Depth Of A Dent

Geometry Level 4

I accidentally dropped a large, heavy metal cube onto my nice wood floor. It landed on one of its corners, and made a huge dent. (Okay, so my wood floor isn't "nice." It's balsa. I knew I shouldn't have just gone with the cheapest option...) The triangular hole it made in the surface of the floor has sides of length 68mm, 75mm, and 77mm. The depth of the dent, in mm, can be written as $\sqrt{n}$ , where n is a positive integer. Find n .

Details & Assumptions:

The depth is measured perpendicular to the floor.

The answer is 864.

2 solutions

Anish Puthuraya
Feb 20, 2014

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It is clear that the cube here represent nothing but the 3-D space.

Let the Triangular face be represented by the points $\displaystyle A,B,C$ with the respective coordinates as mentioned in the figure.

For now, let us forget the $\displaystyle mm$ , i.e the units.

From the given conditions,

$\displaystyle \sqrt{x^2+y^2} = 68$
$\displaystyle \sqrt{y^2+z^2} = 75$
$\displaystyle \sqrt{z^2+x^2} = 77$

Solving these equations,

$\displaystyle x = \sqrt{2464} = l$

$\displaystyle y = \sqrt{2160} = m$

$\displaystyle z = \sqrt{3465} = n$

Now, let us find the equation of the plane formed by these 3 points. I shall focus on the methods rather than the calculation.

First, we find the vectors $\displaystyle \vec{CA}$ and $\displaystyle \vec{CB}$ .
We take their cross product to find the Normal vector.

$\displaystyle \vec{CA} = l\hat{i}-n\hat{k}$
$\displaystyle \vec{CB} = m\hat{j}-n\hat{k}$

Performing $\displaystyle \vec{CA}\times\vec{CB}$ , we get,
$\displaystyle \vec{n} = mn\hat{i} + nl\hat{j} + lm\hat{k}$

Thus, the equation of the plane is,
$\displaystyle (mn)x + (nl)y + (lm)z = d$

We find $\displaystyle d$ by substituting any of the points...It turns out to be,
$\displaystyle d = lmn$

Now, the problem asks us to find the depth of the dent.
In our case, the equivalent question is the distance of the plane from Origin O , which is,

$\displaystyle h = \frac{lmn}{\sqrt{(mn)^2+(nl)^2+(lm)^2}}$

On substituting the values of $\displaystyle l,m,n$ , we get,
$\displaystyle h = 29.393 units$

Squaring both sides,
$\displaystyle h^2 = 864$

Thus,
$\displaystyle\boxed{n = 864}$

I found your three sides as well, and then I computed the volume of the pyramid in two different ways: by using that it was $\frac16 lmn$ (thinking of the right triangles as supplying the base and the height), and that it was $\frac13 Ah$ where $A$ is the area of the $68-75-77$ triangle and $h$ is the thing we're looking for. Setting them equal and using Heron's formula gave me $h$ .

Patrick Corn - 7 years, 3 months ago

You can also use the well-known fact that the perpendicular from O to the plane of ABC (in this case) crosses the triangle ABC at it's orthocentre. This can be proved using properties of the orthic triangles and plane theory (basically 3-dimensional angle chasing).

TheKnee OfJustice - 7 years, 3 months ago

That is awesome...Where do you find these types of facts?? Please tell me.!

Anish Puthuraya - 7 years, 3 months ago

very good

Aditya Raj - 7 years, 3 months ago

I didn't get why u did CA X CB, why it gave equation of plane and obviously whatever u did after that can anyone explain plz........ i know vectors quite well but haven't been taught all this eqn of plane etc. etc.

Ak Sharma - 7 years, 3 months ago

Note: I shall assume that you know how to perform cross products.

Can we agree on the fact that the cross product of $\displaystyle\vec{CA}$ and $\displaystyle\vec{CB}$ gives us the Normal Vector to the given plane? (if not, please tell me and I shall explain it to you).

Moving on to the Equation of the Plane,
Consider the following situation.

Let $\displaystyle \vec{r_o}$ be the position vector of some point $\displaystyle P_o = (x_o,y_o,z_o)$ , which lies on the plane.
Let $\displaystyle \vec{n} = (a,b,c)$ be the Normal vector to the plane.

If $\displaystyle P = (x,y,z)$ is a general point on the plane, then it is clear that,
The vector drawn from $\displaystyle P_o$ to $\displaystyle P$ is perpendiculat to $\displaystyle \vec{n}$ (that is why $\displaystyle n$ is called the Normal vector, right?)

Hence,
$\displaystyle \vec{PP_o}\perp \vec{n}$

$\displaystyle\Rightarrow \vec{n}\cdot\vec{PP_o} = 0$

Now,

$\displaystyle \vec{PP_o} = (x-x_o)\hat{i}+(y-y_o)\hat{j}+(z-z_o)\hat{k}$ and,
$\displaystyle \vec{n} = a\hat{i}+b\hat{j}+c\hat{k}$

Substituting these values,

$\displaystyle a(x-x_o)+b(y-y_o)+c(z-z_o) = 0$

Thus,

$ax+by+cz = d$ ,where $\displaystyle d = (ax_o+by_o+cz_o)$

Therefore, this is the Equation of the Plane. This is what I used in my solution if you look carefully.

For More Information, check out this link

Anish Puthuraya - 7 years, 3 months ago

Angelito Matibag
Apr 16, 2014

LET: A, B, C - be the corner of the triangle on the floor O - be the edge of the cube that landed on the floor h -be the height perpendicular to the floor

so,A B = 68, AC = 75 & BC = 77 and since it is from a cube, triangles AOB, AOC & BOC right triangles, hence:

(AO)^2 + (BO)^2 = 68^2 EQ. 1 (AO)^2 + (CO)^2 = 75^2 EQ. 2 (BO)^2 + (CO)^2 = 77^2 EQ. 3

solving these equations simultaneously, we get:

AO = √2160 BO = √2464 CO = √3465

solving for the area of ABC by Heron's formula,

S = (AB + AC + BC)/2 = (68 + 75 + 77)/2 = 110

[ABC] = √S(S-AB)(S-AC)(S-BC) = √110(110-68)(110-75)(110-77) = 2310

computing the volume of the pyramid,

V =( 1/3)[(1/2)(AO)(BO)(CO)] V = (1/3)[(1/2)(√2160)(√2464)(√3465)] V = 22633.28522

also we can compute the volume by this formula:

V = (1/3)([ABC])(h)

since "h" is the only variable that is unknown, we now easily get its value by substituting all the known values to the formula above.

22633.28522 = (1/3)(2310)(h) h = 22633.28522/[(1/3)(2310)] h=29.39387691

so,

h^2 = \boxed{864}

Superb solution! Appreciate it

Krishna Ar - 7 years, 1 month ago

Depth Of A Dent

The answer is 864.

2 solutions

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