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ABCD is a unit square. Construct a point E on DC (extended) such that AE intersects BC at F with EF = 1.

enter image description here

After trying for long, the only solution I could find was to solve for the length of $CE$ algebraically (it turns out to be $\frac{1}{2} (-1 + \sqrt{2} + \sqrt{2 \sqrt{2} - 1})$ ) and then construct that length, which can be done because it is possible to take square root of any arbitrary number via geometric construction.

However, this approach is more algebraic than geometric, and I think that it undermines the style of geometric constructions. I am looking for a better solution to this problem.

EDIT:

The length of $CE$ can be found (by using triangle similarity) to be $\frac{1}{2} (-1 + \sqrt{2} + \sqrt{2 \sqrt{2} - 1})$. This can easily be constructed, thus solving the problem. However, to find out the value of $CE$, I had to solve a 4-degree polynomial (using Wolfram|Alpha).

This method seems (to me), to be kind of brute-forcing a geometrical construction problem. Generally, such problems can be solved by using innovative constructions and using patterns in the question. I am looking for a solution where geometric reasoning is used, instead of resorting to algebraic techniques.

EDIT #2:

After reading @heropup's comment below his answer, I think I can specify my question a little better. I am looking for a synthetic geometric proof, as opposed to the analytic method of solving for the length of $CE$.

Jam
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Soham Saha
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  • The automated system tells me that my choice of the title is not good enough, but I can’t think of anything better. Feel free to edit! – Soham Saha Jun 04 '24 at 07:38
  • Here is an attempt at a better title. – Semiclassical Jun 04 '24 at 08:29
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    I couldn't solve this problem yet. But using $AE= \frac1{FC}$ would help. As there is a construction for reciprocal of a segment. (see this). Really, not sure if this actually helps. – User Jun 04 '24 at 08:36
  • @Semiclassical actually I know how to construct the length $\frac{1}{2} (-1 + \sqrt{2} + \sqrt{2 \sqrt{2} - 1})$. Please read the post, I am looking for another, more geometrical way. Still thanks for trying to help. – Soham Saha Jun 04 '24 at 09:05
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    For posterity, the same figure is drawn in question 3614337, but no geometric solution is given there. – Jam Jun 05 '24 at 08:08

1 Answers1

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I have a construction, but I'm afraid it's not much more than a thinly disguised algebraic solution. It might be interesting if one could retrospectively devise a purely geometric (i.e., in the style of Euclid) argument from this construction.

enter image description here

  1. Given square $ABCD$:
  2. Draw a circle $D$ with radius $BD$ and center $D$.
  3. Locate the intersection of circle $D$ with the line $\overline{CD}$, to the right of point $C$; call this intersection $G$.
  4. Erect a perpendicular to $\overline{DG}$ at $G$, and locate the point $H$ on this perpendicular that intersects the line $\overline{AC}$.
  5. Construct the circle with diameter $BH$.
  6. The rightmost intersection of this circle with line $\overline{CD}$ is the desired point $E$ satisfying $EF = AB$.
heropup
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  • +1 for this simplified construction. I understood upto the construction of $BH=\sqrt{5-2\sqrt{2}}$. Could you explain (or just a hint would be OK) why the last part works? – Soham Saha Jun 04 '24 at 12:52
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    @SohamSaha See https://en.wikipedia.org/wiki/Carlyle_circle for more information. – heropup Jun 04 '24 at 17:38
  • Ah, so you’re taking $C$ as the origin, $B$ as $(0,1)$ and $H$ as $(s,p)$, right? Thanks for introducing me to the Carlyle circle :) – Soham Saha Jun 05 '24 at 05:58
  • And as for motivation for your method, did you go back from $x=\frac{1}{2} (-1 + \sqrt{2} + \sqrt{2 \sqrt{2} - 1})$ to get $x(x-(\sqrt2-1))=\sqrt2-1$? – Soham Saha Jun 05 '24 at 06:04
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    @SohamSaha When I recognized that the length $CE$ corresponded to the solution of a quadratic with relatively simple coefficients, it occurred to me to use a Carlyle circle to perform the computation. All that remained was to find an efficient way to incorporate the coordinates of the circle's diameter into the existing diagram. So it's still very much an algebraically motivated construction at present, although (as I stated earlier) it may be quite interesting to see if this construction could lead to a synthetic geometric proof. – heropup Jun 05 '24 at 06:19