*Answers and discussion of problems below are now available in Comments*.

Some of these questions are reprinted from copyrighted materials from the College Board. In some cases, I've modified the questions for instructional purposes. These questions are linked to more advanced topics involving polyhedra and college geometry but they are appropriate for middle and secondary students as well. When is it valuable for students to actually enumerate the objects asked for? These questions also stress the importance of the phrase 'determined by.'

1. What is the total number of right angles formed by the edges of a cube?

Modified version for classroom use: Show that there are 24 right angles formed by the edges of a cube. You and your partner must find at least TWO different methods. [Note: By giving students the 'answer', the focus is then on process.]

2. How many distinct pairs of parallel edges are there in a cube (or rectangular solid)?

3. How many different planes are determined by the vertices of a cube (or rectangular solid)?

4. How many equilateral triangles are determined by the vertices of a cube?

## Thursday, February 22, 2007

### Geometry SAT Problems - Do These Questions Help Students Develop Spatial Sense and Combinatorial Thinking?

Posted by Dave Marain at 9:36 AM

Labels: combinatorial math, cubes, geometry, SAT-type problems, spatial sense

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## 19 comments:

#3 is tricky. My first (quick) answer was not correct. Ways of counting this might be interesting to compare.

Something (potentially) useful for all parts of this question: introduce some sort of notation for the vertices.

Something else (potentially) useful: choose a naming system for the directions that segments or planes can run within the cube. Certainly the external faces need a collective name (x- y- or z- planes? But I am biased in favor of coordinates)

jonathan--

i agree! #3 is definitely the hardest because there are so many to consider! Most of the planes are easy to visualize in the first 30 seconds but some are not. Students and most everyone else struggle without a good model and way to keep track. One might think it makes sense to use 8C3 = 56, since any combination of 3 vertices determines a plane but then you have to remove the duplicates. Note that 8C3 also determines how many triangles and there should not be duplicates, so does this mean there are 56 possible triangles! What is the nature of these triangles? This could have been queston 4, leaving the equilateral triangles for #5. There's a lot more to say here...

Labeling the vertices is an excellent strategy. I usually use A,B,C,D for the upper face going clockwise and E,F,G,H for the bottom. A naming procedure for systematic counting of planes is critical.

I will post answers later but by then our regular viewers will have solved it and be ready to move on!

Answers:

1) 24 = 6x4 (4 on each of 6 faces)

= 8x3 (3 at each of 8 vertices)

2) 18 (3 x 4C2)

3) 28 (I believe this is correct but you may disagree!)

4) 8 (one for each vertex - why?!?)

PLS LET ME KNOW IF YOU DISAGREE OR AGREE. I WOULD LOVE A SIMPLE COUNTING METHOD FOR #3. I NEEDED TO LABEL THE VERTICES AND CHECK SEVERAL COMBINATIONS. THE HARDEST FOR ME WERE THE EIGHT PLANES CONTAINING THE TRIANGLES FORMED BY AN EDGE, FACE DIAGONAL AND CUBE DIAGONAL. VERY EASY TO MISS THESE. USUALLY SOMEONE SEES A SIMPLE WAY TO ENUMERATE!

For question three, I like your idea of doing 8C3 = 56 and subtracting the duplicates. However, I end up with the answer being just 20 unique planes.

The first set of duplicates is easy to visualize because each face of the cube has four vertices and 4C3 = 4, so we've counted each of those 6 planes 4 times (that is, 3 extra times). So 3*6 = 18 duplicates. The next set of duplicates occur when we consider the planes formed by the planes through the diagonals on opposite faces. There are six of these. (You can imagine the "X" on each of the faces of the cube and see that there are three sets of two planes). These six planes also have four vertices each, are thus also being counted four times instead of once. So 56 - 18 - 18 = 20.

How did you count your 28? If I was adding them up positively, I would count 6 faces, the 6 opposite face diagonals, and the 8 tricky ones you mentioned. That's 20 also. Which ones am I missing?

I like the thinking that's required here, but I'm not sure that my students will even be able to approach problems 3 or 4.

mrc--

i think there are 8 other 'triangles' that are formed from an edge, face diagonal and cube diagonal. I believe that the planes containing these triangles are not duplicated from the others you mentioned and distinct from the planes containing those 8 equilateral triangle from question 4. I will have to list these using the vertices. Of course, I could be wrong and they really are part of the 20. By the way, i got 20 the first time I tried this and then suspected I missed a few so I looked at other combinations of vertices.

Do you think it would help students to use plastic (transparent) models for these slices or some excellent geometry software to show this in full color and 3-D. I found an excellent site that shows animations of planes slicing a cube (actually the plane is fixed and the cube passes through it at different angles) but it doesn't exactly solve this problem. What it does do is to demonstrate how it is possible for a cube to have an hexagonal cross-section! I may post the link to this.

Here are a couple of websites that help one to visualize some of the triangles and plane cross-sections in a cube:

Equilateral Triangles in a Cube

Slices in a Cube

(check out the 3rd movie - it's awesome!)

Ok, now for the attempt to label the 8 OTHER planes...

Label the vertices clockwise on the top face A,B,C,D where A is the vertex at the top left (looking down on the cube), B is top right, C is bottom right, etc. Do the same for the base of the cube using E,F,G,H wjere E is directly below A, F below B, G below C, etc.

Here's an example of one of the EIGHT planes containing the equilateral triangles:

The plane containing BDG, formed by the 3 face diagonals BD, DG, BG.

Here's an example of TWO of the 'other' EIGHT:

The planes containing the triangles DFG and DFA.

Now I could be mistaken and these have already been counted among the 20 but I don't think so.

To summarize, the 28 break down as follows:

6 faces

2 diagonal planes ACGE and BDHF

4 diagonal planes connecting any top edge to opposite bottom edge like ABGH

(Note: mrc, you combined the 4 & 2)

8 forming those equilateral triangles like ACH

8 more (the killer ones) like DFG, unless these were previously counted!

Total: 28 - ok, correct my errors!!

Oops! The more I look at this darn cube, the more i think those 8 other 'triangles' are part of the planes connecting a top edge to the opposite lower edge! Thus, my DFG is really part of ADFG. If that's the case, we're back to 20 and I have egg all over my

FACE!LOOKS LIKE IT'S 20 AFTER ALL! Guess I could use some work on my spatial reasoning!

Well, I asked this set of questions to the other math teachers at our department meeting today. We love puzzle time! They did fine on the first two questions, but were lost in approaching the third. They were reaching for a paper cube model right away.

Thinking about the mental gymnastics we've been going through, I was looking for a better way to explain the "tricky 8" planes. I came up with this: There are exactly two unique tetrahedra that share all their vertices with a cube but share no faces with that cube. In your notation they would be ACFH and BDEG. Each tetrahedron has four faces, and thus we have 8 planes.

Now any grouping of three vertices you can pick from the cube will define a plane that falls into one of these three categories:

1. A face of the cube (6 unique planes)

2. A plane that cuts the cube in half, into two right isosceles triangular prisms (6 unique planes)

3. A face of one of our two tetrahedra (8 unique planes)

Now I'm confident that there are exactly 20 unique planes through these 8 vertices.

mrc--

no doubt about it - 20 planes determined by the vertices! Thank you for exploring this problem further and carefully verifying the result! You make writing these problems worthwhile. One of my colleagues asked what my sources are for these questions. Well, I've been writing problems for 3+ decades so I do have a lot of material (if i could find them all!). However, many of the questions I'm posting are new variations of these or brand-new problems. It's a labor of love, but I'm having labor pains!

I allowed three students to play with the planes question (was not a class). I really did not intervene. It was a freshman, a sophomore, and a junior. The freshman labeled the vertices and then systematically listed all combinations of 3. The sophomore took charge, crossing out duplicates. The junior participated a little, supervised a little. They got 20. Then I counted, and matched my results against theirs. 6 faces. 6 planes joining opposite parallel sides. And then I started counting the odd ones out, and realized that each remaining plane "cuts off" one vertex. 8 vertices, 8 of these. And I listed the 20 with their labels, a perfect match (esp since the slightly bossy sophomore ruled that 1,3,5,7 must be listed as 1,3,5 and not 1,5,7)

In case not everyone knows, there's a geometrical SAT problem that made the news around 1980:

Take two pyramids, one with a triangular base and the other with a square base, with all edges equal. All the triangular faces are congruent, so we may attach the two by aligning respective triangular faces and gluing them together.

The original pyramids had four and five faces, respectively. How many faces does the compound polyhedron have?

Oops, I forgot to sign my comment about the compound of two pyramids. I'm

Dan Hoey, haoyuep@aol.com

you are very confusing and i think the answer is deffantly 8

#3 is very tricky, but i have come to the conclusion that the answer is possibly 10

There are no planes defined by vertices of a cube. The vertices only define faces. A cube has no internal faces.

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