Lesson 0-8: Perspective (Day 8)

Today on her Mathematics Calendar 2019, Theoni Pappas writes:

Segment BE is a median for trapezoid ACDF, whose area is 128 square units. Find the length of AF.

(Here is the given info from the diagram: D, F right angles, CD = 9, DE = 4, with E on DF.)

Let's use the trapezoid area formula as given in Lesson 8-6 of the U of Chicago text:

A = (1/2)h(b_1 + b_2)

The right angles at D and F imply that DF is the height of the trapezoid, and since BE is a median, it means that DE = 4 must be half of the height. So we can plug 4 in directly to (1/2)h. Then CD = 9 must be one of the bases, and our task is to find the other base.

128 = 4(9 + b_2)
32 = 9 + b_2
23 = b_2

Therefore the missing base is 23 -- and of course, today's date is the 23rd.

This is what I wrote last year about today's lesson:

Lesson 0.8 of Serra's Discovering Geometry is called "Perspective." This is the second of two sections appearing in the Second Edition but not in the modern editions. Serra begins:

"Many of the paintings created by European artists during the Middle Ages were commissioned by the Roman Catholic Church. The art was symbolic; that is, people and objects in the paintings were symbols representing religious ideas."

Unlike Lesson 0.5 on mandalas, which we choose to include on the blog even though it's "missing" from the modern editions, Lesson 0.8 can just be left out altogether. This is because we'll be starting the U of Chicago text next week, and that text already has a lesson on perspective (Lesson 1-5), so Day 15 would just be a repeat of Day 8.

Then again, we recall that in my class last year, the students in all grades had trouble drawing cubes even though those were on isometric paper rather than in true perspective). So we might wish to teach perspective on both Day 8 and Day 15. True perspective drawings should most likely be completed on plain unlined white paper, with a straightedge to draw lines toward the vanishing point. Lined notebook paper for one-point perspective drawing may also be acceptable -- but not for two-point perspective (the subject of today's worksheet).

The worksheet below comes from "marcandersonarts" and "Daisuke Motogi."

Here is the Blaugust prompt for today:

Time Capsule - revisit an old post and reflect. If you are new to blogging - find a post on this day from the past on someone else's blog-read, share, reflect.

Well, technically I've already reblogged last year's post for Lesson 0.8. There were some other things I wrote about in that post that I could add to my reblogging that's not directly related to 0.8:

This is what Theoni Pappas writes on page 235 of her Magic of Mathematics:

"This famous drawing by Leonardo da Vinci appeared in the book De Divina Proportione, which Leonardo illustrated for mathematician Luca Paoli in 1509. Leonardo wrote an extensive section on the proportions of the human body in one of his notebooks."

This is the first and only page of the section "Secrets of the Renaissance Man." Of course you can't see what drawing Pappas is referring to here, so let me provide a link:

https://leonardodavinci.stanford.edu/submissions/clabaugh/history/leonardo.html

Pappas explains:

"In his book, he also made reference to the works of Vitruvius, the Roman architect (circa 30 B.C.) who also dealt with the proportions of the human body."

The title, De Divina Proportione, refers to the "divine proportion," which is also known as the golden ratio or Phi. Both Vitruvius and Leonardo believed that Phi = (1 + sqrt(5))/2 appeared in certain ratios of the human body. This is explained at the following link:

https://www.goldennumber.net/leonardo-da-vinci-golden-ratio-art/

• In the distance from the Da Vinci’s guide line drawn at the hairline to the guide line at the foot, the following are all at golden ratio points:
  •  the navel, which is most often associated with the golden ratio of the total height and not the height of the hairline
  • the guidelines for the pectoral nipples
  • the guidelines for the collar bone
• In the distance from the Da Vinci’s guide line drawn at the elbow to the guideline at the fingertips
  • the base of the hand is at a golden ratio point.

Pappas concludes:

"Leonardo adds, The length of a man's outspread arms is equal to his height."

This estimation appears in a Square One TV song, "Rule of Thumb" by Kid 'n Play. The rappers are trying to measure the length of the floor. One member knows that his height is about six feet, so he concludes that the length of his outspread arms from fingertip to fingertip is also six feet. There isn't a separate video on YouTube for this song, but it does appear at the start of this YouTube video on Math Talk, a spin off of Square One TV. This was recently posted a few months ago:

How can I connect this back to the science class I taught last year? Well, the actual ratios of the human body isn't part of the curriculum, although the human body itself is.

Under the old California standards, here's how a seventh grade life science class was organized. It began with a little chemistry with an emphasize on the elements required for life (hydrogen, carbon, oxygen, and so on). Then the lessons focus on cell structure, DNA, and genes. Next would be evolution and the history of life on earth. This is usually followed by biodiversity, with lessons first on microbes and fungi, plants, and then animals. Within the animal unit, typically invertebrates are covered first, then the various orders of vertebrates -- fish, amphibians, reptiles, birds, mammals. So humans appear last in this section -- but then this is followed immediately by the human body, also known as anatomy. Since I should have followed the California standards for seventh grade, this meant that the unit on the human body should have appeared at the end of the year.

And if you prefer me to reblog a 2016 post (my year in the classroom) rather than 2018, then let me do so right here. I don't want another August post, so instead I'll choose a September 2016 post, halfway through the first trimester:

Today the students take a test. This is for all three grade levels, now that I've changed my original assessment schedule. The eighth grade test is on rational approximations. So far, many students fare well on the test, since much of it involves approximating a square root on the calculator and rounding it off from zero to three decimal places.

Day 30 marks the midpoint of the trimester. In the past I've referred to half of a trimester by a special name -- the "hexter."

I've written about my plans to give four tests this trimester. Therefore, I ought to have two tests during each hexter. But as it turns out, I ended up giving only one test the first hexter, since I printed up the progress reports before grading the tests. Furthermore, the last major grade before printing the first hexter progress reports was a Dren Quiz, which was easy.

As it turns out, all of my eighth graders are earning a C or better. But there are a few students who were failing until the Dren Quiz raised their grades to a C. This might make the progress reports misleading, since the grades were artificially inflated by a Dren Quiz -- oops! As it turns out, most of my failing students are seventh graders. The first test was difficult, and no Dren Quiz can erase all the 10% and 20% scores received on the test.

Here is the song for today:

UNIT RATES

If you want to find unit rates,
There's one thing you must know.
To find a unit rate,
All you do is divide!
To see if it's proportional,
All you do is divide!
Write it as a fraction,
Reduce it then you're fine.
Graph it at (0, 0),
Then just draw a line.

If you want to find square roots,
There's one thing you must know.
To find an estimate,
4 and below, round down!
To find an estimate,
5 and above, round up!
1 place for tenths, 2 for hundredths,
3 for thousandths, you're fine.
Graph it between two values,
Right on the number line.

Tomorrow is also the beginning of a new module. Learning Module 3 of the Illinois State text is called "What's the Best Advantage?" In this module, students will finish the mousetrap cars that they started back in Module 1.

For my eighth graders, this will be an excellent opportunity to integrate science in the lesson. As I wrote earlier, the next NGSS science lesson on the computer is on motion and force. I've been delaying it until the students can learn about force and Newton's Laws. Well, as it just so happens, the students are supposed to measure the force used to launch the mousetrap cars -- in Newtons! So the idea is to have the eighth graders use the first hour to use the mousetrap cars and then the second hour to complete the online assignment.

But the problem is that our mixed-up Wednesday schedule might finally be changing. Here's how the old schedule worked: first period I would have sixth grade, then second period I'd have the eighth graders for "science" (the online assignment), and then third period I'd keep the eighth graders for STEM (which I'd use for either math or an Illinois State project). The problem with the old schedule is trying to fit music into the schedule. According to the music teacher's schedule, eighth grade music started near the end of first period and was intended to last into second period. So the eighth graders began the day in the history classroom and switched to music when the music teacher arrived -- only to have it end 15 minutes later when the sixth graders arrived to the history classroom.

The new, more logical schedule has eighth grade music line up with second period. This means, among other things, that the eighth graders won't be in my room for both an online science lesson and a STEM project, since they'll still be in music. I assume that I will begin the day with the sixth graders in my room -- but I can't send them to the history classroom after first period, since the eighth graders will still be in there for the music lesson. So I'd either keep the sixth graders an extra period or have them go to English and have the seventh graders come to my room.

Under the old schedule, the seventh graders came to my room for fourth period -- but then their music lesson took place in my room, and I didn't see them for math or science at all! Frankly, I wouldn't mind seeing the seventh graders tomorrow, and I bet they'll enjoy beginning the project. If they're in my room only for music as usual, then I'll just have them do the project on Thursday -- indeed, I suspect the project will bleed into Thursday for all the grades no matter what.

And if I do lose an hour with my eighth graders, then I'll just do the project today and have them do the online science assignment tomorrow after lunch -- that time on Thursday is usually for online math assignments, but I'll just have them do science instead.

But this is a two-day post, and I won't know what happens until tomorrow. That's right -- we received an email informing us of the new music schedule, but we were never told what to do with the students outside of music time! The English teacher has given up trying to figure out the schedule and says that she'll just give an online English assignment to whatever kids show up in her classroom! So this is what Wednesdays are like at our middle school, even one hexter into the year!

Returning to 2019, the Blaugust prompt is to reflect on these old posts. Note that in 2016 I mentioned a planned science lesson, then in 2018 I wrote about my failure to teach science. Yes, in 2016 I'd made so many plans to teach science properly, but none of them turned out well.

In that old 2016 post, I seemed to give the uncertainty of the Wednesday music lessons as an excuse not to teach science properly. Of course, the decision to make Wednesday the day for science was, in hindsight, a rather poor choice. I probably should have made Thursday the science day -- after all, I was guaranteed to see all three grades on Thursday. So I could have taught science to all three grades that day without any problems.

Today we return to Beth Ferguson, the only Blaugust participant who posted today:

http://algebrasfriend.blogspot.com/
http://algebrasfriend.blogspot.com/2019/08/relearning-elementary-math-eye-opening.html

Once again, Ferguson is a former teacher who's now an elementary math coach, so once again this week we turn back to basic arithmetic:

I’ve learned a lot by working with these children.  The math curriculum in our elementary schools is excellent.  The curriculum emphasizes finding solutions using multiple methods.  Typical algorithms are introduced after exploring a variety of other strategies designed to help students internalize place value and the meaning of each operation.

And of course, by "other strategies" she means "not the standard algorithm" -- so there's nothing in this post that traditionalists will like. Indeed, she includes two pictures here -- the "plus-minus" method for addition and "chunk out" for division.

Whatever "number talks" are, they sure don't look like anything traditionalists would approve of.

Instead of dwelling on elementary math again, let's get back to Benjamin Leis and the Geometry problem that he posted yesterday. Even though we arrived at a correct solution, I doubt it's the method that Leis intends for us to use.

Indeed, this reminds me of the tweeter CCSSIMath. Every once in a while, CCSSIMath would tweet a problem and say that Japanese seventh (or eighth, or whatever) graders can solve the problem, but Americans at that age can't -- because Common Core is holding them back. Many of these are Geometry problems that can be solved with some clever insight -- but I can't figure out that clever insight even after staring at this problem for a few hours. And if I can't figure it out, I wonder how seventh (or whatever) graders are supposed to solve it!

Anyway, Leis labels this as a Geometry problem, not a Trig problem. The solution I posted yesterday requires the difference formula for tangent plus advanced algebraic manipulation that most Geometry students haven't mastered yet. Therefore what I posted isn't the intended purely geometric solution.

Let me restate the problem from yesterday along with what we demonstrated:

ABCD is a square. P is on side CD, Q is on side BC, Angle PAQ = 45, and AB = 1 cm. What is the perimeter of Triangle PCQ?

Solution: The desired perimeter is 2 cm.

The fact that the perimeter works out to be such a simple value suggests that a purely geometric solution is possible. And in fact, the perimeter of the triangle is exactly half that of the square -- that is, the triangle perimeter is exactly twice the side length of the square. This is significant because two of the triangle sides overlap parts of the square sides.

Working backward, we wish to prove that CP + CQ + PQ = 2, which we know equals BC + CD. And so we subtract out the sides that overlap. So we're now left to prove that BQ + DP = PQ.

Continuing to work backward, we notice that there might exist a point R on PQ such that BQ = QR, as well as DP = PQ. If such a point exists, then it would be easy to show that BQ + DP = PQ.

And in fact, if we then drew in AR, we form triangles that just might be congruent -- Triangle ABQ to be congruent to ARQ, and Triangle ADP to ARP. This seems to be the way to go. So let's try writing out a two-column proof:

Given: information from Leis as stated above
Prove: CP + CQ + PQ = 2

Proof:
Statements                         Reasons
1. info from Leis               1. Given
2. AQ = AQ, AP = AP       2. Reflexive Property of Congruence
3. Draw R such that ???    3. ???
4. ???                                 4. ???
5. Triangle ABQ = ARQ,   5. ???
    Triangle ADP = ARP
6. BQ = RQ, DP = RP       6. CPCTC
7. RP + RQ = DP + BQ     7. Addition Property of Equality
8. CP + CQ + RQ + RP =  8. Addition Property of Equality
    CP + DP + CQ + BQ
9. CP + CQ + PQ =           9. Segment Addition/Betweenness Theorem
    CD + BC
10. CP + CQ + PQ = 2      10. Given Properties of a Square

In Step 5, we must prove that both pairs of triangles are congruent. All of our congruence theorems require knowing three pairs of congruent parts (ASA, SAS, and so on) -- and these are to be filled in to Steps 2-4. One congruence is already known -- Step 2 from the Reflexive Property.

Now R is to be chosen at a specific location so that a second congruence is guaranteed. But my problem is that, after hours of looking at this problem, I found no way to get the third congruence.

One possibility is to let R be the foot of a perpendicular from A to PQ. This makes two right angles at both ARQ and ARP, so that Angle B = ARQ and Angle D = APQ (as ABCD is a square). But I found no way to get a third congruence out of this.

Another possibility is to draw the ray AR such that Angle BAQ = RAQ and DAP = RAP. Such a ray exists from the given fact that Angle PAQ = 45. Since Angle BAD is 90, it means that the sum of the two angles RAQ and RAP (which is 45) equals the sum of the angles (BAQ and DAP), which makes BAQ = RAQ and DAP = RAP possible. But once again, I found no way to get a third congruence out of this.

A third possibility comes from transformations. Let lines AQ and AP be mirrors, and consider the composite of this two reflections. Since the mirrors intersect at A, the composite must be a rotation centered at A. Since the angle between the mirrors is 45, the magnitude of the rotation must be twice this, or 90. The rotation maps ray AB to ray AD -- and since AB = AD, the image point of B must be exactly the point D.

Now let R be the image of B under the first reflection. (Therefore D, of course, is the image of R via the second reflection.) Then we automatically have the congruent triangles -- ABQ = ARQ as well as ADP = ARP, because in each case, a reflection maps one triangle to the other. The problem here is that there's no reason for this point R to be anywhere near PQ, much less on PQ. (That R lies on that segment is needed for the Segment Addition of Step 9.)

It's possible to use trig to show that R lies on PQ. (Since trig is used, this can't be the original proof intended by Leis.)

To do so, let's set up coordinate axes. We'll place A at the origin and AB on the x-axis, so AD must be on the y-axis. (From the picture from Leis, AD is horizontal. But it's actually more convenient for me to place AB on the x-axis, which keeps everything in Quadrant I.)

Now from yesterday's work, we find the coordinate of Q to be (1, tan t) where t is theta, the angle we defined yesterday as the measure of BAQ. Then P becomes (tan (45 - t), 1).

We want to find the equation of line PQ. Since we have two points, we use Point-Slope. The slope of this line is easily found to be -CQ/CP, which we worked to find yesterday. Simplifying gives us:

m = (tan^2 t - 1)/(2 tan t)

So by Point-Slope we obtain the equation of the line:

y - tan t = (tan^2 t - 1)/(2 tan t) * (x - 1)

Now we wish to show that R must lie on this line. The coordinates of R are easy to find -- because reflections preserve distance, AR = AB = 1, so R lies on the unit circle. Its argument is Angle BAR, and since BAR = BAQ + RAQ and BAQ = t and BAQ = RAQ (reflections preserve angles), we conclude that BAR = 2t. Thus the coordinates of R must be (cos 2t, sin 2t).

Plugging this into our equation, we obtain:

sin 2t - tan t = (tan^2 t - 1)/(2 tan t) * (cos 2t - 1)

Our goal now is to show that R always lies on PQ -- that is, that the above equation is a trig identity:

sin 2t - tan t = (tan^2 t - 1)/(2 tan t) * (cos 2t - 1)
                    = (sin^2 t - cos^2 t)/(2 sin t cos t) * (cos 2t - 1) (multiplying fraction by cos^2 t/cos^2 t)
                    = -cos 2t/sin 2t * (cos 2t - 1)
                    = (-cos^2 2t + cos 2t)/sin 2t
                    = (sin^2 2t - 1 + cos 2t)/sin 2t
                    = sin 2t + (cos 2t - 1)/sin 2t
                    = sin 2t + (1 - 2 sin^2 t - 1)/(2 sin t cos t) (using alternate double-angle formula for cos)
                    = sin 2t - 2 sin^2 t/(2 sin t cos t)
                    = sin 2t - sin t/cos t
                    = sin 2t - tan t
identity proved

Thus R lies on PQ. This means that the triangles ABQ = ARQ and ADP = ARP are always congruent, and the above proof works to show that the perimeter of PCQ is 2. QED

But once again, we needed trig to save the day. Once again, this can't be intended by Leis.-- but at least I know that the triangles are supposed to be congruent (so that we're not wasting all our effort only to find out that the triangles aren't really congruent).

I just feel that there must be some way to complete the two-column proof using pure geometry without any trig. Since we should use the fact that PAQ = 45 somewhere, the second method by selecting the ray that makes Angles ABQ = ARQ and ADP = ARP is most likely the correct Step 3 in the proof. But what eludes me is Step 4 -- what other congruence can be made so that the triangles are indeed congruent. So far, all we have is AS, and we need an additional A or S to complete the proof.