Difference between revisions of "2000 AMC 10 Problems/Problem 5"
Dairyqueenxd (talk | contribs) (→Problem) |
(→Solution) |
||
| Line 26: | Line 26: | ||
==Solution== | ==Solution== | ||
| − | (a) | + | (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) Obviously, the perimeter changes. For example, imagine if P was extremely far to the left. | ||
Revision as of 10:17, 29 May 2023
Problem
Points 
and 
are the midpoints of sides 
and 
of 
. As 
moves along a line that is parallel to side 
, how many of the four quantities listed below change?
(a) the length of the segment 
(b) the perimeter of
(c) the area of
(d) the area of trapezoid 
![[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]](http://latex.artofproblemsolving.com/0/b/6/0b66755a2c42a83a93ad849254ee5df3266bb45c.png)
Solution
(a) Triangles 
and 
are similar, and since 
, 
.
(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 and its corresponding height remain the same.
(d) The bases and do not change, and neither does the height, so the area of the trapezoid remains the same.
Only 
quantity changes, so the correct answer is 
.
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. 
