# Difference between revisions of "1964 AHSME Problems/Problem 24"

## Problem

Let $y=(x-a)^2+(x-b)^2, a, b$ constants. For what value of $x$ is $y$ a minimum? $\textbf{(A)}\ \frac{a+b}{2} \qquad \textbf{(B)}\ a+b \qquad \textbf{(C)}\ \sqrt{ab} \qquad \textbf{(D)}\ \sqrt{\frac{a^2+b^2}{2}}\qquad \textbf{(E)}\ \frac{a+b}{2ab}$

## Solution 1

Expanding the quadratic and collecting terms gives $y = 2x^2 - (2a + 2b)x + (a^2+b^2)$. For a quadratic of the form $y = Ax^2 + Bx + C$ with $A>0$, $y$ is minimized when $x = -\frac{B}{2A}$, which is the average of the roots.

Thus, the quadratic is minimized when $x = \frac{2a+2b}{2\cdot 2} = \frac{a+b}{2}$, which is answer $\boxed{\textbf{(A)}}$.

## Solution 2

The problem should return real values for $ab = 0$ and $ab < 0$, which eliminates $E$ and $C$. We want to distinguish between options $A, B, D$, and testing $(a, b) = (2, 0)$ should do that, as answers $A, B, D$ will turn into $2, 1, \sqrt{2}$, respectively.

PLugging in $(a, b) = (2, 0)$ gives $y = (x - 2)^2 + x^2$, or $y = 2x^2 - 4x + 4$. This has a minimum at $x = -\frac{-4}{2}$, or at $x=2$. This is answer $\boxed{\textbf{(A)}}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 