Adopting Different Point of View

The typical training that students receive in school basically prepares them to solve problems in a single, straightforward fashion. This leads to a solution, but not always in the most efficient way. It is sometimes beneficial to the problem solver to adopt a different point of view than that to which he or she was led initially by the problem. This sort of thinking was used when we discussed the working backwards strategy. The problem was inspected from a different point of view, namely, from the end back to the beginning. We use this sort of thinking automatically in everyday life. Consider the problem of being separated from and trying to find a friend in a crowd. It would be logical to stand on some sort of elevated platform so that your overview of the crowd is different (more advantageous); that is, so that you can adopt a different point of view (in this case, quite literally!)

Such a different point of view might be one where instead of focusing on the winners of a contest, we look at the losers. For example, suppose you were asked to determine how many games would have to be played in a single-elimination tennis tournament that began with 25 players. You would find it very cumbersome to work directly by setting up a diagram of the tournament, counting the winners in each round, following them through the hypothetical tournament, and arriving at a single winner. You would then have to count the number of games that had been played. Yes, the problem could be solved in this manner! It would be much more efficient, however, to look at this from a different point of view; that is, to ask yourself how many losers this tournament must have. The answer is simply 24. Since 1 player is eliminated each time a game is played, there would have to be 24 games to have 24 losers!

We can look at this problem from still another point of view. Consider (hypothetically, of course), that you know who the winner will be. Then suppose that this “winner” plays each of the other contestants and defeats them (of course). Because this is a single-elimination tournament, players do not need to play each other. This “winner” must play 24 games to get 24 losers!

Many times the complexity of the setting in which a problem is placed impedes the students’ ability to even understand what must be done. As a result, experienced problem solvers often adopt a slightly different point of view or setting in an attempt to simplify their understanding.

Now, solve these problems.

Problem 1

Find the area of the shaded region between triangle ABC and triangle GHI, if the corresponding sides of the three triangles are parallel and 1 unit apart. Triangle DEF lies midway between the other two triangles. The lengths of the three sides of triangle DEF are 5, 6, and 7 units. (See Figure 1.1)

Figure 1.1

Problem 2

What is the greatest value of the expression

ab + be + cd + ad

if a, b, c, and d have values 1, 2, 3, and 4, but not necessarily in that order?

Problem 3

In Figure 2.1, ABCD is a square, and P and Q are the midpoints of the sides. What is the ratio of the area of triangle DPQ to the area of the square?

Figure 2.1

to get the answer for these questions and others cases, please click in here

Quoted from “Problem-Solving Strategies for Efficient and Elegant Solutions

by Alfred S Posamentier and Stephen Krulik


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