MHB Geometry and Trigonometry Challenge

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The area of a circumscribing rectangle, denoted as \( A^2 \), is derived from the dimensions \( x \) and \( y \) of the inner rectangle and the angle \( \theta \). The formula for the area is \( A^2 = (x\cos\theta + y\sin\theta)(x\sin\theta + y\cos\theta) \), which simplifies to \( A^2 = xy + \frac{1}{2}(x^2 + y^2)\sin 2\theta \). The minimum area occurs when \( \theta = 0 \) or \( \pi/2 \), yielding \( A^2 = xy \), while the maximum area occurs at \( \theta = \pi/4 \), resulting in \( A^2 = \frac{1}{2}(x+y)^2 \). Consequently, \( A \) can take any value from \( \sqrt{xy} \) to \( \frac{1}{\sqrt{2}}(x+y) \) inclusively. This analysis provides a comprehensive understanding of the relationship between the dimensions of the rectangle and the area of the circumscribing rectangle.
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A rectangle with sides $x$ and $y$ is circumscribed by another rectangle of area $A^2$. Find all possible values of $A$ in terms of $x$ and $y$.
 
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[sp]Let $ABCD$ be the given rectangle and $PQRS$ a circumscribing rectangle, as in the diagram, where the angle $\theta = \angle PBA$ lies between $0$ and $\pi/2.$


Then $PQ = PB+BQ = x\cos\theta + y\sin\theta$ and $PS = x\sin\theta + y\cos\theta$. So the area of $PQRS$ is $$A^2 = ( x\cos\theta + y\sin\theta)(x\sin\theta + y\cos\theta) = xy(\sin^2\theta + \cos^2\theta) + (x^2+y^2)\sin\theta\cos\theta = xy + \tfrac12(x^2+y^2)\sin2\theta.$$ Thus the minimum value of A^2 is $xy$ (when $\theta=0$ or $\pi/2$) and the maximum value is $xy + \frac12(x^2+y^2) = \frac12(x+y)^2$ (when $\theta = \pi/4$). Finally, $A$ takes all values from $\sqrt{xy}$ to $\frac1{\sqrt2}(x+y)$ inclusive.[/sp]
 

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Opalg said:
[sp]Let $ABCD$ be the given rectangle and $PQRS$ a circumscribing rectangle, as in the diagram, where the angle $\theta = \angle PBA$ lies between $0$ and $\pi/2.$


Then $PQ = PB+BQ = x\cos\theta + y\sin\theta$ and $PS = x\sin\theta + y\cos\theta$. So the area of $PQRS$ is $$A^2 = ( x\cos\theta + y\sin\theta)(x\sin\theta + y\cos\theta) = xy(\sin^2\theta + \cos^2\theta) + (x^2+y^2)\sin\theta\cos\theta = xy + \tfrac12(x^2+y^2)\sin2\theta.$$ Thus the minimum value of A^2 is $xy$ (when $\theta=0$ or $\pi/2$) and the maximum value is $xy + \frac12(x^2+y^2) = \frac12(x+y)^2$ (when $\theta = \pi/4$). Finally, $A$ takes all values from $\sqrt{xy}$ to $\frac1{\sqrt2}(x+y)$ inclusive.[/sp]

Well done, Opalg and thanks for participating!:)
 
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