2000 AMC 8 竞赛试题/第25题
Problem
The area of rectangle 
is 
units squared. If point 
and the midpoints of 
and 
are joined to form a triangle, the area of that triangle is
![[asy] pair A,B,C,D; A = (0,8); B = (9,8); C = (9,0); D = (0,0); draw(A--B--C--D--A--(9,4)--(4.5,0)--cycle); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW);[/asy]](/ueditor/php/upload/image/20180523/1527047432714194.png)
Solution 1
To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that can have any dimension. Give the rectangle dimensions of 
and 
, which is the easiest way to avoid fractions. Labelling the right midpoint as 
, and the bottom midpoint as 
, we know that 
, and 
.
![$[\triangle ADN] = \frac{1}{2}\cdot 6\cdot 6 = 18$](/ueditor/php/upload/image/20180523/1527047444531918.png)
![$[\triangle MNC] = \frac{1}{2}\cdot 3\cdot 6 = 9$](/ueditor/php/upload/image/20180523/1527047445937502.png)
![$[\triangle ABM] = \frac{1}{2}\cdot 12\cdot 3 = 18$](/ueditor/php/upload/image/20180523/1527047446129380.png)
![$[\triangle AMN] = [\square ABCD] - [\triangle ADN] - [\triangle MNC] - [\triangle ABM]$](/ueditor/php/upload/image/20180523/1527047448630476.png)
![$[\triangle AMN] = 72 - 18 - 9 - 18$](/ueditor/php/upload/image/20180523/1527047449506752.png)
![$[\triangle AMN] = 27$](/ueditor/php/upload/image/20180523/1527047450496579.png)
, and the answer is 
Solution 2
The above answer is fast, but satisfying, and assumes that the area of 
is independent of the dimensions of the rectangle. All in all, it's a very good answer though. However this is an alternative if you don't get the above answer. Label 
and 
Labelling and as the right and lower midpoints respectively, and redoing all the work above, we get:
![$[\triangle ADN] = \frac{1}{2}\cdot h\cdot \frac{l}{2} = \frac{lh}{4}$](/ueditor/php/upload/image/20180523/1527047459343785.png)
![$[\triangle MNC] = \frac{1}{2}\cdot \frac{l}{2}\cdot \frac{w}{2} = \frac{lh}{8}$](/ueditor/php/upload/image/20180523/1527047460657562.png)
![$[\triangle ABM] = \frac{1}{2}\cdot l\cdot \frac{h}{2} = \frac{lh}{4}$](/ueditor/php/upload/image/20180523/1527047462967328.png)
![$[\triangle AMN] = lh - \frac{lh}{4} - \frac{lh}{8} - \frac{lh}{4}$](/ueditor/php/upload/image/20180523/1527047465216301.png)
![$[\triangle AMN] = \frac{3}{8}lh = \frac{3}{8}\cdot 72 = 27$](/ueditor/php/upload/image/20180523/1527047466866942.png)
, and the answer is
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