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.
“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.
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 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 + n2, m2 – 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
Now, if we form three triangles from the following pairs of generators, a and b, a and c, a 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.
The answer to our puzzle in smallest possible numbers is as follows:
The area in every case is 341,880 square furlongs.
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.
|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.
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