Week 14

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. 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 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◦ .