MHB Solving a Challenging Math Problem | Expert Tips & Strategies

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The discussion focuses on solving a complex math problem involving a rectangle inscribed within a reduced ellipse. The original rectangle has dimensions defined by width "w" and height "h," while the ellipse is characterized by semi-axes of length "pw/2" and "ph/2." A coordinate system is established with the ellipse's center as the origin, and the upper left corner of the rectangle is positioned at (Bwa/2, Bha/2). The key equation to solve is \frac{Bwa^2}{p^2w^2}+ \frac{Bha^2}{p^2h^2}= 1, which ensures that the rectangle fits within the ellipse. Participants are encouraged to ask questions for further clarification and assistance.
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Take the original rectangle to have width "w" and height "h". Then the "reduced" ellipse has axe of length pw and ph, semi-axes pw/2 and ph/2, and is given by the equation \frac{x^2}{\left(\frac{pw}{2}\right)^2}+ \frac{y^2}{\left(\frac{ph}{2}\right)^2}= \frac{4x^2}{p^2w^2}+ \frac{4y^2}{p^2h^2}= 1.

Now, we have a rectangle Bwa by Bha that fits inside that reduced ellipse. Setting up a coordinate system with origin at the center of the ellipse, coordinate axes along the axes of the ellipse. In that coordinate system, the upper left corner of the rectangle has coordinates (Bwa/2, Bha/2) and, since that corner lies on the ellipse, must satisfy the equation of the ellipse. That is, we must have \frac{Bwa^2}{p^2w^2}+ \frac{Bha^2}{p^2h^2}= 1
 
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