Showing posts with label puzzles. Show all posts
Showing posts with label puzzles. Show all posts

Wednesday, 1 June 2011

Two Ladders Puzzle: Answer

Previously I posted the Two Ladders Puzzle in which two ladders lean against opposite walls. The length of the ladders is given and so is the height of the point where they intersect. The task was to find the distance between the walls. The problem can be solved using Pythagoras and the law of similar triangles, so really it can classify as an easy problem. Nonetheless, the answer requires the solution of a quartic equation. We begin by naming the lengths of the ladders a and b and the, as yet unknown heights at which the ladders lean against the walls by x and y. Also let's divide the distance between the two walls into the parts m and n. We now have 5 unknowns, so we need 5 equations. The easy one is

m + n = d

Then we have Pythagoras for the two ladders

a2 = d2 + x2

b2 = d2 + y2

Finally we have the similar triangles

\frac{d}{x} = \frac{n}{h}

\frac{d}{y} = \frac{m}{h}

If we add these two equations, we get the sum (m + n) / h on the right hand side. But m + n = d and thus we can eliminate d to obtain the nice intermediate result

\frac{1}{h} = \frac{1}{x} + \frac{1}{y}

This means that h is the harmonic mean of x and y, independent of the values of a,b or d.

The rest of the solution will become a tiny bit messy. Instead of looking for a solution of d, we can look first for the values of x or y. Then we can use Pythagoras to find d. To do this we subtract the two Pythagorean equations to eliminate d

x2y2 = a2b2

Solving the harmonic equation for y we get

y = \frac{hx}{x-h}

With this we can eliminate y from the above difference of squares

x^2-\frac{h^2 x^2}{(x-h)^2} = a^2-b^2

Obviously this is a fourth order equation for x. I know this is not really a nice result so we will have to get some help for solving it. So let's insert the values and stick it into WolframAlpha. Of course there are four possible solutions. We get one positive, one negative and two complex solutions. Only the positive solution of x \approx 3.03692 is relevant. Now we can use the first Pythagoras to find the value for d

d \approx 2.60329

Wednesday, 25 May 2011

Two Ladders Puzzle

This puzzle is an old favourite of mine. It looks quite straightforward but it's not as easy as one might expect.

Two ladders are leaning against opposite walls in an alleyway. The bottom of each ladder is placed on the ground in the opposite corner. One ladder is 4m long, the other is 3m long. The height at which the two ladders meet is 1m.


Question: How far are the walls apart?

Wednesday, 11 May 2011

The Three Wise Men Puzzle: Answer

This is the answer to the "Three Wise Men Puzzle" posted earlier.

To understand the reason the men stop laughing we follow the thoughts of one of the philosophers, let's call him A. A sees the other two men, B and C laughing. Assuming that he doesn't have a mark on his forehead, he thinks that B is laughing at C and vice versa. But he must also assume that B is unaware of his own mark on the forehead. If A had no mark, what does B think C is laughing about? In other words, if A's assumption was true, B should quickly realise that C is laughing at B and therefore B should stop laughing. But after enough time has passed, and B hasn't stopped, A must assume that he himself has a spot on his forehead.

Monday, 9 May 2011

Three Philosophers Puzzle (Three Wise Men Puzzle)

Here is a logic problem which I was told some years ago and I found it quite interesting at the time. I tried finding a reference to this problem on the internet but wasn't really successful.

EDIT: Thanks to one of the readers, who pointed out that the puzzle is actually well known and is called the "Three Wise Men Puzzle". A variation of the puzzle is the "Muddy Children Problem".

The problem goes like this:

The Three Philosophers  -  by Giorgione
Three philosophers are sitting in the shade of a tree discussing deep philosophical problems. They talk and talk and, of course, they don't reach an agreement or come any closer to a solution. After some hours of talking the afternoon sun makes the three men grow tired and they finally decide to take a nap. While the philosophers are sleeping a young boy from the local village passes by. He looks at the men and decides to play a prank on them. The boy takes some white paint and on each of their foreheads he paints a white spot that resembles bird droppings. After some time the philosophers wake up again and, looking at the others, they all burst out into laughter. Of course each of them thinks that a bird has relieved itself on their colleagues foreheads but none of them is aware that, he too, has a white spot on their forehead. In this way they keep on laughing, but they don't let the others know of their plight. But suddenly they all stop laughing, suddenly realising that they too must have the mark on their forehead.

Question: How, just by using logic, did they arrive at this conclusion?

The answer can be found here.