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given a line segment with length a. pick a random point within that line segment, would generate two line segments with length b and a-b where 0<b<a. what kind of geometric operation would generate the value of a/(b*(a-b)) ? or is it even possible?

peng yu
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    What is your work on the subject ? – Jean Marie Sep 09 '19 at 21:09
  • By geometric operation, do you mean a geometric construction that has length, area, or volume etc. equal to $1/(b(a-b))$? – Charles Hudgins Sep 09 '19 at 21:10
  • @CharlesHudgins yes. – peng yu Sep 09 '19 at 21:12
  • @JeanMarie purely for fun and trying to understand more about primitive geometry and math. actually i have some decision tree induction method related to this. trying to make a mapping from a ellipse boundary to splitting criteria – peng yu Sep 09 '19 at 21:13
  • Clearly we can form $b(a-b)$ by building a rectangle with sides $b$ and $a-b$. The question then is just how to form $1/x$ given area $x$. – Charles Hudgins Sep 09 '19 at 21:13
  • Take a look at https://math.stackexchange.com/q/1275768 – Jean Marie Sep 09 '19 at 21:16
  • @JeanMarie thanks for the link, i figured that i have missed a very important definition of 1, in my case, the 1 is a, not sure the solution would apply – peng yu Sep 09 '19 at 21:26
  • It's not hard to construct a length that represents your number, but finding a nice/simple geometrical interpretation seems harder. If only the $b$ was in the numerator... there are some very simple and nice geometrical constructions for $\frac{ab}{b-a}$ (e.g. make a right-angled triangle with lengths $x$ and $y$ and inscribe a square which has a side $z$ then $1/x+1/y = 1/z$). – Winther Sep 09 '19 at 21:51
  • Thanks ! I’m interested in both constructing the length and simple interpretation if possible! – peng yu Sep 09 '19 at 21:58
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    As the numerator and denominator of $c:=a/(b(a-b))$ are not homogeneous expressions of the same degree, constructing a segment of length $c$ from segments of length $a$ and $b$ requires a segment of length $1$. Otherwise, you could scale the resulting figure by, say, $2$, and the same construction would yield a segment of length $c':=2c$ from segments of length $a':=2a$ and $b':=2b$; but $c'\neq a'/(b'(a'-b'))$. ... Now, with segments $p$, $q$, $1$, it's not difficult to construct $pq$, then $1/(pq)$, and then $(p+q)/pq$. Can you see how? – Blue Sep 10 '19 at 03:46
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    @Blue Fundamental remark (I had overlooked this aspect). You should give it as an answer that I would be happy to upvote. – Jean Marie Sep 10 '19 at 07:44

1 Answers1

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(Converting comment to answer, by request.)

As the numerator and denominator of $c:=\dfrac{a}{b(a−b)}$ are not homogeneous expressions of the same degree, constructing a segment of length $c$ from segments of length $a$ and $b$ requires an auxiliary segment of length $1$. (Otherwise, you could scale the resulting figure by, say, $2$, and the same construction would yield a segment of length $c':=2c$ from segments of length $a':=2a$ and $b':=2b$; but $c'\neq \dfrac{a'}{b'(a'−b')}$.)

Now, with segments of length $p$, $q$, $1$, it's not difficult to construct a segment of length $pq$, then one of length $1/(pq)$, and then one of length $(p+q)/pq$:

enter image description here

  • First, arrange $\overline{OP}$, $\overline{OQ}$, $\overline{OR}$ (of lengths $p$, $q$, $1$, respectively) with the first two on opposite sides of $O$ and the last perpendicular to them. Construct $\bigcirc PQR$ (via well-known methods), and extend $\overline{OR}$ to meet this circle at $S$. Then, by the Power of a Point Theorem, $|OP||OQ|=|OR||OS|$; that is, $|OS|=pq$.

enter image description here

  • Next, construct $\overline{OT}$ (of length $1$) perpendicular to $\overline{OS}$. Construct the perpendicular to $\overline{ST}$ at $T$, and let it meet the extension of $\overline{OS}$ at $U$. Since $\triangle SOT\sim \triangle TOU$, we have $|OU|/|OT| = |OT|/|OS|$; that is, $|OU| = 1/(pq)$.

  • Finally, construct a perpendicular to $\overline{PQ}$ at $P$, and a parallel to $\overline{TU}$ at $Q$; let these lines meet at $V$. Since $\triangle UOT \sim \triangle VPQ$, we have $|PV|/|PQ|=|UO|/|OT|$, so that $|PV|=(p+q)/(pq)$.

This completes the construction. $\square$

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