- Tools for the Common Core Standards. Bill McCallum, math team coordinator of the Common Core State Standards, provides expert answers to questions about the implementation of the mathematics standards. If only the responses were a bit more timely.
- Progressions Documents for the Common Core State Standards. As of 8/10/2013, there's no high school geometry document, but one has been promised soon.
- Teaching Geometry According to the Common Core Standards. Dr. Hung-Hsi Wu provides detailed guidance on the implementation of the CCSS. The document will be an intellectual work-out for most teachers of high school geometry, but it is the best single resource as of late 2013.
- Beauty, Rigor, Surprise: Elementary Euclidean Geometry. A complete CCSS-aligned course in geometry. Classroom-ready PowerPoints and worksheets.
- Illustrative Mathematics. An initiative of the Institute for Mathematics & Education funded by the Bill & Melinda Gates Foundation. A source for CCSS-aligned tasks, organized by standard. Sometimes classroom-ready, sometimes not.
- Mathematics Vision Project. A set of CCSS-aligned texts released under the Creative Commons License. I'm deeply impressed (and I don't impress easily).

## Friday, August 16, 2013

### The Common Core Standards for Geometry: Resources

I'm a Common Core convert. Here in Indiana, they're a real improvement over the standards they replace. I've compiled a list of resources for Common Core geometry. All are of use to a teacher as she plans her CC geometry. If you know of something not on my list, please let me know.

## Saturday, August 3, 2013

### Parallels: The Triangle Exterior Angle Inequality

The Triangle Exterior Angle Inequality (TEAI) does much of the work in the proofs to come. Let us prove it first.

Triangle ABC is given. We extend side AC through C and mark point D on the extension. Angle DCB is by definition an exterior angle for triangle ABC. The angles of ABC to which it is not adjacent - angles A and B - are its remotes.

We will prove that the exterior angle DCB is greater than the remote angle B.

Mark the midpoint of BC; name it M. Construct the segment AN through M such that AM = MN. Connect N to C.

Angles BMA and NMC are vertical and thus congruent. M is the midpoint of segment BC and so BM = MC. AN was constructed so that AM = MN. Thus triangles BMA and CMN are congruent by SAS.

From this it follows that angles ABM and NCM are congruent. But angle DCB is the sum of angles DCN and NCM, and so angle DCB is greater than angle NCM. Thus angle DCB is greater than angle ABM, as was to be shown.

This technique can be used to prove that any exterior angle of any triangle is greater than its two remotes.

The result proven holds in both Euclidean and hyperbolic geometry. But it does not hold in elliptic geometry. Where does the proof break down in elliptic geometry? We assumed that the segment from N to C is unique and that it lies in the interior of angle DCB. This cannot be assumed in elliptic geometry.

Triangle ABC is given. We extend side AC through C and mark point D on the extension. Angle DCB is by definition an exterior angle for triangle ABC. The angles of ABC to which it is not adjacent - angles A and B - are its remotes.

We will prove that the exterior angle DCB is greater than the remote angle B.

Mark the midpoint of BC; name it M. Construct the segment AN through M such that AM = MN. Connect N to C.

Angles BMA and NMC are vertical and thus congruent. M is the midpoint of segment BC and so BM = MC. AN was constructed so that AM = MN. Thus triangles BMA and CMN are congruent by SAS.

From this it follows that angles ABM and NCM are congruent. But angle DCB is the sum of angles DCN and NCM, and so angle DCB is greater than angle NCM. Thus angle DCB is greater than angle ABM, as was to be shown.

This technique can be used to prove that any exterior angle of any triangle is greater than its two remotes.

**Commentary**The result proven holds in both Euclidean and hyperbolic geometry. But it does not hold in elliptic geometry. Where does the proof break down in elliptic geometry? We assumed that the segment from N to C is unique and that it lies in the interior of angle DCB. This cannot be assumed in elliptic geometry.

## Friday, August 2, 2013

### Parallels: Assumptions

I've had the intent for some time now to do a little non-Euclidean geometry with my students.

I didn't want it to be mere history. A bit of history is fine, but mostly it should be mathematics. This means that we should prove a set of results.

But what results? I'd like to shock a bit, and so I decided to seek out a proof that, in the variety of non-Euclidean geometry that is called hyperbolic, the sum of the angles of a triangle is less than 180 degrees.

Of course the proof must be elementary. My students are quite bright, but they're only beginners. They have only the resources of that part of Euclidean geometry that we've developed. Thus my task was to find such a proof.

I think I have it. The technique comes from Saccheri. In a set of posts titled

Today I'll list those assumptions on which I'll draw. Some are definitions, some are postulates, some are theorems. Which is which is irrelevant. All that matters is that they'll be in place when on that day late in the second semester I begin.

I wish my assumptions to be, as it were, geometry-neutral. I wish them to hold in both Euclidean and hyperbolic, non-Euclidean geometry. Thus I do not include the Parallel Postulate (or any proposition equivalent to it) in the list.

The concept of congruence is key, and so I'll begin there.

I didn't want it to be mere history. A bit of history is fine, but mostly it should be mathematics. This means that we should prove a set of results.

But what results? I'd like to shock a bit, and so I decided to seek out a proof that, in the variety of non-Euclidean geometry that is called hyperbolic, the sum of the angles of a triangle is less than 180 degrees.

Of course the proof must be elementary. My students are quite bright, but they're only beginners. They have only the resources of that part of Euclidean geometry that we've developed. Thus my task was to find such a proof.

I think I have it. The technique comes from Saccheri. In a set of posts titled

*Parallels*, I'll outline the proof.Today I'll list those assumptions on which I'll draw. Some are definitions, some are postulates, some are theorems. Which is which is irrelevant. All that matters is that they'll be in place when on that day late in the second semester I begin.

I wish my assumptions to be, as it were, geometry-neutral. I wish them to hold in both Euclidean and hyperbolic, non-Euclidean geometry. Thus I do not include the Parallel Postulate (or any proposition equivalent to it) in the list.

The concept of congruence is key, and so I'll begin there.

**Assumptions**- Vertical angles are congruent
- In congruent polygons, sides and angles can be paired up in such a way that sides which correspond and angles which correspond have the same measure.
- In congruent triangles, side which correspond lie opposite angles which correspond.
- SAS. If two sides and an included angle of one triangle are congruent to two sides and an included angle of a second triangle, then those triangles are congruent.
- SSS. If the three sides of one triangle are congruent to the three sides of another, then those triangles are congruent.
- ASA. If two angles and an included side of one triangle are congruent to two angles and an included side of a second triangle, then those triangles are congruent.
- HL. If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg, respectively, of a second right triangle, then those triangles are congruent.
- Through a pair of given points a line may be constructed. It is unique.
- Lines are infinite in extent.

On one variety of non-Euclidean geometry - elliptic - lines are finite in extent. On the other - hyperbolic - lines are infinite, as they are in Euclidean geometry.

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