Let S be a square of unit area. Consider any quadrilateral which has one vertex on each side of S. I

Q: Wait, why do we use coordinate geometry instead of just pure Euclidean geometry?

While pure geometry is elegant, coordinate geometry is the 'brute force' tool that guarantees we don't miss any configurations. By parameterizing the vertices, we turn a geometric problem into an algebraic one, allowing us to use the power of calculus and inequalities to find the exact bounds.

Q: Why does the sum of squares simplify so nicely? Is it a coincidence?

It is not a coincidence! It is the beauty of the unit square. Because the sides are length 1, the displacements are always related to $x$ and $1-x$. When you square these, the linear terms $-2x$ and $-2(1-x)$ interact perfectly to create the symmetric structure we see. It is a testament to the underlying symmetry of the square.

Q: How do we know the maximum is 4?

The maximum occurs at the boundaries of our domain, where $x, y, z, w \in [0, 1]$. Since each term $(v-\frac{1}{2})^2$ is maximized at the endpoints ($0$ or $1$), the entire sum reaches its peak when the quadrilateral vertices coincide with the square's corners, effectively making the quadrilateral the square itself.

Question

Let $S$ be a square of unit area. Consider any quadrilateral which has one vertex on each side of $S$. If $a, b, c$, and $d$ denote the lengths of the sides of the quadrilateral, prove that $2 \le a^2+b^2+c^2+d^2 \le 4$.

Sigma Percentile · Accepted Answer

### The Geometry of Constraints: Unlocking the Square

Welcome, future engineer. Today, we are not just solving a problem; we are exploring the hidden architecture of a square.

You have been given a unit square, and inside it, a quadrilateral is dancing, its vertices tethered to the sides of the square. We want to prove that the sum of the squares of its sides, $a^2 + b^2 + c^2 + d^2$, is trapped between $2$ and $4$.

This is a classic JEE Advanced problem that tests your ability to bridge the gap between visual geometry and algebraic rigor.

### Phase 1: The Cartesian Bridge

Imagine standing on a coordinate plane. We place our unit square $S$ with its vertices at $(0,0), (1,0), (1,1),$ and $(0,1)$.

By placing the square here, we have turned a vague geometric shape into a set of precise equations. The sides of the square are now defined by the lines $x=0, x=1, y=0,$ and $y=1$.

Now, let us place our quadrilateral. Let the vertices be $P, Q, R,$ and $S$. Since each vertex lies on a side, we can parameterize them: $P(x,0), Q(1,y), R(z,1),$ and $S(0,w)$.

Here, $x, y, z,$ and $w$ are variables constrained between $0$ and $1$. We have successfully mapped the physical constraints of the problem into the domain of real numbers.

### Phase 2: The Distance Formula

Now, we need the lengths of the sides. Let us look at side $a$, which connects $P(x,0)$ and $Q(1,y)$.

By the Pythagorean theorem, the square of the side length is:

$$a^2 = (1-x)^2 + y^2$$

We repeat this for all four sides. For side $b$ (between $Q$ and $R$), we get $b^2 = (1-y)^2 + (1-z)^2$.

For side $c$ (between $R$ and $S$), we get $c^2 = z^2 + (1-w)^2$. Finally, for side $d$ (between $S$ and $P$), we get $d^2 = x^2 + w^2$.

### Phase 3: The Algebraic Transformation

Now, let us sum them up. Let $L = a^2 + b^2 + c^2 + d^2$.

When we expand these terms, we find:

$$L = (2x^2 - 2x) + (2y^2 - 2y) + (2z^2 - 2z) + (2w^2 - 2w) + 4$$

This is where the magic happens. We use the technique of 'completing the square'. For any variable $v$, we know that $2v^2 - 2v = 2(v - \frac{1}{2})^2 - \frac{1}{2}$.

Applying this to all four variables, our expression transforms into:

$$L = 2\left[(x-\frac{1}{2})^2 + (y-\frac{1}{2})^2 + (z-\frac{1}{2})^2 + (w-\frac{1}{2})^2ight] + 2$$

### Phase 4: The Final Verdict

This form is the key to the kingdom. Since a squared real number is always non-negative, the minimum value of each term $(v-\frac{1}{2})^2$ is $0$.

This happens when $x=y=z=w=\frac{1}{2}$. Thus, the minimum value of $L$ is $2$.

Conversely, since $0 \le v \le 1$, the maximum value of $(v-\frac{1}{2})^2$ is $(\frac{1}{2})^2 = \frac{1}{4}$. If we set all variables to the boundaries ($0$ or $1$), the sum of the four squared terms becomes $4 	imes \frac{1}{4} = 1$.

Plugging this back into our equation, $L = 2(1) + 2 = 4$.

We have proven that **$2 \le a^2 + b^2 + c^2 + d^2 \le 4$**. You have just navigated from a simple square to a profound inequality.

Animated Solution for Mathematics - Straight Lines: Let $S$ be a square of unit area. Consider any quadrilateral which has one vertex on each side of $S$ . If $a, b, c$ , and $d$ denote the lengths of the sides of the quadrilateral, prove that $2 \leq a^{2} + b^{2} + c^{2} + d^{2} \leq 4$ .

Visualized Solution

Visualizing the Problem

Setting Up the Coordinate System

Parameterizing the Quadrilateral Vertices

Applying the Distance Formula for Side $a$

Distance Formula for Sides $b, c, d$

Summing the Squares

Grouping and Simplifying the Expression

Completing the Square

The Transformed Sum

Finding the Lower Bound (Minimum)

Finding the Upper Bound (Maximum)

Conclusion and Geometric Intuition

The Sigma Insight: Distance and Section Formulas

The Geometry of Constraints

Phase 1

Phase 2

Phase 3

Phase 4

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