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 "2000 AMC 10 Problems/Problem 5"

m (→‎Solution)
(→‎Problem)
Line 1: Line 1:
 
==Problem==
 
==Problem==
 +
 +
Points <math>M</math> and <math>N</math> are the midpoints of sides <math>PA</math> and <math>PB</math> of <math>\triangle PAB</math>. As <math>P</math> moves along a line that is parallel to side <math>AB</math>, how many of the four quantities listed below change?
 +
 +
(a) the length of the segment <math>MN</math>
 +
 +
(b) the perimeter of <math>\triangle PAB</math>
 +
 +
(c) the area of <math>\triangle PAB</math>
 +
 +
(d) the area of trapezoid <math>ABNM</math>
 +
 +
<asy>
 +
draw((2,0)--(8,0)--(6,4)--cycle);
 +
draw((4,2)--(7,2));
 +
draw((1,4)--(9,4),Arrows);
 +
label("$A$",(2,0),SW);
 +
label("$B$",(8,0),SE);
 +
label("$M$",(4,2),W);
 +
label("$N$",(7,2),E);
 +
label("$P$",(6,4),N);
 +
</asy>
 +
 +
<math>\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 2 \qquad\mathrm{(D)}\ 3 \qquad\mathrm{(E)}\ 4</math>
  
 
==Solution==
 
==Solution==

Revision as of 21:28, 8 January 2009

Problem

Points $M$ and $N$ are the midpoints of sides $PA$ and $PB$ of $\triangle PAB$. As $P$ moves along a line that is parallel to side $AB$, how many of the four quantities listed below change?

(a) the length of the segment $MN$

(b) the perimeter of $\triangle PAB$

(c) the area of $\triangle PAB$

(d) the area of trapezoid $ABNM$

[asy] draw((2,0)--(8,0)--(6,4)--cycle); draw((4,2)--(7,2)); draw((1,4)--(9,4),Arrows); label("$A$",(2,0),SW); label("$B$",(8,0),SE); label("$M$",(4,2),W); label("$N$",(7,2),E); label("$P$",(6,4),N); [/asy]

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

Solution

(a) Clearly does not change, as $MN=\frac{1}{2}AB$. Since $AB$ does not change, neither does $MN$.

(b) Obviously, the perimeter changes.

(c) The area clearly doesn't change, as the base and height remain the same.

(d) The bases $AB$ and $MN$ do not change, and neither does the height, so the trapezoid remains the same.

Only $1$ changes, so $\boxed{\text{B}}$.

See Also

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_10_Problems/Problem_5&oldid=29311"