MaxWong

Simplifying, we have \((3y - 11)^4\).

So the coefficient of \(y^4\) is \(3^4 = 81\)

Extract the useful information: M is the midpoint of BC. 

Then triangle AMC has half the base of triangle ABC, and the same height. 

The area of triangle AMC would be half that of triangle ABC, so the answer is 90.

Label the diagram as shown. Let \(\angle FCE = \theta\) and \(EC = x\).

Let H be a point on AC such that FH is perpendicular to AC. Then \(FH = 8\).

Note that \(CE = FD = HA = x\). Then \(CH = 10 - x\). 

\(CF = \sqrt{8^2 + (10 - x)^2} = \sqrt{x^2 - 20x + 164}\) by using Pythagoras theorem in \(\triangle CFH\).

In \(\triangle ECG\), \(\sin \theta = \dfrac{EG}{EC} = \dfrac1x\).

In \(\triangle FHC\), \(\sin \theta = \dfrac{FH}{CF} = \dfrac8{\sqrt{x^2 -20x+164}}\).

So we have \(\dfrac1x = \dfrac8{\sqrt{x^2 - 20x + 164}}\). Squaring gives \(\dfrac1{x^2} = \dfrac{64}{x^2 - 20x + 164}\).

Rearranging, 

\(64x^2 = x^2 - 20x + 164\\ 63x^2 + 20x - 164 = 0\\ x = \dfrac{-20 \pm \sqrt{20^2 - 4(63)(-164)}}{2(63)}\\ x = \dfrac{8 \sqrt{163} - 10}{63}\text{ (reject negative root)} \)

Hence, the area is \(\dfrac{8x}2 = \dfrac{32 \sqrt{163} - 40}{63}\).

This is wrong, but I can't edit it, see CPhill's answer below

... 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.