Geometry, Common Core Style: 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.