MaxWong

... and let $X$ be a point such that $MH = MX$, as shown below

Did you forget to attach a diagram?

Suppose O is the centre of the circle. Then \(\Delta OP_1 P_2\) is an isosceles triangle with \(OP_1 = OP_2 = 1\) and \(\angle P_1OP_2 = \dfrac{360^\circ}{10} = 36^\circ\).

Then \(P_1P_2 ^2 = 1^2 + 1^2 - 2(1)(1) \cos 36^\circ = 2 - \dfrac{1 + \sqrt 5}2 = \dfrac{3 - \sqrt 5}2\\ \). Note that \( \left(\dfrac{\sqrt 5 - 1}2\right)^2 = \dfrac{3 - \sqrt 5}2\). Then \(P_1P_2 = \dfrac{\sqrt 5 - 1}2\).

Repeating the same process 10 times gives \(P_1 P_2 + P_2 P_3 + P_3 P_4 + \cdots + P_9 P_{10} + P_{10} P_1 = 10\cdot \dfrac{\sqrt 5 -1}2 = 5(\sqrt 5 - 1)\).

Divide by 5 on both sides.

\(125x = \dfrac 95\)

Note that \((\sin (a))^5 < 1\) and \((\cos(a))^5 < 1\).

Then, we have

\(\begin{array}{rcl} (\sin(a))^7 + (\cos(a))^7 &=& (\sin(a))^2(\sin(a))^5 + (\cos(a))^2 (\cos(a))^5\\ &<& (\sin(a))^2 + (\cos(a))^2\\ &=& 1 \end{array}\)

The inequality is shown.

Suppose the radius of the circle = r units 

\(\pi r^2 = 25\pi\\ r = 5\)

Then, the radius of the sphere is \(\sqrt{r^2 + 5^2} = 5 \sqrt 2\) units.

Then the surface area is \(4 \pi (5 \sqrt 2)^2 = 200\pi\).

Using Pythagoras theorem in \(\triangle ABX\) gives \(AX^2 + BX^2 = AB^2\), substituting AX = 15 and AB = 25 gives

\(15^2 + BX^2 = 25^2\\ BX^2 = 400\\ BX = 20\)

Note that \(\angle BAD = \angle CAD = \dfrac12 \cdot 30^\circ = 15^\circ\).

Considering the interior angle sum of \(\triangle ADC\) gives \(\angle ADC = 120^\circ\).

Using Sine Law in \(\triangle ADC \) gives 

\(\dfrac{10}{\sin 120^\circ} = \dfrac{AD}{\sin 45^\circ}\\ AD = \dfrac{10 \sqrt 6}3\)

Considering the angles around point D gives \(\angle ADB = 60^\circ\). Considering the interior angle sum of \(\triangle ABD\) gives \(\angle ABD = 105^\circ\).

Using Sine Law in \(\triangle ABC\) gives 

\(\dfrac{10}{\sin 105^\circ} = \dfrac{AB}{\sin 45^\circ}\\ AB = 10(\sqrt 3 - 1)\)

Then, the area of \(\triangle ABD\) is \(\dfrac12 (AB)(AD) \sin 15^\circ = \dfrac12 (10(\sqrt 3 - 1))\cdot \dfrac{10 \sqrt 6}3 \cdot \dfrac{\sqrt 6 - \sqrt 2}4 = \dfrac{50(2 \sqrt 3 - 3)}{3} \).

Each of the 8 pyramids being cut off has a right isosceles base with equal sides 3, and height 3.

Using the formula for volume of cube and volume of pyramid, we have

\(\text{Volume} = 6^3 - 8 \cdot \dfrac13 \left(\dfrac{3 \cdot 3}2\right)\cdot 3 = 180\)

\(\displaystyle \frac{1}{\sqrt{36} + \sqrt{42}} + \frac{1}{\sqrt{39} + \sqrt{45}}\\ =\dfrac{\sqrt{42} - \sqrt{36}}{42 - 36} + \dfrac{\sqrt{45} - \sqrt{39}}{45 - 39}\\ = \dfrac16 \left(\sqrt{45} + \sqrt{42} - \sqrt{39} - \sqrt{36}\right)\\ = \dfrac{3\sqrt 5 + \sqrt{42} - \sqrt{39} - 6}6\)

\(\dfrac1{1 + \dfrac1{4 + \dfrac12}}\\ = \dfrac1{1 + \dfrac1{\dfrac92}}\\ = \dfrac1{1 + \dfrac29}\\ = \dfrac1{\dfrac{11}{9}}\\ = \dfrac9{11}\)

Recall that 

If the roots of the quadratic equation \(x^2 + bx + c = 0\) are \(\alpha\) and \(\beta\), then \(\alpha + \beta = -b\) and \(\alpha \beta = c\).

(Vieta's formula)

By this fact, we have \(b = -(5 + 3i) - (1 - i) = -6 - 2i\), \(c = (5 + 3i)(1 - i) = 5 - 2i - 3i^2 = 8 - 2i\).

Then, \(b + c = -6-2i + (8-2i) = \boxed{2 - 4i}\)

Suppose \(xy = a\). Then,

\(x^2 + y^2 = (x^2 + 2xy + y^2) - 2xy = (x + y)^2 - 2xy = 100 - 2a\)

\(x^3 + y^3 = (x + y)^3 - 3xy(x + y) = 1000 - 30a\)

So the second equation is equivalent to 

\(100 - 2a = 1000 - 30a\\ 28a = 900\\ a = \dfrac{225}7\\ xy = \dfrac{225}7\\ x = \dfrac{225}{7y}\)

Substituting this into x + y = 10, we have

\(y + \dfrac{225}{7y} = 10\\ 7y^2 - 70y + 225 = 0\)

Checking the discriminant of this equation, \(\Delta = (-70)^2 - 4(7)(225) = -1400 < 0\).

Therefore, there are no real solutions.