise a strategy that guarantees the robber will be spotted by a police officer at some point, no matter how he tries to avoid them?

Note: The officers can see infinitely far, but their running speeds are lower than the speed of the robber.

Let the police officers are located at points with coordinates (100N, 0) for N = 0, ±1, ±2… First, we fix the positions of all officers stationed at points (±200N, 0), then repeatedly perform the following procedure, step by step:

On step M, we let the non-fixed officers who are closest to the center move to the free points with coordinates (K, 0) and (0, K) for K = 0, ±1, ±2, … ±M. Then we fix their positions.

Since there are fixed officers at points (200N, 0) at all times, the robber is contained within some vertical strip the entire time. Therefore, at some point there will be two fixed officers that will restrict the robber within a horizontal segment of size 1, at coordinates (x, T) for x ∈ (S, S+1) and some T. Finally, at some point an officer will move to the point (0, T) and will spot the robber.

alex-lvovsky
Alex Lvovsky June 13, 2023
0 Comments

Numbers on a Dodecahedron

Section: Brain Teasers Category: Mathematics
Difficulty:

We have written the numbers from 1 to 12 on the faces of a regular dodecahedron. Then, we have written on each vertex the sum of the five numbers on the faces incident with it. Is it possible that 16 of these 20 sums are the same?

We color the vertices in five colors as shown in the image below, such that each face of the dodecahedron has 1 vertex of each color. Then, the sum of the four numbers from each color must be equal to the sum of all numbers written on the faces: 1 + 2 + … + 12 = 78.

If 16 of the vertices have the same number written on them, then by the pigeonhole principle there will be 4 vertices with identical colors and identical numbers. Since 78 is not divisible by 4, we conclude that this is impossible.

unknown-author
Unknown Author April 13, 2023
0 Comments

Sum Up to 999

Section: Brain Teasers Category: Mathematics
Difficulty: Tags:

Can you find a triple of three-digit numbers that sum up to 999 and collectively contain all digits from 1 to 9 exactly once? How many such triples are there? What if the sum was 1000?

There are exactly 180 such triples that sum up to 999 and none that sum up to 1000.

In order to see that, notice that the sum of the first digits of the numbers can be no more than 9. Since the sum of all digits is 45, the sum of the middle and the sum of the last digits should be both no more than 9+8+7=24, and no less than 45-9-24=12. We then see that the sum of the last digits should be exactly 19 and the sum of the middle digits should be exactly 18. The sum of the first digits should be 45-19-18=8.

There are 2 ways to get 8 using unique digits from 1 to 9: 1+2+5 and 1+3+4.

  • If the first digits are {1, 2, 5}, the options for the middle digits are {3, 6, 9}, {3, 7, 8}, and {4, 6, 8}. The last digits end up {4, 7, 8}, {4, 6, 9}, and {3, 7, 9} respectively.
  • If the first digits are {1, 3, 4}, the options for the middle digits are {2, 7, 9} and {5, 6, 7}. The last digits end up {5, 6, 8} and {2, 8, 9} respectively.

Since the set of the first digits, the set of the middle digits, and the set of the last digits of the numbers can be permuted in 6 ways each, we get a total of 5×6×6×6=1080 solutions, or 180 up to permutation of the 3 three-digit numbers.

In order to see that we cannot get a sum of 1000, we note that since the sum of the digits from 1 to 9 is divisible by 9, then the sum of the 3 three-digit numbers should be divisible by 9 as well. Since 1000 is not divisible by 9, the statement follows.

unknown-author
Unknown Author January 16, 2023
17 Comments

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