April 1st, 2017, 12:27 PM  #1 
Newbie Joined: Apr 2017 From: New York Posts: 1 Thanks: 0  Math game
King Arthur wishes to construct a new shield by attaching square pieces with sizes 10×10 and 8×8 so that the corner of the smaller square is placed on the diagonal of the larger square and 3 units away from its center. King Arthur wants the smaller square to be sloped so that the area of overlap between the two squares is as small as possible since this will give a shield of the largest possible area. What is the area of King Arthur’s new shield? (Give a full explanation for your answer.)
Last edited by skipjack; April 1st, 2017 at 07:25 PM. 
April 3rd, 2017, 01:12 AM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 1,264 Thanks: 650 
I'm showing the overlap area is independent of the rotation angle for angles $\pi/4 \leq \theta \leq \pi/4$ Surprised by it myself but that's what it's looking like. 
April 3rd, 2017, 02:45 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 17,466 Thanks: 1312  Squares.jpg There are two diagrams that would be relevant. One is given above. In the other, PQ intersects the square ABCD. 
April 3rd, 2017, 03:24 AM  #4  
Senior Member Joined: Sep 2015 From: CA Posts: 1,264 Thanks: 650  Quote:
One bit of ambiguity in the OP is that it doesn't specify whether the smaller corner is pinned to 3 in the nw direction or the se direction of the center of the larger. I assumed the se direction.  
April 3rd, 2017, 04:04 AM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 9,746 Thanks: 651  
April 3rd, 2017, 04:28 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 17,466 Thanks: 1312  
April 3rd, 2017, 01:09 PM  #7  
Senior Member Joined: Sep 2015 From: CA Posts: 1,264 Thanks: 650  Quote:
there are admittedly 4 interpretations as to where this pivot point might be but they will all end up having equivalent answers as they are all the same problem but in a rotated coordinate system. Do you have mathematica? I'd love a 2nd pair of eyes on this sheet.  
April 3rd, 2017, 07:56 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 17,466 Thanks: 1312  Squares2.PNG You're right. I was interrupted when I first read the question, having got as far as just making the larger square. A few days later, I came across the unfinished diagram and decided to finish it, but I misread the description. 
April 3rd, 2017, 07:59 PM  #9 
Senior Member Joined: Sep 2015 From: CA Posts: 1,264 Thanks: 650  The area of the quadrilateral outlined in black is $\dfrac{59}{2}  15 \sqrt{2}$ which is independent of $\theta$ 
April 3rd, 2017, 08:10 PM  #10 
Global Moderator Joined: Dec 2006 Posts: 17,466 Thanks: 1312 
It's easy to see that a tilt angle of magnitude up to 45° doesn't affect the area of overlap. Hence the minimum overlap area is (5  3/√2)² (square units). 

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