Unfortunately it looks like the question has changed in from the original solutions video, so I can’t look up the solution.
Can someone help me explain why the quantity a is greater? My understanding was that the greater angle rotation of 60 corresponds to a greater diagonal and therefore greater side lengths, increasing the area of the square. So quantity b should be greater.
What am I missing?
Thanks for your help!
Try reviewing the explanation of the last question of the PrepSwift quiz for “Inscribed Polygon Problems II”.
I did that, but still not clear how this one compares. My understanding was that the greater angle rotation of 60 corresponds to a greater diagonal and therefore greater side lengths, increasing the area of the square. So quantity b should be greater.
In the event you missed it, Greg talked about it somewhat at the 21:15 mark in the solution video. I know he didn’t get into the details as much and he’s talking about a modified question, but what helped me was drawing the following list of θs on a sketch paper.
What happens to the shape and area of the smaller square inscribed in the larger square? And what happens to angle ∠RCB? I think if you can answer these questions, you’ll likely get it.
If the above tip is still too challenging, click here for a bigger tip: At which θ is the area of the smaller square getting maximised? And which θ is the area getting minimised? 45’ or 0’? .
