Two-tone Pinwheel

Let the intersection of the diagonals be $O$ .

Note that $AOB$ and $COD$ are similar ( $\angle ABO = \angle CDO, \angle AOB = \angle COD$ ), and so as $BOC$ and $DOA$ . Let $AO = x, BO = y, CO = z$ , then $DO = z \cdot \frac{y}{x} = \frac{yz}{x}$ by similarity, and furthermore $AO = \frac{yz}{x} \cdot \frac{z}{y} = \frac{z^2}{x}$ by similarity. Thus $x = \frac{z^2}{x}$ ; since $x, z > 0$ , we have $x = z$ , so $AO = CO$ . Thus $AOB$ and $COD$ are in fact congruent, and so as $BOC$ and $DOA$ . Thus $CO = x, DO = y$ .

Let $\angle BAO = \alpha, \angle CBO = \beta$ . Note that $\alpha + \beta = 95^\circ$ because the sum of all internal angles must be $360^\circ$ . By sine rule,

$\frac{AO}{\sin \angle ABO} = \frac{BO}{\sin \angle BAO} \implies \frac{x}{\sin 40^\circ} = \frac{y}{\sin \alpha}$

Similarly, from $BOC$ we can conclude $\frac{x}{\sin \beta} = \frac{y}{\sin 45^\circ}$ . Thus $\frac{x}{y} = \frac{\sin 40^\circ}{\sin \alpha} = \frac{\sin \beta}{\sin 45^\circ}$ . Cross-multiplying, we have $\sin \alpha \sin \beta = \sin 40^\circ \sin 45^\circ$

Remember that $\alpha + \beta = 95^\circ$ . Thus,

$\begin{aligned} \cos 95^\circ &= \cos (\alpha + \beta) \\ &= \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ &= \cos \alpha \cos \beta - \sin 40^\circ \sin 45^\circ \\ \cos \alpha \cos \beta &= \cos 95^\circ + \sin 40^\circ \sin 45^\circ \end{aligned}$

Thus, we can compute $\alpha - \beta$ :

$\begin{aligned} \cos (\alpha - \beta) &= \cos \alpha \cos \beta + \sin \alpha \sin \beta \\ &= (\cos 95^\circ + \sin 40^\circ \sin 45^\circ) + \sin 40^\circ \sin 45^\circ \\ &= \cos 95^\circ + 2 \sin 40^\circ \sin 45^\circ \end{aligned}$

Let's go back to the problem. We want to find the difference between the two solutions. We know that $\angle AOB = 140^\circ - \alpha, \angle BOC = 135^\circ - \beta$ . We want the minimum of these two. Their sum is $275^\circ - (\alpha + \beta) = 275^\circ - 95^\circ = 180^\circ$ , so whichever is less than $90^\circ$ , we take it. In other words, if $\alpha > 50^\circ$ (and thus $\beta < 45^\circ$ ), then we take $\angle AOB$ , otherwise we take $\angle BOC$ .

The trick is to realize that the two solutions might make $\angle AOB$ smaller in one solution, and $\angle BOC$ smaller in the other! In the above, we see that if $\alpha > 50^\circ$ , we take $\angle AOB$ . But $\alpha - \beta = 2 \alpha - (\alpha + \beta) = 2 \alpha - 95^\circ$ . Thus if $\alpha - \beta > 5^\circ$ , we take $\angle AOB$ , otherwise we take $\angle BOC$ . We know the value of $\cos (\alpha - \beta)$ as above; it just suffices to show that $\cos (\alpha - \beta) = \cos 95^\circ + 2 \sin 40^\circ \sin 45^\circ < \cos 5^\circ$ , which would imply that $|\alpha - \beta| > 5^\circ$ , so one of its two values will be greater than $5^\circ$ . (The other will be negative and hence obviously less than $5^\circ$ .)

This can be shown in a few ways. One that comes to my mind is this:

We know

$\begin{aligned} \cos 95^\circ &= -\cos 85^\circ \\ &= -\cos (45^\circ + 40^\circ) \\ &= -\cos 40^\circ \cos 45^\circ + \sin 40^\circ \sin 45^\circ \end{aligned}$

Also, we know

$\begin{aligned} \cos 5^\circ &= \cos (45^\circ - 40^\circ) \\ &= \cos 40^\circ \cos 45^\circ + \sin 40^\circ \sin 45^\circ \end{aligned}$

Thus,

$\begin{aligned} (\cos 95^\circ + 2 \sin 40^\circ \sin 45^\circ) - (\cos 5^\circ) &= (-\cos 40^\circ \cos 45^\circ + 3 \sin 40^\circ \sin 45^\circ) - (\cos 40^\circ \cos 45^\circ + \sin 40^\circ \sin 45^\circ) \\ &= -2 (\cos 40^\circ \cos 45^\circ - \sin 40^\circ \sin 45^\circ) \\ &= -2 \cos 85^\circ \\ &< 0 \end{aligned}$

This proves the claim. Since $|\alpha - \beta| > 5^\circ$ , the two solutions have $\angle AOB$ smaller in one solution, and $\angle BOC$ smaller in the other.

Let the two solutions for $\alpha, \beta$ be $(\alpha_1, \beta_1), (\alpha_2, \beta_2)$ such that $\alpha_1 - \beta_1 > 0$ (and thus $\alpha_2 - \beta_2 < 0$ . From the above, we know that $\alpha_1 - \beta_1 > 5^\circ$ , so this solution makes $\angle AOB = 140^\circ - \alpha_1$ as the smaller angle. The second solution has $\alpha_2 - \beta_2 < 0 < 5^\circ$ , so this solution makes $\angle BOC = 135^\circ - \beta_2$ as the smaller angle. We're looking for their difference; that is, $|(140^\circ - \alpha_1) - (135^\circ - \beta_2)|$ .

But we know that $\alpha_1 - \beta_1 = - (\alpha_2 - \beta_2)$ , and $\alpha_1 + \beta_1 = \alpha_2 + \beta_2$ (both are $95^\circ$ ). Thus we can in fact conclude that $\alpha_1 = \beta_2$ , and so their difference is $|(140^\circ - \alpha_1) - (135^\circ - \beta_2)| = 140^\circ - 135^\circ = \boxed{5^\circ}$ .

Since there are two pairs of equal alternate interior angles, quadrilateral $ABCD$ is a parallelogram. See the figure above; let $OP=$ $PA=$ $OQ=$ $QC=1$ and $\angle DQP = \theta$ . We note that:

$\begin{aligned} \tan (90^\circ - \theta + 40^\circ) & = \frac {EC}{DE} \\ & = \frac 2{DP-EP} \\ \implies \color{#3D99F6} \tan (130^\circ - \theta) & = \frac 2{\tan \theta -1} & \small \color{#3D99F6} \text{As }\tan (180^\circ - x) = - \tan x \\ \color{#3D99F6} \tan (50^\circ + \theta) & = \frac 2{1-\tan \theta} \\ \frac {\tan 50^\circ + \tan \theta}{1-\tan 50^\circ \tan \theta} & = \frac 2{1-\tan \theta} \end{aligned}$

$\begin{aligned} \tan 50^\circ + \tan \theta - \tan 50^\circ \tan \theta - \tan^2 \theta & = 2-2\tan 50^\circ \tan \theta \\ \tan^2 \theta - (1+ \tan 50^\circ)\tan \theta + 2 - \tan 50^\circ & = 0 \end{aligned}$

Solving the quadratic equation we get $\theta = \begin{cases} 59.863^\circ \\ 25.137^\circ \end{cases}$ , then the angle between the diagonals $= \begin{cases} 59.863^\circ + 45^\circ = 104.863^\circ \\ 25.137^\circ + 45^\circ = 70.137^\circ \end{cases}$ ; the smaller ones $= \begin{cases} 180^\circ - 104.863^\circ = 75.137^\circ \\ 70.137^\circ \end{cases}$ and the difference $= 75.137^\circ - 70.137^\circ = \boxed{5}^\circ$ .

We'll start constructing the solution by drawing points DB and the directions, $40^\circ$ off from the diagonal $DB$ , of the sides DC and AB. The second diagonal will go through point E midway between D and B, due to the symmetry of the problem. $\angle ECB$ should be $45^\circ$ . We can think of this as the inscribed angle inside a circle going through E, B, and C. Corresping central angle of $90^\circ$ makes locating the center F trivial.

Let's set the radius of the circle arbitrarily at 1. This will give us $\overline{EB}=\overline{ED}=\sqrt{2}$ . In the $\triangle DEF$ we know also $\angle DEF=135^\circ$ and $\overline{EF}=1$ . From the law of cosines we get the remaining side as $\overline{DF}=\sqrt{5}$ , and the law of sines will then give us angle $\alpha=18.435^\circ$ .

Switching to $\triangle DFC$ we know angle $\beta=40-\alpha=21.565^\circ, \overline{DF}=\sqrt{5}$ , and $\overline{FC}=1$ . We can look for angle $\gamma$ using the law of sines.

$sin(\gamma)=\sqrt{5}\times sin(21.565)$

There are two solutions: $\gamma_{1}=124.726^\circ$ and $\gamma_{2}=55.274^\circ$ . Law of cosines applied to $\triangle DFC$ will give us $\overline{DC_{1}}=1.5099$ and $\overline{DC_{2}}=2.6492$ .DEC

We now switch to the $\triangle DEC$ as it is the $\angle DEC$ , or its supplementary angle, that we need to find. We know $\angle EDC=40^\circ, \overline{ED}=\sqrt{2}$ and the side $\overline{DC}$ has one of the above values. Law of cosines provides $\overline{EC_{1}}=1.00413$ and $\overline{EC_{2}}=1.81059$ . Law of sines applied to the $\triangle DEC$ will then give us the angle between the diagonals. Using inverse sine function will automatically provide the smaller of those two angles.

$\angle DEC_{1}=arcsin(sin(40)\times 1.5099/1.00413)=75.1369$

$\angle DEC_{2}=arcsin(sin(40)\times 2.6492/1.81059)=70.1369$

The difference is 5.

45° - 40*° = 5°