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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 \(a^5 = (-1)^5 = -1\).

To find the distance between two opposite corners of the rectangular prism, we can visualize a diagonal line going through the entire prism. This diagonal will form a right triangle with sides that correspond to the lengths, width, and height of the prism.

We can use the Pythagorean Theorem to find the length of this diagonal.

Pythagorean Theorem:

a^2 + b^2 = c^2

where:

a and b are the lengths of the two legs of a right triangle.

c is the length of the hypotenuse (the diagonal in this case).

Applying the theorem:

In our case:

a = length of the prism = 3 units

b = width of the prism = 5 units

c = diagonal length (what we want to find)

Solve for c (diagonal length):

c^2 = a^2 + b^2

c^2 = 3^2 + 5^2

c^2 = 9 + 25

c^2 = 34

Taking the square root of both sides (remembering there might be positive and negative square roots), we get:

c = ±√34

Since c represents a distance (and cannot be negative), we take the positive square root:

c = √34 units

Therefore, the distance between two opposite corners of the rectangular prism is √34 units. (This is the exact answer, but you can approximate it to a decimal value if needed.)

To find the probability that intervals \(I\) and \(J\) overlap, we need to consider the possible positions of the four points \(x_1, x_2, x_3,\) and \(x_4\) within the interval \([0, 1]\).

Without loss of generality, let's assume that \(x_1 < x_2\) and \(x_3 < x_4\).

For \(I\) and \(J\) to overlap, one of the following conditions must be true:

1.  \(x_1 < x_3 < x_2 < x_4\)

2.  \(x_3 < x_1 < x_4 < x_2\)

Let's calculate the probability for each condition:

1.  Probability of Condition 1:

   - The probability that \(x_1\) falls in the interval \([0, 1]\) is \(1\).

   - The probability that \(x_3\) falls in the interval \([x_1, 1]\) is \(1 - x_1\).

   - Given \(x_1\) and \(x_3\), the probability that \(x_2\) falls in the interval \((x_1, 1]\) is \(1 - x_1\).

   - Given \(x_1\), \(x_3\), and \(x_2\), the probability that \(x_4\) falls in the interval \((x_3, 1]\) is \(1 - x_3\).

   - So, the probability of Condition 1 is \(1 \times (1 - x_1) \times (1 - x_1) \times (1 - x_3) = (1 - x_1)^2 (1 - x_3)\).

2.  Probability of Condition 2:

   - The probability that \(x_3\) falls in the interval \([0, 1]\) is \(1\).

   - The probability that \(x_1\) falls in the interval \([0, x_3]\) is \(x_3\).

   - Given \(x_3\) and \(x_1\), the probability that \(x_4\) falls in the interval \((x_3, 1]\) is \(1 - x_3\).

   - Given \(x_3\), \(x_1\), and \(x_4\), the probability that \(x_2\) falls in the interval \((x_1, 1]\) is \(1 - x_1\).

   - So, the probability of Condition 2 is \(1 \times x_3 \times (1 - x_1) \times (1 - x_3) = x_3 (1 - x_1) (1 - x_3)\).

Now, since \(x_1, x_2, x_3,\) and \(x_4\) are chosen independently and uniformly at random in the interval \([0, 1]\), we can find the probability that intervals \(I\) and \(J\) overlap by summing the probabilities of Condition 1 and Condition 2:

\[ \text{Total probability} = (1 - x_1)^2 (1 - x_3) + x_3 (1 - x_1) (1 - x_3) \]

\[ = (1 - x_1) (1 - x_3) (1 - x_1 + x_3) \]

\[ = (1 - x_1) (1 - x_3) \]

Now, since \(x_1\) and \(x_3\) are chosen uniformly at random in the interval \([0, 1]\), we can find the expected value of the probability by integrating over the joint distribution of \(x_1\) and \(x_3\):

\[ \text{Expected probability} = \int_{0}^{1} \int_{0}^{1} (1 - x_1) (1 - x_3) \, dx_1 \, dx_3 \]

\[ = \int_{0}^{1} \left( \int_{0}^{1} (1 - x_1) (1 - x_3) \, dx_3 \right) dx_1 \]

\[ = \int_{0}^{1} \left( 1 - x_1 - \frac{1}{2} + \frac{x_1^2}{2} \right) dx_1 \]

\[ = \left[ x_1 - \frac{x_1^2}{2} - x_1 + \frac{x_1^3}{6} \right]_{0}^{1} \]

\[ = \left( 1 - \frac{1}{2} - 1 + \frac{1}{6} \right) - \left( 0 - 0 - 0 + 0 \right) \]

\[ = \frac{1}{3} - \frac{1}{2} = \frac{1}{6} \]

Therefore, the expected probability that intervals \(I\) and \(J\) overlap is \( \frac{1}{6} \).

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

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

@MaxWong, you're right about the radius is in fact 16. 

For part b, 

Like you already said, the dimater of the halfcircle is 34. We already know the height of the trapezoid since PQDC is a square. We also know DC is 16. 

Now, we have everything needed for the area of the trapezoid, so let's begin!

We know the area of the trapezoid is \(\frac{(b1+b2)h}{2}\) where b1 and b2 are the bases and the h is the height. 

We have \(\frac{(25+16)16}{2} = (31)8 = 248.\)

So the area of the trapezoid is 248!

Thanks!

EDIT: Yes, Max Wong is right about the fact that PQDC cannot be a square. Plugging the new numbers in gets us an area of 232.5

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 \(-4,-2,0,2,4\).