## Thursday, May 18, 2017

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

This is what Theoni Pappas writes on page 138 of her Magic of Mathematics:

"If you're a surfer, you know it's sometimes difficult to know in advance when the surf will be up. Sometimes the surf appears perfect from the beach, but when you get in the water it's died down, so you wait for that perfect wave for what sometimes seems like hours."

This is a new section, "Mathematics Rides the Crest of the Wave." In this section, Pappas writes about many of the factors influencing wave size -- the wind, an earthquake, the wake of a boat, and the moon and sun controlling the tides. But she doesn't reach the math part until the next page.

Before we get to today's lesson, let me point out that our two weeks of testing are almost over. If I had been still participating in Tina Cardone's "Day in the Life" project, today would have been my post for the special day "A Testing Day," because it would have also doubled as my monthly posting date of the 18th.

Lesson 15-7 of the U of Chicago text is on "Lengths of Chords, Secants, and Tangents." 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!