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?**

#### Solution

If we call * to* at Ana's age,

*to Blanca's and*

**b***to that of Carla, the first information they offer us directly provides the equation*

**c***.*

**to**^{2}+ b^{2}+ c = c2For 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

*. That Carla is four times the age Ana is now, means that*

**a + (c - a) = c***, Or what is the same,*

**c + c - a = 4a***.*

**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

*so that the first one is*

**b**^{2}= 4th^{2}*, Or what is the same,*

**to**^{2}+ 4th^{2}+ c = c2*.*

**5th**^{2}+ c = c^{2}From the other equation we get * a = 2c / 5*, so that

*. Substituting in the first equation, we have to*

**to**^{2}= 4c^{2}/25*, from where*

**5 * 4c**^{2}/ 25 + c = c^{2}*. Removing denominators, we have to*

**4c**^{2}/ 5 + c = c^{2}*. This equation is equivalent to*

**4c**^{2}+ 5c = 5c^{2}*, which is a second degree equation with two solutions.*

**0 = c**^{2}- 5cThe 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

*Y*

**a = 2***. We can verify that this solution fits the statement.*

**b = 4**