Difference between revisions of "2012 AMC 10B Problems/Problem 12"
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== 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> | ||
− | |||
== 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>. | ||
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{{AMC10 box|year=2012|ab=B|num-b=11|num-a=13}} | {{AMC10 box|year=2012|ab=B|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | [[Category: Introductory Geometry Problems]] |
Latest revision as of 20:22, 3 September 2021
Problem
Point is due east of point
. Point
is due north of point
. The distance between points
and
is
, and
. Point
is
meters due north of point
. The distance
is between which two integers?
Solution
If point B is due east of point A and point C is due north of point B,
is a right angle. And if
,
is a 45-45-90 triangle. Thus, the lengths of sides
,
, and
are in the ratio
, and
is
.
is clearly a right triangle with
on the side
.
is 20, so
.
By the Pythagorean Theorem, .
, and
. Thus,
must be between
and
. The answer is
.
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.