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)
m
 
(29 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== 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 <math>10\sqrt 2</math>, and <math>\angle BAC= 45^\circ</math>. Point D is 20 meters due north of point C. The distance AD is between which two integers?
+
Point <math>B</math> is due east of point <math>A</math>. Point <math>C</math> is due north of point <math>B</math>. The distance between points <math>A</math> and <math>C</math> is <math>10\sqrt 2</math>, and <math>\angle BAC = 45^\circ</math>. Point <math>D</math> is <math>20</math> meters due north of point <math>C</math>. The distance <math>AD</math> is between which two integers?
−
 
 
  
 
<math>\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</math>
 
<math>\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</math>
Line 8: Line 7:
 
== Solution ==
 
== 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);
 +
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, <math>\angle CBA</math> is a right angle. And if <math>\angle BAC = 45^\circ</math>, <math>\triangle CBA</math> is a 45-45-90 triangle. Thus, the lengths of sides <math>CB</math>, <math>BA</math>, and <math>AC</math> are in the ratio <math>1:1:\sqrt 2</math>, and <math>CB</math> is <math>10 \sqrt 2 \div \sqrt 2 = 10</math>.
 
If point B is due east of point A and point C is due north of point B, <math>\angle CBA</math> is a right angle. And if <math>\angle BAC = 45^\circ</math>, <math>\triangle CBA</math> is a 45-45-90 triangle. Thus, the lengths of sides <math>CB</math>, <math>BA</math>, and <math>AC</math> are in the ratio <math>1:1:\sqrt 2</math>, and <math>CB</math> is <math>10 \sqrt 2 \div \sqrt 2 = 10</math>.
 +
 +
<math>\triangle DBA</math> is clearly a right triangle with <math>C</math> on the side <math>DB</math>. <math>DC</math> is 20, so <math>DB = DC + CB = 20 + 10 = 30</math>.
 +
 +
By the Pythagorean Theorem, <math>DA = \sqrt {DB^2 + BA^2} = \sqrt {30^2 + 10 ^2} = \sqrt {1000}</math>.
 +
 +
<math>31^2 = 961</math>, and <math>32^2 = 1024</math>. Thus, <math>\sqrt {1000}</math> must be between <math>31</math> and <math>32</math>. The answer is <math>\boxed {B}</math>.
 +
 +
==See Also==
 +
 +
{{AMC10 box|year=2012|ab=B|num-b=11|num-a=13}}
 +
{{MAA Notice}}
 +
[[Category: Introductory Geometry Problems]]

Latest revision as of 20:22, 3 September 2021

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); 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=161520"