Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC 10A/12A testing, the AoPS Wiki is in read-only mode and no edits can be made.

1978 AHSME Problems/Problem 11: Difference between revisions

Created page with "The circle <math>x^2 + y^2 = r</math> has center <math>(0,0)</math> and radius <math>\sqrt{r}</math>. Therefore, if the line <math>x + y = r</math> is tangent to the circle <m..."
 
Coolmath34 (talk | contribs)
editor
376 edits
No edit summary
 
Line 1: Line 1:
== Problem 11 ==
If <math>r</math> is positive and the line whose equation is <math>x + y = r</math> is tangent to the circle whose equation is <math>x^2 + y ^2 = r</math>, then <math>r</math> equals
<math>\textbf{(A) }\frac{1}{2}\qquad
\textbf{(B) }1\qquad
\textbf{(C) }2\qquad
\textbf{(D) }\sqrt{2}\qquad
\textbf{(E) }2\sqrt{2}    </math>
== Solution ==
The circle <math>x^2 + y^2 = r</math> has center <math>(0,0)</math> and radius <math>\sqrt{r}</math>. Therefore, if the line <math>x + y = r</math> is tangent to the circle <math>x^2 + y^2 = r</math>, then the distance between <math>(0,0)</math> and the line <math>x + y = r</math> is <math>\sqrt{r}</math>.
The circle <math>x^2 + y^2 = r</math> has center <math>(0,0)</math> and radius <math>\sqrt{r}</math>. Therefore, if the line <math>x + y = r</math> is tangent to the circle <math>x^2 + y^2 = r</math>, then the distance between <math>(0,0)</math> and the line <math>x + y = r</math> is <math>\sqrt{r}</math>.


Line 6: Line 17:
<cmath>\frac{r}{\sqrt{2}} = \sqrt{r}.</cmath>
<cmath>\frac{r}{\sqrt{2}} = \sqrt{r}.</cmath>
Then <math>r = \sqrt{r} \cdot \sqrt{2}</math>, so <math>\sqrt{r} = \sqrt{2}</math>, which means <math>r = \boxed{2}</math> or (B), <math>2</math>.
Then <math>r = \sqrt{r} \cdot \sqrt{2}</math>, so <math>\sqrt{r} = \sqrt{2}</math>, which means <math>r = \boxed{2}</math> or (B), <math>2</math>.
==See Also==
{{AHSME box|year=1978|num-b=10|num-a=12}}
{{MAA Notice}}

Latest revision as of 11:26, 13 February 2021

Problem 11

If $r$ is positive and the line whose equation is $x + y = r$ is tangent to the circle whose equation is $x^2 + y ^2 = r$, then $r$ equals

$\textbf{(A) }\frac{1}{2}\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }\sqrt{2}\qquad \textbf{(E) }2\sqrt{2}$

Solution

The circle $x^2 + y^2 = r$ has center $(0,0)$ and radius $\sqrt{r}$. Therefore, if the line $x + y = r$ is tangent to the circle $x^2 + y^2 = r$, then the distance between $(0,0)$ and the line $x + y = r$ is $\sqrt{r}$.

The distance between $(0,0)$ and the line $x + y = r$ is \[\frac{|0 + 0 - r|}{\sqrt{1^2 + 1^2}} = \frac{r}{\sqrt{2}}.\] Hence, \[\frac{r}{\sqrt{2}} = \sqrt{r}.\] Then $r = \sqrt{r} \cdot \sqrt{2}$, so $\sqrt{r} = \sqrt{2}$, which means $r = \boxed{2}$ or (B), $2$.

See Also

1978 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1978_AHSME_Problems/Problem_11&oldid=146444"