MaxWong

You can check your answer by substituting \(x = 0\) into the original equation:

\(\begin{array}{rcl} \text{LHS} &=& 400 + 0\\ &=& 400\\ \text{RHS} &=& 200x\\ &=& 200(0)\\ &=& 0\\ &\neq& \text{LHS} \end{array}\)

Then \(x= 0\) does not satisfy the equation, and hence your answer is not correct.

\(\begin{array}{rcl|c} 400+x&=&200x\\ 400+x-x&=&200x-x&\text{subtract }x\\ 400 &=& 199x\\ \dfrac{400}{199} &=& x&\text{divide by }199\\ x &=& \dfrac{400}{199} &\text{exchange sides} \end{array}\)

Note that 

\(\begin{array}{rcl}f(x) &=& (x^2 + 6x + 9)^{50} - 4x + 3\\ &=&\left( (x + 3)^2\right)^{50} - 4x + 3 \\&=&(x+3)^{100}-4x+3\end{array}\)

For each of \(i=1,2,\cdots,100\), since \(x = r_i\) is a root of \(f(x)\):

\(\begin{array}{rcl} f(r_i) &=& 0\\ (r_i+ 3)^{100} &=& 4r_i - 3 \end{array}\)

Now, using the equation above,

\(\begin{array}{cl} &(r_1 + 3)^{100} + (r_2 + 3)^{100} + \cdots + (r_{100} + 3)^{100}\\ =&(4r_1 - 3) + (4r_2 - 3) + \cdots + (4r_{100} - 3)\\ =&4(r_1 + r_2 + \cdots + r_{100}) - 300 \end{array}\)

By Vieta's formula, we have:

\(r_1 + r_2 + \cdots + r_{100} = -\dfrac{\text{coefficient of }x^{99}\text{ in }f(x)}{\text{coefficient of }x^{100}\text{ in }f(x)}\)

Using the binomial theorem, we can expand \((x + 3)^{100}\).

\((x + 3)^{100} -4x+3 = x^{100} + 300x^{99}+\cdots\)

Therefore, 

\(r_1 + r_2 + \cdots + r_{100} = -\dfrac{300}1 = -300\)

Recall that \((r_1 + 3)^{100} + (r_2 + 3)^{100} + \cdots + (r_{100} + 3)^{100}=4(r_1 + r_2 + \cdots + r_{100}) - 300\). Then:

\(\begin{array}{cl} &(r_1 + 3)^{100} + (r_2 + 3)^{100} + \cdots + (r_{100} + 3)^{100}\\ =&4(-300) -300\\ =& -1500 \end{array}\)

\(\quad\dfrac{\sqrt 2}{\sqrt 2 + \sqrt 3 - \sqrt 5} - \dfrac12\\ = \dfrac{\sqrt2 \left(\sqrt2 + \sqrt 3 + \sqrt 5\right)}{(\sqrt 2 + \sqrt 3)^2 - \sqrt5^2} - \dfrac12\\ =\dfrac{\sqrt 2 \left(\sqrt 2 + \sqrt 3 + \sqrt 5\right)}{2\sqrt 6} - \dfrac12\\ = \dfrac{\sqrt 2 + \sqrt 3 + \sqrt 5}{2 \sqrt 3} - \dfrac{\sqrt 3}{2\sqrt 3}\\ =\dfrac{\sqrt 2 + \sqrt 5}{2 \sqrt 3}\\ = \dfrac{\sqrt 6 + \sqrt{15}}{6}\)

\((f\cdot g)(x) = f(x) \cdot g(x) = 6x(x + 5) = 6x^2 + 30x\)

But if you meant \((f\circ g)(x)\):

\((f\circ g)(x) = f(g(x)) = 6 g(x) = 6(x + 5) = 6x + 30\)

Let r be the radius.

Then OA = OB = OC = r.

By Pythagorean theorem, 

\(CP^2 = OP^2 + OC^2\\ CP^2 = 20^2 + r^2\\ CP = \sqrt{r^2 + 400}\)

Then,

\(AP = AO - OP = r - 20\\ BP = BO + OP = r + 20\)

Consider the power of point P.

\(CP \cdot PQ = AP \cdot PB\\ 7\sqrt{r^2 + 400} = (r - 20)(r + 20)\\ 49(r^2 + 400) = (r^2 - 400)^2\\ (r^2)^2 - 800r^2 + 160000 - 49r^2 - 19600 = 0\\ (r^2)^2 - 849r^2 + 140400 = 0\)

Using quadratic formula,

\(r^2 = \dfrac{849 \pm \sqrt{849^2 - 4(140400)}}{2} \\ r^2 = 225\text{(rejected) or }\boxed{r^2 = 624}\)

Let r be the radius of the circle.

Obviously, OC = r.

Let M be the mid-point of CD.

Since \(\angle ABD = 90^\circ\), ABMO is a rectangle. Therefore, OM = 12.

Since M is the midpoint of CD, CM = 18/2 = 9.

Then, by Pythagorean theorem, 

\(OC^2 = OM^2 + MC^2\\ r^2 = 12^2 + 9^2 = 225\\ r = 15\)

Consider the power of point M with respect to each of the circles.

\(MT^2 = MU^2 = MB \cdot MA = 3 \cdot 12 = 36\)

This implies \(MT = MU = 6\)

Therefore, \(TU = MT + MU = 6 +6 = 12\)

By mid-point theorem, 

\(KL = \dfrac12 GJ\\ 75 - 10 x = \dfrac{x + 3}{2}\\ 150 - 20x = x + 3\\ 21x = 147\\ x = 7\\GJ = x + 3 = 10\)

Both Melody and CPhill are offline now. I will try to answer that.

This means \(F_{3n} = aF_{3n - 3} + bF_{3n - 6}\) by the definition of \(G\).

Then, we try some \(F_{3n}, F_{3n - 3}, F_{3n - 6}\). For this to be possible, we need the values of \(F_n\) from n = 0 up to n = 9.

\(F_0 = 0\\ F_1 = 1\\ F_2 = 1\\ F_3 = 2\\ F_4 = 3\\ F_5 = 5\\ F_6 = 8\\ F_7 = 13\\ F_8 = 21\\ F_9 = 34\)

Then, using the equation on the first line of my solution and substituting n = 2 and n = 3,

\(\begin{cases}F_6 = aF_3 + bF_0\\F_9=aF_6+bF_3\end{cases}\)

Simplifying,

\(\begin{cases}2a = 8\\8a + 2b = 34\end{cases}\)

Solving gives \(a = 4, b = 1\).

Therefore, \((a, b) = (4, 1)\).