**Example 1**. Three children and two adults want to cross river using a row boat. The boat can carry no more than a single adult or, in case there is no adult in the boat, a maximum of two children. Show that the least number of times the boat needs to cross the river to transport all five people is 11.

**Example 2**. In the following figure, *ABC* is an isosceles triangle with an inscribed circle, whose centre is at *O*. Let *P* be the mid-point of *BC*. If *AB* = *AC* = 15 and *BC* = 10, find the value of *OP*.

**Example 3**. The numbers 4 and 52 share the following features:

- both are sums of two squares, i.e., 4 = 0
^{2}+ 2^{2}, 52 = 4^{2}+ 6^{2}; - both exceed another square by 3, i.e., 4 – 3 = 1
^{2}; 52 – 3 = 7^{2}.

Show that there are infinitely many numbers that have these two characteristics.

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**. This is a slight variation of a classic puzzle in which there are two adults and only two children. Common sense is the best approach.

The crucial task is to get the two adults across. There is no point in an adult bringing the boat back. Nor is there any point in a single child rowing the boat forward except possibly at the end. Note also that the last forward trip cannot be by an adult, because for this to be possible, somebody else must have brought the boat backward in the last but one trip, and this person will not be able to go with the adult in the last forward trip.

These observations serve to design a scheme of rowing and also show that it is optimal, i.e., has least number of trips.

Call the two adults as *A* and *B* and the three children as *C*, *D* and *E*. It is foolish to begin with only one person in the very first trip because then he/she will also have to bring the boat back and thus the first two trips are a waste. So, first send two children, say *C* and *D*, across and let *C* bring the boat back. Then *A* can go across and *D* will bring the boat back.

So a minimum of four crossings is needed to get an adult across and bring the boat back. After two such sessions of four trips each for *A* and *B*, only *C*, *D* and *E* will be left. Any two of them can go forward, either one of them can bring the boat backward and take the third child with him/her in the last forward trip.

Hence the minimum number of trips needed is 4 + 4 + 3 = 11.

**Solution 2**. Denote the length of *OP* by *r*.

Since *O* is the radius of the inscribed circle, in the usual terminology, *O* is the in-centre and *r* is the in-radius of Δ*ABC*.

From the given data, semi-perimeter *s* of Δ*ABC* equals (15 + 15 + 10) / 2 = 20.

Hence the area of the triangle *ABC* equals *sr* = 20*r*. But the area also equals (*h* × *BC*) / 2, where *h* (= *AP*) is the altitude through *A*.

As *AB* = *AC*, *AP* ⊥ *BC*. Also, *P* is the mid-point of *BC*. From the right-angled triangle *APC*, we have

It follows that

**Solution 3**. We need to show that there are infinitely many integer values of *k* such that *k *= *a*^{2} + *b*^{2} and *k* = *c*^{2 }+ 3 for non-negative integers *a*, *b* and *c*. Here is a way of generating such values.

Let *a* = 2*n*, *b* = 2*n*^{2} − 2 and *c* = 2*n*^{2} – 1 for a positive integer *n*.

Then

a^{2}+b^{2}= 4

n^{2}+ 4(n^{2}− 1)^{2}= 4

n^{2}+ 4(n^{4}− 2n^{2}+ 1)= 4(

n^{4}−n^{2}+1);

and

c^{2}+ 3= (2

n^{2}− 1)^{2}+ 3= 4

n^{4}− 4n^{2}+ 1 + 3= 4(

n^{4}−n^{2}+ 1).

It follows that, *a*^{2} + *b*^{2} = *c*^{2} + 3.

Moreover, there is an infinite sequence of numbers *k* = 4(*n*^{4} − *n*^{2} + 1) with these two characteristics, where *n* = 1, 2, 3, . . .

*Note*. The numbers 4 and 52 correspond to *n* = 0 and *n* = 2 respectively.