Materials for language lessons

Posts tagged ‘medieval’

Medieval brain teasers

I have found these medieval brain teasers on pedagonet, maybe you will like them, too:
The Amulet

A strange man was one day found loitering in the courtyard of the castle, and the retainers, noticing that his speech had a foreign accent, suspected him of being a spy.

So the fellow was brought before Sir Hugh, who could make nothing of him. 

He ordered the varlet to be removed and examined, in order to discover whether any secret letters were concealed about him. 

All they found was a piece of parchment securely suspended from the neck, bearing this mysterious inscription:

To-day we know that Abracadabra was the supreme deity of the Assyrians, and this curious arrangement of the letters of the word was commonly worn in Europe as an amulet or charm against diseases. 

But Sir Hugh had never heard of it, and, regarding the document rather seriously, he sent for a learned priest.
“I pray you, Sir Clerk,” said he, “show me the true intent of this strange writing.”

“Sir Hugh,” replied the holy man, after he had spoken in a foreign tongue with the stranger, “it is but an amulet that this poor wight doth wear upon his breast to ward off the ague, the toothache, and such other afflictions of the body.”

“Then give the varlet food and raiment and set him on his way,” said Sir Hugh.
“Meanwhile, Sir Clerk, canst thou tell me in how many ways this word ‘Abracadabra’ may be read on the amulet, always starting from the A at the top thereof?”

Place your pencil on the A at the top and count in how many different ways you can trace out the word downwards, always passing from a letter to an adjoining one.

The Amulet
The puzzle was to place your pencil on the A at the top of the amulet and count in how many different ways you could trace out the word “Abracadabra” downwards, always passing from a letter to an adjoining one.

“Now, mark ye, fine fellows,” said Sir Hugh to some who had besought him to explain, “that at the very first start there be two ways open: whichever B ye select, there will be two several ways of proceeding (twice times two are four); whichever R ye select, there be two ways of going on (twice times four are eight); and so on until the end. 

Each letter in order from A downwards may so be reached in 2, 4, 8, 16, 32, etc., ways. 

Therefore, as there be ten lines or steps in all from A to the bottom, all ye need do is to multiply ten 2’s together, and truly the result,1024, is the answer thou dost seek.”


The Four Princes

The dominions of a certain Eastern monarch formed a perfectly square tract of country.
It happened that the king one day discovered that his four sons were not only plotting against each other, but were in secret rebellion against himself. After consulting with his advisers he decided not to exile the princes, but to confine them to the four corners of the country, where each should be given a triangular territory of equal area, beyond the boundaries of which they would pass at the costof their lives.Now, the royal surveyor found himself confronted by great natural difficulties, owing to the wild character of the country.
The result was that while each was given exactly the same area, the four triangular districts were all of different shapes, somewhat in the manner shown in the illustration.

The puzzle is to give the three measurements for each of the four districts in the smallest possible numbers—all whole furlongs.
In other words, it is required to find (in the smallest possible numbers) four rational right-angled triangles of equal area.

The Four Princes
Answer :When Montucla, in his edition of Ozanam’s Recreations in Mathematics, declared that “No more than three right-angled triangles, equal to each other, can be found in whole numbers, but we may find as many as we choose in fractions,” he curiously overlooked the obvious fact that if you give all your sides a common denominator and then cancel that denominator you have the required answer in integers!Every reader should know that if we take any two numbers, m and n, then m2 + n2m2 – n2, and 2mn will be the three sides of a rational right-angled triangle.
Here m and n are called generating numbers.
To form three such triangles of equal area, we use the following simple formula, where m is the greater number:

mn + m2 + n2 = a
m2 – n2 = b
2mn + n2 = c

Now, if we form three triangles from the following pairs of generators, a and ba and ca and b + c, they will all be of equal area. This is the little problem respecting which Lewis Carroll says in his diary (see his Life and Letters by Collingwood, p. 343), “Sat up last night till 4 a.m., over a tempting problem, sent me from New York, ‘to find three equal rational-sided right-angled triangles.’ I found two … but could not find three!”

The following is a subtle formula by means of which we may always find a R.A.T. equal in area to any given R.A.T.
Let z = hypotenuse, b = base, h = height, a = area of the given triangle; then all we have to do is to form a R.A.T. from the generators z2 and 4a, and give each side the denominator 2z (b2 – h2), and we get the required answer in fractions.
If we multiply all three sides of the original triangle by the denominator, we shall get at once a solution in whole numbers.

The answer to our puzzle in smallest possible numbers is as follows:

First Prince 518 1320 1418
Second Prince 280 2442 2458
Third Prince 231 2960 2969
Fourth Prince 111 6160 6161

The area in every case is 341,880 square furlongs.
I must here refrain from showing fully how I get these figures. I will explain, however, that the first three triangles are obtained, in the manner shown, from the numbers 3 and 4, which give the generators 37, 7; 37, 33; 37, 40.
These three pairs of numbers solve the indeterminate equation, a3b – b3a = 341,880.
If we can find another pair of values, the thing is done. These values are 56, 55, which generators give the last triangle.
The next best answer that I have found is derived from 5 and 6, which give the generators 91, 11; 91, 85; 91, 96.
The fourth pair of values is 63, 42.

The reader will understand from what I have written above that there is no limit to the number of rational-sided R.A.T.’s of equal area that may be found in whole numbers.

The Crescent and the Cross

When Sir Hugh’s kinsman, Sir John de Collingham, came back from the Holy Land, he brought with him a flag bearing the sign of a crescent, as shown in the illustration.
It was noticed that De Fortibus spent much time in examining this crescent and comparing it with the cross borne by the Crusaders on their own banners.One day, in the presence of a goodly company, he made the following striking announcement:

“I have thought much of late, friends and masters, of the conversion of the crescent to the cross, and this has led me to the finding of matters at which I marvel greatly, for that which I shall now make known is mystical and deep.

Truly it was shown to me in a dream that this crescent of the enemy may be exactly converted into the cross of our own banner. Herein is a sign that bodes good for our wars in the Holy Land.”

Sir Hugh de Fortibus then explained that the crescent in one banner might be cut into pieces that would exactly form the perfect cross in the other.

It is certainly rather curious; and I show how the conversion from crescent to cross may be made in ten pieces, using every part of the crescent.
The flag was alike on both sides, so pieces may be turned over where required.

The Crescent and The Cross
Answer :”By the toes of St. Moden,” exclaimed Sir Hugh de Fortibus when this puzzle was brought up, “my poor wit hath never shaped a more cunning artifice or any more bewitching to look upon.
It came to me as in a vision, and ofttimes have I marvelled at the thing, seeing its exceeding difficulty.
My masters and kinsmen, it is done in this wise.”

The worthy knight then pointed out that the crescent was of a particular and somewhat irregular form—the two distances a to b andc to d being straight lines, and the arcs ac and bd being precisely similar.
He showed that if the cuts be made as in Figure 1, the four pieces will fit together and form a perfect square, as shown in Figure 2, if we there only regard the three curved lines.
By now making the straight cuts also shown in Figure 2, we get the ten pieces that fit together, as in Figure 3, and form a perfectly symmetrical Greek cross.
The proportions of the crescent and cross in the original illustration were correct, and the solution can be demonstrated to be absolutely exact and not merely approximate.

I have a solution in considerably fewer pieces, but it is far more difficult to understand than the above method, in which the problem is simplified by introducing the intermediate square


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