MaxWong

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.

Go from bottom to top:

5a = 20

a = 4

4a + 5b = 16

4(4) + 5b = 16

5b = 0

b = 0

3a + 4b + 5c = 15

3(4) + 4(0) + 5c = 15

5c = 3

c = 3/5

2a + 3b + 4c + 5d = 34

2(4) + 3(0) + 4(3/5) + 5d = 34

5d = 118/5

d = 118/25

a + 2b + 3c + 4d + 5e = 41

4 + 2(0) + 3(3/5) + 4(118/25) + 5e = 41

5e = 408/25

e = 408/125

\(a^2 + b^2 = 6 + 2ab\\ a^2 + 2ab + b^2 = 6 + 4ab\\ (a + b)^2 = 6 + 4ab\)

But a + b = 4, so

\(4^2 = 6 + 4 ab\\ ab = \dfrac 52\)

Then consequently \(a^3 + b^3 = (a + b)^3 - 3ab(a + b) = 4^3 -3\cdot \dfrac52 \cdot 4 = 34\).

\(\dfrac1{1\times 7} + \dfrac1{4\times 10} = \dfrac17 + \dfrac1{40} = \dfrac{40 + 7}{280} = \dfrac{47}{280}\)

Suppose f(x) is an integer. Consequently sqrt(x) is an integer. Suppose N = sqrt(x). Then sqrt(x - sqrt(x)) = sqrt(N^2 - N) = sqrt(N(N-1)).

x is 3 digits, so the range of N is \(\sqrt{100} \leq N \leq \sqrt{961} \), which is \(10 \leq N\leq 31\). There are no integer in this range where N(N - 1) is a square number.

Hence, there are no solution.

Repeatedly using the recurrence formula with n = 3, 4, 5, 6 gives respectively:

\(a_3 = a_2 - a_1 = -4\\ a_4 = a_3 - a_2 = -11\\ a_5 = a_4 - a_3 = -7\\ a_6 = a_5 - a_4 = \boxed{4} \)

Multiplying all terms by a^2 gives 

\(a^4 + a^3 + a + 1 = 0\\ (a^3 + 1)(a + 1) = 0\\ (a + 1)^2(a^2 - a + 1) = 0\\ a = -1\text{ or }a^2 - a + 1 = 0\)

Note that the equation \(a^2 - a +1 = 0\) has determinant \(\Delta = (-1)^2 - 4(1)(1) = -3 < 0\), so \(a^2 - a +1 = 0\) has no real solutions.

Hence, a = -1 is the only possibility. Then \(a^5 = (-1)^5 = -1\).

Suppose P = (x, y). Since the slope of OP is 4, by slope formula, we have \(\dfrac{y - 0}{x - 0} = 4\), which means \(y = 4x\).

So, P = (x, 4x) for some x. By the same argument, Q = (w, 5w) for some w. 

Now, we use the condition \(OP = OQ\) to find a connection between w and x. Using distance formula to calculate OP and OQ, we have

\(\sqrt{x^2 + (4x)^2} =\sqrt{w^2 + (5w)^2}\)

\(17x^2 = 26w^2\)

\(x = \sqrt{\dfrac{26}{17}}w\)

Now, the slope of PQ is \(\dfrac{5w - 4x}{w - x} = \dfrac{5w - 4\cdot \sqrt{\frac{26}{17}}w}{w - \sqrt{\frac{26}{17}} w} = \dfrac{5 \sqrt{17} - 4 \sqrt{26}}{\sqrt{17} - \sqrt{26}}\)

Rationalizing gives \(m_{PQ} = \dfrac{(5 \sqrt{17} - 4\sqrt{26})(\sqrt{17} + \sqrt{26})}{17 - 26} = \dfrac{-19 + \sqrt{442}}{-9} = \dfrac{19 - \sqrt{442}}9\).

If you just need an approximate value, \(\dfrac{19 - \sqrt{442}}{9} \approx -0.2249\).

Note that \(\triangle TMP \sim \triangle TUS\). Then \(\angle TMP = \angle TUS = 180^\circ - \angle TSU - \angle STU = 89^\circ\).

Considering the interior angle sum in triangle TUZ gives 

\(90^\circ + \angle STU + \angle TUZ = 180^\circ\\ \angle TUZ = 90^\circ - \angle STU = 61^\circ\)

Therefore,

\(\angle TMP + \angle TUZ = 89^\circ + 61^\circ = \boxed{150^\circ}\)

Using the slope formula gives \(\dfrac{b^2 - a^2}{b - a} = 2\).

Note that \(b^2 - a^2 = (b - a)(a + b)\) by difference of squares formula. Then, simplifying, we have

\(\dfrac{(b - a)(a + b)}{b - a} = 2\\ a + b = \boxed{2}\)

Note that for any non-zero number a, the quantity \(\dfrac a{|a|}\) can only take two values: -1 when a is negative, and 1 when is positive.

Therefore, \(\dfrac a{|a|}\) serves as an indicator of whether a is positive. So, there are only 5 cases:

Case 1: x, y, z are all positive, making x + y + z also positive. Then \(\displaystyle {\frac{x}{|x|} + \frac{y}{|y|} + \frac{z}{|z|} + \frac{x + y + z}{|x + y + z|}} = 4\).

Case 2: One of x, y, z, x + y + z is negative (for example, x = -1, y = 3, z = 4). Then \(\displaystyle\frac{x}{|x|} + \frac{y}{|y|} + \frac{z}{|z|} + \frac{x + y + z}{|x + y + z|} = 2\).

Case 3: Two of x, y, z, x + y + z are negative (for example, x = -1, y = -2, z = 4). Then \(\displaystyle\frac{x}{|x|} + \frac{y}{|y|} + \frac{z}{|z|} + \frac{x + y + z}{|x + y + z|} = 0\).

Case 4: Three of x, y, z x + y + z are negative (for example, x = -7, y = -5, z = 1). Then \(\displaystyle\frac{x}{|x|} + \frac{y}{|y|} + \frac{z}{|z|} + \frac{x + y + z}{|x + y + z|} = -2\).

Case 5: x, y, z are all negative, making x + y + z also negative. Then \(\displaystyle\frac{x}{|x|} + \frac{y}{|y|} + \frac{z}{|z|} + \frac{x + y + z}{|x + y + z|} = -4\).

Therefore, the possible values of \(\displaystyle\frac{x}{|x|} + \frac{y}{|y|} + \frac{z}{|z|} + \frac{x + y + z}{|x + y + z|}\) are \(-4,-2,0,2,4\).