What is distinctive about the approach to geometry laid out in the Common Core State Standards for Mathematics (CCSSM)? Let’s look to the Introduction to HS Geometry to find out:

The concepts of congruence, similarity, and symmetry can be understood from the perspective of geometric transformation. Fundamental are the rigid motions: translations, rotations, reflections, and combinations of these, all of which are here assumed to preserve distance and angles (and therefore shapes generally). Reflections and rotations each explain a particular type of symmetry, and the symmetries of an object offer insight into its attributes—as when the reflective symmetry of an isosceles triangle assures that its base angles are congruent. In the approach taken here, two geometric figures are defined to be congruent if there is a sequence of rigid motions that carries one onto the other.  This is the principle of superposition.

Symmetry. Rigid motions. Translations, rotations, reflections. Superposition. Most recent approaches to the study of geometry tinker with these ideas, but the CCSSM places them squarely at the center of the discussion. In this series of posts, let’s explore the potential advantages of this approach, while also owning up to the challenges it poses.

Benefit #1: Hands-on Exploration

The use of rigid motions allows students to explore figures in a concrete, tactile way. Students can trace a figure onto a transparency or patty paper, then spin it, slide it, or flip it over.

They can also manipulate figures dynamically in a virtual environment like GeoGebra. For instance, this sketch allows students to engage with corresponding angles on parallel lines.

Benefit #2: Visual Intuition

One of the key skills involved in thinking geometrically is the ability to engage in visual reasoning. Students are well-positioned to master the subject when they can see the reflection symmetry of a rectangle, which explains why it has congruent diagonals, or the rotation symmetry of a parallelogram, which explains why its diagonals bisect each other (and many other things as well!).

Benefit #3: Direction for Proofs

Proofs involving rigid motions fall into a consistent pattern: 1) determine which rigid motions take one figure onto another, 2) set about showing that the rigid motions do what you say they’ll do. This is not to say that there is anything easy about developing the proofs, but this approach does provide a clear sense of direction for students about the goal towards which they are supposed to be working.

For instance, to show that the isosceles triangle above has reflection symmetry (which proves that the base angles are congruent), we just need to prove that the reflection over line $b$ exchanges $A$ and $C$ while keeping $B$ in place. The burden in developing the proof falls on linking these claims to the definition of a reflection and the things we assume are true of reflections (namely, that they preserve distance and angles).

Are you convinced that rigid motions provide a powerful and coherent framework for the study of congruence? Even if you are, it still remains to see how the program plays out. In this series of posts, my aim is to show how students can live up to the vision outlined in the Introduction to Geometry:

During high school, students begin to formalize their geometry experiences from elementary and middle school, using more precise definitions and developing careful proofs.

I hope you find this series valuable! Please feel free to add your thoughts to the comments.

We talk a lot about degree measure in geometry: a right angle is 90 degrees, the angles in a triangle make 180 degrees, and on and on. But just what is a degree? Let’s give this term a precise definition, and then let’s use the symmetry of the circle to show that the measure of a straight angle must be 180 degrees.

The degree is a unit of measure that can be used to describe the magnitude of a rotation. We’d like to be able to say that a turn through 360 degrees is a full turn; this will be the basis for our definition.

Beginning in Grade 4, students learn to work with the degree as a unit of measure for angles. In Standard 4.MD.5, we read this:

An angle is measured with reference to a circle with its center at the common endpoint of the rays, by considering the fraction of the circular arc between the points where the two rays intersect the circle. An angle that turns through 1/360 of a circle is called a “one-degree angle,” and can be used to measure angles.

[Note that the distance around a circular arc is an undefined notion in the CCSS, per Standard G-CO.1.]

As an example, the angle below cuts an arc that is 40/360 as long as the complete circle, and so the degree measure of $\angle AOB$ is 40.

We apply this understanding each time we use a protractor to measure an angle, as the sides of the angle cut part of the circle at the boundary of the protractor.

This brings us to an interesting question: if two students use protractors of different sizes to measure the same angle, will they always get the same measurement? For instance, if the purple arc in the figure below is 40/360 of the circle it lives on, does that necessarily mean that the red arc is 40/360 of the circle it lives on? This question will be taken up in the lessons on size transformations.

Now that we have a clear idea of what it means to measure an angle, let’s state an important theorem.

Straight Angle Theorem. Let $A$ and $B$ be points on a circle with center $O$. Then the measure of arc $AB$ is $180^\circ$ if and only if points $A$, $O$, and $B$ are collinear.

This theorem can be proven by showing that the reflection over line $AB$ maps the arc above the line to the arc below. Since a reflection preserves angles, the two arcs must be equal, and so each arc must be $180^\circ$.

Next up: let’s use the Straight Angle Theorem to establish the Vertical Angles Theorem using a rotation.

Rotations are powerful. They can be used as a means of understanding and explaining a whole host of important results in geometry, from vertical angles, to alternate interior angles, to properties of parallelograms. In order for students to reason with rotations, the first essential task is to formulate a precise definition of rotation. The Common Core State Standards for Mathematics call for students to be able to do exactly this in Standard G-CO.4.

Standard G-CO.2 requires students to be able to “describe transformations as functions that take points in the plane as inputs and give other points as outputs.” To define a rotation $R$ is to be able to describe precisely how each point in the plane responds to the action of $R$: where does it end up? Suppose that $R$ is the rotation of the plane about point $O$ through 50 degrees. Here are the key features that define $R$:

1. The center of rotation spins in place. In other words, the center is a fixed point of $R$.
• $R(O)=O$.
2. Every other point in the plane moves along a circular path through a specified number of degrees.
• $OA=OA'$
• $\angle AOA'=50^{\circ}$

That’s it! When we start to write proofs that involve rotations, we’ll see that each of the conditions above is absolutely essential. In the next few posts, I’ll give some examples of such proofs.

The task of defining rotations is also described at Illustrative Mathematics.

In the figure above, $\angle AOB$ and $\angle A'OB'$ are vertical angles.

• How do we prove that they are congruent?
• What does it even mean to prove that two angles are congruent?
• Which proofs of this theorem are most accessible to students?
• Is it reasonable to expect students to independently produce a proof of this theorem on an assessment?
• The CCSSM call for students to use rigid motions to prove that two figures are congruent. Is this a viable approach for proving this particular theorem?

Please weigh in with your thoughts on these questions in the comments. In this post, I’ll present two proofs of this theorem. Please feel free to critique my proofs:

• Is the level of rigor appropriate for a typical high school student?
• If you feel there is too much detail, what would you leave out?
• If you feel there is not enough detail, what would you add?
• Are you aware of alternative proofs not listed here that are more accessible to students?

Approach #1: Show that $\alpha = \alpha'$ using algebra.

1. Since points $A$, $O$, and $A'$ are on the same line, $\angle AOA'=180$. It follows that $\alpha +\beta =180$.
2. Since points $B$, $O$, and $B'$ are on the same line, $\angle BOB'=180$. It follows that $\beta +\alpha' =180$.
3. From steps 1 and 2, we get $\alpha + \beta = \beta + \alpha'$, and thus $\alpha =\alpha'$.

The proof above shows that $\angle AOB$ has the same degree measure as $\angle A'OB'$. Does this prove that these two angles are congruent? It depends on how you define congruence.

• In some approaches to geometry, the link between equality of angles and congruence is established through a definition: two angles are congruent if and only if their degree measures are equal.
• In the approach taken in the CCSSM, congruence is defined in terms of rigid motions. So the idea that “two angles are congruent if and only if their degree measures are equal” is its own theorem and therefore requires a separate proof. In that sense, the argument above is not enough to prove that vertical angles are congruent.

Approach #2: Show that a rotation maps $\angle AOB$ onto $\angle A'OB'$.

1. Let $r$ be a 180-degree rotation about point $O$. We’ll show that $r(\angle AOB)=\angle A'OB'$.
2. Circle $K$ meets line $n$ at points $A$ and $A'$, so $\angle AOA'=180$ and $r(A)=A'$.
3. Circle $K$ meets line $m$ at points $B$ and $B'$, so $\angle BOB'=180$ and $r(B)=B'$.
4. A rotation about point $O$ fixes point $O$, so $r(O)=O$.
5. From steps 2, 3, and 4, we have $r(\angle AOB)=\angle A'OB'$. Since a rotation is a rigid motion, $\angle AOB \cong \angle A'OB'$, which was to be proved.

Which approach do you prefer? Does thinking about the theorem and its proof in terms of rigid motions add value? What are the strengths and limitations of the two approaches? Please add your thoughts to the comments.

Trig students are expected to be able to compute things like the sine and cosine of 5π/6, 3π/4, and the like. How do we locate points on the unit circle corresponding to these angles, and how do we find their coordinates? This post explains how.

[AP students: the course description for AP Calculus has this sentence under the “Prerequisites” heading: “Students must…know the values of the trigonometric functions at the numbers 0, π/6, π/4, π/3, π/2, and their multiples.”  If you need a brush-up on trig, this post is for you!]

Let’s begin at the beginning: we can define the radian measure of an angle in terms of arc lengths in a unit circle (that is, a circle with radius r = 1).  How much arc is in a full circle?  Since the circumference of a circle is 2πr, we get 2π(1) = 2π radians for the full unit circle.  So one complete rotation encompasses 2π radians.  That means a half-rotation is π radians.

Consequence #1: If a “straight angle” is 1π radians, then a right angle is (1/2)π radians.

Proof: 1/2 + 1/2 = 1.

Consequence #2: If you bisect a right angle, you get (1/4)π radians.

Proof: 1/4 + 1/4 = 2/4 = 1/2.

Consequence #3: If you trisect a right angle, you get (1/6)π radians.

Proof: 1/6 + 1/6 + 1/6 = 3/6 = 1/2.

Now for some theorems about triangles (or trigons!), the figure of interest in trigonometry.

Theorem #1: Since the angles in a triangle can be arranged in a straight line, they must make π radians altogether.

Theorem #2: Since the angles in an equilateral triangle are the same, each angle must be (1/3)π radians.

Proof: 1/3 + 1/3 + 1/3 = 1.

Theorem #3: If you bisect an angle in an equilateral triangle, you get two angles with measure (1/6)π radians.  Proof: 1/6 + 1/6 = 2/6 = 1/3.

Theorem #4: The acute angles in an isosceles right triangle are both (1/4)π radians.

Proof: 1/4 + 1/4 + 1/2 = 2/4 + 1/2 = 1/2 + 1/2 = 1.

Now let’s actually answer some trig questions: what are the sine and cosine associated with an angle of π/4 radians?

Per Theorem #4 above, we are dealing with an isosceles right triangle.

The vertical and horizontal legs must both be sqrt(1/2).  Proof: if you square each side in the triangle, you get (1/2) + (1/2) = 1, which is obviously correct.  A more detailed treatment of this type of triangle is found here.

Note: many folks write the sqrt(1/2) as sqrt(2)/2, which is the same thing (how so?).

Thus sine(π/4) is sqrt(1/2) and cosine(π/4) is sqrt(1/2).  From this one can easily work out the sine and cosine of (3/4)π, (5/4)π, (7/4)π, etc.  We just have to pay attention to whether the triangle is “pointing up/down/left/right” to get the sign correct.

On to the next major question: what are the sine and cosine associated with an angle of π/6 radians?

Theorem #3 above tells us we’re really dealing with half of an equilateral triangle, so the answer can be found via symmetry:

Reflecting the triangle across the x-axis gives an equilateral triangle with side length 1.  So clearly the vertical leg is 1/2.  This means the horizontal leg must be sqrt(3/4).  Proof: squaring each side gives 3/4 + 1/4 = 1, which is obviously true.

A more detailed treatment of this type of triangle can be found here.

Note: many folks write sqrt(3/4) as sqrt(3)/2, which is clearly the same thing, since both quantities are equal to sqrt(3)/sqrt(4).

Thus sine(π/6) is 1/2 and cosine(π/6) is sqrt(3/4).  From this it’s easy to work out the sine and cosine of (5/6)π, (7/6)π, (11/6)π, etc.  Again we just have to look to see which way the triangle is pointing to know whether to assign a positive or negative value to these lengths.

That’s pretty much it!  The only thing left to do is to write down all possible ratios of the sides of a right triangle.  We have 3 sides to work with, one of which is to be called “top” and the other “bottom,” so there are 6 ratios altogether.

Ratio #1: a/b.  We call this ratio the “tangent” of x.

Ratio #2: b/a.  This one is known as the “cotangent” of x.

Ratio #3: a/c.  The sine of x!

Ratio #4: c/a.  The “cosecant” of x.

Ratio #5: b/c.  The cosine of x!

Ratio #6: c/b.  The “secant” of x.

Learning the names of the ratios above is truly a task for flash cards.  If there is any intuition to the names, it has surely been lost on the likes of me!  Someone out there can probably unpack the Latin origins of these terms, but that someone is not me!

As an example, let’s work out the secant of π/4 radians.  The relevant triangle has legs of length sqrt(1/2) and hypotenuse 1, so the ratio must be 1/(sqrt(1/2)).  So we just need to find the reciprocal of sqrt(1/2).  But this is easy: sqrt(1/2) = sqrt(1)/sqrt(2), which is 1/sqrt(2), so the reciprocal must simply be sqrt(2).

Exercise 1: work out all 6 of the fundamental trig ratios for π/6.  Repeat for each of the first 12 multiples of π/6 (i.e. through one full rotation).

Exercise 2: work out all 6 of the fundamental trig ratios for π/4.  Repeat for each of the first 8 multiples of π/4 (which is, again, one full rotation).

Calculus students: if you can do the above 2 exercise sets fluidly, confidently, and correctly, you’ve officially mastered half of one sentence from the “Prerequisites” section of the AP Course Description!  Congrats.

Algebra 2 students all over the country are being taught to use “synthetic division” to divide a polynomial by a linear expression. It goes something like this:

We need to do some algebraic manipulation next:

Here we have a factorization of the original polynomial, so the expression on the left must be the quotient we were seeking. But how exactly does this work???

A curious person might wonder: where does 3/2 come from?  And why, through repeated cycles of multiplying and adding, does this algorithm produce the quotient polynomial we were seeking?  It seems to operate by magic!

Let’s try to take the mystery out of synthetic division.  To do so, let’s go back to an old friend: multiplication.  We’ll begin by multiplying (2x^2 + 5x -1) by (2x – 3).  We choose a box with an appropriate number of “rows” and “columns.”  With a small stretch of the imagination, we allow for the number of rows to be a variable (e.g. 2x rows), and we also allow rows or columns to have negative values (e.g. -1 columns).  So we’re looking at this picture:

We proceed by finding the area of each box:

Next, we notice that terms with like exponents lie on the diagonals of the box.  (To see why this is true, try drawing this diagram with two standard polynomials where all the coefficients are just 1.)

Adding like terms gives us an expression for the total area of the box, which is the product we were seeking.

Grand!  We have succeeded at multiplying two polynomials.  Now it’s time to put on our work gloves and attempt a division problem.  Why don’t we try doing the same problem backwards?  That would certainly give us added confidence in our solution.  Here goes: let’s divide 2x^3 + 7x^2 – 17x + 3 by 2x – 3.

Now — how to get started?  Well, since the like terms are on the diagonals, it must be true that the corner boxes contain the x^3-term and the constant term.

What next?  Well, the corner box has area 2x^3 and height 2x, so its width must be x^2.  And so the area of the other box in the first column must be (x^2)(-3).

How about the box that is catty-corner from the last one?  We know that its area, together with -3x^2, makes 7x^2.  Therefore, its area must be 10x^2.

We can continue reasoning in this way to work out the rest of the diagram:

Voila!  The width of the box is x^2 + 5x – 1, so this must be the quotient you get when you divide 2x^3 + 7x^2 – 17x + 3 by 2x – 3.

Reflection: in the first go-round, we used “length time width gives area;” in the second go-round, we used “area divided by length gives width.”

Now — all of that seems crystal-clear and wonderful.  Students can master this with a bit of practice.  Why mess with a good thing?  Some people apparently want to abstract the above diagram *even further* and produce this following picture with “just numbers.”

This is, in fact, the synthetic division algorithm at work.  Indeed, if you compare this diagram to the one at the top of the post, you’ll see that the 6 intermediate values are the same (except for their signs), and the 3 final values on the top row are also the same.

So what are we to do with all of this?  My personal opinion: stick with the box diagram, showing the powers of x explicitly.  The box diagram works when the divisor is linear, and it also works when it isn’t.  And it’s perfectly clear at all stages what is going on.  Students can clearly see that the process of division is merely multiplying in reverse.

“Synthetic division,” however, requires a linear divisor, requires somewhat strange manipulations before and after you start it, and is generally *not transparent.*  Its sole advantage is speed.

As for me, I’ll take clear understanding over speed 7 days of the week.

Here’s my take on how to “see” the sides of a 30-60-90 triangle, without “memorizing” anything.  Students learning trigonometry are expected to know the side ratios for “special angles,” and so it’s useful to know how to work them out quickly.

First, students should learn to recognize that a 30-60-90 triangle is half of an equilateral triangle.  Thus, if the hypotenuse is 1 (thinking of a unit circle here), then the shorter leg must be 1/2.

Now let’s deal with that pesky longer leg.  Assuming we’ve worked out this length using the Pythagorean theorem before, it won’t come as a surprise that it’s “the square root of something.”  But what is that “something?”

Applying the Pythagorean theorem, we get this:

Clearly there is only room for one quantity here: if “something” plus 1/4 is 1, then that something had better be 3/4!

Thus the long leg is the square root of 3/4.  The advantage of writing the leg in this way is that we can do a quick mental check: the square root of 3/4 squared, plus 1/2 squared, is indeed equal to 1 squared, as 3/4 + 1/4 = 1.  Done!