Example 1. When 98 is added to a perfect square, another perfect square is obtained. How many such pairs of perfect squares exist?
Example 2. There are 1000 doors named D1, D2, D3, . . . , D1000. There are 1000 persons named P1, P2, P3, . . . , P1000. At first all the doors are closed. P1 goes and opens all the doors. Then P2 goes and closes even numbered doors i.e., D2, D4, D6, . . . , D1000 and leaves the odd numbered doors i.e., D1, D3, D5, . . . , D999 as it is. Then P3 goes and changes the state of the doors (if open then closes and if closed then opens) which are multiple of 3 i.e., the doors D3, D6, . . . Then P4 goes and changes the state of the doors which are multiple of 4. In this way 1000 persons goes and changes the states of the doors accordingly. At the end how many doors will be open and how many doors will be closed?
Example 3. Let ABC be a triangle and D be the mid-point of BC. Suppose the angle bisector of ∠ADC is tangent to the circumcircle of triangle ABD at D. Prove that ∠A = 90◦.
Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own.
Just remember the words of Paul Halmos, who says “the only way to learn mathematics is to do mathematics”.
Solution 1. According to the symmetry of the given question, we have
- A2 = B2 + 98
- ⇒ A2 – B2 = 98
- ⇒ (A – B) (A + B) = 98
Different possibilities for A, A2and B, B2 are given in the following table:
|A||A2||B||B2||A2 − B2||Remarks|
|Even||Even||Even||Even||Even||May be possible|
|Odd||Odd||Odd||Odd||Even||May be possible|
|Even||Even||Odd||Odd||Odd||Not possible (as the difference = 98)|
|Odd||Odd||Even||Even||Odd||Not possible (as the difference = 98)|
Following cases we need to consider:
Case 1. Let both A and B are even.
Then both A2 and B2 are going to be even. In fact, they are all multiples of 4.
Hence A2 – B2 is also a multiple of four, say 4k.
But 98 is not divisible by 4.
So, both A and B are even is not possible.
Case 2. Let both A and B are odd.
Then (A – B) and (A + B) both will be even.
Hence the product (A – B) (A + B) looks like an even number × an even number.
This is definitely a multiple of four, say 4m.
But 98 is not divisible by 4.
So, both A and B are odd is not possible.
Case 3. Let any one of A and B is even and the other being odd.
Clearly, square of an even number is even and that of an odd number is odd.
Hence A2 – B2 is always an odd number.
But 98 is an even number.
So, any one of A and B is even and the other being odd is impossible.
Hence we conclude now that no such set exists.
Solution 2. Let us take an example.
Let us see how many times and which persons are operating on D28.
The persons who are operating on D28 are P1, P2, P4, P7, P14, P28.
Now, 1, 2, 4, 7, 14, 28 are factors of 28.
So, any door Di is getting operated by the persons Pj where j’s are the factors of i.
In this case we are not interested to find the factors of every number up to 1000. Rather we are interested in which doors are getting operated odd number of times and which doors are getting operated even number of times.
In other words, which numbers have odd number of factors and which numbers have even number of factors.
If there are odd number of factors, then odd number of operations and the door will be open. On the other hand, if there are even number of factors then even number of operations and the door will be closed.
We have seen that every number has even number of factors except the square numbers, which have odd number of factors.
So, D1, D4, D9, D16, . . . , D961 these 31 doors will stay open and rest of the doors will stay closed.
Solution 3. Let P be the centre of the circumcircle T of triangle ABC.
Let the tangent at D to T intersect AC in E.
Then PD ⊥ DE. Since DE bisects ∠ADC, this implies that DP bisects ∠ADB.
Hence the circumcentre and the incentre of triangle ABD lies on the same line DP.
This implies that DA = DB. Thus DA = DB = DC and hence D is the circumcentre of triangle ABC.
This gives ∠A = 90◦ .