Sum of angles in a $1$-by-$3$ rectangle

Question

This problem was in a competition for a job. It seems simple BUT the challenge is you cannot use trigonometry.

Let there be 3 squares with side length of $\ell$ arranged in such a way that it forms a rectangle with length $3\ell$ and width $\ell$. So $ABCD$, $CEHD$, and $EFGH$ are squares. (This is notation because it's easier to tell you this way.)

If $\alpha = m(\angle AED)$ and $\beta = m(\angle AFD)$, then what is $\alpha + \beta$?

I tried to use angles of quadrilaterals but so far I didn't find anything useful.

Have fun solving it!

Non-OP edit: added diagram to confirm construct or clarify

Answer 1

The best answer I can give not using trigonometry is: $\alpha + \beta$ is equal to the smallest angle of the $1, 2, \sqrt{5} $ right triangle.

Proof:

Translate the angle $\alpha$ from AED to DFH.

Draw a diagonal from H to C

Let's call K the point where the segment HC intersects the diagonal AF.

In this way we get the triangle FHK which is a right triangle because the angle CHF is a right angle. (see Proof 1)

The length of the side FH must be $\sqrt{2}ℓ$, because it is the diagonal of a square with sides of length $ℓ$. The length of the side HK must be $\sqrt{2}\frac{ℓ}{2}$, because it is the diagonal of a square with sides of length $\frac{ℓ}{2}$, because K is the center of the center square DCEH. (see Proof 2)

Consider a similar triangle with sides of length R, S and T, scaled by a factor $x=\frac{\sqrt{2}}{ℓ}$, such as:

$R=x|HK|=\frac{\sqrt{2}}{ℓ}\sqrt{2}\frac{ℓ}{2}=1$

$S=x|FH|=\frac{\sqrt{2}}{ℓ}\sqrt{2}ℓ=2$

These are the two sides that form the right angle.

By the Pythagorean theorem, $R^2+S^2=T^2=1^2+2^2=5$, so $T=\sqrt{5}$

Hence, the triangle FHK is a right triangle similar to one with sides $1, 2, \sqrt{5}$, and the angle KFH, which is $\alpha+\beta$, is its the smallest angle.

Proof 1:

FH is the diagonal of a square so it forms an angle $\frac{\pi}{4}$ with the sides, so the angle GHF is $\frac{\pi}{4}$.

HC is the diagonal of a square so it forms an angle $\frac{\pi}{4}$ with the sides, so the angle CHD is $\frac{\pi}{4}$.

The angles CHD + CHF + GHF add up to $\pi$, and since CHD and GHF are $\frac{\pi}{4}$, CHF must be $\frac{\pi}{2}$

Proof 2:

The midpoint of the diagonal HC must be the center of the square DCEH.

The midpoint of the diagonal AF must be the center of the rectangle ABFG.

The center of the square DCEH and the center of the rectangle ABFG coincide, so the midpoints of diagonals HC and AF coincide which means that they intersect at that point.

The point where HC and AF intersect is K, by definition. Therefore K is the center of the square DCEH.

Answer 2

Construct the "$3\times 3$-square" with side $AG$ in the same half-plane w.r.t to the line $AG$ where the given three squares are already constructed.

Let $O$ be the point $O=CD\cap EG$. Reflection w.r.t. $EG$ is denoted by a prime. Then by a simple lattice argument a (chess) knight in $O$ can equally jump in $A,H,F=H',A'$. So these points are on the same circle centered in $O$. Then the wanted angle $\alpha+\beta$ is $\widehat{AFH}$, and in the figure we can write many equalities... $$ \alpha+\beta= \widehat{AFH}= \widehat{AA'H}= \frac 12 \widehat{AOH}= \widehat{AOD}= \widehat{DOH}= \widehat{FDG}= \widehat{EAH}=\arctan\frac 12\ . $$ $\square$

Note: Using $\arctan x-\arctan y=\arctan\frac {x-y}{1+xy}$ (a relation parallel to the one for the tangent of a difference of angles, apply $\tan$ on both sides to check), we can trigonometrically check: $$ \small \alpha+\beta= \widehat{AFG}- \widehat{HFG}= \arctan 3-\arctan 1=\arctan\frac{3-1}{1+3\cdot 1} =\arctan\frac24 =\arctan\frac12 =\widehat{FDG}\ . $$

Answer 3

$$\alpha = \widehat{AED} = \widehat{DFH}$$ $$\beta + \alpha = \widehat{AFD} + \widehat{DFH} = \widehat{AFH}$$ $$S_{\triangle{AFH}} = \frac12 FG.AH = \frac12 AF.HF \sin{\widehat{AFH}}$$ $$l^2 = \frac12 \sqrt{10}l.\sqrt2l.\sin{\widehat{AFH}}$$ $$\widehat{AFH} = \alpha + \beta = \arcsin{\frac{1}{\sqrt5}} = \arccos{\frac{2}{\sqrt5}} = \arctan{\frac{1}{2}}$$