Given the three line segments below, of lengths a, b and 1, respectively:enter image description here

construct the following length using a compass and ruler: $$\frac{1}{\sqrt{b+\sqrt{a}}} \ \ \text{and} \ \ \ \sqrt[4]{a} $$

Make sure to draw the appropriate diagram(s) and describe your process in words. We are also to use the following axioms and state where they are used:

  1. Any two points can be connected by a line segment,
  2. Any line segment can be extended to a line,
  3. Any point and a line segment define a circle,
  4. Points are born as intersection of lines, circles and lines and circles
  • $\begingroup$ Addition is easy enough. Two more difficult items you need are square root and reciprocal. These require a length defined to be $1.$ Similar triangles and right triangles are used over and over. And semicircles. Main theorem is that a triangle with one edge the diameter of a circle and another vertex on the circle is a right triangle. $\endgroup$ – Will Jagy Apr 7 '17 at 3:22
  • $\begingroup$ Are you okay with contracting right angles and replicating angles? If so you can create a right triangle with hypotenuse a and a leg of 1. So... triangle witbone side has a similar triangle with a coresponding leg of 1 and... $\endgroup$ – fleablood Apr 7 '17 at 6:42

I don't want to answer for you but I'll give you hints.

Forget constructions for a moment. How would we represent $\frac 1k$ and $\sqrt{k}$ geometrically?

As $k$ won't necessarily correspond to any integer or rational value, the only way I see off the top of my head to do ratios is via similar triangles.

Can you construct two similar triangles $ABC$ and $XYZ$ ($A$ corresponds to $X$, $B$ to $Y$ and $C$ to $Y$) so that $ABC$ has sides of length $BC =a$ and $XYZ$ has a side of length $XY = 1$ any $YZ = AB$?

That is $AB$ ~ $XY= 1$ and $BC = a$ ~$YZ=AB$ and $AC$ ~ $XZ$.

If so that would mean $\frac {AB}{BC=a} = \frac {XY = 1}{YZ=AB}$ so $AB^2 = a$. So $AB = \sqrt{a}$.

So how can we construct such similar triangles? Well, maybe $YZ$ should be $AB$. Thus two triangles $ABC$ ~ $XBA$ where $BC = a$ and $XB = 1$

So construct a line $CBX$ so that $CB=a$ and $BX = 1$. Can you construct a point $A$ so that $ABC$ ~ $XBA$? What if the line $\overline{BA}\perp\overline {CB}$ and $\overline{CA}\perp \overline {AX}$?

Finding a line $\perp$ to $\overline {CX}$ and $B$ is straightforward because we have the points. But finding an unknown point $A$ so that $\overline{CA}\perp \overline {AX}$ may not be obvious. But Hint: given a semi-circle any angle from diameter endpoint, to point in the semi-circle, to the other endpoint will be a right angle.

So that $\sqrt{a}$.

To find $\frac 1a$ is easier. Find similar triangles $ABC$ and $XBA$ as above where $BC = a$, $AB= 1$ and Then $\frac {AB=1}{BC=a} = \frac {XB}{BA = 1}$ so $XB =\frac 1a$.

To construct those triangle just create $ABC$ be making a $\perp$ to a line 1 at point $B$. Mark off $C$ one line so $BC = a$ and $A$ on the other. At $A$ construct a $\perp$ to $BC$. Where this perpendicular line intersects the extended $\overline {BC}$, mark as $X$.

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  • $\begingroup$ Can you provide a picture or drawing of the first two similar triangles for reference? $\endgroup$ – HKT Apr 7 '17 at 20:58
  • $\begingroup$ Some parts of your answer are quite vague.. $\endgroup$ – HKT Apr 7 '17 at 20:59
  • $\begingroup$ Can I provide a picture. No. I can not. But Will Jagy did. His illustrations were exactly what I was trying to describe and he add a way to get ab which you didn't ask for. $\endgroup$ – fleablood Apr 7 '17 at 22:14
  • $\begingroup$ flea, I bought an inexpensive scanner, the first diagrams I posted were for math.stackexchange.com/questions/173016/… Worthwhile investment, uses no ink. I got the simple type, just a single page per scan. Helps with keeping various types of records, I keep copies as jpegs. $\endgroup$ – Will Jagy Apr 8 '17 at 0:45
  • $\begingroup$ those drawings I made are at answer math.stackexchange.com/questions/173016/… $\endgroup$ – Will Jagy Apr 8 '17 at 0:50

enter image description here


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