# Origin Triangle Tetrahedron Volume

I have a problem that goes like this:

Triangle $ABC$, with sides of length $5$, $6$, and $7$, has one vertex on the positive $x$-axis, one on the positive $y$-axis, and one on the positive $z$-axis. Let $O$ be the origin. What is the volume of tetrahedron $OABC$?

I really can't wrap my head around this problem. I thought I could find the area of the triangle base by using Heron's Formula, and then multiply that by $1/3$ of the height, but I've got no way of finding the height.

The posted solution for this question was very confusing.

It says:

wlog use $A(a,0,0), B(0,b,0), C(0,0,c)$

(i) wlog set $5 = \sqrt{a^2+b^2} \qquad 6 = \sqrt{b^2+c^2} \qquad 7 = \sqrt{c^2+a^2}$

(ii) Square for $25 = a^2+b^2 \qquad 36 = b^2+c^2 \qquad 49 = c^2+a^2$

(iii) Adding and dividing by 2 gives $a^2+b^2+c^2 = 55$

(iv) Subtracting each element of (ii) from (iii) gives $a^2 = 19 \qquad b^2 = 6 \qquad c^2 = 30$

(v) Multiplying gives $a^2b^2c^2 = 19 \cdot 6 \cdot 30 = 19 \cdot 5 \cdot 6^2$

(vi) Squareroot and divide by 6 to get $V = \frac{1}{6} abc = \sqrt{95} \implies \boxed{\text{C}}$

What I don't understand about this solution is why they are using these $a, b$, and $c$ coordinates and taking the square roots of two coordinates added together and setting them equal to the triangle's side lengths.

Perhaps I just don't properly understand how tetrahedrons and $x$-$y$-$z$ coordinates work. I'd really appreciate some help wrapping my head around this problem. Thanks :)

• See Distance Formula. Feb 3, 2018 at 2:30
• If you copied this solution from a post on MSE, please give due credit to the original post Feb 3, 2018 at 2:30
• This is 2008 AMC12A problem 18 artofproblemsolving.com/wiki/… Feb 3, 2018 at 2:48 $OA^2 + OB^2 = 5^2\\ OB^2 + OC^2 = 6^2\\ OA^2 + OC^2 = 7^2$

And even though it looks quadratic, it is really a system of linear equations.

If it feels strange to treat variables with squared terms as linear equations, rewrite it as:

$x + y = 25\\ y + z = 36\\ x + z = 49$

And once you have $x, y, z$ finding the volume is $\frac 16 \sqrt {xyz}$

Because $A$ is on the $x$-axis, there exists a number $a$ such that the coordinates of $A$ are $(a,0,0)$. Similarly, there exist numbers $b$ and $c$ such that the coordinates of $B$ and $C$ are $(0,b,0)$ and $(0,0,c)$, respectively.

We know that $AB,BC,$ and $CA$ are $5,6,7$ in some order. We may assume $AB=5,BC=6,$ and $CA=7$, because otherwise we could relabel the tetrahedron so that this is the case.

Note that $\triangle OAB$ is a right triangle, so $OA^2+OB^2=AB^2$. But $OA=a,OB=b,$ and $AB=5$, so the equation becomes $a^2+b^2=25$.

Is the rest of the solution now clear?

• Oh shoot I just noticed that my rep is exactly 1234 right now Feb 3, 2018 at 2:46

I don't understand about this solution is why they are using these a, b, and c coordinates and taking the square roots of two coordinates added together and setting them equal to the triangle's side lengths.

HINT: You've been given the side lengths, so in the solution they are solving for the coordinates of the three points. If your question is, why this method works for solving those coordinates, then have a look at this simple diagram of a tetrahedron: It has three points on each $x$,$y$ and $z$ axes respectively. Can you identify three right triangles in it, coplanar with the $xy$, $yz$ and $xz$ planes? Now, can you apply the Pythagoras theorem...?