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 "1997 AHSME Problems/Problem 25"

(→‎Solution)
(→‎Problem)
 
(3 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
−
Let <math>ABCD</math> be a parallelogram and let <math>\overrightarrow{AA^\prime}</math>, <math>\overrightarrow{BB^\prime}</math>, <math>\overrightarrow{CC^\prime}</math>, and <math>\overrightarrow{DD^\prime}</math> be parallel rays in space on the same side of the plane determined by <math>ABCD</math>. If <math>AA^\prime = 10</math>, <math>BB^\prime = 8</math>, <math>CC^\prime = 18</math>, and <math>DD^\prime = 22</math> and <math>M</math> and <math>N</math> are the midpoints of <math>A^\primeC^\prime</math> and <math>B^\primeD^\prime</math>, respectively, then <math>MN = </math>  
+
Let <math>ABCD</math> be a parallelogram and let <math>\overrightarrow{AA^\prime}</math>, <math>\overrightarrow{BB^\prime}</math>, <math>\overrightarrow{CC^\prime}</math>, and <math>\overrightarrow{DD^\prime}</math> be parallel rays in space on the same side of the plane determined by <math>ABCD</math>. If <math>AA^{\prime} = 10</math>, <math>BB^{\prime}= 8</math>, <math>CC^\prime = 18</math>, and <math>DD^\prime = 22</math> and <math>M</math> and <math>N</math> are the midpoints of <math>A^{\prime} C^{\prime}</math> and <math>B^{\prime}D^{\prime}</math>, respectively, then <math>MN =</math>
  
 
<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4 </math>
 
<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4 </math>
Line 7: Line 7:
 
==Solution==
 
==Solution==
  
−
Let <math>ABCD</math> be a unit square with <math>A(0,0,0)</math>, <math>B(0,1,0)</math>, <math>C(1,1,0)</math>, and <math>D(1,0,0)</math>.  Assume that the rays go in the +z direction.  In this case, <math>A^\prime(0,0,10)</math>, <math>B^\prime(0,1,8)</math>,  <math>C^\prime(1,1,18)</math>, and <math>D^\prime(1,0,22)</math>.  Finding the midpoints of <math>A^\prime C^\prime</math> and <math>B^\prime D^\prime</math> gives <math>M(\frac{1}{2}, \frac{1}{2}, 14)</math> and <math>M(\frac{1}{2}, \frac{1}{2}, 15)</math>.  The distance <math>MN</math> is <math>15 - 14 = 1</math>, and the answer is <math>\boxed{B}</math>.     
+
Let <math>ABCD</math> be a unit square with <math>A(0,0,0)</math>, <math>B(0,1,0)</math>, <math>C(1,1,0)</math>, and <math>D(1,0,0)</math>.  Assume that the rays go in the +z direction.  In this case, <math>A^\prime(0,0,10)</math>, <math>B^\prime(0,1,8)</math>,  <math>C^\prime(1,1,18)</math>, and <math>D^\prime(1,0,22)</math>.  Finding the midpoints of <math>A^\prime C^\prime</math> and <math>B^\prime D^\prime</math> gives <math>M(\frac{1}{2}, \frac{1}{2}, 14)</math> and <math>N(\frac{1}{2}, \frac{1}{2}, 15)</math>.  The distance <math>MN</math> is <math>15 - 14 = 1</math>, and the answer is <math>\boxed{B}</math>.     
 +
 
  
 
While this solution is not a proof, the problem asks for a value <math>MN</math> that is supposed to be an invariant.  Additional assumptions that are consistient with the problem's premise should not change the answer.  The assumptions made in this solution are that <math>ABCD</math> is a square, and that all rays <math>XX^\prime</math> are perpendicular to the plane of the square.  These assumptions are consistient with all facts of the problem.  Additionally, the assumptions allow for a quick solution to the problem.
 
While this solution is not a proof, the problem asks for a value <math>MN</math> that is supposed to be an invariant.  Additional assumptions that are consistient with the problem's premise should not change the answer.  The assumptions made in this solution are that <math>ABCD</math> is a square, and that all rays <math>XX^\prime</math> are perpendicular to the plane of the square.  These assumptions are consistient with all facts of the problem.  Additionally, the assumptions allow for a quick solution to the problem.
Line 13: Line 14:
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1997|num-b=24|num-a=26}}
 
{{AHSME box|year=1997|num-b=24|num-a=26}}
 +
{{MAA Notice}}

Latest revision as of 21:13, 7 December 2023

Problem

Let $ABCD$ be a parallelogram and let $\overrightarrow{AA^\prime}$, $\overrightarrow{BB^\prime}$, $\overrightarrow{CC^\prime}$, and $\overrightarrow{DD^\prime}$ be parallel rays in space on the same side of the plane determined by $ABCD$. If $AA^{\prime} = 10$, $BB^{\prime}= 8$, $CC^\prime = 18$, and $DD^\prime = 22$ and $M$ and $N$ are the midpoints of $A^{\prime} C^{\prime}$ and $B^{\prime}D^{\prime}$, respectively, then $MN =$

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution

Let $ABCD$ be a unit square with $A(0,0,0)$, $B(0,1,0)$, $C(1,1,0)$, and $D(1,0,0)$. Assume that the rays go in the +z direction. In this case, $A^\prime(0,0,10)$, $B^\prime(0,1,8)$, $C^\prime(1,1,18)$, and $D^\prime(1,0,22)$. Finding the midpoints of $A^\prime C^\prime$ and $B^\prime D^\prime$ gives $M(\frac{1}{2}, \frac{1}{2}, 14)$ and $N(\frac{1}{2}, \frac{1}{2}, 15)$. The distance $MN$ is $15 - 14 = 1$, and the answer is $\boxed{B}$.


While this solution is not a proof, the problem asks for a value $MN$ that is supposed to be an invariant. Additional assumptions that are consistient with the problem's premise should not change the answer. The assumptions made in this solution are that $ABCD$ is a square, and that all rays $XX^\prime$ are perpendicular to the plane of the square. These assumptions are consistient with all facts of the problem. Additionally, the assumptions allow for a quick solution to the problem.

See also

1997 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 26
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

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=1997_AHSME_Problems/Problem_25&oldid=206602"