Algebra/geometry question

Factual Questions
1

At least I don’t think resorting to Trigonometry should be necessary; but here’s hoping.

My question is related to what’s known as the right triangle altitude theorem, aka. the geometric mean theorem. Basically, when you have a right triangle with the hypotenuse taken as the base, the theorem relates the relationship between the hypotenuse, the vertical altitude, and the bases of the two resulting smaller similar triangles:

If h denotes the altitude in a right triangle and p and q the segments on the hypotenuse then the theorem can be stated as h = {\sqrt {p×q}}

To give an example with all whole numbers, if you have a 3-4-5 right triangle of hypotenuse 25 with sides of 15 and 20, the altitude will be 12 and p and q will be 9 and 16.

My question is concerned with a rectangle containing the right triangle, with height the same as the altitude and a length of p + q. I want to be able to take the ratio of length to width of the rectangle as a single number, and plug that into a formula that will give the ratio p : q. This should be doable, yet my algebra skills seem insufficient. Anyone?

2

If I understand correctly, here’s the formula you’re looking for. If r = (p+q)/\sqrt{pq} is the ratio of length to width, then

\frac pq = \tfrac 12\left( r^2 - 2 + r\sqrt{r^2-4} \right).

The square root is OK because r\geq 2 by the arithmetic-geometric mean inequality.

Here’s how I derived it: Let x = p/q. Then

r = \frac{p+q}{\sqrt{pq}} = \sqrt{x} + \frac{1}{\sqrt x}

so

r^2 = x + 2 + \frac 1x.

Rearranging,

x^2 + (2-r^2)x + 1 = 0

and now you can write down the solutions using the quadratic formula. One of those will give you x and the other gives you 1/x.

3

I don’t know if I understand the question either, but if

x = \frac{p+q}{\sqrt{pq}}

then

\frac{p}{q} = \frac{x^2+x\sqrt{x^2-4}-2}{2}.
4

It must be right, we both got the same formula!

Last comment while I’m in the edit window: This formula assumes p\geq q. Put another way, it gives the larger of p/q or q/p.

5

Got it (I think). I’ll play around with it and see if some examples give the expected answers. Thanks all!

6

I’m sorry but this left me completely dead in the water. I don’t know how you derive the second half of the equation from the first half. I’m missing a step somewhere.

7

It might be easier to go from the right side to the left?
\sqrt{x}+\frac{1}{\sqrt{x}}
\sqrt\frac{p}{q}+\sqrt\frac{q}{p}
\frac{p}{\sqrt{pq}}+\frac{q}{\sqrt{pq}}
\frac{p+q}{\sqrt{pq}}

8

Right. I left out the intermediate steps. But the key in either direction is realizing that you can rewrite the fraction

\frac{p+q}{\sqrt{pq}} = \frac{p}{\sqrt{pq}} + \frac{q}{\sqrt{pq}}.
9

That much I got; but apparently if I was ever taught how to rework square roots in fractions, I’ve forgotten it.
\sqrt\frac{q}{p} equals \frac{q}{\sqrt{pq}} ??

10

Inside the square root, multiply and divide by q:

\sqrt{\frac qp} = \sqrt{\frac{q^2}{pq}} = \frac{\sqrt{q^2}}{\sqrt{pq}} = \frac{q}{\sqrt{pq}}

Or, you can work it from right to left, first using that q = \sqrt{q^2}.

11

Well it’s apparent why I couldn’t come up with a formula on my own. And yet given concrete numbers for the hypotenuse and the altitude, I was calculating 𝑝 and 𝑞 fairly easily; I just couldn’t figure out how to generalize it.