Euclidea Shenanigans

Looking more at the Euclidea Wiki, I’m conflicted. That website is schizophrenic. On the one hand, the actual solutions themselves are pretty complete and comprehensive. I’m especially impressed by the “E” solutions.

On the other hand, there’s the Explanation sections. Many times they’re incomplete, and as we’ve seen in previous posts, completely missing from later problems.

And then, every once in a while, you stumble upon something truly special - like the Angle of 45° explanation. I’ll reproduce everything here in case it changes.

Alternative 5E Construction

1. Construct the circle with center A and radius AB
2. Construct the circle with center B and radius AB, intersecting circle A at C and D
3. Construct line BC, intersecting circle B at E
4. Construct the circle with center E and radius AE, intersecting line BC at F and G
5. Construct line AF

Beautiful, but also slightly mysterious!

The Explanation

Now check out this explanation from the wiki:

Understanding the alternate 5E solution is a bit more interesting, particularly since it does not directly produce any $90^\circ$ angles until the 2V solution, where Thales’ theorem states that $\overline{AF}$ and $\overline{AG}$ are perpendicular because they inscribe the diameter $\overline{FG}$ of circle $E$. But determining that $\overline{AF}$ and $\overline{AG}$ form a $45^\circ$ angle from ray $\overrightarrow{AB}$ takes more thought. Note that both $ACBD$ and $ABED$ are each a rhombus by construction from two equilateral triangles, with $\angle ABF = \angle BAD = \angle BED = 60^\circ$, and $\angle ABE = \angle ADE = 120^\circ$. Circle $E$ is created using the length of the longer diagonal of that rhombus, $\overline{AE}$, which is $|\overline{AB}| \cdot \sqrt{3}$. The length of $\overline{BF}$ is thus $|\overline{AB}| \cdot (\sqrt{3}-1)$. Now consider point $H$ placed on $\overline{AB}$ as the base of the altitude of $\triangle ABF$ - the sub-triangle on the right, $\triangle BFH$, is a $30^\circ$-$60^\circ$-$90^\circ$ triangle, so the length of $\overline{HB}$ is $|\overline{AB}| \cdot (\sqrt{3}-1)/2$, and the length of $\overline{HF}$ is $|\overline{AB}| \cdot (3/2-\sqrt{3}/2)$. But we also know $|\overline{AH}| = |\overline{AB}| - |\overline{HB}| = |\overline{AB}| \cdot (1 - (\sqrt{3}-1)/2) = |\overline{AB}| \cdot (3/2 - \sqrt{3}/2)$. Thus, since $|\overline{AH}| = |\overline{HF}|$, $\triangle AHF$ is a right isosceles triangle, and we have proven that $\angle BAF = 45^\circ$.

I’m honestly not sure how you could write this up in good faith. While probably correct, alarm bells should be ringing: “This can’t be the best way to show that angle is $45^\circ$!” Do we really need auxiliary points, rhombuses, side length computations, arithmetic, similar triangles, AND angle chasing? Of course not. But most importantly, this explanation completely obscures the actual motivation behind discovering the solution.

The ACTUAL Explanation

Connect $\overline{AE}$ and consider $\triangle AFE$:

We know three things about this triangle:

1. $\overline{EA}$ and $\overline{EF}$ are both radii of circle $E$, so the triangle is isosceles.
2. $\overline{BA}$ and $\overline{BE}$ are both radii of circle $B$.
3. $\angle AEF = 30^\circ$ since $\triangle ACE$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle. Why? $\angle ACE = 60^\circ$ (from the equilateral triangle), and $\angle CAE = 90^\circ$ because it inscribes diameter $\overline{CE}$.

$\triangle AFE$ by itself is just:

Angle chase this triangle and watch the fog lift, revealing the construction in its elegant simplicity: