Well, why don't we work on Lesson 13-6, on Uniqueness? This actually fits here for several reasons:

-- Chapter 13 of the U of Chicago text won't be left intact this year. Instead, I'm breaking it up and including lessons in units where they make more sense. (Earlier this year I covered Lessons 13-1 and 13-2, on logic, along with Chapter 2.)

-- Lesson 13-6 actually mentions glide reflections, so it fits in a glide reflections unit.

-- After today, we can proceed in Chapter 13 with Exterior Angles and the Triangle Inequality.

It makes sense to include the Triangle Inequality in the first semester, just as Dr. Franklin Mason does in his Inequalities chapter (which is his Chapter 5). So much of what I'm writing now follows the same order as Dr. M.

But let's get back to Lesson 13-6. We've mentioned this lesson before, as it mentions the Uniqueness of Parallels Theorem, also known as Playfair's Parallel Postulate. Then here is the glide reflection theorem stated in the text:

In a glide reflection, the midpoint of the segment connecting a point to its image lies on the glide-reflection line.

So now you may wonder, if parallel lines appear in Chapters 3 and 5, and glide reflections in Chapter 6, why does the U of Chicago wait all the way until Chapter 13 to present this material? The answer is that indirect proofs are required. The U of Chicago wanted to protect students from indirect proofs until second semester -- and originally, so did I. But since I dropped the U of Chicago's definition, indirect proofs have already crept into my lessons. So I might as well give Chapter 13 material today.

Last year, I didn't actually cover Lesson 13-6 very well. I had combined it with Lesson 13-5 on a worksheet that ended up being mostly 13-5 with very little 13-6 -- as it turns out, Lesson 13-5, on Tangents to Circles and Spheres, actually does appear on the PARCC (the End-of-Year assessment, but not the Performance Based Assessment).

Recall that the U of Chicago text defines a glide reflection as the composite of a reflection followed by a translation, with the translation parallel to the mirror. Keep that in mind as we see the proof of the above theorem as given by the U of Chicago text:

Given: C = T o r_

*l*(*A*), where T is a translation parallel to*l*.*N*is the midpoint of

Prove:

*l*contains*N*.
Proof:

Let ~~AB~~. From the Midpoint Connector Theorem, line ~~BC~~. But, from the definition of reflection, ~~BC~~. Thus both ~~BC~~ through

*B*= r_*l*(*A*) and let*M*be the midpoint of*MN*is parallel to*l*contains*M*, and because the translation T is parallel to*l*,*l*is parallel to*l*and line*MN*are parallel to*M*. By the Uniqueness of Parallels Theorem,*l*and line*MN*must be the same line. Thus*l*contains*N*. QED
But there's a problem with this proof here -- and it's not that it's an indirect proof (because it isn't -- it's the proof of Playfair itself that's indirect). The problem is that it requires the Midpoint Connector Theorem, which is usually proved using either coordinates or similar triangles (the U of Chicago does the former in Lesson 11-5). But we're not ready to prove that theorem yet.

Fortunately, Dr. Hung-Hsi Wu has provided a proof of the Midpoint Connector Theorem -- he uses it to prove both the Centroid Concurrency Theorem and the Fundamental Theorem of Similarity. This is the version of Dr. Wu's proof that I posted last year:

Given: In triangle

Prove:

Proof:

Statements Reasons

1. bla, bla, bla 1. Given

2. Let F be on ray DE such that DF = 2DE 2. Ruler Postulate (Point-Line)

3. E midpoint of DF 3. Definition of midpoint

4. ADCF is a parallelogram 4. Sufficient Conditions (pgram test), part (c)

5. CF = AD 5. Properties (pgram consequence), part (b)

6. CF = BD 6. Transitive Property of Equality

7. CF | | AD (same as CF | | BD) 7. Definition of parallelogram

8. DBCF is a parallelogram 8. Sufficient Conditions (pgram test), part (d)

9. DF = BC 9. Properties (pgram consequence), part (b)

10. BC = 2DE 10. Transitive Property of Equality

11. DF | | BC (same as DE | | BC) 11. Definition of parallelogram

*ABC*,*D*and*E*are midpoints of*AB*and*AC*, respectivelyProve:

*| |*~~DE~~*,*~~BC~~*BC*= 2*DE*Proof:

Statements Reasons

1. bla, bla, bla 1. Given

2. Let F be on ray DE such that DF = 2DE 2. Ruler Postulate (Point-Line)

3. E midpoint of DF 3. Definition of midpoint

4. ADCF is a parallelogram 4. Sufficient Conditions (pgram test), part (c)

5. CF = AD 5. Properties (pgram consequence), part (b)

6. CF = BD 6. Transitive Property of Equality

7. CF | | AD (same as CF | | BD) 7. Definition of parallelogram

8. DBCF is a parallelogram 8. Sufficient Conditions (pgram test), part (d)

9. DF = BC 9. Properties (pgram consequence), part (b)

10. BC = 2DE 10. Transitive Property of Equality

11. DF | | BC (same as DE | | BC) 11. Definition of parallelogram

(Notice that

It's possible for me to write a special case of this theorem for the special case where

Given: In triangle

Prove:

Proof:

Statements Reasons

1. bla, bla, bla 1. Given

2. Let F be on ray DE such that DF = 2DE 2. Ruler Postulate (Point-Line)

3. Angle

4. Triangle

5.

6.

7.

8.

9.

10.

11.

Of course, this would be the second time this month that we used this "a rectangle is an isosceles trapezoid, which is a trapezoid trick" for the sole purpose of avoiding the parallelogram properties.

We've rewritten some proofs here on the blog in order to avoid the Fifth Postulate, and that makes sense as mathematicians prefer proofs that avoid certain postulates to ones that require them. But avoiding the parallelogram properties just because of something I wrote back in August makes no sense at all.

And so I decided just to post the parallelogram properties after all. This week we cover Lesson 7-6, on the Parallelogram Consequences, and Lesson 7-7, on the Parallelogram Tests.

We've already seen how useful the Parallelogram Consequences really are. The reason that they are delayed until 7-6 in the U of Chicago text is that they are best proved using triangle congruence, but triangle congruence doesn't appear until Chapter 7. I moved triangle congruence up since we knew that this vital topic shouldn't have to wait until the end of the first semester. But I only moved up the triangle congruence Lessons 7-1 to 7-5, and not the parallelogram Lessons 7-6 and 7-7.

It would probably make more sense to cover parallelograms along with the other quadrilaterals in Chapter 5, but it's too late now. Fortunately, I have this opening in the schedule now to cover both Lessons 7-6 and 7-7. This will allow us to cover both the Midpoint Connector Theorem (using Wu's proof) and the Lesson 13-6 Glide Reflection Theorem (using the U of Chicago's proof).

The time after winter break is still scheduled for miscellaneous Common Core topics that are covered either poorly or not at all in the U of Chicago text. This includes Inequalities in Triangles (Chapter 5 in Dr. M, but spread throughout several chapters in the U of Chicago) and the Concurrency Theorems (where the Midpoint Connector Theorem will come in handy).

And today's Lesson 7-6 also includes the Center of a Regular Polygon Theorem, which the U of Chicago text proves using induction. This fits with the Putnam-based lesson that I posted yesterday.

Oh, and by the way, I found the following page about the Fibonacci sequence and generalizations:

http://mrob.com/pub/seq/linrec2.html

(That's right -- this is my third straight post on the

This is what I wrote last year about the Center of a Regular Polygon Theorem:

So how does Dr. Merryfield solve Putnam problems like 2014 B1 or 2015 A2? As it turns out, he uses a proof technique called

So here is the proof of the Center of a Regular Polygon Theorem as given by the U of Chicago -- in paragraph form, just as printed in the text (rather than converted to two columns as I usually do):

Proof:

Analyze: Since the theorem is known to be true for regular polygons of 3 and 4 sides, the cases that need to be dealt with have 5 or more sides. What is done is to show that the circle through three consecutive vertices of the regular polygon contains the next vertex. Then that fourth vertex can be used with two others to obtain the fifth, and so on, as many times as needed.

Given: regular polygon

Prove: There is a point

Draw:

Write: Let

Now the "Analyze" part of this proof contains the induction. If the first three vertices lie on the circle, then so does the fourth. If the fourth vertex lies on the circle, then so does the fifth. If the fifth vertex lies on the circle, then so does the sixth. If the

Every induction proof begins with an initial step, or "base case." In this proof, the base case is that the first three points lie on a circle. This is true because

We have seen how powerful a proof by induction can be. We have proved the Center of a Regular Polygon Theorem. But now we wish to prove another related theorem -- one that is specifically mentioned in the Common Core Standards:

What we wish to derive from the Center of a Regular Polygon Theorem is that we can rotate this polygon a certain number of degrees -- about that aformentioned center, of course -- or reflect it over any angle bisector or perpendicular bisector. But the U of Chicago text, unfortunately, doesn't give us a Regular Polygon Symmetry Theorem or anything like that.

I'm of two minds on this issue. One way would be to take this theorem and use it to prove that when rotating about

*D*and*E*are the points that we are now calling*M*and*N*.) Unfortunately, this proof is quite long just for something that I want to use as a lemma for our Glide Reflection Theorem -- and moreover, it uses the properties of parallelograms, that I was saving for after winter break, according to the unit plans that I posted on the blog back in August.It's possible for me to write a special case of this theorem for the special case where

*ABC*is a right triangle (as*AB*is perpendicular to the glide mirror and*BC*is parallel to it). Then the parallelograms mentioned in this above proof become rectangles. Here is what that proof would look like:Given: In triangle

*ABC*with right angle*B*,*D*and*E*are midpoints of*AB*and*AC*, respectivelyProve:

*| |*~~DE~~*,*~~BC~~*BC*= 2*DE*Proof:

Statements Reasons

1. bla, bla, bla 1. Given

2. Let F be on ray DE such that DF = 2DE 2. Ruler Postulate (Point-Line)

3. Angle

*AED*=*CEF*3. Vertical Angles Theorem4. Triangle

*AED*=*CEF*4. SAS Congruence Theorem (steps 1, 3, 2)5.

*AE*=*CF*, Angle*ADE*=*CFE*5. CPCTC6.

*CF*=*BD*6. Transitive Property of Equality7.

*CF*| |*AD*(same as*CF*| |*BD*) 7. Alternate Interior Angles Test8.

*BC*is perpendicular to*CF*8. Fifth Postulate9.

*BCFD*is an isosceles trapezoid 9. Definition of isosceles trapezoid (sufficient condition)10.

*BCFD*is a trapezoid 10. An isosceles trapezoid is a trapezoid.11.

*DF*| |*BC*(same as*DE*| |*BC*) 11. Definition of trapezoid (meaning)Of course, this would be the second time this month that we used this "a rectangle is an isosceles trapezoid, which is a trapezoid trick" for the sole purpose of avoiding the parallelogram properties.

We've rewritten some proofs here on the blog in order to avoid the Fifth Postulate, and that makes sense as mathematicians prefer proofs that avoid certain postulates to ones that require them. But avoiding the parallelogram properties just because of something I wrote back in August makes no sense at all.

And so I decided just to post the parallelogram properties after all. This week we cover Lesson 7-6, on the Parallelogram Consequences, and Lesson 7-7, on the Parallelogram Tests.

We've already seen how useful the Parallelogram Consequences really are. The reason that they are delayed until 7-6 in the U of Chicago text is that they are best proved using triangle congruence, but triangle congruence doesn't appear until Chapter 7. I moved triangle congruence up since we knew that this vital topic shouldn't have to wait until the end of the first semester. But I only moved up the triangle congruence Lessons 7-1 to 7-5, and not the parallelogram Lessons 7-6 and 7-7.

It would probably make more sense to cover parallelograms along with the other quadrilaterals in Chapter 5, but it's too late now. Fortunately, I have this opening in the schedule now to cover both Lessons 7-6 and 7-7. This will allow us to cover both the Midpoint Connector Theorem (using Wu's proof) and the Lesson 13-6 Glide Reflection Theorem (using the U of Chicago's proof).

The time after winter break is still scheduled for miscellaneous Common Core topics that are covered either poorly or not at all in the U of Chicago text. This includes Inequalities in Triangles (Chapter 5 in Dr. M, but spread throughout several chapters in the U of Chicago) and the Concurrency Theorems (where the Midpoint Connector Theorem will come in handy).

And today's Lesson 7-6 also includes the Center of a Regular Polygon Theorem, which the U of Chicago text proves using induction. This fits with the Putnam-based lesson that I posted yesterday.

Oh, and by the way, I found the following page about the Fibonacci sequence and generalizations:

http://mrob.com/pub/seq/linrec2.html

(That's right -- this is my third straight post on the

*same*Putnam problem.) It shows why numbers of the Fibonacci sequence have other such numbers as factors -- it's because the Fibonacci numbers can be written as polynomials that can be factored to get other Fibonacci polynomials. (Oh, and by the way, I found out by reading this link above that if the Putnam problem had multiplied each term by 3 instead of 4, the sequence would have been 1, 2, 5, 13, 34, 89, ... -- in other words, it's just the Fibonacci sequence with every value skipped!)This is what I wrote last year about the Center of a Regular Polygon Theorem:

So how does Dr. Merryfield solve Putnam problems like 2014 B1 or 2015 A2? As it turns out, he uses a proof technique called

*mathematical induction*. Most proofs in a high school geometry course aren't proved used mathematical induction. Indeed, only one proof in the U of Chicago text is proved this way -- and it just happens to be the Center of a Regular Polygon Theorem! Furthermore, Dr. Wu uses induction in his proofs on similar triangles, so this is a powerful proof technique indeed. (On the other hand, Wu simply defines a regular polygon as an equilateral polygon with a circle through its vertices, so he would have no need to prove the following theorem at all.)So here is the proof of the Center of a Regular Polygon Theorem as given by the U of Chicago -- in paragraph form, just as printed in the text (rather than converted to two columns as I usually do):

Proof:

Analyze: Since the theorem is known to be true for regular polygons of 3 and 4 sides, the cases that need to be dealt with have 5 or more sides. What is done is to show that the circle through three consecutive vertices of the regular polygon contains the next vertex. Then that fourth vertex can be used with two others to obtain the fifth, and so on, as many times as needed.

Given: regular polygon

*ABCD*...Prove: There is a point

*O*equidistant from*A*,*B*,*C*,*D*, ...Draw:

*ABCD*...Write: Let

*O*be the center of the circle containing*A*,*B*, and*C*. Then*OA*=*OB*=*OC*. Since*AB*=*BC*by the definition of regular polygon,*OABC*is a kite with symmetry diagonal*. Thus ray*~~OB~~*BO*bisects angle*ABC*. Let*x*be the common measure of angles*ABO*and*OBC*. Since triangle*OBC*is isosceles, angle*OCB*must have the same measure as angle*OBC*, namely*x*. Now the measure of the angles of the regular polygon are equal to 2*x*, so angle*OCD*has measure*x*also. Then triangles*OCB*and*OCD*are congruent by the SAS Congruence Theorem, and so by CPCTC,*OC*=*OD*. QEDNow the "Analyze" part of this proof contains the induction. If the first three vertices lie on the circle, then so does the fourth. If the fourth vertex lies on the circle, then so does the fifth. If the fifth vertex lies on the circle, then so does the sixth. If the

*n*th vertex lies on the circle, then so does the (*n**+*1)st. I point out that this is induction -- from*n*to*n**+*1.Every induction proof begins with an initial step, or "base case." In this proof, the base case is that the first three points lie on a circle. This is true because

*any*three noncollinear points lie on a circle -- mentioned in Section 4-5 of the U of Chicago (delayed until Chapter 6 on this blog). The induction step allows us to prove that one more point at a time is on the circle, until*all*of the vertices of the regular polygon are on the circle.We have seen how powerful a proof by induction can be. We have proved the Center of a Regular Polygon Theorem. But now we wish to prove another related theorem -- one that is specifically mentioned in the Common Core Standards:

CCSS.Math.Content.HSG.CO.A.3

Given a rectangle, parallelogram, trapezoid, or

Given a rectangle, parallelogram, trapezoid, or

*regular polygon*, describe the rotations and reflections that carry it onto itself.What we wish to derive from the Center of a Regular Polygon Theorem is that we can rotate this polygon a certain number of degrees -- about that aformentioned center, of course -- or reflect it over any angle bisector or perpendicular bisector. But the U of Chicago text, unfortunately, doesn't give us a Regular Polygon Symmetry Theorem or anything like that.

I'm of two minds on this issue. One way would be to take this theorem and use it to prove that when rotating about

*O*, the image of one of those isosceles triangles with vertex*O*and base one side of the polygon is another such triangle. The other way is to do Dr. Wu's trick -- he defines regular polygon so that it's vertices are already on the circle. Then we can perform rotations on the entire circle. (Rotations are easier to see, but it's preferable to do reflections because a rotation is the composite of two reflections.) Notice that the number of degrees of the rotation depends on the number of sides. In particular, for a regular*n*-gon we must rotate it 360/*n*degrees, or any multiple thereof.
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