Let S be a square of unit area....

MathematicsDistance and Section FormulasJEE Advanced 1997Moderate

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 (Hindi)

Visualizing the Problem

Let the square $S$ have vertices at $(0, 0), (1, 0), (1, 1)$ , and $(0, 1)$ .
A quadrilateral is inscribed such that each vertex lies on one side of $S$ .
We need to prove $2 \leq a^{2} + b^{2} + c^{2} + d^{2} \leq 4$ .

Assigning Coordinates

Let the vertices of the quadrilateral be:
Vertex $P$ on $y = 0$ : $(x, 0)$
Vertex $Q$ on $x = 1$ : $(1, y)$
Vertex $R$ on $y = 1$ : $(z, 1)$
Vertex $S$ on $x = 0$ : $(0, w)$
Note that $0 \leq x, y, z, w \leq 1$ .

Applying the Distance Formula

Using the distance formula $d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ :
$a^{2} = (1 - x)^{2} + y^{2}$
$b^{2} = (1 - y)^{2} + (1 - z)^{2}$
$c^{2} = z^{2} + (1 - w)^{2}$
$d^{2} = x^{2} + w^{2}$

Summing the Squares

Sum $L = a^{2} + b^{2} + c^{2} + d^{2}$
$L = (1 - 2 x + x^{2} + y^{2}) + (1 - 2 y + y^{2} + 1 - 2 z + z^{2}) + (z^{2} + 1 - 2 w + w^{2}) + (x^{2} + w^{2})$
$L = 2 x^{2} - 2 x + 2 y^{2} - 2 y + 2 z^{2} - 2 z + 2 w^{2} - 2 w + 4$

Completing the Square

Rewrite $L$ by completing the square for each variable:
$2 x^{2} - 2 x + 1 = 2 (x^{2} - x + \frac{1}{4}) + \frac{1}{2} = 2 (x - \frac{1}{2})^{2} + \frac{1}{2}$
Applying this to all variables:
$L = 2 [(x - \frac{1}{2})^{2} + (y - \frac{1}{2})^{2} + (z - \frac{1}{2})^{2} + (w - \frac{1}{2})^{2}] + 2$

Finding the Minimum and Maximum

Since $(v a r - \frac{1}{2})^{2} \geq 0$ , the minimum value is $L_{min} = 2 (0) + 2 = 2$ .
This occurs when $x = y = z = w = \frac{1}{2}$ (midpoints).
Since $0 \leq v a r \leq 1$ , the maximum value of $(v a r - \frac{1}{2})^{2}$ is $\frac{1}{4}$ (at $0$ or $1$ ).
The maximum value is $L_{ma x} = 2 (\frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4}) + 2 = 2 (1) + 2 = 4$ .
Therefore, $2 \leq a^{2} + b^{2} + c^{2} + d^{2} \leq 4$ .

Key Takeaways

Key Takeaway: Coordinate geometry simplifies geometric constraints into manageable algebraic functions.
Optimization: The sum of squares of distances is minimized at the midpoints and maximized at the vertices of the bounding square.
Next Challenge: Try proving a similar bound for a general rectangle with sides $L$ and $W$ .

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Comprehension Passage

A B C D

is a square of side length 2 units.

C_{1}

is the circle touching all the sides of the square

A B C D

and

C_{2}

is the circumcircle of square

A B C D

L

is a fixed line in the same plane and

R

is a fixed point.

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(A)

0.75

(B)

1.25

(C)

(D)

0.5

Answer:A

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(A)

ellipse

(B)

hyperbola

(C)

parabola

(D)

pair of straight line

Answer:C

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(A)

1/2 sq. units

(B)

2/3 sq. units

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Answer:C

Visualized Solution (Hindi)

Visualizing the Problem

Assigning Coordinates

Applying the Distance Formula

Summing the Squares

Completing the Square

Finding the Minimum and Maximum

Key Takeaways

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