Question

A parallelogram has area 40 units. If $AB = \frac{41}{4}$, then BC, Find the minimum possible value of the square of its longer diagonal; ($d^2 = \frac{a}{b}$) and find $|a-b|$

Answer 1

3208465

Answer 2

Let the sides of the parallelogram be $a$ and $b$, and let $\theta$ be the angle between them. We are given $a = AB = \frac{41}{4}$ and the Area $= 40$. The area of a parallelogram is given by $Area = ab \sin\theta$. So, $(\frac{41}{4}) b \sin\theta = 40$, which implies $b \sin\theta = \frac{160}{41}$.

The squares of the diagonals ($d_1^2, d_2^2$) of a parallelogram are given by: $d_1^2 = a^2 + b^2 - 2ab \cos\theta$ $d_2^2 = a^2 + b^2 + 2ab \cos\theta$

The square of the longer diagonal, $d_{long}^2$, is $a^2 + b^2 + 2ab |\cos\theta|$.

We have $a = \frac{41}{4}$, so $a^2 = \frac{1681}{16}$. From $b \sin\theta = \frac{160}{41}$, we have $b = \frac{160}{41 \sin\theta}$. So, $b^2 = \frac{25600}{1681 \sin^2\theta}$. Also, $ab = (\frac{41}{4}) (\frac{160}{41 \sin\theta}) = \frac{40}{\sin\theta}$.

Substituting these into $d_{long}^2$: $d_{long}^2 = \frac{1681}{16} + \frac{25600}{1681 \sin^2\theta} + 2 \left(\frac{40}{\sin\theta}\right) |\cos\theta|$ $d_{long}^2 = \frac{1681}{16} + \frac{25600}{1681 \sin^2\theta} + \frac{80 |\cos\theta|}{\sin\theta}$

Let $u = |\cot\theta|$. Since $0 < \theta < \pi$, $\sin\theta > 0$. We know $\frac{1}{\sin^2\theta} = 1 + \cot^2\theta = 1+u^2$ and $\frac{|\cos\theta|}{\sin\theta} = |\cot\theta| = u$. Substituting these: $d_{long}^2 = \frac{1681}{16} + \frac{25600}{1681} (1+u^2) + 80u$

Let $f(u) = \frac{25600}{1681} (1+u^2) + 80u = \frac{25600}{1681} u^2 + 80u + \frac{25600}{1681}$. This is a quadratic in $u$. Since $u = |\cot\theta|$, $u \ge 0$. The parabola opens upwards, and its vertex is at $u = -\frac{80}{2 \cdot \frac{25600}{1681}} = -\frac{1681}{640}$. For $u \ge 0$, the minimum value of $f(u)$ occurs at $u=0$. $u=0$ implies $\cot\theta = 0$, so $\theta = \frac{\pi}{2}$. This means the parallelogram is a rectangle.

When $\theta = \frac{\pi}{2}$, $\sin\theta = 1$ and $\cos\theta = 0$. $b = \frac{160}{41 \sin(\frac{\pi}{2})} = \frac{160}{41}$. The minimum square of the longer diagonal is: $d_{min}^2 = a^2 + b^2 = \left(\frac{41}{4}\right)^2 + \left(\frac{160}{41}\right)^2$ $d_{min}^2 = \frac{1681}{16} + \frac{25600}{1681}$ $d_{min}^2 = \frac{1681^2 + 25600 \times 16}{16 \times 1681} = \frac{2825761 + 409600}{26896} = \frac{3235361}{26896}$.

We are given $d^2 = \frac{a}{b}$, so $a = 3235361$ and $b = 26896$. We need to find $|a-b|$. $|a-b| = |3235361 - 26896| = 3208465$.