Always Two

525 /www/nrich/html/content/97/09/six4/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <table border="0"> <tbody> <tr> <td><mdo:image height="136" width="138" src="alwaystwo.jpg" alt="fig"></mdo:image></td> <td>Find all the triples of numbers $a$, $b$, $c$ such that each one of them plus the product of the other two is always $2$.</td> </tr> </tbody> </table> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> The examples below give a feel for the range of solutions we received. They are interesting in their variety of approaches and detail. Well done to all of you. Firstly, Rachael and Katy of Ardingly College Junior School sent a solution obtained through trial and improvement. Some people undervalue this valid mathematical method and it can often lead us to a complete answer or, at the very least, throw light on what to be looking for to find a fuller solution to the problem. For this problem we have found more than one solution. To work this out we used trial and error. We started with $0.5$ giving $0.5 \times 0.5 + 0.5 = 0.25 + 0.5 = 0.75$ The answer was too low so we tried 1 giving $1 \times 1 + 1 = 1 + 1 = 2$. This is a correct solution. The numbers are the same and so equal the same whichever way round they go. Then I remembered that if you square a negative number the answer is positive so I tried $-2$ giving $-2\ \times -2 + (-2) = 4 + (-2) = 2 $. This is another possible solution. James sent us this solution. From the question we can form $3$ simultaneous equations. <div class="math">\begin{eqnarray} ab+ c&=&2 \\ bc+ a&=&2 \\ ca+ b&=&2. \end{eqnarray}</div> So, $$ab + c= ac+ b. $$ Take $b$ and $c$ from each side: $$ a(b-1) = a(c-1) $$ So, if $a$ is not equal to zero, then $b = c$. Since any two of the three equations could have been chosen, by symmetry we must have $a = b= c$. This gives us the quadratic: <div class="math">\begin{eqnarray} a^2 + a&=& 2 \\ a^2 + a- 2 &=& 0 \\ (a-1)(a+2) &=& 0 \end{eqnarray}</div> So the only 2 solutions are: $a =b =c = 1$ and $-2 $. A complete solution, which explains the equality of the three unknowns more fully was supplied by Derek of Tin College. We seek numbers $a, b$ and $c$ such that: $a + b c = 2 \quad (1)$ $b + c a = 2 \quad (2)$ $c + a b = 2 \quad (3)$ By subtracting (1)-(2) and simplifying we get $(a-b)(c-1) = 0 \quad (4)$ and similarly $(b - c)(a - 1)=0 \quad (5)$ $(c - a)(b - 1)=0 \quad (6)$ Notice that if $a=1$ then (from (4)) we have either $b=1$ or $c=1$. Then using (3) or (2) respectively, we deduce that $a=b=c=1$. Similarly if $b=1$ or $c=1$ then it follows that $a=b=c=1$. Now suppose for the moment that none of $a$, $b$ and $c$ are 1 because we are looking for other solutions. Then from (4), (5) and (6), we deduce $a = b = c$. From (1), $a^2 + a - 2 = 0$ and factorising this we get $(a - 1)(a + 2) = 0$. As $a \ne 1$, this gives $a = -2$ and hence $a = b = c = -2$. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why use this problem?</h3> The problem gives practice in writing equations from verbal information and in algebraic manipulation. Learners will experience the value of recognising and making use of the symmetries in the algebraic expressions that occur. As a non-standard problem, it is designed for learners to think for themselves but it does not require any mathematical knowledge beyond knowing how to solve two linear simultaneous equations in two unknowns. <h3>Possible approach</h3> Encourage learners to work in small groups to discuss how they might tackle the problem, then to work out the solutions individually, and finally to check together if their answers agree.This is reassuring for people who are inclined to panic at the unfamiliar and gives practice in communication of mathematical ideas. Learners may find one or both of the solutions by trial and error but then they need to prove that there are no other solutions. <h3>Key questions</h3> Are you using the symmetry of the expressions? If you subtract one equation from another in pairs, what do you notice? What can you deduce if the product of two linear factors in an equation is zero? <h3>Possible support</h3> In order to get some experience of thinking for themselves, and not simply following set procedures to solve a system of equations, the class could first try the problem <a href="http://nrich.maths.org/public/viewer.php?obj_id=543&part=index"> System Speak</a> which is another non-standard problem on simultaneous equations. System Speak is easier in so far as it can be solved by expressing all the letters in terms of one of the letters (eliminating the other variables) and reaching a final equation in one variable. As practice in solving more standard sets of simultaneous equations with unit coefficients in more than 2 unknowns learners could try <a href="http://nrich.maths.org/public/viewer.php?obj_id=5970&part=index"> Equation Sudoko.</a> <h3>Possible Extension</h3> Try <a href="http://nrich.maths.org/public/viewer.php?obj_id=1941&part=index"> Leonardo's Problem</a> where you have first to create the equations then to solve them. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> As some solutions are easily spotted, you may be able to find solutions by trial and error. You still need to prove that the solutions you have found are the only ones. Write down three simultaneous equations from the statement of the problem. Take care to keep the symmetry of the expressions. In pairs subtract one equation from another. What can you deduce if the product of two linear factors is zero? </mdoxml> 2 4 0 0 0 1 1 Always Two Find all the triples of numbers a, b, c such that each one of them plus the product of the other two is always 2. Algebra Simultaneous equations Algebra Factorisation (algebraic) Algebra Quadratic equations Algebra Manipulating algebraic expressions/formulae