# 2000 AMC 8 Problems/Problem 25

## Problem

The area of rectangle $ABCD$ is $72$. If point $A$ and the midpoints of $\overline{BC}$ and $\overline{CD}$ are joined to form a triangle, the area of that triangle is $[asy] pair A,B,C,D; A = (0,8); B = (9,8); C = (9,0); D = (0,0); draw(A--B--C--D--A--(9,4)--(4.5,0)--cycle); label("A",A,NW); label("B",B,NE); label("C",C,SE); label("D",D,SW);[/asy]$ $\text{(A)}\ 21\qquad\text{(B)}\ 27\qquad\text{(C)}\ 30\qquad\text{(D)}\ 36\qquad\text{(E)}\ 40$

## Solution 1

To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that $ABCD$ can have any dimension. Give the rectangle dimensions of $AB = CD = 12$ and $BC = AD= 6$, which is the easiest way to avoid fractions. Labelling the right midpoint as $M$, and the bottom midpoint as $N$, we know that $DN = NC = 6$, and $BM = MC = 3$. $[\triangle ADN] = \frac{1}{2}\cdot 6\cdot 6 = 18$ $[\triangle MNC] = \frac{1}{2}\cdot 3\cdot 6 = 9$ $[\triangle ABM] = \frac{1}{2}\cdot 12\cdot 3 = 18$ $[\triangle AMN] = [\square ABCD] - [\triangle ADN] - [\triangle MNC] - [\triangle ABM]$ $[\triangle AMN] = 72 - 18 - 9 - 18$ $[\triangle AMN] = 27$, and the answer is $\boxed{D}$

## Solution 2

The above answer is fast, but not satisfying, and assumes that the area of $\triangle AMN$ is independent of the dimensions of the rectangle. Label $AB = CD = l$ and $BC = DA = h$

Labelling $M$ and $N$ as the right and lower midpoints respectively, and redoing all the work above, we get: $[\triangle ADN] = \frac{1}{2}\cdot h\cdot \frac{l}{2} = \frac{lh}{4}$ $[\triangle MNC] = \frac{1}{2}\cdot \frac{l}{2}\cdot \frac{w}{2} = \frac{lh}{8}$ $[\triangle ABM] = \frac{1}{2}\cdot l\cdot \frac{h}{2} = \frac{lh}{4}$ $[\triangle AMN] = [\square ABCD] - [\triangle ADN] - [\triangle MNC] - [\triangle ABM]$ $[\triangle AMN] = lh - \frac{lh}{4} - \frac{lh}{8} - \frac{lh}{4}$ $[\triangle AMN] = \frac{3}{8}lh = \frac{3}{8}\cdot 72 = 27$, and the answer is $\boxed{D}$