MaxWong

Assuming the pyramid is a right pyramid and the base is a square:

Surface area = 23² + 4(23)(34)/2 = 2093 cm²

Challenge problem: Find the explicit form of f(n).

This can be done by brute force.

f(3) = f(2) + f(1)

10 = f(2) + 3

f(2) = 7

f(4) = f(3) + f(2)

f(4) = 7 + 10 = 17

f(5) = f(4) + f(3)

f(5) = 17 + 10 = 27

f(6) = f(5) + f(4) = 27 + 17 = 44.

$\quad \sqrt{11\sqrt{11\sqrt{11\sqrt{11}}}}\\ = \sqrt{11} \cdot 11^{1/4} \cdot 11^{1/8} \cdot 11^{16}\\ = 11^{1/2 + 1/4 + 1/8 + 1/16}\\ = 11^{15/16}$

Appendix: Why is $4 \sin x \sin(60^\circ - x) \sin(60^\circ + x) = \sin 3x\\ $?

Proof:

Using compound angle formula,

$4\sin x\sin(60^\circ - x) \sin(60^\circ + x) = 4\sin x \left(\sin 60^\circ \cos x - \cos 60^\circ \sin x\right)\left(\sin 60^\circ \cos x + \cos 60^\circ \sin x\right)$

Simplifying,

$4\sin x\sin(60^\circ - x) \sin(60^\circ + x) = 4\sin x \left(\dfrac{\sqrt 3}2 \cos x - \dfrac12 \sin x\right)\left(\dfrac{\sqrt 3}2 \cos x + \dfrac12 \sin x\right)$

$4\sin x\sin(60^\circ - x) \sin(60^\circ + x) = \sin x \left(\sqrt 3 \cos x - \sin x\right)\left(\sqrt 3 \cos x + \sin x\right)$

Now, by the identity $a^2 - b^2 = (a - b)(a + b)$,

$4\sin x\sin(60^\circ - x) \sin(60^\circ + x) = \sin x (3\cos^2 x - \sin^2 x)$

Starting from the right hand side, expanding using triple angle formula,

(if you haven't learnt that yet, you can try expanding it with compound angle formula and double angle formula.)

$\sin 3x = 3\sin x - 4\sin^3 x$

Now, it suffices to show that $\sin x (3\cos^2 x -\sin^2 x) = 3\sin x - 4\sin^3 x$

To do so, we use the identity $\cos^2 x = 1-\sin^2 x$.

$\sin x (3\cos^2 x -\sin^2 x) = \sin x(3(1 - \sin^2 x) - \sin^2 x) = \sin x (3 - 3\sin^2 x - \sin^2 x) = \sin x (3 - 4\sin^2 x) = 3\sin x - 4\sin^3 x$

Therefore, the original identity $4 \sin x \sin(60^\circ - x) \sin(60^\circ + x) = \sin 3x\\ $ is true.

$\quad\sin 10^\circ \sin 30^\circ \sin 50^\circ \sin 70^\circ\\ = \dfrac12 \sin 10^\circ \sin 50^\circ \sin 70^\circ\\ = \dfrac12 \sin 10^\circ \sin(60^\circ - 10^\circ)\sin(60^\circ + 10^\circ)$

Now, use the identity: $4\sin x \sin (60^\circ - x)\sin(60^\circ + x) = \sin 3x$ and substitute x = 10^o.

$\quad \sin10^\circ \sin30^\circ \sin50^\circ \sin 70^\circ\\ = \dfrac12 \sin 10^\circ \sin(60^\circ - 10^\circ) \sin(60^\circ + 10^\circ)\\ = \dfrac18 \left(4\sin 10^\circ \sin(60^\circ - 10^\circ) \sin(60^\circ + 10^\circ)\right)\\ = \dfrac18 \left(\sin(3\cdot 10^\circ)\right)\\ = \dfrac18 \sin 30^\circ\\ = \dfrac18 \cdot \dfrac12\\ = \dfrac1{16}$

Multiply conjugates.

$\quad \dfrac1{1 + \sqrt 3 + \sqrt 5}\\ = \dfrac1{1 + \sqrt 3 + \sqrt 5} \cdot \dfrac{1 + \sqrt 3 - \sqrt 5}{1 + \sqrt 3 - \sqrt 5}\\ = \dfrac{1 + \sqrt 3 - \sqrt 5}{(1 + \sqrt 3)^2 - \sqrt5^2}\\ = \dfrac{1 + \sqrt 3 - \sqrt 5}{4 + 2\sqrt 3 - 5}\\ = \dfrac{1 + \sqrt 3 - \sqrt 5}{2\sqrt 3 - 1}\\ = \dfrac{1 + \sqrt 3 - \sqrt 5}{2\sqrt 3 - 1}\cdot \dfrac{2\sqrt 3 + 1}{2\sqrt 3 + 1}\\ = \dfrac{(1 + \sqrt 3 - \sqrt 5)(2 \sqrt 3 + 1)}{(2\sqrt 3)^2 - 1^2}\\ = \dfrac{1 + 2\sqrt 3 + 6 + \sqrt 3 - 2\sqrt{15} - \sqrt 5}{11}\\ = \dfrac{7 + 3 \sqrt 3 - \sqrt 5 - 2\sqrt {15}}{11}$

Enumerate possible combinations:

{8}

{4, 4}

{4, 2, 2}

{4, 2, 1, 1}

{4, 1, 1, 1, 1}

{2, 2, 2, 2}

{2, 2, 2, 1, 1}

{2, 2, 1, 1, 1, 1}

{2, 1, 1, 1, 1, 1, 1}

{1, 1, 1, 1, 1, 1, 1, 1}

Total 10 ways.

We calculate side lengths by Pythagorean theorem.

$DE = \sqrt{17}\\ EF = \sqrt{17}\\ DF = \sqrt{34}$

Obviously, $DE^2 + EF^2 = DF^2$

So by converse of Pythagorean theorem, m∠DEF = 90^o.

Let 

$a = x^2 - 10x - 45\\ b = x^2 + 10x - 29\\ c = x^2 - 10x - 69$

$\dfrac1a + \dfrac1b = \dfrac2c\\ 2ab = c(a + b)\\ 2(x^2 - 10x - 45)(x^2 + 10x - 29) = 2(x^2 - 10x - 69)(x^2 - 37)$

$(x^2 - 10x - 45)(x^2 + 10x - 29) = (x^2 - 10x - 69)(x^2 - 37)$

Then expand and simplify.

$5x^3 - 34x^2 - 265x - 624 = 0$

Solving by numerical methods gives $x\approx 12.055$.