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.