Difference between revisions of "2000 AMC 10 Problems/Problem 5"

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<math>\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 2 \qquad\mathrm{(D)}\ 3 \qquad\mathrm{(E)}\ 4</math>
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<math>\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4</math>
  
 
==Solution==
 
==Solution==
  
(a) Clearly does not change, as <math>MN=\frac{1}{2}AB</math>. Since <math>AB</math> does not change, neither does <math>MN</math>.
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(a) Clearly <math>AB</math> does not change, and <math>MN=\frac{1}{2}AB</math>, so <math>MN</math> doesn't change either.
  
(b) Obviously, the perimeter changes.
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(b) Obviously, 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.
  
(d) The bases <math>AB</math> and <math>MN</math> do not change, and neither does the height, so the trapezoid remains the same.
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(d) The bases <math>AB</math> and <math>MN</math> do not change, and neither does the height, so the area of the trapezoid remains the same.
  
 
Only <math>1</math> quantity changes, so the correct answer is <math>\boxed{\text{B}}</math>.
 
Only <math>1</math> quantity changes, so the correct answer is <math>\boxed{\text{B}}</math>.
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{{AMC10 box|year=2000|num-b=4|num-a=6}}
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[[Category:Introductory Geometry Problems]]

Revision as of 09:49, 8 November 2021

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) Clearly $AB$ does not change, and $MN=\frac{1}{2}AB$, so $MN$ doesn't change either.

(b) Obviously, 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 (ProblemsAnswer KeyResources)
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

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