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

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</asy>
 
</asy>
  
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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==
  
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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.
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(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>.
  
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(b) Obviously, the perimeter changes. For example, imagine if A was extremely far to the left.
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(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.
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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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 +
==Video Solution by Daily Dose of Math==
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https://youtu.be/yf4eWFAmmd0?si=voNN42OoTcS8Z3sz
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~Thesmartgreekmathdude
  
 
==See Also==
 
==See Also==
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{{AMC10 box|year=2000|num-b=4|num-a=6}}
 
{{AMC10 box|year=2000|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Geometry Problems]]

Latest revision as of 23:38, 14 July 2024

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 $\boxed{\text{B}}$.

Video Solution by Daily Dose of Math

https://youtu.be/yf4eWFAmmd0?si=voNN42OoTcS8Z3sz

~Thesmartgreekmathdude

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

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