hairyberry

Suppose point A is the first point choosen. To satisfy the contition, the second point has to be on the major arc of BB^'\(\). But the question now is how do we find the value of \(\angle{BO{B}^{'}}\)? Well, we know \(\angle{BO{B}^{'}}=\angle{AO{B}^{'}}*2\). There are many ways to find angle AOB^', but I choose to use law of cosines. I won't get into the details, but

 \(\cos(\angle{AO{B}^{'}})=\frac{1+1-\frac{9}{4}}{2}=-\frac{1}{8}\), so \(\angle{AO{B}^{'}}=\arccos(-\frac{1}{8})\), and \(\angle{BO{B}^{'}}=2*\arccos(-\frac{1}{8})\)\(\).

Supposing you are calculating in radians, the probability is \(\frac{2*\arccos(-\frac{1}{8})}{2\pi}\), or \(\frac{\arccos(-\frac{1}{8})}{\pi}\), which is approximately 0.53989, or 53.989%.

We see there are quadratic terms and linear terms, so Cauchy Schwarz is perfect!

Applying Cauchy Schwarz inequatlity, we get,

\(({x}^{2}+3{y}^2)(1+\frac{1}{3})\ge{(x+y)}^{2}\). Substituting our values in,

\(18*\frac{4}{3}\ge{(x+y)}^{2}\), or \(24\ge{(x+y)}^{2}\), and because x and y are nonnegative, or \(x+y\ge0\), we have \(\sqrt{24} \ge x+y\).

But we need to satisfy the cauchy schwarz equality condition, to know for sure this can be achieved. This condition is, satisfied when x = r, \(\frac{\sqrt{3}}{3}r=\sqrt{3}y\), or \(y=\frac{r}{3}\), so \({r}^{2}+\frac{{r}^{2}}{3}=18\), or \({r}^{2}=27/2, r = \frac{3\sqrt{6}}{2}\), this puts x and y as \(x = \frac{3\sqrt{6}}{2}, y=\frac{\sqrt{6}}{2}\). So we know that this solution works, so we know the maximum value is \(\sqrt{24}\).\(\)

Here is a different solution, which is a generalization of proyaop's method. Here is the graph below, labeled with some extra points and lines. P.S. - In a math competition, or any other place where you don't have good access to resources, I suggest not using this method, as Stewart's theorem is just a generalization of the double law of cosines, proyaop's method. Stewart's theorem is also notoriously hard to memorize. But, it does shorten the amount of work significantly.

Stewart's theorem is perfect for this problem:

First, we know \(\overline{C1C3} = 3, \overline{C2C3}=2\).

We use a mnemonic to memorize Stewart's theorem, man + dad = bmb + cnc, where these all recommend certain side lengths of the triangle. A man and his dad put a bomb in the sink, though this does not exactly make it easier to memorize, as we also have to memorize the configuration of the sides.

So onto the solution,

Substituting the values into the equation we get \(30+5{(5-x)}^{2}=3{(3+x)}^{2}+2{(2+x)}^{2}\).

When solving, we get a convient cancellation of the quadratic terms, and we are left with x = 120/76, or x = 30/19.

Note while this is still easier, I would mostly still recommend proyaop's method.

Between the 5 different mathematics textbooks, there is 5! ways to order them.

Between the 4 different psychology textbooks, there is 4! ways to order them.

Treating each of these as a group, we can order the group of of mathematics textbooks, and psychology textbooks in \(\frac{2!}{1!1!}\) = 2 different ways. (this is because either the group of mathematics textbooks can be first, or the group of psychology textbooks can be first).

There are 5!*4!*2 = 5760 different ways to order the textbooks.

We simplify, into -8x²-kx+47=0.

If the equation has no real solutions then, b²-4ac < 0.

So, k²-4(-8)(47)<0. or k²+1184<0

k² must be greater than or equal to zero, so k²+1184 must be greater than zero, so there is no possibility of the discriminant being negative, so k can be infinitely big.

This is a classic case of Simon's favorite factoring trick.

Supposed the mystery number is m.

We group like terms, getting ab-18a+8b+m.

We can take the a out of the first two terms, a(b-18) + 8b + m. We know b - 18 should also factor into 8b + m.

So, 8b+m = 8(b-18), and m = -144. The mystery number is -144.

This can be factored into (a+8)(b-18).

A lot of casework, but relatively simple.

We find the digits for each number, find the number of ways to arrange them, to represent the different integers.

Here are the possible cases:

5 = 

C1) 1, 1, 1, 1, 1

C2) 1, 1, 1, 2

C3) 1, 2, 2

C4) 1, 1, 3

C5) 2, 3

For C1, there is 1 way to order this.

For C2, there is \(\frac{4!}{3!1!}\), or 4 ways to order this.

For C3, there is \(\frac{3!}{2!1!} \), or 3 ways to order this.

For C4, there is \(\frac{3!}{2!1!}\), or 3 ways to order this.

For C5, there is 2 ways to order this.

There are a total of 1+4+3+3+2 = 13 numbers.

Looking at the terms of both sides, we see that a quadratic * f(x) is a quartic, so f(x) must be a quadratic. 

We additionally know that the coefficient of the quadratic term of f(x) is 2, because the only way 2x⁴ can be formed, is the multiplying of the quadratic terms, and since on quadratic term is x², the other term must be 2x⁴/x² = 2x².

Similarly, applying this same trick, we know that the constant term of f(x) must be 1/5, because only constant terms * constant terms can create constant terms, and on constant term is -5, so the other must be 1/5, so that -5*1/5 = -1. Another way to visualize this is by plugging in 0 as x.

Now we need the linear term.

We know f(x) so far as 2x²+ax+1/5.

To find the linear term, we consider that (x²-2x-5)*f(x) is the quartic.

(x²-2x-5)*(2x²+ax+1/5)=2x⁴+19x³ ... 

From here we can directly get a. We see that the cubic term 19x³ on the right, should also be equal to the cubic term of the expansion on the left.

We notice that cubic terms in the expansion can only be formed from quadratic terms * linear terms.

The cubic terms on the expansion on the left is -2x*(2x²) = -4x^3 and x²*ax, so the cubic term in the expansion is (-4+a)x³,

We equate the two cubic terms getting, -4+a = 19. a = 23.

But we see that (x²-2x-5)(2x²+23x+1/5) is not the correct expansion, and polynomials can't have negative exponents, I believe that this question is not possible. 

We treat the 4 freshmens as a group. 

In the group of freshman there are 4! ways to order the freshmans.

To order the remaining objects, 4 sophomores, one freshman, and one group of freshmans, there is 6! ways to order this.

So there are a total of 4! * 6! ways to order

But we must be careful, because if there is a group of 5 freshmanm, we overcount. The conbination S₂F₁F₄F₃F₅F₂S₃S₄S₁, can be formed as S₂F₁ + a group + S₃S₄S₁, or S₁ + a group + F₂S₃S₄S₁. 

So we overcount each possiblity of five people in a group twice, so we need to subtract the number of possibilities with 5 people in a row.

The possibilities for 5 people in a row, treating the 5 freshmen as a group is 5! (the arrangments within the group) * (5!) (the ways to order 1 group + 4 sophomores). 

So the answer should be 4!*6!-5!*5! = 2880 ways