Three sisters fond of riddles always answer the following when asked about age: The square of Ana's age plus the square of Blanca's age plus Carla's age is equal to the square of Carla's age.
When Ana is the age of Carla now, Carla will be 4 times the age Ana is now and twice the age of Blanca now.
How old is each one?
If we call to at Ana's age, b to Blanca's and c to that of Carla, the first information they offer us directly provides the equation to2 + b2 + c = c2.
For the second information you have to think that for Ana to be the age of Carla now, they will have to pass c - a years because it's the way that a + (c - a) = c. That Carla is four times the age Ana is now, means that c + c - a = 4a, Or what is the same, 2c = 5a.
Obviously, the last relationship can be expressed as 4a = 2b, Or what is the same,
b = 2a.
To solve the system of three equations with three unknowns, we have to substitute in the first, which is second grade. For this it is convenient to clear the unknowns in the others that appear less, so we have to b = 2a by the last equation, that is to say b2 = 4th2 so that the first one is to2 + 4th2 + c = c2, Or what is the same, 5th2 + c = c2.
From the other equation we get a = 2c / 5, so that to2 = 4c2/25. Substituting in the first equation, we have to 5 * 4c2/ 25 + c = c2, from where 4c2/ 5 + c = c2. Removing denominators, we have to 4c2 + 5c = 5c2. This equation is equivalent to 0 = c2 - 5c, which is a second degree equation with two solutions.
The solution c = 0 It makes no sense in this problem (if Carla is 0 years old, and also all her sisters, Ana is already the same age as Carla and the statement itself would not make sense).
The other solution is c = 5, and in that case a = 2 Y b = 4. We can verify that this solution fits the statement.