Triangles and Cubes

How many distinct triangles can be constructed by connecting three different vertices of a cube?
Source: mathcontest.olemiss.edu 11/1/2010

Solution

STEP 1 Try something simpler first.

Consider ADEH a square in a plane (it doesn’t look square because we look at it from space). Connect vertex A to vertex E and we have constructed 2 distinct triangles ADE and AEH. Similarly, connect vertex D to vertex H to form 2 more distinct triangles ADH and DEH. Thus, with 4 vertices we can construct 4 distinct triangles. We need 3 vertices to make a triangle. So how many ways can we choose 3 different vertices out of 4 vertices? The answer is _4\textup{C}_3=4.

STEP 2 Now try a square ADEH and one point B in space above the square.

By looking carefully at the above figure, we can see that there are a total of 10 distinct triangles:
ADE,AEH,ADH,DEH
ABD,BDE,BEH,ABH
ABE,BDH

Alternatively, we can think of choosing 3 different vertices out of 5 vertices to build our distinct triangles, _5\textup{C}_3=10.

STEP 3 One square ADEH and two points B and G in space above the square.

It is getting much harder to count the triangles visually, so we are going to use a different method to identify them. We know that there are _6\textup{C}_3=20 triangles. Let’s write down the 6 vertices in alphabetical order and without looking at the figure pick out 3 different vertices at a time as follows.

ABCDG                                List the 6 vertices in alphabetical order
ABD,ABE,ABH,ABG      pick out 3 different vertices at a time
ADE,ADH,ADG
AEH,AEG,AHG
BDE,BDH,BDG
BEH,BEG,BHG
DEH,DEG,DHG,EHG

Now, look back at the above figure and verify to your satisfaction that indeed there are exactly 20 distinct triangles.

STEP 4 Now consider a cube.
How many distinct triangles can we construct by joining three different vertices?
From the patterns we have seen before, we know that there are _8\textup{C}_3=56 distinct triangles. We can verify this fact by looking at the figure of the cube and list out the triangles as follows.


Front face ABCD ABC ABD ACD BCD
Rear face EFGH EFG EFH EGH FGH
Top face BCFG BCF BCG BFG CFG
Bottom face ADEH ADE ADH AEH DEH
Left face CDEF CDE CDF CEF DEF
Right face ABGH ABG ABH AGH BGH
Front face ABCD and vertex E ABE ACE BCE BDE
Front face ABCD and vertex F ABF ACF ADF BDF
Front face ABCD and vertex G ACG ADG BDG CDG
Front face ABCD and vertex H ACH BCH BDH CDH
Rear face EFGH and vertex A AEF AEG AFG AFH
Rear face EFGH and vertex B BEF BEG BEH BFH
Rear face EFGH andvertex C CEG CEH CFH CGH
Rear face EFGH and vertex D DEG DFG DFH DGH

Be mindful that there are duplicate triangles that were deleted from the above list. For example, in the case of front face ABCD and vertex E, there are 6 duplicate triangles out of 10 triangles. They are listed underlined as follows:
\underline{ABC},\underline{ABD},ABE,\underline{ACD},ACE,\underline{ADE},\underline{BCD},BCE,BDE,\underline{CDE}

Also, there is no need to consider the left and right faces and their opposite vertices, because the triangles will be duplicated.

Thus, we can construct 56 distinct triangles by joining three different vertices of a cube.

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , . Bookmark the permalink.

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