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"

(→‎Solution)
m (→‎Solution)
Line 28: Line 28:
 
(a) Triangles <math>ABP</math> and <math>MNP</math> are similar, and since <math>PM=\frac{1}{2}AP</math>, <math>MN=\frac{1}{2}AB</math>.
 
(a) Triangles <math>ABP</math> and <math>MNP</math> are similar, and since <math>PM=\frac{1}{2}AP</math>, <math>MN=\frac{1}{2}AB</math>.
  
−
(b) Obviously, the perimeter changes. For example, imagine if P was extremely far to the left.
+
(b) We see the perimeter changes. For example, imagine if P was extremely far to the left.
  
 
(c) The area clearly doesn't change, as both the base <math>AB</math> and its corresponding height remain the same.
 
(c) The area clearly doesn't change, as both the base <math>AB</math> and its corresponding height remain the same.

Revision as of 00:24, 27 July 2023

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]

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

Solution

(a) Triangles $ABP$ and $MNP$ are similar, and since $PM=\frac{1}{2}AP$, $MN=\frac{1}{2}AB$.

(b) We see the perimeter changes. For example, imagine if P was extremely far to the left.

(c) The area clearly doesn't change, as both the base $AB$ and its corresponding height remain the same.

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

Only $1$ quantity changes, so the correct answer is $\boxed{\text{B}}$.

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

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=2000_AMC_10_Problems/Problem_5&oldid=196119"