The puzzle question is : On Bagshot Island, there is an airport. The airport is the homebase of an unlimited number of identical airplanes. Each airplane has a fuel capacity to allow it to fly exactly 1/2 way around the world, along a great circle. The planes have the ability to refuel in flight without loss of speed or spillage of fuel. Though the fuel is unlimited, the island is the only source of fuel.
What is the fewest number of aircraft necessary to get one plane all the way around the world assuming that all of the aircraft must return safely to the airport? How did you get to your answer?
Notes:
(a) Each airplane must depart and return to the same airport, and that is the only airport they can land and refuel on ground.
(b) Each airplane must have enough fuel to return to airport.
(c) The time and fuel consumption of refueling can be ignored. (so we can also assume that one airplane can refuel more than one airplanes in air at the same time.)
(d) The amount of fuel airplanes carrying can be zero as long as the other airplane is refueling these airplanes. What is the fewest number of airplanes and number of tanks of fuel needed to accomplish this work? (we only need airplane to go around the world)
As per the puzzle given ablove The fewest number of aircraft is 3! Imagine 3 aircraft (A, B and C). A is going to fly round the world. All three aircraft start at the same time in the same direction. After 1/6 of the circumference, B passes 1/3 of its fuel to C and returns home, where it is refuelled and starts immediately again to follow A and C.
C continues to fly alongside A until they are 1/4 of the distance around the world. At this point C completely fills the tank of A which is now able to fly to a point 3/4 of the way around the world. C has now only 1/3 of its full fuel capacity left, not enough to get back to the home base. But the first "auxiliary" aircraft reaches it in time in order to refuel it, and both "auxiliary" aircraft are the able to return safely to the home base.
Now in the same manner as before both B and C fully refuelled fly towards A. Again B refuels C and returns home to be refuelled. C reaches A at the point where it has flown 3/4 around the world. All 3 aircraft can safely return to the home base, if the refuelling process is applied analogously as for the first phase of the flight.
Thursday, March 24, 2011
Tuesday, March 22, 2011
Given the cube below, find the number of acute triangles formed by connecting three vertices and the number of right triangles formed by connecting 3
Given the cube below, find the number of acute triangles formed by connecting three vertices and the number of right triangles formed by connecting three vertices.
For example, the vertices 1-6-8 form an acute triangle while the vertices 1-2-3 form a right triangle.
Solution:
One approach is to look at each vertex and see how many acute and right triangles can be formed and then remove duplications at the end. This is not the most efficient way.
Another approach is as follows: To determine the number of acute triangles, we shall first find the probability that at a triangle formed will be acute and multiply that number by the total number of triangles that can be formed on the cube. Similarly, we can find the number of right triangles.
Let us start with an arbitrary vertex, say vertex 1. (Notice that we could have started with any vertex.) There are a total of 21 triangles that can be formed using this vertex.
Why are there 21? From the vertex 1, we can choose any of the other 7 vertices for the next vertex of the triangle. For the last vertex, we can choose any of the remaining 6 vertices, but now we have double-counted, since the triangle with vertices 1-2-3 is the same as the triangle with vertices 1-3-2. Thus, we have (7*6)/2 = 21 triangles.
There are three acute triangles, namely 1-6-8, 1-3-6, and 1-3-8.
There are eighteen right triangles, namely 1-2-3, 1-2-4, 1-2-5, 1-2-6, 1-2-7, 1-2-8, 1-3-4, 1-3-5, 1-3-7, 1-4-5, 1-4-6, 1-4-7, 1-4-8, 1-5-6, 1-5-7, 1-5-8, 1-6-7, and 1-7-8.
From that, we see that the probability of picking an acute triangle is 3/21 = 1/7 while the probability of picking a right triangle is 18/21 = 6/7.
In total, there are 56 different triangles that are possible. This comes from the formula for the binomial coefficient. In this case, we have 8 vertices and we are choosing 3 of them.
Thus, there are a total of (1/7)*56 = 8 acute triangles and (6/7)*56 = 48 right triangles that can be formed by connecting three vertices of a cube.
For example, the vertices 1-6-8 form an acute triangle while the vertices 1-2-3 form a right triangle.
Solution:
One approach is to look at each vertex and see how many acute and right triangles can be formed and then remove duplications at the end. This is not the most efficient way.
Another approach is as follows: To determine the number of acute triangles, we shall first find the probability that at a triangle formed will be acute and multiply that number by the total number of triangles that can be formed on the cube. Similarly, we can find the number of right triangles.
Let us start with an arbitrary vertex, say vertex 1. (Notice that we could have started with any vertex.) There are a total of 21 triangles that can be formed using this vertex.
Why are there 21? From the vertex 1, we can choose any of the other 7 vertices for the next vertex of the triangle. For the last vertex, we can choose any of the remaining 6 vertices, but now we have double-counted, since the triangle with vertices 1-2-3 is the same as the triangle with vertices 1-3-2. Thus, we have (7*6)/2 = 21 triangles.
There are three acute triangles, namely 1-6-8, 1-3-6, and 1-3-8.
There are eighteen right triangles, namely 1-2-3, 1-2-4, 1-2-5, 1-2-6, 1-2-7, 1-2-8, 1-3-4, 1-3-5, 1-3-7, 1-4-5, 1-4-6, 1-4-7, 1-4-8, 1-5-6, 1-5-7, 1-5-8, 1-6-7, and 1-7-8.
From that, we see that the probability of picking an acute triangle is 3/21 = 1/7 while the probability of picking a right triangle is 18/21 = 6/7.
In total, there are 56 different triangles that are possible. This comes from the formula for the binomial coefficient. In this case, we have 8 vertices and we are choosing 3 of them.
Thus, there are a total of (1/7)*56 = 8 acute triangles and (6/7)*56 = 48 right triangles that can be formed by connecting three vertices of a cube.
Arrange 20 cubes in 4 piles using these clues:
All piles contain an even number of cubes.there are twice as many cubes in the first pile as in the 2nd pile. the largest number of cubes are in the first pile. all piles have different number of cubes. each pile has at least one cube.how many cubes in pile1,pile2,pile3,pile4????
Pile 2: x number of cubes
Pile 1: 2x
Each pile must have at least 2 cubes to have an even number.
x + 2x = 3x
3x < 20 so x < 6.67
Using trial and error, if pile 2 had 6 cubes and pile 3 had 12 cubes, there would be 2 cubes left. You can't make two piles with 2 cubes, because one is not an even number. If pile 2 had 5 cubes and pile 1 had 10 cubes, there would be 5 cubes left. This is an odd number and can't be divided into two even numbered piles. If pile 2 had 4 cubes and pile 1 had 8 cubes, there would be 8 cubes left. This can then be split into a piles of 2 and 6.
Pile 1: 8
Pile 2: 4
Pile 3: 6
Pile 4: 2
Pile 2: x number of cubes
Pile 1: 2x
Each pile must have at least 2 cubes to have an even number.
x + 2x = 3x
3x < 20 so x < 6.67
Using trial and error, if pile 2 had 6 cubes and pile 3 had 12 cubes, there would be 2 cubes left. You can't make two piles with 2 cubes, because one is not an even number. If pile 2 had 5 cubes and pile 1 had 10 cubes, there would be 5 cubes left. This is an odd number and can't be divided into two even numbered piles. If pile 2 had 4 cubes and pile 1 had 8 cubes, there would be 8 cubes left. This can then be split into a piles of 2 and 6.
Pile 1: 8
Pile 2: 4
Pile 3: 6
Pile 4: 2
Measure ( Minutes With two SandGlasses of 4 MInutes & 7 Minutes
If you have two sand glasses One of them is for the 4 minute another one - is for 7 minutes If you have to measure 9 minutes with these two sand glasses in 9 minutes?
At 0 minutes, turn both the hourglasses.
After 4 minutes:
The 4 minute hourglass: 0 minutes left.
The 7 minute hourglass: 3 minutes left.
Turn the 4 minute hourglass.
After 7 minutes:
The 4 minute hourglass: 1 minute left.
The 7 minute hourglass: 0 minutes left.
Turn the 7 minute hourglass.
After 8 minutes:
The 4 minute hourglass: 0 minutes left.
The 7 minute hourglass: 1 minute left.
Turn the 7 minute hourglass.
After 9 minutes:
The 4 minute hourglass: 0 minute left.
The 7 minute hourglass: 0 minutes left.
Now you know you are finished.
At 0 minutes, turn both the hourglasses.
After 4 minutes:
The 4 minute hourglass: 0 minutes left.
The 7 minute hourglass: 3 minutes left.
Turn the 4 minute hourglass.
After 7 minutes:
The 4 minute hourglass: 1 minute left.
The 7 minute hourglass: 0 minutes left.
Turn the 7 minute hourglass.
After 8 minutes:
The 4 minute hourglass: 0 minutes left.
The 7 minute hourglass: 1 minute left.
Turn the 7 minute hourglass.
After 9 minutes:
The 4 minute hourglass: 0 minute left.
The 7 minute hourglass: 0 minutes left.
Now you know you are finished.
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