MaxWong

Moving some terms yields: \(2x^2 + 2y^2 - x - y - 8 = 0\)

Note that geometrically, this is the equation of a circle, because we can rewrite the equation as follows:

\(\begin{array}{rcl} 2x^2 + 2y^2 - x - y - 8 &=& 0\\ x^2 + y^2 - \dfrac12 x - \dfrac12 y - 4&=&0\\ \left(x - \dfrac14\right)^2 + \left(y - \dfrac14\right)^2 &=& \dfrac{33}8 \end{array}\)

Consider the straight line x - y = c, where c is varying through real numbers.

You can find the maximum value of c such that the line and the circle intersects, i.e., \(\begin{cases}2x^2 + 2y^2 - x - y - 8 = 0\\x - y = c​​​​\end{cases}\) has solutions.

To do so, you need to substitute y = x - c into the circle equation, then consider the discriminant of the resulting equation. Since the system has solutions, the discriminant of the resulting equation must be nonnegative. Solving the equation \(\Delta \geq 0\) will give you the suitable range of values of c.

Please tell me if you want more explanations or you need more help.

Applying the Law of Sines yields:

\(\dfrac{CA}{\sin B} = \dfrac{CB}{\sin A}\)

After some simplification, 

\(\sin B = \dfrac{CA \cdot \sin A}{CB} = \dfrac{4 \sqrt 2 \sin 45^\circ}{4\sqrt 3}\)

Now you can find the value of sin B, and hence the value of B using inverse trigonometric function.

Cross-multiply. You will get: 

\(c(c - 1) = 4(c - 5)\)

Now expand both sides, and move all terms to the left-hand side. You will get a quadratic equation. Then you can solve it with the quadratic formula.

Okay! 

Contour map is a nice method! I didn't know it was called "contour map" before.

Here's another probability problem solved with contour maps, which shows the power of contour maps. It makes the problem much more easier.

Since \(\alpha\) is inversely proportional to \(\beta\), \(\alpha = \dfrac k\beta\) for some nonzero real number \(k\), which we call the proportionality constant.

We find k by substituting a pair of values of \(\alpha\) and \(\beta\) into the equation.

Substituting \(\alpha = -3\) and \(\beta = -6\), \(-3 = \dfrac k{-6}\). Solving gives \(k = 18\).

Now, substitute \(\beta = 18\) and \(k = 18\) to get the corresponding value of \(\alpha\) when \(\beta = 18\).

\(\alpha = \dfrac{18}{18} = \boxed{1}\).

You have assumed \(\alpha\) is directly proportional to \(\beta\) instead of inversely proportional to \(\beta\).

You can register for an email account and then register here.

If you don't have an email account, it is going to be useful for other purposes too.

Guests can post at most 3 questions daily, I believe.

Note that

\(\begin{array}{cl} &\displaystyle\frac{(y-x)^2}{(y-z)(z-x)}+\frac{(z-y)^2}{(z-x)(x-y)}+\frac{(x-z)^2}{(x-y)(y-z)} \\ =& \dfrac{-(y - x)^3-(z - y)^3 -(x - z)^3}{(y - z)(z - x)(x - y)}\\ =& \dfrac{(x - y)^3 + (y - z)^3 + (z - x)^3}{(y - z)(z - x)(x - y)} \end{array}\)

Now, we use the identity \(a^3 + b^3 + c^3 - 3abc = (a + b +c)(a^2 + b^2 + c^2 - ab - bc -ca)\), with \(a = x - y, b =y - z, c = z - x\). 

(If you don't believe me, you can try to expand the right-hand side of the identity. It should take some effort, but you will eventually arrive at the left-hand side.)

\(\begin{array}{rcl} (x - y)^3 + (y - z)^3 + (z - x)^3 - 3(x - y)(y - z)(z - x) &=& ((x - y) + (y - z) + (z - x))(\text{something})\\ &=& (x - y + y - z + z - x)(\text{something})\\ &=& (0)(\text{something})\\ &=& 0\\ (x - y)^3 + (y - z)^3 + (z - x)^3 &=& 3(x - y)(y - z)(z - x) \end{array}\)

Therefore, the numerator is exactly 3 times the denominator. 

We have, for any distinct real numbers x, y, z, \(\displaystyle\frac{(y-x)^2}{(y-z)(z-x)}+\frac{(z-y)^2}{(z-x)(x-y)}+\frac{(x-z)^2}{(x-y)(y-z)} = 3\).

Since it is constantly 3, the minimum value must be 3.