Here's a problem from this year's Intermediate Maths Challenge (a UK competition for, I think, 17 to 18 year olds). They had 25 multiple choice questions to do in an hour. My wife, who is a teacher, did the paper at the same time so that she could answer her pupils' questions afterwards. However she couldn't solve the following one and I felt very pleased with myself when I worked it out in just over a minute.
The six stacked squares in this diagram are all equal in size and they are stacked such that there is a vertical line of symmetry.
A straight line is drawn from point A and going through BC such that the total area of the squares and part squares on one side of the line is equal to that on the other side.
Does this line cut BC at:
a) B
b) C
c) 1/3 of the way down BC
d) 1/2 way down BC
e) 2/3 of the way down BC
Originally posted by Fat Lady Here's a problem from this year's Intermediate Maths Challenge (a UK competition for, I think, 17 to 18 year olds). They had 25 multiple choice questions to do in an hour. My wife, who is a teacher, did the paper at the same time so that she could answer her pupils' questions afterwards. However she couldn't solve the following one and I felt very pleased w ...[text shortened]... at:
a) B
b) C
c) 1/3 of the way down BC
d) 1/2 way down BC
e) 2/3 of the way down BC
If we substract one square from one side of the line and another from the other side of the line, namely the top square and the bottom right one, doesn't this make the problem easier so solve with your eyes only, concidering symmetry?
Originally posted by FabianFnas If we substract one square from one side of the line and another from the other side of the line, namely the top square and the bottom right one, doesn't this make the problem easier so solve with your eyes only, concidering symmetry?
The answer is alternative (a) - B.
I guess that's how the person setting the question wanted them to solve it, just by matching up the bits of square.
My solution was more rigorous:
If you say that the line cuts BC at a point which is h vertically above A, then the area of the bottom half can easily be seen to be a triangle plus half a square:
(0.5 * 2.5 * h) + 0.5
We know the area must be equal to 3 (half of six), so
Originally posted by Fat Lady I guess that's how the person setting the question wanted them to solve it, just by matching up the bits of square.
My solution was more rigorous:
If you say that the line cuts BC at a point which is h vertically above A, then the area of the bottom half can easily be seen to be a triangle plus half a square:
(0.5 * 2.5 * h) + 0.5
We know the ar ...[text shortened]... 3 (half of six), so
3 = (0.5 * 2.5 * h) + 0.5
=> h = 2, so the line goes through BC at B.
Why calculate when resoning is enough? Quicker and simpler.
Take away one square of either side and see the symmetry. Easy peasy.
Originally posted by FabianFnas Why calculate when resoning is enough? Quicker and simpler.
Take away one square of either side and see the symmetry. Easy peasy.
You can't always rely on it 'looking' right, we all know that optical illusions can trick the eye.
Originally posted by Fat Lady You can't always rely on it 'looking' right, we all know that optical illusions can trick the eye.
This doesn't apply here. We're told that the squares are the same and there is a line of symmetry. Fabian's solution is not about "eye-balling" but about reasoning.
Originally posted by Palynka This doesn't apply here. We're told that the squares are the same and there is a line of symmetry. Fabian's solution is not about "eye-balling" but about reasoning.
AB is not a line of symmetry - the shape of the top half has nine sides whereas the bottom half only has five sides! If you remove the two uncut squares then each shape has five sides, but AB is clearly not a line of symmetry.
The question states that there is a vertical line of symmetry so that you know the squares are stacked 'neatly', i.e. the second row is in the middle of the bottom row and the top row is in the middle of the second row.
Fabian's solution relies on us guessing that the bits of square on either side are the same area and hence cancel each other out.
Originally posted by Fat Lady Fabian's solution relies on us guessing that the bits of square on either side are the same area and hence cancel each other out.
No guesswork needed. No risque of any optical illusion here. Symmetry is obvious once you remove the two squares.
In geometry, and mathematics in general, when a symmetry is found, calculation can be very simplified, sometimes even not necessary (as in this case).
Symmetry is every mathematicians best friend.
Edit: AB is actually a line of symmetry, when two well chosen sqares are remove.
Originally posted by Fat Lady AB is not a line of symmetry - the shape of the top half has nine sides whereas the bottom half only has five sides! If you remove the two uncut squares then each shape has five sides, but AB is clearly not a line of symmetry.
The question states that there is a vertical line of symmetry so that you know the squares are stacked 'neatly', i.e. the second ...[text shortened]... ing that the bits of square on either side are the same area and hence cancel each other out.
We are told that there is a vertical line of symmetry and the squares are the same. From this you can immediately conclude that C is in the midpoint of the side of the lower-right square and also that the vertex of the middle-left square is in the midpoint of the top side of the lower left square.
Fabian's solution only relies on these two things.
Originally posted by Palynka We are told that there is a [b]vertical line of symmetry and the squares are the same. From this you can immediately conclude that C is in the midpoint of the side of the lower-right square and also that the vertex of the middle-left square is in the midpoint of the top side of the lower left square.[/b]
I'm not sure I understand what you mean here, but you seem to be saying that the top right corner of the left hand square in the middle row is the mirror image of point C when reflected through the line AB. If so this is clearly incorrect.
Originally posted by Fat Lady I'm not sure I understand what you mean here, but you seem to be saying that the top right corner of the left hand square in the middle row is the mirror image of point C when reflected through the line AB. If so this is clearly incorrect.
Originally posted by Fat Lady I can't see the line of symmetry.
http://i19.photobucket.com/albums/b177/gallicrow/stacked_squares_2.jpg
Removing those two squares helps in that it becomes clear that if a half square is added to the top and bottom shapes you end up with equal triangles.
I think you would agree that each half of the object is congruent to the other. Although they aren't "mirror symmetrical" to each other, they are "rotationally symmetrical". However I solved this problem the same way you did, with an area formula.
I'm not sure if this is the proper name for this type of solution, but to me it seemed "constructive" (i.e. building up the area by pieces until you've described the whole). Continuing along these lines, I think the cleverness behind the geometric solution is that it is "reductive" (i.e. starting with the whole, then removing parts until you reach a workable analogue). The shift in thinking between these two approaches, especially from "constructive" where you know what all the pieces are and where they fit on the playing field, to "reductive" where a whole new world of possibilities opens up from inside the puzzle with only a deft removal of some key pieces, gives one a rush of new information and the feeling of glee at having discovered a "secret", resulting in that AHA!! moment we all relish so much. 🙂
Stacked squares
10 Feb '09 10:501 edit
Back to Top