Let $ A(4, -2), B(1, 1) $ and $ C(9, -3) $ be the vertices of a triangle ABC. Then the maximum area of the parallelogram AFDE, formed with vertices D, E and F on the sides BC, CA and AB of the triangle ABC respectively, is __________.
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Correct Answer: 3
Approach Solution - 1
Approach Solution -2
We have triangle \( \triangle ABC \) with vertices \( A(4,-2), B(1,1), C(9,-3) \). A parallelogram \( AFDE \) is formed with \( D \in BC,\; E \in CA,\; F \in AB \) and one vertex at \( A \). We need to find the maximum possible area of such a parallelogram.
Concept Used:
Parametrize points on sides through vectors from \( A \): let \( \vec{u}=\overrightarrow{AB} \) and \( \vec{v}=\overrightarrow{AC} \). If \( F=A+\alpha \vec{u} \) and \( E=A+\beta \vec{v} \) with \( \alpha,\beta \in [0,1] \), then the fourth vertex of the parallelogram is \( D = F + E - A = A + \alpha\vec{u} + \beta\vec{v} \). For \( D \) to lie on \( BC \), we must have coefficients that are convex on \( \vec{u},\vec{v} \), i.e., \( \alpha,\beta \ge 0 \) and \( \alpha + \beta = 1 \). The area of the parallelogram spanned by sides \( \overrightarrow{AF}=\alpha\vec{u} \) and \( \overrightarrow{AE}=\beta\vec{v} \) is
\[ [\text{Parallelogram}] = \left\lvert \overrightarrow{AF} \times \overrightarrow{AE} \right\rvert = \alpha\beta\, \left\lvert \vec{u} \times \vec{v} \right\rvert. \]
Thus the problem reduces to maximizing \( \alpha\beta \) subject to \( \alpha,\beta \ge 0 \) and \( \alpha+\beta=1 \), and computing \( \lvert \vec{u} \times \vec{v} \rvert \) from the given coordinates.
Step-by-Step Solution:
Step 1: Compute the side vectors from \( A \):
\[ \vec{u}=\overrightarrow{AB}=B-A=(1-4,\; 1-(-2))=(-3,\,3),\quad \vec{v}=\overrightarrow{AC}=C-A=(9-4,\; -3-(-2))=(5,\,-1). \]
Step 2: Compute the magnitude of the 2D cross product (determinant) \( \lvert \vec{u}\times\vec{v}\rvert \):
\[ \lvert \vec{u}\times\vec{v}\rvert=\left| \det\begin{pmatrix} -3 & 5\\ 3 & -1\end{pmatrix}\right| =\left|(-3)(-1)-3\cdot 5\right|=\left|3-15\right|=12. \]
The area of \( \triangle ABC \) is \( \dfrac{1}{2}\lvert \vec{u}\times\vec{v}\rvert = \dfrac{1}{2}\cdot 12 = 6 \).
Step 3: Impose the parallelogram-on-sides condition. With \( F=A+\alpha\vec{u} \) and \( E=A+\beta\vec{v} \), the fourth vertex is \( D=A+\alpha\vec{u}+\beta\vec{v} \). For \( D \in BC \), points on \( BC \) can be written as \( A + s\vec{u} + t\vec{v} \) with \( s,t \ge 0 \) and \( s+t=1 \). Hence the necessary and sufficient condition is
\[ \alpha \ge 0,\quad \beta \ge 0,\quad \alpha+\beta=1. \]
Step 4: Express the area of \( AFDE \) in terms of \( \alpha,\beta \):
\[ [\text{Parallelogram }AFDE] = \left\lvert \overrightarrow{AF}\times\overrightarrow{AE} \right\rvert = \left\lvert (\alpha\vec{u})\times(\beta\vec{v}) \right\rvert = \alpha\beta\, \lvert \vec{u}\times\vec{v} \rvert = \alpha\beta \cdot 12. \]
Step 5: Maximize \( \alpha\beta \) under \( \alpha,\beta \ge 0 \) and \( \alpha+\beta=1 \). Using AM-GM or a quadratic, the maximum occurs at \( \alpha=\beta=\dfrac{1}{2} \), giving
\[ \max \alpha\beta = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4}. \]
Final Computation & Result
Thus the maximum area is
\[ [\text{AFDE}]_{\max} = 12 \cdot \frac{1}{4} = 3 \]
Equivalently, it is half the area of \( \triangle ABC \) (since \( [\triangle ABC]=6 \)). Hence the maximum area is 3 square units.
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