Lesson 15-7: Lengths of Chords, Secants, and Tangents (Day 157)

Today I subbed in a seventh grade P.E. class. It's in my first OC district -- in fact, I've subbed in this particular class before. I mentioned it in my March 5th post. Since it's P.E., it's too unrepresentative for me to do "A Day in the Life."

Notice that during the entire month of April, I subbed only twice for middle school classes -- and those were on the first and last days of the month! The reason for this drought is obvious -- my LA County district is a high-school only district. So once it finally reopened, I've obviously been subbing at many more high schools than middle schools.

Even then, I've found myself in more Orange County high schools lately as well. In fact, the last time I subbed in a middle school in this district was -- you guessed it! -- right here on March 5th.

Just like that March morning, each class begins with laps -- except now it's three laps, not two. Both March 5th and today are Fridays, and so there's a Friday workout. This time, all ten types of stretches are timed, so instead of doing a certain number of jumping jacks, they do them for thirty seconds. The class is still learning about Track and Field, except now the emphasis is on "Field." They watch a video about the high jump.

The song I perform during the walk today is "Plug It In." Yes -- yesterday I wrote that "Mousetrap Car Song" would be my new walking song, and now I sing a different song. Oh well -- I have several songs that work well on walks, and this is one of them.

Today is Sixday on the Eleven Calendar:

Resolution #6: We ask, what would our heroes do?

Once again, the only heroes the students see today are Olympians. The regular teacher tried to leave a video on the computer so all I have to do is play it (in the locker room, that is). But this computer has an automatic screensaver which can only be turned off with a password. Instead, I go to YouTube, search for high jump, and play the first video I see -- the 2016 Olympic finals. The winner was Canadian Derek Drouin with a mark of 2.38 meters (nearly eight feet). Again, I remind them that champions like Drouin don't slack off during the laps or stretches just because it's hot (about ninety degrees in SoCal).

Today on her Daily Epsilon on Math 2021, Rebecca Rapoport writes:

Find the volume of the parallelepiped.

[Here's the given info from the diagram: The solid is more like a right "parallelogram prism" -- its height is 2, and its base is a parallelogram with sides 5, 6 and a 30-degree angle.]

Notice that the area of the base here is (5)(6)sin 30 = (5)(6)/2 = 15 square units. Multiplying by the height of the prism gives us a volume of 30 cubic units -- and of course, today's date is the thirtieth.

I know -- today's my first middle school in some time, and it's also my first Rapoport problem posted to the blog in some time. While I might have missed a suitable problem or two this month on a weekend or other non-posting day (including spring break), the truth is that there simply weren't that many Geometry problems at all this month.

Lesson 15-7 of the U of Chicago text is called "Lengths of Chords, Secants, and Tangents." In the modern Third Edition of the text, lengths of chords, secants, and tangents appear in Lesson 14-7. As for the new Lesson 14-6, this is a lesson that doesn't appear in the old text. It's all about three circles associated with a triangle -- the circumcircle, incircle, and nine-point circle. The first two of these circles are directly related to the concurrency theorems that appear in the Common Core Standards, so it's good that the authors include this lesson in the third edition.

This is what I wrote two years ago about today's lesson:

Here are the two big theorems of this lesson:

Secant Length Theorem:
Suppose one secant intersects a circle at A and B, and a second secant intersects the circle at C and D. If the secants intersect at P, then AP * BP = CP * DP.

Given: Circle O, secants AB and CD intersect at P.
Prove: AP * BP = CP * DP.

There are two figures, depending on whether P is inside or outside the circle, but proofs are the same.

Proof:
Statements                    Reasons
1. Draw DA and BC.     1. Two points determine a line.
2. Angle BAD = BCD,  2. In a circle, inscribed angles intercepting
    Angle ADC = ABC       the same arc are congruent.
3. Triangle DPA ~ BPC 3. AA~ Theorem (steps 2 and 3)
4. AP / CP = DP / BP    4. Corresponding sides of similar
                                           figures are proportional.
5. AP * BP = CP * DP   5. Means-Extremes Property

This leads, of course, to the definition of power of a point.

Tangent Square Theorem:
The power of point P for Circle O is the square of the length of a segment tangent to Circle O from P.

Given: Point P outside Circle O and Line PX tangent to Circle O at T.
Prove: The power of point P for Circle O is PT^2.

Proof:
Statements                    Reasons
1. Draw Ray TO which 1. Two points determine a line.
intersects Circle O at B.
2. Let PB intersect         2. A line and a circle intersect in at most
Circle O at A and B.           two points.
3. PT perpendicular TB 3. Radius-Tangent Theorem and def, of semicircle
   and TAB in semicircle     
4. PTB right triangle     4. Definitions of right angle, right triangle,
with altitude TA,                and altitude
5. PT^2 = PA * PB        5. Right Triangle Altitude Theorem
6. The power of point P 6. Definition of power of a point
for Circle O is PT^2

By the way, we can now finally prove the Bonus Question from Lesson 15-4. I think I'll dispense with two-column proofs here and give a paragraph proof. We begin with a lemma:

Lemma:
Suppose two circles intersect in two points. Then for each point on their common secant line, the power of that point for first circle equals the power of that point for the second circle.

Given: Circles A and B intersect at C and D, Point P on secant CD
Prove: The power of point P for Circle A equals the power of point P for Circle B

Proof:
For both circles, the power of P is CP * DP, no matter whether P is inside or outside the circle. This common secant has a name -- the radical (or power) axis of the two circles. QED

Theorem:
Suppose each of three circles, with noncollinear centers, overlaps the other two. Then the three chords common to each pair of circles are concurrent.

Proof:
The proof of this is similar to that of the concurrency of perpendicular bisectors of a triangle (which I'll compare to this proof in parentheses). Let A, B, and C be the three circle centers. Every point on the radical axis of A and B has the same power for both circles. (Compare how every point on the perpendicular bisector of XY is equidistant from the points X and Y.) Every point on the radical axis of B and C has the same power for both circles. (Each point on the perpendicular bisector of YZ is equidistant from Y and Z.) So the point where these chords intersect has the same power for all three circles, and thus must lie on the radical axis of Circles A and C. (So the point where the perpendicular bisectors of XY and YZ must be equidistant from all three points, and so must lie on the perpendicular bisector of XZ.) The point where all three radical axes intersect is called the radical center (for perpendicular bisectors, it's called the circumcenter.) QED

Notice that if three centers are collinear, then the three radical axes are parallel (just as if three points are collinear, then the perpendicular bisectors are parallel).

I decided to make this the activity day for this week, since the Exploration section includes two questions rather than one. This is more like a puzzle, since the key to both questions isn't Geometry but arithmetic (or Algebra) and the properties of multiplication!

Hmm, you might wonder whether I should post a pandemic-friendly Desmos today -- or should I wait until Monday, which has suddenly become my Desmos day lately? I will do Desmos today -- it's much easier to find Desmos for today's lesson than for Lessons 15-8 or 15-9.

The following activity was created by Rachel Saunders:

https://teacher.desmos.com/activitybuilder/custom/5eb0c7da10eaf1096e463fdc

But even this activity focuses more on angles (Lesson 15-5 and 15-6 stuff) and doesn't really get into Lesson 15-7 until a video in Slide 9.

Because of this, I'll retain the old activity from two years ago in addition to Desmos.

Lesson 15-6: Angles Formed by Tangents (Day 156)

Today I subbed in a high school self-contained special ed class. It's in my new district. This is obviously an unrepresentative class, and so there's no need to do "A Day in the Life" today.

While several of the students are working on math, it's mostly on a very familiar website, IXL. None of them seem to need my help with their math IXL assignments.

Most of the day is spend preparing for an upcoming field trip to Catalina Island. For non-Californian blog readers, I refer here to the 1950's song "26 Miles (Santa Catalina)" by the Four Preps. Yes, Catalina is an island that's approximately that distance from the mainland. I've never been there, but apparently, these students are going there soon. They will take a ferry that takes about an hour to traverse that distance. One guy, an Honors Biology student with an interest in marine life, is especially looking forward to the trip.

Another guy definitely has issues, though. (As usual, I don't disclose these issues on the blog.) So a pair of aides and I take him out for a walk. On this walk, I sing "Mousetrap Car Song" again -- I didn't plan on this becoming my new walking song now that the Big March is over, but the guy seems to enjoy it and tries to sing part of it with me.

Today is Fiveday on the Eleven Calendar:

Resolution #5: We treat people who are great at math as heroes.

Of course, I don't have much opportunity to fulfill this resolution today. The original version of this resolution referred to the year 1955. Well, the Catalina song I mentioned earlier is from the 1950's, albeit a few years after 1955.

Today is Thursday, and so it's time for my weekly series on COVID-97 and high school Track. This week, my alma mater participated in the third dual meet of the season -- the second league meet. It was held on our home track.

But as I mentioned in last Friday's post, the league combines the two smallest schools -- and those two were our opponents last week. As expected, neither of those school has very many participants -- indeed not in the distance races. (They might have had some sprinters, though.)

And so our school runs the 1600 races unopposed. There is only one girls and one boys 1600 race (as opposed to having separate Varsity and Junior Varsity races). The top boys appear to run just under five minutes, with some runners closer to my own best times.

With fewer races, the meet progresses quickly, and so I decided to stay through the 800's. The top boys in this race are just over two minutes. I don't remember my own 800 times as well as my 1600 times, but I believe that my time was closer to the mid-2:00 range. This would have placed me much closer to the middle of the pack.

On April 29th, 1999 (third dual meet, second league meet), my 1600 time would have been 5:12.

This would have put me in fourth place in the combined 1600 race.

Lesson 15-6 of the U of Chicago text is called "Angles Formed by Tangents." In the modern Third Edition of the text, angles formed by tangents appear in Lesson 14-5.

This is what I wrote two years ago about today's lesson:

The theorems in this lesson are similar to those in yesterday's lesson.

Tangent-Chord Theorem:
The measure of an angle formed by a tangent and a chord is half the measure of the intercepted arc.

Given: AB chord of Circle O, Line BC tangent to Circle O
Prove: Angle ABC = Arc AB/2

Proof:
Statements                                     Reasons
1. Draw diameter BD.                    1. Through any two points there is exactly one line.
2. Arc AD = 180 - AB                    2. Arc Addition Postulate
3. CB perpendicular BD                 3. Radius-Tangent Theorem
4. Angle ABC = 90 - ABD             4. Angle Addition Postulate
5. Angle ABC = 180/2 - Arc AD/2 5. Inscribed Angle Theorem
6. Angle ABC = (180 - Arc AD)/2 6. Distributive Property
7. Angle ABC = Arc AB/2              7. Substitution Property of Equality

Tangent-Secant Theorem:
The measure of the angle between two tangents, or between a tangent and a secant, is half the difference of the intercepted arcs.

Given: Line AB secant, Ray EC tangent at point C, forming Angle E,
Arc AC = x, Arc BC = y
Prove: Angle E = (x - y)/2

Proof ("between a tangent and a secant"):
Statements                                     Reasons
1. Draw AC.                                   1. Through any two points there is exactly one line.
2. Angle DCA = x/2, EAC = y/2     2. Inscribed Angle Theorem
3. Angle DCA = EAC + E              3. Exterior Angle Theorem
4. Angle E = DCA - EAC               4. Subtraction Property of Equality
5. Angle E = x/2 - y/2                     5. Substitution Property of Equality
6. Angle E = (x - y)/2                     6. Distributive Property

In the text, the "between two tangents" is given as an exercise. The Given part of this proof with the way the points are labeled is completely different from the first part.

Given: Ray PV tangent at Q, Ray PU tangent at R
S on Circle O (same side of QR as P), T on Circle O (opposite side of QR as P)
Prove: Angle P = (Arc QTR - QSR)/2

Proof ("between two tangents"):
Statements                                     Reasons
1. Draw QR.                                   1. Through any two points there is exactly one line.
2. Angle VQR = Arc QTR/2,          2. Inscribed Angle Theorem
    Angle PQR = Arc QSR/2
3. Angle VQR = PQR + P              3. Exterior Angle Theorem
4. Angle P = VQR - PQR               4. Subtraction Property of Equality
5. Angle P = Arc QTR/2 - QSR/2  5. Substitution Property of Equality
6. Angle P = (Arc QTR - QSR)/2   6. Distributive Property

In some ways, the Tangent-Chord Theorem is just like yesterday's Angle-Chord Theorem, except that one of the intercepted arcs is 0 degrees. The bonus question concerns a solar eclipse.

Lesson 15-5: Angles Formed by Chords or Secants (Day 155)

Today I subbed in a special ed English class. It's in my first OC district -- indeed, it's the class that I've subbed for several times this year, most recently in my April 12th post.

Many of those visits to this classroom have been on Tuesdays, when odd periods meet. One of those classes was fifth period Business Math. But since today is an even day, there's no Business Math -- it's just junior English, senior English, and conference period. With no math class, there's no reason for me to do "A Day in the Life" today. (Hey, at least there was a Business Math class at yesterday's school.)

This class has finally finished The Great Gatsby, so the students watch the movie version of the novel. We aren't able to complete the movie today -- we end up right after the car crash. (That's right -- the English class I covered at another school last week was reading that part of the novel.)

Since the film takes the entire period, there's no time for any song today. Indeed, I rarely get to sing on movie days.

Today marks the end of the seventh quaver. This time, I don't hear the regular teacher discuss any progress reports (although he does plan on giving the students a day to make up missing assignments).

Today is Fourday on the Eleven Calendar:

Resolution #4: We need to inflate the wheels of our bike.

Obviously, I don't have any opportunity at all to fulfill this resolution today.

This is what I wrote two years ago about today's lesson:

Lesson 15-5 of the U of Chicago text is on "Angles Formed by Chords or Secants." There is one vocabulary term as well as two theorems to learn.

The vocabulary word to learn is secant. The U of Chicago defines a secant as a line that intersects a circle in two points. This is in contrast with a tangent, a line that intersects the circle in one point.

At this point, I often wonder why we have tangent and secant lines as well as tangent and secant functions in trig. Well, here's an old (nearly 20 years!) Dr. Math post with the explanation:

[2021 update: That link is now dead. Last week, I linked to the blog of James Tanton, who also discussed this in more detail. Here I retain the old Dr. Math discussion.]

Now, the tangent and the secant trigonometric functions are related to 
the tangent and secant of a circle in the following way.

Consider a UNIT circle centered at point O, and a point Q outside the 
unit circle. Construct a line tangent to the circle from point Q and 
call the intersection of the tangent line and the circle point P. Also 
construct a secant line that goes through the center O of the circle 
from point Q. The line segment OQ will intersect the circle at some 
point A. Next draw a line segment from the center O to point P. You 
should now have a right triangle OPQ.

A little thought will reveal that the length of line segment QP on the 
tangent line is nothing more but the tangent (trig function) of angle 
POQ (or POA, same thing). Also, the length of the line segment QO on 
the secant line is, not surprisingly, the secant (trig function) of 
angle POA.

And now let's look at the theorems:

Angle-Chord Theorem:
The measure of an angle formed by two intersecting chords is one-half the sum of the measures of the arcs intercepted by it and its vertical angle.

Given: Chords AB and CD intersect at E.
Prove: Angle CEB = (Arc AD + Arc BC)/2

Proof:
Statements                                            Reasons
1. Draw AC.                                          1. Through any two points there is exactly one segment.
2. Angle C = Arc AD/2,                        2. Inscribed Angle Theorem
    Angle A = Arc BC/2
3. Angle CEB = Angle C + Angle A    3. Exterior Angle Theorem
4. Angle CEB = Arc AD/2 + Arc BC/2 4. Substitution

Angle-Secant Theorem:
The measure of an angle formed by two secants intersecting outside the circle is half the difference of the arcs intercepted by it.

Given: Secants AB and CD intersect at E
Prove: Angle E = (Arc AC - Arc BD)/2

Proof:
Statements                                            Reasons
1. Draw AD.                                          1. Through any two points there is exactly one segment.
2. Angle ADC = Arc AC/2,                   2. Inscribed Angle Theorem
    Angle A = Arc BD/2
3. Angle A + Angle E = Angle ADC     3, Exterior Angle Theorem
4. Angle E = Angle ADC - Angle A      4. Subtraction Property of Equality
5. Angle E = Arc AC/2 - Arc BD/2       5. Substitution

In the end, I must admit that of all the theorems in the text, I have trouble recalling circle theorems the most.

I decided to include another Exploration Question as a bonus:

The sides of an inscribed pentagon ABCDE are extended to form a pentagram, or five-pointed star.
a. What is the sum of the measures of angles, F, G, H, I, and J, if the pentagon is regular?

Notice that each angle satisfies the Angle-Secant Theorem. So Angle F is half the difference between CE (which is two-fifths of the circle, Arc CD + Arc DE = Arc CE = 144) and AB (which is one-fifth of the circle, Arc AB = 72). So Angle F = (144 - 72)/2 = 36 degrees. All five angles are measured the same way, so their sum is 36(5) = 180 degrees.

b. What is the largest and smallest this sum can be if the inscribed polygon is not regular.

Well, let's write out the Angle-Secant Theorem in full:

Angle F + G + H + I + J
= Arc (CD + DE - AB + DE + EA - BC + EA + AB - BC + AB + BC - DE + BC + CD - EA)/2
= Arc (CD + DE + EA + AB + BC)/2
= (360)/2 (since the five arcs comprise the entire circle)
= 180

So the largest and smallest this sum can be is 180. The sum of the five angles is a constant.