Let A(4, -2), B(1, 1) and C(9, -3) be the vertices of a triangle ABC. Then the maximum area of the p

Q: Wait, why is the area of the parallelogram always half the triangle's area?

This is a beautiful result of affine geometry. When you place the vertices of the parallelogram at the midpoints of the triangle's sides, you effectively divide the original triangle into four smaller, congruent triangles. The parallelogram occupies two of these four triangles, which means its area is exactly $\frac{2}{4} = \frac{1}{2}$ of the total area of the original triangle.

Q: Can I solve this using calculus by setting variables for the sides?

Yes, you absolutely can! You could define the position of point $F$ on $AB$ using a parameter $t$ and then express the coordinates of $D$ and $E$ in terms of $t$. Then, you would write the area of the parallelogram as a function of $t$, differentiate it, and set the derivative to zero to find the maximum. However, the geometric theorem is a much faster and more elegant way to reach the same result.

Q: Does the position of the vertices $D, E, F$ matter for the maximum area?

Yes, it matters significantly. The theorem states that the maximum area is achieved specifically when the vertices $D, E, F$ are the midpoints of the sides $BC, CA, AB$. If you move these points away from the midpoints, the area of the parallelogram will decrease. The midpoint configuration is the unique optimal solution.

Question

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 ______ .

Options:

(A) 3

(B) 4

(C) 6

(D) 8

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast coordinate plane, looking at a triangle defined by three points: $A(4, -2)$, $B(1, 1)$, and $C(9, -3)$. We are tasked with finding the maximum area of a parallelogram $AFDE$ inscribed within this triangle, where $F$ lies on $AB$, $D$ on $BC$, and $E$ on $AC$.

This problem is a classic example of how coordinate geometry and pure geometry dance together. Before we dive into the algebra, let's appreciate the structure of the canvas.

### The Canvas: Calculating the Area of $	riangle ABC$

To find the area of the parallelogram, we first need to know the area of the container—the triangle itself. We use the classic coordinate area formula:

$$\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

Substituting our vertices $A(4, -2)$, $B(1, 1)$, and $C(9, -3)$, we get:

$$\Delta = \frac{1}{2} |4(1 - (-3)) + 1(-3 - (-2)) + 9(-2 - 1)|$$

Let's be meticulous with the signs. Inside the modulus, we have $4(4) + 1(-1) + 9(-3)$, which simplifies to $16 - 1 - 27$. This gives us $\frac{1}{2} |-12|$, which is $6$ square units.

### The Hidden Symmetry

Now, we introduce the parallelogram $AFDE$. It shares vertex $A$ with the triangle, with $F$ on $AB$, $D$ on $BC$, and $E$ on $AC$.

You might be tempted to start writing complex equations for the lines $AB$, $BC$, and $AC$ and then parameterizing the coordinates of $D, E,$ and $F$. While that is a valid path, it is the long road. There is a deeper, more elegant truth here.

In any triangle, if you inscribe a parallelogram that shares a vertex with the triangle, the area of that parallelogram is maximized when its other three vertices are the midpoints of the triangle's sides. This is a powerful theorem in geometry.

When you connect the midpoints of the sides of a triangle, you create four smaller triangles, each with an area equal to one-fourth of the original triangle's area. The parallelogram formed by these midpoints covers two of these smaller triangles. Thus, the area of the parallelogram is:

$$	ext{Area} = 2 	imes \frac{1}{4} \Delta = \frac{1}{2} \Delta$$

### Final Calculation

With this theorem in our toolkit, the rest of the problem collapses into a simple calculation. We know the area of $	riangle ABC$ is $6$ square units.

Therefore, the maximum area of the parallelogram $AFDE$ is simply:

$$	ext{Area} = \frac{1}{2} 	imes 6 = 3 	ext{ square units}$$

It is truly remarkable how a problem that seems to require complex optimization can be solved with such geometric insight. The **maximum area is 3 square units**. Keep this property in your arsenal—it is a shortcut that will save you precious time in the heat of an exam.

Select Answer:

Visualized Solution

Visualizing the Triangle $A B C$

Area of a Triangle

Substituting the Coordinates

Simplifying the Expression

Calculating the Area

Final Area of $△ A B C$

The Inscribed Parallelogram

The Maximum Area Theorem

Calculating Maximum Area

Final Answer

The Sigma Insight: Area of Triangle

Analyzing the Setup

The Canvas

The Hidden Symmetry

Final Calculation

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