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Cutting Corners (Posted on 2003-11-20) Difficulty: 4 of 5
Suppose you have a circle that is perfectly inscribed in a rectangle.
A smaller rectangle is placed on top of the first one, such that one corner is on the edge of the circle and the opposite corner matches a corner of the larger rectangle.

If the smaller rectangle is twice as long as it is high, how many of them will fit into of the larger one (without overlapping, of course)?

  Submitted by DJ    
Rating: 3.6364 (11 votes)
Solution: (Hide)
50 (or 2).

The larger rectangle, of course, is a square, since it circumscribes a circle. It sides are equal in length to the diameter of the inscribed circle.
Let's call the sides of the small rectangle x and 2x.

First, draw the radius from the center of the circle to the corner of the small rectangle. Call this length r (naturally). Then, extend one of the legs of the triangle, say the length (2x), into the circle. Then draw the right triangle that has that and the radius for sides.

The legs of this right triangle run parallel to the sides of the square (actually, to both rectangles). The shorter leg (the extension of the rectangle) has a length of r-2x, while the longer leg has a length of r-x. The hypotenuse of the triangle is the radius of the circle, or r.

The only right triangle in which the sides differ in length by a value of x is a 3x, 4x, 5x right triangle. So, the hypotenuse of the triangle (and the radius of the circle is 5x.

Since the circle is inscribed in the square, the side of the square is the diameter of the circle, which we now know must be 10x.

Therefore, 50 of the smaller rectangles will fit into the square.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
No SubjectK Sengupta2005-07-05 07:27:16
Some ThoughtsRectanglesbrianjn2003-11-23 23:06:08
Solution2 solutionsTristan2003-11-20 20:22:08
hm..Victor Zapana2003-11-20 19:08:12
solutionVictor Zapana2003-11-20 19:07:17
Solutionre(2): What it looks like...SilverKnight2003-11-20 18:59:50
Solutionre: What it looks like...Popstar Dave2003-11-20 18:56:17
stupid question, or is it?Victor Zapana2003-11-20 18:49:43
Hints/TipsWhat it looks like...Popstar Dave2003-11-20 18:26:44
SolutionNo SubjectC.B.2003-11-20 16:54:08
re(2): sdrawkcabbackwards?wonshot2003-11-20 16:05:31
re: sdrawkcabbackwards?wonshot2003-11-20 15:50:16
Another possibility--being picky.Charlie2003-11-20 15:45:07
sdrawkcabbackwards?Eric2003-11-20 15:35:46
solutionwonshot2003-11-20 15:16:34
Some Thoughtsplease clarifySilverKnight2003-11-20 15:09:31
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