MaxWong

The diameter AB is 9 + 16 + 9 = 34. Therefore, the radius is half the diameter, which is \(\dfrac{34}2 = 17\).

EDIT: There is one detail I missed, that is, it is not possible for PQDC to be a square. Simple calculations gives \(QD = \sqrt{9(9 + 16)} = 15\), not 16. The quadrilateral PQDC is a rectangle with side lengths 16 and 15.

I assume you mean \(3x^3\) and not \(3x^2\), because it is impractical to write polynomials in an un-simplified way. (I hope you understand what I meant, my English is not that good)

We state the factor theorem as follows:

Let \(p(x)\) be a polynomial. If \(p(x)\) is divisible by \(x - k\), then \(p(k) = 0\).

This is a well-known result in algebra. We prove it using division algorithm: Suppose that when p(x) is divided by x - k, the quotient is q(x) and the remainder is r. (The remainder is a constant because it must have lower degree than the divisor x - k.) By division algorithm, we must have \(p(x) = (x - k)q(x) + r\), substituting x = k gives \(r = p(k) + 0q(k) = p(k)\). Therefore the factor theorem is equivalent to saying "If p(x) is divisible by x - k, then the remainder is 0 when p(x) is divided by by x - k", which is obviously true.

We can use this result to solve the problem. From the factor theorem, we know that \(f(3) = 0\). Then, substituting x = 3 into the definition of f(x) gives \(3k -87 = 0\), which gives k = 29. 

Suppose \(f(x) = (x - 3)(ax^2 + bx + c)\). Then \(ax^3 + (b - 3a)x^2 + (c - 3b)x - 3c = 3x^2 - 20x^2 + 29x + 12\). Comparing coefficients gives \(a = 3, b = -11, c = -4\). Then we have \(f(x) = (x - 3)(3x^2 - 11x - 4)\). Factorizing the quadratic part further, we have \(f(x) = (x - 3)(3x + 1)(x - 4)\).

For the bonus part, simple inspection gives the roots x = 3, x = 4, and x = -1/3.

In fact, you have more information than enough to determine the angle. 

Note that \(\angle ZXW, \angle XWZ, \angle XZW\) are the 3 interior angles of \(\triangle WXZ\), so they should add up to \(180^\circ\).

Then,

\(90^\circ + 60^\circ + \angle XZW = 180^\circ\)

\(\angle XZW = 30^\circ\)

The information about Y can be arbitrary, it does not affect the answer.

This particular condition on Y (i.e., \(\angle XYZ = 120^\circ\)) just makes the diagram easy to draw because it is an isosceles trapezoid, i.e., WX = YZ. 

Draw the diagram. You will notice that \(\triangle MPF \sim \triangle GPH\) and \(\triangle HNQ \sim \triangle FGQ\). 

Using the ratio of corresponding sides in these pairs of similar triangles, we have

\(\dfrac{HQ}{QF} = \dfrac{2}{2 + 5} = \dfrac27\)

and

\(\dfrac{HP}{PF} = \dfrac{3 + 7}7 = \dfrac{10}7\)

We suppose \(HQ : QP : PF = 1 : a : b\) (if not, we just divide through the ratio until the first component is 1). 

Then,

\(\begin{cases} \frac1{a + b} = \frac27\\ \frac{1+a}b = \frac{10}7 \end{cases}\)

Cross-multiplying,

\(\begin{cases} 2a + 2b = 7\\ -7a +10b = 7 \end{cases}\)

Solving gives \(a = \dfrac{28}{17}\) and \(b = \dfrac{63}{34}\). Hence, \(HQ : QP : PF = 34 : 56 : 63\).

In particular, \(\dfrac{PQ}{FH} = \dfrac{56}{34 + 56 + 63} = \dfrac{56}{153}\).

This is the diagram. You can try to rotate A_3 around A_4 in your mind. Notice the angle in question is just an interior angle of the regular polygon.

We use the formula for an interior angle of a regular n-gon: \(\alpha = \dfrac{(n-2)\times 180^\circ}n\). Here, n is the number of sides, which is 15 in this case.

We get \(\alpha = \dfrac{(15 - 2) \times 180^\circ}{15} = 156^\circ\).

You can think of a dilation as multiplying lengths by the dilation factor.

The scale factor is 5, and the image (end result) of the dilation is 30. 

To get the original length OA, simply ask: What multiplied by 5 is 30?

The answer is 6.

Note that PX bisects \(\angle QPR\). Then, we can suppose \(\angle XPQ = \angle XPR = \theta\).

Recall the formula:

If two sides of a triangle are of length a and b respectively, and the angle included by these two sides is \(\phi\), then the area of the triangle is \(\dfrac12 ab \sin \phi\).

Now, draw the diagram and look at triangle PMY. Its height is MY (which is 8) and its base is PM (which is 18). So the area of triangle PMY is \(\dfrac12 \cdot 8 \cdot 18 = 72\). But on the other hand, using the formula stated above, the same area is also equal to \(\dfrac12 (PM)(PY) \sin \theta = 9\cdot PY \sin \theta\).

Therefore, we know that \(PY \sin \theta = 8\) by comparing.

Now, we look at triangle PYR. By the formula, the area is \(\dfrac12 (PY)(PR) \sin \theta\), but we know PR = 22. Using \(PY \sin \theta = 8\), the required area is \(\dfrac12 (22) (PY \sin \theta) = 11(8) = \boxed{88}\).

Recall the following formula:

The area of a sector with radius r and central angle \(\theta^\circ\) is \(\dfrac{\theta}{360} \cdot \pi r^2\).

Using the formula, we have

\(\dfrac{83}{360} \cdot \pi r^2 = 60 \pi\)

This means 

\(r^2 = \dfrac{360 \cdot 60}{83} = \dfrac{21600}{83}\)

\(r = \sqrt{\dfrac{21600}{83}} \approx 16.13\)