 Problem of the Week

Problem D and Solution

To the Other Side

Points $$A$$ and $$C$$ are vertices of a cube with side length 2 cm, and $$B$$ is the point of intersection of the diagonals of one face of the cube, as shown below. Determine the length of $$CB$$. Solution

Solution 1

Label vertices $$D$$, $$E$$ and $$G$$, as shown.
Drop a perpendicular from $$B$$ to $$AD$$. Let $$F$$ be the point where the perpendicular meets $$AD$$. Join $$B$$ to $$F$$ and $$C$$ to $$F$$.

The faces of a cube are squares. The diagonals of a square meet at the centre of the square. Therefore, $$BF = 1$$ and $$AF =1$$. Now, $$\triangle CAF$$ is right-angled.

Using the Pythagorean Theorem in $$\triangle CAF$$,
$$CF^2=CA^2+AF^2=2^2+1^2=5$$.
Therefore, $$CF=\sqrt{5}$$, since $$CF>0$$.

Looking at $$\triangle CFB$$, we know from above that $$CF~=~\sqrt{5}$$ and $$BF=1$$. We also know that $$\angle CFB = 90^\circ$$. Because of the three-dimensional nature of the problem, it may not be obvious to all that $$\angle CFB = 90^{\circ}$$. To help visualize this, notice that $$CF$$ and $$BF$$ lie along faces of the cube that meet at $$90^{\circ}$$.
Using the Pythagorean Theorem in $$\triangle CFB$$, $$CB^2=CF^2+BF^2=\sqrt{5}^2+1^2=5+1= 6$$. Since $$CB>0$$, we have $$CB=\sqrt{6}$$.
Therefore, the length of $$CB$$ is $$\sqrt{6}$$ cm.
Solution 2

Label vertices $$D$$ and $$E$$, as shown. The faces of a cube are squares. The diagonals of a square right bisect each other. It follows that $$AB=BE=\frac{1}{2}AE$$. Since the face is a square, $$\angle ADE=90^{\circ}$$ and $$\triangle ADE$$ is right-angled. Using the Pythagorean Theorem in $$\triangle ADE$$, $$AE^2=AD^2+DE^2=2^2+2^2=8$$. Since $$AE>0$$, we have $$AE=\sqrt{8}$$. Then $$AB=\frac{1}{2}AE=\frac{\sqrt{8}}{2}$$.
Because of the three-dimensional nature of the problem, it may not be obvious to all that $$\angle CAB=90^{\circ}$$. To help visualize this, note that $$\angle CAD=90^{\circ}$$ because the face of the cube is a square. Rotate $$AD$$ counterclockwise about point $$A$$ on the side face of the cube so that the image of $$AD$$ lies along $$AB$$. The corner angle will not change as a result of the rotation, so $$\angle CAD=\angle CAB=90^{\circ}$$.
We can now use the Pythagorean Theorem in $$\triangle CAB$$ to find the length $$CB$$. $CB^2=CA^2+AB^2=2^2 + \left(\frac{\sqrt{8}}{2}\right)^2=4+\frac{8}{4}=4+2=6$ Since $$CB>0$$, we have $$CB=\sqrt{6}$$ cm.
Therefore, the length of $$CB$$ is $$\sqrt{6}$$ cm.
Note, we could have simplified $$AB=\frac{1}{2}AE=\frac{\sqrt{8}}{2}$$ to $$\sqrt{2}$$ as follows: $\frac{\sqrt{8}}{2}=\frac{\sqrt{4 \times 2}}{2}=\frac{\sqrt{4} \times \sqrt{2}}{2}=\frac{2\sqrt{2}}{2}=\sqrt{2}.$ The calculation of $$CB$$ would have been simpler using $$AB=\sqrt{2}$$. Often simplifying radicals is not a part of the curriculum at the grade 9 or 10 level.