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

Difference between revisions of "2012 AMC 10B Problems/Problem 12"

(→‎Solution)
(→‎Solution)
Line 19: Line 19:
 
pair D=(10,30);
 
pair D=(10,30);
 
label ("D",(10,30),E);
 
label ("D",(10,30),E);
−
label (10,(A--B),S);   
+
label ("10",(A--B),S);   
 
dot(A);
 
dot(A);
 
dot(B);
 
dot(B);

Revision as of 13:28, 31 December 2019

Problem

Point B is due east of point A. Point C is due north of point B. The distance between points A and C is $10\sqrt 2$, and $\angle BAC= 45^\circ$. Point D is 20 meters due north of point C. The distance AD is between which two integers?


$\textbf{(A)}\ 30\ \text{and}\ 31 \qquad\textbf{(B)}\ 31\ \text{and}\ 32 \qquad\textbf{(C)}\ 32\ \text{and}\ 33 \qquad\textbf{(D)}\ 33\ \text{and}\ 34 \qquad\textbf{(E)}\ 34\ \text{and}\ 35$

Solution

[asy] unitsize(4); pair A=(0,0); label ("A",(0,0),W); pair B=(10,0); label ("B",(10,0),E); pair C=(10,10); label ("C",(10,10),E); pair D=(10,30); label ("D",(10,30),E); label ("10",(A--B),S); dot(A); dot(B); dot(C); dot(D); draw(A--B); draw(A--C); draw(A--D); draw(C--D); draw(B--C); [/asy] If point B is due east of point A and point C is due north of point B, $\angle CBA$ is a right angle. And if $\angle BAC = 45^\circ$, $\triangle CBA$ is a 45-45-90 triangle. Thus, the lengths of sides $CB$, $BA$, and $AC$ are in the ratio $1:1:\sqrt 2$, and $CB$ is $10 \sqrt 2 \div \sqrt 2 = 10$.

$\triangle DBA$ is clearly a right triangle with $C$ on the side $DB$. $DC$ is 20, so $DB = DC + CB = 20 + 10 = 30$.

By the Pythagorean Theorem, $DA = \sqrt {DB^2 + BA^2} = \sqrt {30^2 + 10 ^2} = \sqrt {1000}$.

$31^2 = 961$, and $32^2 = 1024$. Thus, $\sqrt {1000}$ must be between $31$ and $32$. The answer is $\boxed {B}$.

See Also

2012 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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
All AMC 10 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_12&oldid=113846"