Choosing Correct Variables

When reviewing solutions presented by others, we often see how taking a different perspective allows us to greatly reduce the apparent difficulty of the question. This allows us to find an easier approach to the problem, because we have identified the key aspects. Determining the correct perspective is a skill that is slowly gained over time and part of it involves choosing the correct variables to interpret the question.

When a problem seems to have insufficient information, this often requires us to approach the problem from another viewpoint. If we can identify the stumbling blocks, it is possible for us to work on it from another direction. Let's start with a simple example.

If a+c=7a+c = 7 and b+c=11 b + c = 11 ,what is the value of ab+bc+ca+c2 ab + bc + ca + c^2 ?

First approach: Starting with 2 linear equations, we might be tempted to solve it in order to find the variables. However, since there are 3 variables, this system of linear equations is underdetermined, and we cannot compute the exact values of aa, bb and cc. This is our stumbling block.

Second approach: Since we do not have enough information, let's rewrite all the variables in terms of aa. In this case, we have c=7a c = 7-a and b=11c=11(7a)=4+a b = 11 - c = 11 - (7-a) = 4 + a . Hence,

ab+bc+ca+c2=a(4+a)+(4+a)(7a)+(7a)a+(7a)2=77. ab + bc + ca + c^2 = a(4+a) + (4+a)(7-a) + (7-a)a + (7-a)^2 = 77.

Amazingly, the variable aa isn't involved in the final expression! _\square

Third approach: Looking at ab+bc+ca+c2ab+bc+ca+c^2 , since the variable cc is involved several times, it might be better for us to expand in terms of cc. In this case, we have a=7c a = 7-c and b=11c b = 11 - c . Hence,

ab+bc+ca+c2=(7c)(11c)+(11c)c+c(7c)+c2=77. ab + bc + ca + c^2 = (7-c)(11-c)+(11-c)c + c(7-c) + c^2 = 77.

We have one less expansion to do, which could save us on careless mistakes. _\square

Fourth approach: Staring at ab+bc+ca+c2 ab+bc+ca+c^2, we realize that it can be factorized into (a+c)(b+c) (a+c) ( b+c) . Hence, if we set x=a+c x = a+c and y=b+c y = b+c , then the question is essentially asking for xy xy , and the answer is 7×11=77 7 \times 11 = 77 . _\square

Worked Examples

1. The tortoise and the rabbit race in a 21 kilometer half marathon. The rabbit can run 4 times as fast as the tortoise. Less than halfway into the race, the rabbit dropped his lucky rabbit's foot, and so had to stop to find it. After finding it, he started running to catch up with the tortoise. When the tortoise reached the finishing line, the rabbit was 1 kilometer away. How far (in km) did the tortoise crawl while the rabbit was looking for his lucky rabbit's foot?

There are many variables defined in the question, including
1. Speed of rabbit
2. Speed of tortoise
3. Distance that rabbit initially ran
4. Time that rabbit was searching for his foot
5. Distance that rabbit ran after finding foot
6. Distance that the tortoise was crawling while the rabbit was looking for his foot.

However, very few equations are given, and it would thus be hopeless to try and solve for each of them.

Solution: Let's focus on the distances that the tortoise crawled while the rabbit was running, and while the rabbit was stationary. We know that the rabbit ran 211=20 21 - 1 = 20 kilometers. While the rabbit was running this distance, the tortoise can only crawl 204=5 \frac{20}{4} = 5 kilometers. Hence, when the rabbit stooped, the tortoise must have crawled the remaining distance, which is 215=16 21 - 5 = 16. _\square


2. x1x_1, x2x_2, \ldots x7x_7 are real numbers that satisfy:

x1+4x2+9x3+16x4+25x5+36x6+49x7=14x1+9x2+16x3+25x4+36x5+49x6+64x7=1016x1+25x2+36x3+49x4+64x5+81x6+100x7=100.\begin{array} { r r r r r r r r l} x_1 &+ 4x_2 &+ 9x_3 &+ 16x_4 &+25x_5 &+ 36x_6& + 49x_7 &=& 1 \\ 4x_1 &+ 9x_2 &+ 16x_3 &+ 25x_4 &+ 36x_5 &+ 49x_6 &+ 64x_7 &=& 10 \\ 16x_1 &+ 25x_2 &+ 36x_3 &+ 49x_4 &+ 64x_5 &+ 81x_6 &+ 100x_7 &=& 100. \\ \end{array}

What is the value of 9x1+16x2+25x3+36x4+49x5+64x6+81x79x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7?

Almost certainly, with 7 variables and 3 equations, the system is under determined. Also, it seems difficult to express variables in terms of each other, and hope that the calculations simplify.

Solution: Consider the polynomial f(n)=(n)2(x1)+(n+1)2(x2)++(n+6)2(x7), f(n) = (n)^2 (x_1) + (n+1)^2 (x_2) + \ldots + (n+6)^2(x_7),

which is quadratic in nn. We are given that f(1)=1,f(2)=10,f(4)=100 f(1) = 1, f(2) = 10, f(4) = 100 , hence by Lagrange Interpolation Formula, the quadratic must be f(n)=12n227n+16f(n) = 12n^2 - 27n + 16. Thus, the answer is given by evaluating ff at n=3n=3, which gives f(3)=12(9)27(3)+16=43 f(3) = 12(9) - 27(3) + 16 = 43 . _\square

Note by Calvin Lin
7 years, 3 months ago

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Is Lagrange interpolation important for jee advance

Chaitya Shah - 7 years ago

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i dont think so.

Rishi Sharma - 6 years, 6 months ago

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Not at all. Its not even in prescribed syllabus. Though you may want to apply it in question based on something else.

Pranjal Jain - 6 years, 6 months ago

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Could you please explain how you got the desired quadratic.

Aayush Patni - 6 years, 5 months ago

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Which quadratic are you referring to? Are you talking about worked example 2?

Calvin Lin Staff - 6 years, 5 months ago

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Yes the 'n' one

Aayush Patni - 6 years, 4 months ago

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