Which statement is not true?
1) 4 noncoplanar points determine a plane.
2) {2, 6, 8} is a Pythagorean triplet.
3) If the lengths of 3 sides of a triangle are given, then Heron's formula can always be used to find the triangle's area.
4) The intersection of two distinct Euclidean lines is always a point.
We already know why multiple choice questions always occur during the first week of a month. The question tells us that three of these statements are true and one is false, and we are asked to identify the false one. But unfortunately, more than one statement is false:
1) Four noncoplanar points clearly can't define a plane. How can noncoplanar things define a plane when by definition, noncoplanar things don't lie in a plane?
2) We check the Pythagorean formula: 2^2 + 6^2 = 4 + 36 = 40, which clearly isn't 8^2.
And arguably:
4) The two distinct lines might be parallel. The intersection of two parallel lines obviously isn't a point, as we found out in yesterday's lesson.
Here Pappas most likely means two intersecting lines -- that is, it was implied, but not directly stated, that the lines intersect in at least one point. I would have cut her some slack on this question had we not already had two false statements in 1) and 2) -- now I might as well point out all the errors.
Today's date is the second, so 2) is the intended false statement. We notice here that 2 + 6 = 8 without squaring anything. Thus they can't be the sides of any triangle at all -- much less a right triangle -- since they violate the Triangle Inequality.
And so there must be an error in 1). We think back to the Point-Line-Postulate, with the following part introduced in Lesson 9-1:
f. Through three noncollinear points, there is exactly one plane.
And part f is sometimes restated: Three noncollinear points determine a plane. In other words, by "four" Pappas meant "three," and by "noncoplanar" Pappas meant "noncollinear."
There's actually another statement that Pappas might have had in mind:
Four noncollinear points determine space.
I've discussed this on the blog before, though it's a bit awkward. But if we have three noncollinear points, we can only derive the existence of the plane (that is, we can't rule out that the entire universe is two-dimensional). Once we have a fourth noncoplanar point, we have enough to conclude the existence of (at least) three dimensions.
The mathematician Hilbert, who made Euclid's postulates more rigorous, wrote the following:
I.8. There exist at least four points not lying in a plane.
Finally, notice that 3) above is arguably false as well (which would make all four false), since Heron's formula fails if the three sides don't satisfy the Triangle Inequality -- for example, (2, 6, 8} in the formula gives an area of zero. But Pappas did say that we are given three sides (of a triangle), not three numbers. In the case of {2, 6, 8} we aren't given three sides. If the three sides satisfy the inequality -- that is, if they actually form a triangle -- then the formula gives a positive area. So let's give Pappas some credit -- she did include one true statement.
Let's rewrite the problem fixing the errors:
Which statement is not true?
1) 4 noncoplanar points determine space.
2) {2, 6, 8} is a Pythagorean triplet.
3) If the lengths of 3 sides of a triangle are given, then Heron's formula can always be used to find the triangle's area.
4) If two distinct Euclidean lines intersect, then that intersection is always a point.
Statement 1) is true by Hilbert I.8. Statement 3) is true -- Heron's formula appears as an exercise in Lesson 10-6 of the U of Chicago text. Statement 4) is true by the Line Intersection Theorem as given in Lesson 1-7. But Statement 2) is false, as it fails the Pythagorean Theorem of Lesson 8-7.
Therefore the desired statement that is not true is 2) -- and of course, today's date is the second.
This is what I wrote last year about today's lesson:
Lesson 3-5 of the U of Chicago text is called "Perpendicular Lines." This corresponds to Lesson 3-8 in the modern Third Edition of the text. Meanwhile, Lesson 3-7 of the new edition is actually an introduction to "size transformations" (or dilations). This introduction is basically the same as Lesson 12-1 in my old version, and dilations are studied in more detail in Chapter 12 of both editions.
In past years, I changed the way I covered Lesson 3-5, and so I had a long discussion about my old and new ways of teaching the lesson. I preserve some of this discussion here, if only because I'm still posting those old worksheets. But first, let me post a YouTube video from Square One TV on perpendicular lines.
Officially, I'm doing Section 3-5 now, but then again, not really. Let's consider the contents of this particular section:
- The definition of perpendicular has already been covered. I'm moved it to Section 3-2 when I defined right angles, because I wanted to get it in before jumping to Chapter 4 on reflections, since reflections are defined in terms of the perpendicular bisector.
Let's look at Playfair's Parallel Postulate again:
Through a point not on a line, there is at most one line parallel to the given line.
This is a straightforward, easy to understand rendering of a Parallel Postulate, and Dr. M uses this postulate to derive his Parallel Consequences. But let's look at Euclid's Fifth Postulate, as written on David Joyce's website:
http://aleph0.clarku.edu/~djoyce/java/elements/bookI/post5.html
That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.
No modern geometry text would word its Parallel Postulate in this manner. For one thing, even though the use of degrees to measure angles dates back to the ancient Babylonians, Euclid never uses degrees in his Elements. So the phrase "less than two right angles" is really just Euclid's way of writing "less than 180 degrees." Indeed, in Section 13-6, the U of Chicago text phrases Euclid's Fifth Postulate as:
If two lines are cut by a transversal, and the interior angles on the same side of the transversal have a total measure of less than 180, then the lines will intersect on that side of the transversal.
But let's go back to the Perpendicular to Parallels Theorem as stated in Section 3-5:
In a plane, if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other.
Now count the number of right angles mentioned in this theorem. The transversal being perpendicular to the first line gives us our first right angle, and the conclusion that the transversal is perpendicular to the second line gives us our second right angle. So we have two right angles -- just like Euclid! So in some ways, making the Perpendicular to Parallels Theorem into our Parallel Postulate is making our postulate more like Euclid's Fifth Postulate, not less.
Of course, if we wanted to make our postulate even more like Euclid's, we could write:
If a plane, if a transversal is perpendicular to one line and form an acute angle (that is, less than right) with another, then those two lines intersect on the same side of the transversal as the acute angle.
But this would set us up for many indirect proofs, which I want to avoid. So our Perpendicular to Parallels Postulate is the closest we can get to Euclid without confusing students with indirect proofs.
So this is exactly what I plan on doing. Since we're in Section 3-5, the section that has Perpendicular to Parallels given, I could include it here -- we don't need to worry about how to prove it since I want to make it a postulate. But I already said that I want to wait until Chapter 5 before including any sort of Parallel Postulate. And so the new postulate will be given in that chapter.
Returning to Lesson 3-5:
- The Perpendicular Lines and Slopes Theorem must also wait. Common Core Geometry gives an interesting way to prove this theorem, but the proof depends on similar triangles, which I don't plan on covering until second semester.
If two coplanar lines l and m are each perpendicular to the same line, then they are perpendicular to each other.
This theorem doesn't require any Parallel Postulate to prove. Indeed, even though I just wrote that I don't want to use indirect proof, in some ways this theorem is just begging for an indirect proof:
Indirect Proof:
Assume that lines l and m are both perpendicular to line n, yet aren't parallel. Then the lines must intersect (as they can't be skew, since we said "coplanar") at some point P. So l and m are two lines passing through point P perpendicular to n. But the Uniqueness of Perpendiculars Theorem (stated on this about two weeks ago) states that there is only one line passing through point P perpendicular to n, a blatant contradiction. Therefore l and m must be parallel. QED
But this would be a very light lesson indeed if all I included is this one theorem. Because of this, I decided to include a theorem that I mentioned back in July -- the Line Parallel to Mirror Theorem (companion to the Line Perpendicular to Mirror Theorem mentioned a few weeks ago):
Line Parallel to Mirror Theorem:
If a line l is reflected over a parallel line m, then l is parallel to its image l'.
Lemma:
Suppose T is a transformation with the following properties:
- The image of a line is a line.
- Through every point P in the plane, there exists a line L passing through P such that L is invariant with respect to T -- that is, T maps L to itself.
Now all Common Core transformations satisfy the first property -- that the image of a line is itself some line. (Forget about that Geogebra "circle reflection" where the image of a line can be a circle, since that's not a Common Core transformation.) As it turns out, there are five types of Common Core transformations that satisfy the second property:
- Any reflection
- Any translation
- Any glide reflection
- Any dilation
- A rotation of 180 degrees
Here is a proof of the lemma. (By the way, "lemma" means a short theorem that is mainly used to prove another theorem.) Let l be the original line and l' its image, and let P be any point on l. Since l doesn't contain any fixed points, the image of P can't be P itself -- so instead, it must be some point distinct from P, which we'll call P'. So of course P' lies on l'. Now the point P lies on some invariant line L -- and by invariant, we mean that P' lies on it. Now through the two points P and P', there is exactly one line, and that line is L, not l. Since P lies on l, it means that P' can't lie on l. But this is true for every point P on l. For every point P on l, P' is not on l. So l can't intersect its image l', since every point on l fails to have an image on l. In other words, l and its image l' are parallel. QED
OK, that's enough from four years ago -- we don't need to go any further because we haven't actually learned reflections this year. So here are all the old worksheets from that year -- except that I replaced Line Parallel to Mirror with a worksheet that is more nearly aligned with Lesson 3-5. There are now questions referring to Perpendicular to Parallels, and Perpendicular Lines and Slopes.
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