bonanova Posted August 19, 2009 Report Share Posted August 19, 2009 Draw a sketch of a square with three-inch sides. On one side, mark off 1-inch distances - i.e., trisect one of the sides. Draw a four-inch line segment from the center of the square through one of these 1-inch marks. Use that line as a side of a four-inch square that has one of its corners at the center of the three-inch square. Complete the sketch of the four-inch square, and shade in the region of overlap. This is much simpler than I've probably made it sound. Anyway, the challenge is to quickly determine the area of the overlapping region. Please use spoilers Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 19, 2009 Report Share Posted August 19, 2009 9/4 or have I missed something? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 19, 2009 Report Share Posted August 19, 2009 Looking at it I'd expect it'd always be a quarter of the 3-inch cube, so 2.25 square inches. Did some rudimentary geometry calculations and get the same. So long as the second square has sides longer than.... uhh 2.12 inches it's always the same. I think. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 19, 2009 Report Share Posted August 19, 2009 2.295 (without a calc so mite be a bit out) Quote Link to comment Share on other sites More sharing options...
0 plainglazed Posted August 19, 2009 Report Share Posted August 19, 2009 With the assumption that the OP is correct (pretty solid in this case) that there is a unique solution and realizing that there are two such larger 4" squares that can be drawn to fit the solution so both must be equal and it is obvious that together they take up half the smaller 3" square so each is one fourth of the 3 x 3 area or 9/4. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 19, 2009 Report Share Posted August 19, 2009 The area is 1/4 of the smaller square, this can be seen by swapping the two areas shown. Sorry I don't know how to put this image in the spoiler. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 19, 2009 Report Share Posted August 19, 2009 The area is 1/4 of the smaller square, this can be seen by swapping the two areas shown. Sorry I don't know how to put this image in the spoiler. The problem with this solution is the start point of you 4 inch square isn't centered to the 3 inch square. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 19, 2009 Report Share Posted August 19, 2009 The problem with this solution is the start point of you 4 inch square isn't centered to the 3 inch square. It's not to scale. The red square it the 3" square. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 19, 2009 Report Share Posted August 19, 2009 If you realize that as you rotate the 4" square 90 degrees around the center point of the three inch square, each rotation covers a new area of the three inch square while keeping the area of overlap the same. You can create 4 overlap areas by doing this, thus the overlap of one of the squares is 1/4 the area of the 3" square = 3x3/4 = 9/4 in2 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 19, 2009 Report Share Posted August 19, 2009 looks to me like the sum of the area of 2 right triangles with sides of 0.5 and 1.5 inches long, and a rectangle of 1 inch by 1.5 inches, for a total of 2.25 sq. inches. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 19, 2009 Report Share Posted August 19, 2009 Basically, you have a 4" square pivoting around the center of a 3" square. Suppose it intersects at right angles, producing a common square with sides of 1.5". This means the area is 1.5^2, or 2.25, or 9/4. Consider another case, where the larger square is rotated by 45˚. This produces 2 isosceles right triangles, two legs of 1.5". The area of one triangle is (1.5^2)/2, so the area of both is 1.5^2, or 2.25, or 9/4. If the area changed, it would oscillate between these two cases, which are equal, so the area is constant. So long as the rotating square is bigger than the small square, which has a side of s, the overlapping area = (s/2)^2 . Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 19, 2009 Report Share Posted August 19, 2009 2.25 Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted August 20, 2009 Author Report Share Posted August 20, 2009 Yeah ... you all got this one. Extend the lines at the centered corner of the 4-inch square. Both lines bisect the 3-inch square. Quote Link to comment Share on other sites More sharing options...
Question
bonanova
Draw a sketch of a square with three-inch sides.
On one side, mark off 1-inch distances - i.e., trisect one of the sides.
Draw a four-inch line segment from the center of the square through one of these 1-inch marks.
Use that line as a side of a four-inch square that has one of its corners at the center of the three-inch square.
Complete the sketch of the four-inch square, and shade in the region of overlap.
This is much simpler than I've probably made it sound.
Anyway, the challenge is to quickly determine the area of the overlapping region.
Please use spoilers
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