2010 AMC 10B Problem 4

P1-10~1 minPrint

Functions Set up the equation/formula and compute; no special trick needed

For a real number $x$ , define $\heartsuit(x)$ to be the average of $x$ and $x^2$ . What is $\heartsuit(1)+\heartsuit(2)+\heartsuit(3)$ ?

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Key insight

The heart of a number is the number plus its square, divided by 2. For 1, 2 and 3 that gives 1, 3 and 6.

Solution

The average of two numbers is their sum divided by $2$ , so the heart of $x$ is
$\heartsuit(x) = \frac{x + x^2}{2}.$

$x = 1$ : the square is $1$ , so $\heartsuit(1) = \frac{1 + 1}{2} = 1$ .
$x = 2$ : the square is $4$ , so $\heartsuit(2) = \frac{2 + 4}{2} = 3$ .
$x = 3$ : the square is $9$ , so $\heartsuit(3) = \frac{3 + 9}{2} = 6$ .
Add them: $1 + 3 + 6 = 10$ .

The answer is $\boxed{\textbf{(C)}\ 10}$ .

Why this works

A made-up symbol like $\heartsuit$ is just a rule. Write the rule as a formula once, then plug in each number. (The values $1, 3, 6$ are the triangular numbers $1$ , $1+2$ , $1+2+3$ , because $\frac{x + x^2}{2} = \frac{x(x+1)}{2}$ .)

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

By definition $\heartsuit(x) = \dfrac{x + x^2}{2}$ . Evaluate at the three inputs:
$\heartsuit(1) = \frac{1+1}{2} = 1,\qquad \heartsuit(2) = \frac{2+4}{2} = 3,\qquad \heartsuit(3) = \frac{3+9}{2} = 6.$
Their sum is $1 + 3 + 6 = 10$ .

The answer is $\boxed{\textbf{(C)}\ 10}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 16: Functions
Introduction to Algebra, Chapter 17: Graphing Functions (§17.2)
Intermediate Algebra, Chapter 19: Functional Equations (§19.1, §19.2)