Paper Fortune Teller

Let's look at the paper before last unfolding:

As we can see, $\triangle ACD,\triangle BCD,\triangle ACE,\triangle BCE$ are right isosceles triangles. Let us assume $AC=1$ . Naturally, $AC = BC = AD = BD= AE = BE = 1$ and $CD = CE = \sqrt { 2 }$ .

Another observation we need to make is that the fortune teller consists of four equal parts/petals. That means that the angles between the planes separating these parts are right. Thus, $\angle A(CE)B=\frac { \pi }{ 2 }$ .

Let point M be the middlepoint of CE. Since $\triangle ACE$ and $\triangle BCE$ are isosceles, $AM$ and $BM$ are also their heights and perpendicular to $CE$ . Knowing that we get $AM = BM = \frac { \sqrt { 2 } }{ 2 }$ and $\angle AMB=\frac { \pi }{ 2 }$ . That means that $\triangle ABM$ is also right isosceles triangle and $AB = 1$ .

We now know that $\triangle ABC, \triangle ABD$ and $\triangle ABE$ are equilateral triangles, $\angle ACB = \angle ADB = \angle AEB = \frac { \pi }{ 3 }$ .

Now let's consider one petal alone. We can see that it consists of four right equilateral triangles but also we know that $AB=1$ . This means we have two equal tetrahedrons, $ABCD$ and $ABCE$ , which are also symmetrical relative to plane CDE.

Let point O be the midpoint of AB. Since $\triangle ABC, \triangle ABD$ and $\triangle ABE$ are equilateral triangles and $AB = 1$ , $CO = DO = EO = \frac { \sqrt { 3 } }{ 2 }$ . Point O is in one plane with points C, D and E due to symmetry. Let's take a look at this plane:

$\triangle CDO,\quad \triangle CEO$ are equal isosceles triangles, let's consider only one of them:

$\angle DCE=\angle DCO+\angle OCE=\angle DCO+\angle OCE=\pi -\angle COD$

Now let CH be penperdicular to DO and point N be a middlepoint of CD:

By Pythagoras' theorem, $NO = \frac { 1 } { 2 }$ . Triangles $\triangle CNO,\quad \triangle CDH$ are similar, because $\angle CNO=\angle CHO=\frac { \pi }{ 2 }$ and $\angle DCO=\angle NDO$ due to $\triangle CDO$ being isosceles. That means:

$\frac { CO }{ CD } =\frac { CN }{ DH } =\frac { NO }{ CH } \\ \frac { \frac { \sqrt { 3 } }{ 2 } }{ \sqrt { 2 } } =\frac { \frac { \sqrt { 2 } }{ 2 } }{ DH } =\frac { \frac { 1 }{ 2 } }{ CH } \\ DH=\frac { 2 }{ 3 } \sqrt { 3 } ;CH=\frac { 1 }{ 3 } \sqrt { 6 }$

$OH=DH-DO=\frac { 1 }{ 6 } \sqrt { 3 } ;CH=\frac { 1 }{ 3 } \sqrt { 6 }$

Now we can find our tan value:

$\tan { \angle DCE } =\tan { \left( \pi -\angle COD \right) } =\tan { \angle COH } =\frac { CH }{ OH } =2\sqrt { 2 }$

And finally get the final answer:

$\cos { \angle ACB } =\cos { \frac { \pi }{ 3 } } =\frac { 1 }{ 2 }$

$\cos { \angle ACB } \cdot \tan { \angle DCE } =\frac { 1 }{ 2 } \cdot 2\sqrt { 2 } =\boxed{\sqrt { 2 } }$