This problem is from Ancient Rome.
A rich senator died, leaving his wife pregnant. After the senator’s death it was found out that he left a property of 210 talents (an Ancient Roman currency) in his will as follows: “In the case of the birth of a son, give the boy two thirds of my property (i.e. 140 talents) and the other third (i.e. 70 talents) to the mother. In the case of the birth of a daughter, give the girl one third of my property (i.e. 70 talents) and the other two thirds (i.e. 140 talents) to the mother.”
The senator’s widow gave birth to twins: one boy and one girl. This possibility was not foreseen by the late senator. How can the property be divided between three inheritors so that it is as close as possible to the instructions of the will?
Let \(n\) be some positive number. It is obvious that \[2n-1<2n.\] Take another positive number \(a\), and multiply both sides of the inequality by \((-a)\) \[-2na +a< -2na.\] Now, subtracting \((-2na)\) from both sides of the inequality we get \(a<0\).
Thus, ALL POSITIVE NUMBERS ARE NEGATIVE!
Suppose \(a \neq b\). We can write \[-a = b - (a+b)\] and \[-b = a - (a+b)\] Since \((-a)b = a(-b)\), then \[( b - (a+b))b = a(a - (a+b))\] Removing the brackets, we have \[b^2 - (a+b)b = a^2 - a(a+b)\] Adding \(\left(\frac{a+b}{2}\right)^2\) to each member of the equality we may complete the square of the differences of two numbers \[\left(b - \frac{a+b}{2}\right)^2 = \left(a - \frac{a+b}{2}\right)^2\] From the equality of the squares we conclude the equality of the bases \[b - \frac{a+b}{2} = a - \frac{a+b}{2}.\] Adding \(\frac{a+b}{2}\) to both sides of equality we get \(a=b\). Therefore, WE HAVE SHOWED THAT FROM \(a\neq b\) IT FOLLOWS \(a=b\).
Let \(x\) be equal to 1. Then we can write \(x^2=1\), or putting it differently \(x^2 -1 =0\). By using the difference of two squares formula we get \[(x+1)(x-1)=0\] Dividing both sides of the equality by \(x-1\) we obtain \[x+1=0,\] in other words \(x=-1\). But earlier we assumed \(x=1\). THUS \[-1=1\ !\]
In every right-angled triangle the arm is greater than the hypotenuse. Consider a triangle \(ABC\) with right angle at \(C\).
The difference of the squares of the hypothenuse and one of the arms is \(AB^2 -BC^2\). This expression can be represented in the form of a product \[AB^2 -BC^2 = (AB - BC)(AB+BC)\] or \[AB^2 -BC^2 = -(BC - AB)(AB+BC)\] Dividing the right hand sides by the product \(-(AB-BC)(AB+BC)\), we obtain the proportion \[\frac{AB+BC}{-(AB+BC)} = \frac{BC-AB}{AB-BC}.\] Since the positive quantity is greater than the negative one we have \(AB+BC>-(AB+BC)\). But then also \(BC-AB>AB-BC\), and therefore \(2BC>2AB\), or \(BC>AB\), i.e. THE ARM IS GREATER THAN THE HYPOTENUSE!
If you are on a boat and toss a suitcase overboard, will the water level rise or fall?
Philip and Denis cut a watermelon into four parts. When they finished eating watermelon (they ate the whole thing), they discovered that there were five watermelon rinds left. How is it possible, if no rind was cut after the initial cutting?
a) You have a \(10\times20\) chocolate bar and 19 friends. Since you are good at maths they ask you to split this bar into 19 pieces (always breaking along the lines between squares). All the pieces have to be of a rectangular shape. Your friends don’t really care how much they will get, they just want to be special, so you need to split the bar in such way that no two pieces are the same.
(b) The friends are quite impressed by your problem solving skills. But one of them is not that happy with the fact you didn’t get a single piece of the chocolate bar. He thinks you might feel that you are too special, therefore he convinces the others that you should get another \(10\times20\) chocolate bar and now split it into 20 different pieces, all of rectangular shapes (and still you need to break along the lines between squares). Can you do it now?
After having lots of practice with cutting different hexagons with a single cut Jennifer thinks she found a special one. She found a hexagon which cannot be cut into two quadrilaterals. Provide an example of such a hexagon.
In the magical land of Anchuria there is a drafts championship made up of several rounds. The days and cities in which the rounds are carried out are determined by a draw. According to the rules of the championship, no two rounds can take place in one city, and no two rounds can take place on one day. Among the fans, a lottery is arranged: the main prize is given to those who correctly guess, before the start of the championship, in which cities and on which days all of the round will take place. If no one guesses, then the main prize will go to the organising committee of the championship. In total, there are eight cities in Anchuria, and the championship is only allotted eight days. How many rounds should there be in the championship, so that the organising committee is most likely to receive the main prize?