"Sam Loyd's Get off the Earth Puzzle...."
This is the first page of the subsection "A Few Oldies." We may be done with the checkerboard, but we're just getting started with Sam Loyd, the famous puzzlist.
This is visual and also requires a rotation, so here's a link to this classic puzzle. Notice that the puzzle rotates automatically on the screen:
http://www.murderousmaths.co.uk/games/getofftheearth.htm
Yes, a rotation is an isometry, and isometries preserve angle measure, betweenness, collinearity, and distance, but apparently they don't preserve population. When we perform the rotation, somehow 13 people become 12, so apparently one person has "got off the earth."
I've seen this puzzle before -- not Loyd's, but versions of it. Indeed, the link above gives the same example where an extra player appears on a football team. Apparently, like rotations, translations fail to preserve population as well.
You'll have to wait until I post the solution tomorrow -- because Pappas doesn't provide an answer, and neither does the link above! You should be glad that I've seen this problem before....
Meanwhile, I did say yesterday that I'd post the solution to Sam Loyd's Mrs. Pythagoras puzzle as given by Pappas. The link I provided yesterday gives a different solution from Pappas. As I describe this solution, notice that Pappas connects the 5 * 5 square to the lower left corner of the 12 * 12 square -- the reflection image of Loyd's drawing, with the small square at the lower right corner.
For our first cut, we begin where the 5 * 5 and 12 * 12 squares meet. We cut down one unit, then cut four units to the right (so we're now in the "interior" of the 12 * 12 square). Then we cut down two units, then right four units, and finally down two units. This piece we cut out already has a length of 13 units -- so of course this will be the base of our final 13 * 13 square.
The second cut begins at the point which I labeled as "interior" of the square above (that is, where the first 4-unit cut ended and the first 2-unit cut began). We now cut in a zigzag pattern -- one unit right, then one unit up, and then continue until we reach the right edge of the original figure (which should be one unit from the top).
We now have the three pieces, and it should be easy to put them together to form the 13 * 13 square. I don't post the solution to the Battle Royal problem, since that's provided at yesterday's link.
Lesson 5-5 of the U of Chicago text is called "Properties of Trapezoids." In the modern Third Edition of the text, trapezoids appear in Lesson 6-6.
Unfortunately, I made many changes to Lesson 5-5 both three years and two years ago. I did so for several reasons -- one of which is the lack of a "Same Side Interior Angles Theorem" in my version of the U of Chicago text (where a "Trapezoid Angle Theorem" appears instead). Actually, the modern Third Edition now includes a "Same Side Interior Angles Theorem," but again, my worksheets were based on the old Second Edition, which I then modified.
To rectify this situation, I'm actually posting the so-called "Lesson 5-6" worksheet today, since this is actually more aligned with the real Lesson 5-5 from my text. Tomorrow I'll have to post something else for Lesson 5-6 instead.
This is what I wrote three years ago in describing this worksheet:
The first theorem that we have is the Isosceles Trapezoid Symmetry Theorem:
The perpendicular bisector of one base of an isosceles trapezoid is the perpendicular bisector of the other base and is a symmetry line for the trapezoid.
Notice that this theorem is similar to the Kite Symmetry Theorem. In many ways, there is a sort of dualism between the kite and the isosceles trapezoid. A kite is defined by having consecutive equal sides, while an isosceles trapezoid has consecutive equal angles. The symmetry line for the kite bisects two of the angles, and the symmetry line for the isosceles trapezoid bisects two of the sides.
Here is the proof of the Isosceles Trapezoid Symmetry Theorem as given by the U of Chicago. This time, since Section 5-5 gives a paragraph proof, let me post a two-column proof here on the blog.
Given: ZOID is an isosceles trapezoid with angles I and D equal in measure.
m is the perpendicular bisector of
Prove: m is the perpendicular bisector of
m is a symmetry line for ZOID
Proof:
Statements Reasons
1. m is the perp. bis. of
2. D' = I, I' = D 2. Definition of reflection
3. angle I = angle D 3. Given
4. ray DZ reflected is IO 4. Reflections preserve angle measure.
5. Z' lies on ray IO 5. Figure Reflection Theorem
6. ZOID is a trapezoid 6. Given
7.
8.
9. Z' lies on line ZO 9. Definition of reflection
10. Z' = O 10. Line Intersection Theorem
11. O' = Z 11. Flip-Flop Theorem
12. ZOID reflected is OZDI 12. Figure Reflection Theorem
13. m symm. line of ZOID 13. Definition of symmetry line
14. m is the perp. bis. of OZ 14. Definition of reflection
This proof is quite long and can be intimidating for students. A teacher can break it down by actually folding the isosceles trapezoid along line m. The teacher can ask, "Where does D fold to?" The students will probably answer I, only to have the teacher ask "Why?"
Then the tougher part is to show that Z folds to O. We do it one step at a time -- first we show that since reflections preserve angle measure, Z' must be somewhere on ray OI -- although not necessarily O (but of course students will want to jump to that conclusion).
Now we have another theorem, the Isosceles Trapezoid Theorem:
In an isosceles trapezoid, the non-base sides are equal in measure.
In other words, our definition of isosceles trapezoid implies what we usually think of when we hear the word isosceles. The text states that this is merely a corollary of the previous theorem -- since the reflection of one of the non-base sides is the other.
Returning as a text, another corollary given is the Rectangle Symmetry Theorem:
Every rectangle has two symmetry lines, the perpendicular bisectors of its sides.
This corollary follows from the Isosceles Trapezoid Symmetry Theorem in the same way that the Rhombus Symmetry Theorem (which shows that a rhombus also has two symmetry lines) follows from the Kite Symmetry Theorem. It shows again the beauty of using inclusive definitions.
So far in this section, we've mentioned the symmetries of the kite, rhombus, isosceles trapezoid, as well as the rectangle. A general trapezoid, as we've mentioned before, has no symmetry. But there's one special quadrilateral that's missing -- the parallelogram. So does a parallelogram have symmetry?
Also, so far in this section we've mentioned reflections and reflection symmetry, but there's another transformation that we've learned about -- rotations and rotational symmetry. So do any of our special quadrilaterals have rotational symmetry.
We can answer both of these questions at once: A parallelogram has rotational symmetry!
Notice that the U of Chicago text doesn't discuss rotational symmetry of any figure, much less that of the parallelogram. Yet we see in the Common Core Standards:
CCSS.MATH.CONTENT.HSG.CO.A.3
Given a rectangle, parallelogram, trapezoid, or regular polygon, describe therotations and reflections that carry it onto itself.
(emphasis mine)
[2017 update: The rotational symmetry of the parallelogram appears in the Third Edition. Therefore, the many of the changes I made three years ago can also be viewed as "fixing" my Second Edition to include material from the Third Edition. The tricky part is that in adhering to the Second Edition order, the students don't actually see rotations until Chapter 6. So my old worksheet mentions rotations, but that means there's Chapter 6 material on a Chapter 5 worksheet. I'll keep this discussion of rotations here on the blog, but teachers may want to save rotations for Chapter 6.]
And so we should discuss rotational symmetry in general and that of the parallelogram in particular.
But let's begin with the rhombus and rectangle before going on to the general parallelogram. We recall that our definition of rotation is the composite of reflections in intersecting lines -- and if those lines are symmetry lines of a figure, each one maps the figure to itself. Therefore, the entire rotation must map the figure to itself. It follows that any figure that contains two lines of symmetry must automatically have rotational symmetry as well! And the Two Reflection Theorem for Rotations tells us exactly what the center and magnitude of the rotation are. The center is where the two symmetry lines intersect, and the magnitude is double the angle between the intersecting lines.
Corollary to Two Reflection Theorem for Rotations:
If l and m are intersecting symmetry lines of a figure, then it also has rotational symmetry, where the magnitude is twice the non-obtuse angle between l and m and the center is the point of intersection of l and m.
Applying this to the rhombus, we see that both its diagonals are symmetry lines. Because a rhombus is a kite, its diagonals are perpendicular. So the magnitude of the rotation is twice 90, or 180 degrees, about the point where the diagonals intersect.
And now let's look at the rectangle. Its symmetry lines are the perpendicular bisectors of its sides, and we can use the Two Perpendiculars Theorem and the Fifth Postulate to show that these symmetry lines are also perpendicular. So the magnitude of the rotation is twice 90, or 180 degrees, about the point where the diagonals intersect.
But as it turns out, all parallelograms -- not just the rhombuses and rectangles -- can be shown to have 180-degree rotational symmetry. The U of Chicago doesn't prove this, but Hung-Hsi Wu does. In his Theorem 4, he proves that the opposite sides of a parallelogram are equal, by showing that the parallelogram has rotational symmetry.
Here is Wu's proof of his Theorem 4, in paragraph form (as Wu himself gives it):
Given: ABCD is a parallelogram
Prove: AD = BC
Proof:
Let M be the midpoint of the diagonal
Because MA = MC and rotations preserve distance, we have C" = A, so that R maps line BC to a line passing through A and parallel to line BC. Since the line AD has exactly the same two properties by assumption, Playfair implies that R maps line BC to line AD. Similarly, R maps line AB to CD. Thus since B lies on both AD and CD, its image B" is a point that lies on both AD and CD. By the Line Intersection Theorem, this point is exactly D.
Recall we also have C" = A. Therefore R maps the segment
This proof is very similar to the Isosceles Trapezoid Symmetry Theorem. Both theorems depend on what Wu calls "Lemma 6" -- identifying the reflection or rotation image of a point by finding two lines that intersect at the preimage point and noting that the images of these two lines must intersect at the image of the point.
Like the Isosceles Trapezoid Symmetry Theorem and its corollary the Isosceles Trapezoid Theorem, we should call part of the proof the Parallelogram (Rotational) Symmetry Theorem and then derive that opposite sites are equal as a corollary. Wu also derives as a corollary that opposite angles are equal -- but we can also derive this by applying the Trapezoid Angle Theorem twice -- since a parallelogram is a trapezoid. Even the third major property of parallelograms -- that their diagonals bisect each other -- can be derived as a corollary (since rotations preserve distance, we must also have MB = MD in the above proof).
So we can prove all the major properties of parallelograms using only rotations and not using triangle congruence at all. The U of Chicago derives the properties of other figures using reflection symmetry but reverts to triangle congruence for the parallelogram properties (in Chapter 7).
Finally, notice that since rhombuses and rectangles are parallelograms, we can show that they have 180-degree rotational symmetry without looking at their lines of symmetry. But I still like using the Corollary to the Two Reflection Theorem for Rotations because it can be used to determine the rotational symmetry for figures other than quadrilaterals. We earlier proved that an equilateral triangle has three lines of symmetry -- so it also has rotational symmetry.
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