# Tag Archives: box

## Three Little Geometry Problems

Here are three variations of geometry problems I got from @jamestanton on Twitter.

1. The numerical measure of a rectangle’s area and perimeter are equal (P=A) (obviously the units are different).  If the rectangle’s sides have integer lengths, what are the dimensions of the rectangle?
2. The numerical measure of a box’s surface area and volume are equal (V=SA).  If the box’s sides have integer lengths, what are its dimensions?
3. The numerical measure of a right triangle’s area and perimeter are equal (P=A).  If the triangle’s sides have integer lengths, what are its dimensions?

To give a complete solution to a math problem, remember that you must

1. show that your proffered solution(s) is (are) correct, and
2. shows that no other solutions exist.

Find convincing arguments that you have found all of the solutions for each.  While my solutions are shown below, I eagerly welcome suggestions for any other approaches.

STOP!!!
DO NOT READ ANY FURTHER IF YOU WANT TO SOLVE THESE PROBLEMS ON YOUR OWN.

Problem 1:  $P=A$ for a rectangle
Let the a=length and b=width.  Without loss of generality, assume $a\le b$.  Then, $a\cdot b=2(a+b) \Longrightarrow \frac{1}{2}=\frac{1}{a}+\frac{1}{b}$ which implies $2.

If $a=3, b=6$.  If $a=4, b=4$.  Thus, the only rectangles for which $P=A$ are a 3×6 and a square with side 4.

Problem 2:  $P=SA$ for a box
Let the a=length, b=width, and c=height.  Without loss of generality, assume $a\le b\le c$.  Then, $a\cdot b\cdot c=2(a\cdot b+a\cdot c+b\cdot c) \Longrightarrow \frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ which implies $2.

It wasn’t worth it to find all these by hand, so I wrote a quick spreadsheet to find all c values for given a and b values under the condition $a\le b\le c$.

Therefore, there are 8 such boxes with integer dimensions: 3x7x42, 3x8x24, 3x9x18, 3x10x15, 4x5x20, 4x6x12, 5x5x10, and 6x6x6.

Problem 3:  $P=A$ for a right triangle
Let the legs be a and b and the hypotenuse be c.  Perhaps there is a way to employ the technique I used on the first two problems, but my first successful solution invoked a variation on Euclid’s formula:  For some integer values of k, m, and n with $m\le n$, $a=k\cdot (m^2-n^2), b=2k\cdot m\cdot n,$ and $c=k\cdot (m^2+n^2)$ will form all Pythagorean triples (although not uniquely).

Because $\frac{1}{2} a\cdot b=a+b+c$, Euclid’s formula gives

$\begin{tabular}{ r c l } $$2$$ & $$=$$ & $$\frac{\displaystyle a\cdot b}{\displaystyle a+b+c}$$ \\ & $$=$$ & $$\frac{\displaystyle [k\cdot (m^2-n^2)]\cdot [2k\cdot m\cdot n]}{\displaystyle [k\cdot (m^2-n^2)]+[2k\cdot m\cdot n]+[k\cdot (m^2+n^2)]}$$ \\ & $$=$$ & $$\frac{\displaystyle 2k^2mn(m^2-n^2)}{\displaystyle 2km(m+n)}$$ \\ & $$=$$ & $$kn(m-n)$$ \end{tabular}$

What started out feeling like a difficult search for unknown side lengths has been dramatically simplified.  $2=kn(m-n)$ can only happen if

1. $k=2, n=1, m-n=1 \rightarrow m=2$ which Euclid’s formula converts to $a=6, b=8, c=10$ .
2. $k=1, n=2, m-n=1 \rightarrow m=3$ which Euclid’s formula converts to $a=5, b=12, c=13$ .
3. $k=1, n=1, m-n=2 \rightarrow m=3$, but this leads to a repeat of the first solution.

Therefore, there are only two right triangles with the property $P=A$:  the 6-8-10 and the 5-12-13 right triangles.

Again, any other solution approaches are encouraged and will be posted.