# A cube is dropped on the floor, and the triangular hole created has sides of lengths $68,$ $75,$ and $77.$ Find the depth of the hole.

I dropped a heavy cube the other day, creating a dent in my floor. The triangular hole in the surface of the floor has sides of lengths $$68,$$ $$75,$$ and $$77.$$ Find the depth of the hole.

I drew a diagram, but I'm really not sure what to do next. I also tried taking some cross sections of the diagram, but that also doesn't do much. I'm really not sure how to continue from here. :(

• HINT...you can calculate the sides of the triangle surface hole and work out the area, then use the volume to work out the height Mar 11 at 17:10
• A hole is a 3-dimensional object: in this case, in the form of a triangular pyramid, which has six edges. Given that the pyramid has a cuboidal apex, its dimensions can be determined by three edges. Here, presumably either (A) the apical or (B) the basal edges are specified. Three of the answers (including one deleted answer) deal with case A, while the other (accepted) answer is for case B. The question does not specify which case is intended. Mar 12 at 10:19

First, you can calculate the sides of the tetrahedron whose apex is the deepest point of the dent. We already know three of its sides which are $$68, 75, 77$$, and since we have perpendicular planes, then if these side lengths are $$x, y, z$$, then,

$$x^2 + y^2 = 68^2 , x^2 + z^2 = 75^2 , y^2 + z^2 = 77^2$$ and these three equations solve to

$$x = 12 \sqrt{15} , y = 4 \sqrt{154} , z = 3 \sqrt{385}$$

Now placing the origin of a coordinate frame at the intersection of the three perpendicular planes, with the axes along the edges, then the equation of the plane of the triangular hole is

$$\dfrac{x}{12 \sqrt{15} } + \dfrac{ y }{4 \sqrt{154}} + \dfrac{z}{3 \sqrt{385}} = 1$$

Hence the distance of the origin (which is the deepest point) to the plane of the hole is

$$d = \dfrac{1 }{\sqrt{ \dfrac{ 1 }{ 144(15) } + \dfrac{1}{16(154)} + \dfrac{1}{9(385) } }}$$

And this comes to

$$d = 12 \sqrt{6}$$

Take the bottom of the hole as the origin $$O$$, and the recessed edges from $$O$$ as the axes, with unit vectors $$\pmb i$$, $$\pmb j$$, and $$\pmb k$$ respectively along the edges of length $$a$$, $$b$$, and $$c$$. The vector $$x\pmb i+y\pmb j+z\pmb k$$ to the point $$A$$ vertically above $$O$$ in the plane of the floor is perpendicular to the vectors from $$A$$ to the points on the axes with respective vector positions $$\pmb i$$, $$\pmb j$$, and $$\pmb j$$ from $$O$$. This perpendicularity gives us three zero scalar products: $$x(a-x)-y^2-z^2=0,$$ $$-x^2+y(b-y)-z(c-z)=0,$$ $$-x^2-y^2+z(c-z)=0.$$ These equations may be written $$ax=by=cz=x^2+y^2+z^2.$$ Substituting $$y=(a/b)x$$ and $$z=(a/c)x$$ gives $$x=\frac{r^2}a,\qquad y=\frac{r^2}b\qquad z=\frac{r^2}c,$$ where $$\dfrac1{r^2}=\dfrac1{a^2}+\dfrac1{b^2}+\dfrac1{c^2}.$$ In conclusion, we get $$|OA|=\sqrt{x^2+y^2+z^2}=r.$$ For $$a=68$$, $$b=75$$, and $$c=78$$, the corresponding value of $$r$$ is $$\frac1{\sqrt{\dfrac1{68^2}+\dfrac1{75^2}+\dfrac1{78^2}}}.$$

Coordinate-free Solution

Consider a cuboid with sides $$68$$, $$75$$, and $$78$$. Now the "hole" is a sixth of that cuboid. Hence its volume equals $$68\cdot75\cdot77/6=65\,450$$.

(Let$$ABCDEFGH$$ be the cuboid. The pyramid $$ABCDE$$ is third of the cuboid and the tetraeder $$ABDE$$ is half of that pyramid.)

Due to Heron, see https://en.wikipedia.org/wiki/Heron%27s_formula, (after some easy calculations) the square of the area of the triangle is $$77^2\cdot75^2+68^2\cdot77^2+68^2\cdot75^2$$.

Alternative solution:

Let $$a$$, $$b$$, and $$c$$ be the given lengths. Then verify that the plane through $$(a,0,0)$$, $$(0,b,c)$$ and $$(0,0,c)$$ is given by the equation $$\left\langle\begin{pmatrix} bc\\ ac\\ ab \end{pmatrix},\vec x\right\rangle-abc=0,$$ hence its distance from the origin is $$\frac{abc}{\sqrt{a^2b^2+a^2b^2+b^2c^2}}$$

• I've continued. Mar 11 at 18:57