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

(Solution 1)
(Solution 2)
 
Line 17: Line 17:
 
==Solution 2==
 
==Solution 2==
  
The problem should return real values for <math>ab = 0</math> and <math>ab < 0</math>, which eliminates <math>E</math> and <math>C</math>.  We want to distinguish between options <math>A, B, D</math>, and testing <math>(a, b) = (2, 0)</math> should do that, as answers <math>A, B, D</math> will turn into <math>2, 1, \sqrt{2}</math>, respectively.
+
The problem should return real values for <math>ab = 0</math> and <math>ab < 0</math>, which eliminates <math>E</math> and <math>C</math>.  We want to distinguish between options <math>A, B, D</math>, and testing <math>(a, b) = (2, 0)</math> should do that, as answers <math>A, B, D</math> will turn into <math>1, 2, \sqrt{2}</math>, respectively.
  
PLugging in <math>(a, b) = (2, 0)</math> gives <math>y = (x - 2)^2 + x^2</math>, or <math>y = 2x^2 - 4x + 4</math>.  This has a minimum at <math>x = -\frac{-4}{2}</math>, or at <math>x=2</math>.  This is answer <math>\boxed{\textbf{(A)}}</math>.
+
PLugging in <math>(a, b) = (2, 0)</math> gives <math>y = (x - 2)^2 + x^2</math>, or <math>y = 2x^2 - 4x + 4</math>.  This has a minimum at <math>x = -\frac{-4}{4}</math>, or at <math>x=1</math>.  This is answer <math>\boxed{\textbf{(A)}}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 03:19, 24 July 2019

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 $1, 2, \sqrt{2}$, 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 $x = -\frac{-4}{4}$, or at $x=1$. This is answer $\boxed{\textbf{(A)}}$.

See Also

1964 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
All AHSME Problems and Solutions

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

Invalid username
Login to AoPS