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Given a line segment with length $ a $, a line segment with length $ b $, a compass, and a straightedge (you can only measure line segments with lengths $a $ or $ b $), is it possible to construct a line segment with length $ ab $ or $ \frac{b}{a} $?

Motivation: In one of the geometric construction problems that I solved, I noticed that I constructed a line segment with length $ \frac{ab}{a+b} $. The construction is as follows: Assume $ b > a $. Draw a right-angle triangle $ABC$ with $ CA = a $ and $ BC = b $, where $ \angle A = 90^\circ $. Now draw the angle bisector of $ \angle C $. Let the point of intersection between the angle bisector of $ \angle C $ and $ \overline{AB} $ be $ E $, and draw a line parallel to $ \overline{AC} $ intersecting $ BC $ at $ F $. Let $ x := EF $; it is easy to see that $ DC = x $.

$$ \frac{x}{a} = \frac{b - x}{b} $$ $$ x = \frac{ab}{a + b} $$ enter image description here

I attempted to find a geometric construction for $ ab $ but was not successful.

pie
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    Once a fixed segment is defined as length $1,$ given segments $a,b$ we can construct lengths $a+b, |a-b|, ab, \frac{a}{b}.$ Finally, $\sqrt a.$ – Will Jagy May 14 '24 at 00:00
  • @WillJagy the segment with length $1$ is not given, so one should construct it. and $a+b, |a-b|$ are can be trivially done. – pie May 14 '24 at 00:06
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    It's not possible to construct a segment of length $1$. And without a segment of length $1$ it's not possible to construct $ab$ or $\frac{a}{b}$. – jjagmath May 14 '24 at 00:08
  • @jjagmath I have two questions: the first is how to construct these segments with knowing the unit length, second question is: if the segment with length $1$ is not given why it is impossible to construct $ab$ or $\frac a b $ ? – pie May 14 '24 at 00:11
  • For the first question: Did you try by yourself? You asked just 3 minutes after I post my comment... – jjagmath May 14 '24 at 00:16
  • https://en.wikipedia.org/wiki/Constructible_number#Equivalence_of_algebraic_and_geometric_definitions – Will Jagy May 14 '24 at 00:26
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    For the second question: Imagine you have two segments $a$ and $b$ and you have a construction that gives you $a b$. If $1<a<2$ and $1<b<2$ then the segment $a b$ will be bigger than both $a$ and $b$. But now repeat the exact same construction with a segments of length $a/2$ and $b/2$, then it should give you a segment that is bigger than $a/2$ and $b/2$, which is absurd because $(a/2)(b/2) < a/2$ – jjagmath May 14 '24 at 00:26
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    Regarding multiplication: Suppose you have a segment of length $d$ and a construction that builds $d^2$. Whether the second segment is longer, shorter, or equal to the first segment depends entirely upon how $d$ compares to "$1$". So, you have to know what "$1$" means before you begin. ... The thing is, due to similarity in Euclidean geometry, anything you can construct at some scale you can construct at any scale. There's no intrinsic sense of size; no inherent "$1$". You pick a segment you like, and declare it "$1$". (FYI: Non-Euclidean geometry doesn't have this problem. Intrigued?) – Blue May 14 '24 at 01:20
  • @Blue That seems very interesting. – pie May 14 '24 at 02:06

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In this answer, an elegant method to construct $xy$ was given, assume $1$ is known. In the same image, if $x, y$ and $xy$ are all known, we would be able to construct $1$. Therefore if you cannot construct $1$ given $x$ and $y$, you cannot construct $xy$. Similarly, you cannot construct $\frac{x}{y}$, because if so, we have $\frac{x}{y} \cdot y = x$, so replace $x$ by $\frac{x}{y}$ and $y$ by $y$, we would be able to construct $1$.

Intuitively, if $1$ is not known, then $xy$ is not fixed, because of the change of units. For example, if $x=1$ kilo meter and $y=2$ kilo meter, then $xy$ is $2$ kilo meter squared, but for the same length, $x=1000$ meter and $y=2000$ meter, then $xy$ is $2\cdot 10^6$ meter squared. So is $xy=2$ or $xy=2,000,000$?

As for division, even if $x=y$, then $\frac{x}{y}=1$, but without specifying $1$, there is not even a way to know if $x=1$ or not. That is, you can map division to arithmetic (when $x/y$ is rational), but you cannot map arithmetic to geometry without knowing $1$. The problem is still $x,y$ may have units of length, but $\frac{x}{y}$ is dimensionless, so not comparable to $x$ or $y$ or any length on the plane.

This also explains somehow why $\frac{ab}{a+b}$ can be constructed: it has the same dimension as $a, b$. Similarly, you can also construct $\frac{a^2}{b}, \frac{b^2}{a}$, etc. The idea is to take an arbitrary length as $1$, and construct multiplication and division with the usual method, then it can be shown that if the quantity has the same dimension as $a$ and $b$, it's independent of the choice of $1$.

Just a user
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  • But my question is given a line segment with length $a$ , a line segment with length $b$, so $1$ is defined (twice), so that there is omly one $ab$. – pie May 14 '24 at 02:26
  • @Pie My argument is this is an empty promise. If you give the question to two different people and give them two different lengths as $1$, they will return different but correct answers to you. Therefore, there isn't enough enough information to get the answer. This is like given $x^2=y$, what is $x$? Even if in your mind, $y$ is fixed, if it's not given, there is not unique answer. – Just a user May 14 '24 at 02:31
  • To put another way, what's given are not two lenghts, but two segments, unless you define lengths as the equivalence classes of segments, then the addition of "lengths" is well-defined, but the multiplication and division are not. Unless you translate the lengths into numbers, but this is equivalent to fix $1$, and if it's only fixed in your mind, not revealed to the problem solver, it doesn't count. Or, there are many different "isomorphisms" from the geometric plane to $\mathbb R^2$, and they don't agree with each other about multiplications. – Just a user May 14 '24 at 02:37