MaxWong

Using Law of Cosine in \(\triangle PQR\) gives

\(8^2 = 5^2 + 11^2 - 2(5)(11) \cos \angle Q\\ \cos \angle Q = \dfrac{5^2 + 11^2 - 8^2}{2(5)(11)}\\ \cos \angle Q = \dfrac{41}{55}\)

Using Law of Cosine in \(\triangle PQM\) gives

\(PM^2 = 5^2 + \left(\dfrac{11}2\right)^2 - 2(5)\left(\dfrac{11}2\right) \cos \angle Q\\ PM^2 = 5^2 + \left(\dfrac{11}2\right)^2 - 2(5)\left(\dfrac{11}2\right)\left(\dfrac{41}{55}\right)\\ PM^2 = \dfrac{57}4\\ PM = \dfrac{\sqrt{57}}2\)

Suppose \(\overline{AC} = x\) and \(\overline{BC} = y\). From \(\overline{AM} = \overline{BN} = 1\), we have

\(\begin{cases} \left(\dfrac x2\right)^2 + y^2 = 1\\ x^2 + \left(\dfrac y2\right)^2 = 1 \end{cases}\)

Adding, we have 

\(\dfrac 54 \left(x^2 + y^2\right) = 2\\ x^2 + y^2 = \dfrac 85\\ \sqrt{x^2 + y^2} = \dfrac{2 \sqrt {10}}5\)

Hence, length of the hypotenuse is \(\dfrac{2 \sqrt {10}}5\)

The most important thing is the unit of n. Simply compute by law of indices:

\(n = \dfrac ky = \dfrac{1\text{ kg m}^{1}}{1\text{ kg m}^{-1}} = 1\,\text{m}^2\)

\(\begin{cases} x^2 + y^2 = 5\\ y= \dfrac{3x+2}4 \end{cases}\)

Substituting, we have

\(x^2 + \left(\dfrac{3x + 2}4\right)^2 = 5\\ \dfrac{25}{16}x^2 + \dfrac34x - \dfrac{19}4 = 0\\ 25x^2 + 12x - 76 = 0\\ (x + 2)(25x - 38) = 0\\ x = -2 \text{ or }x = \dfrac{38}{25}\)

When x = -2, y = (3(-2) + 2)/4 = -1.

When x = 38/25, y = (3(38/25) + 2)/4 = 41/25.

The required distance is:

\(\sqrt{\left(-2 - \dfrac{38}{25}\right)^2 +\left(-1 - \dfrac{41}{25}\right)^2 } = \dfrac{22}5\) units.

We tackle the problem in the following manner: First rewrite f(x) = f(f(x)) as the system of equations \(\begin{cases} f(k) = k\\ f(x) = k \end{cases}\).

(This system is obtained by letting k = f(x) and substituting)

Simplifying f(x) gives \(f(x) = x^3 - 6x^2 + 2x\). For f(k) = k, that means:

\(k^3 - 6k^2 +2k = k\\ k^3 - 6k^2 + k = 0\\ k(k^2 - 6k + 1) = 0\\ k = 0 \text{ or }k^2 - 6k + 1 = 0\)

For the remaining part of the problem, we solve it by quadratic formula to get \(k = 3 \pm 2 \sqrt 2\).

For each value of k, we solve the equation f(x) = k. We first start with the easy case k = 0. For k = 0, we have

\(x^3 - 6x^2 + 2x = 0\\ x(x^2 - 6x + 2) = 0\\ x = 0 \text{ or }x^2 - 6x + 2 = 0\\ x = 0\text{ or }x = 3\pm \sqrt 7\)

For k = 3 + 2 sqrt(2), we know that x = 3 + 2 sqrt(2) is a root of f(x) = k because f(k) = k. We have, by long division,

\(x^3 - 6x^2 + 2x = 3 + 2 \sqrt 2\\ (x - (3 + 2 \sqrt 2))(x^2 + (2 \sqrt 2 - 3)x + 1) = 0\)

The quadratic factor has negative discriminant. That means no real root there. The only real root from this case is \(x = 3 + 2 \sqrt 2\).

Similarly, for the case k = 3 - 2 sqrt(2), we have

\((x - (3 - 2 \sqrt 2))(x^2 - (3 + 2 \sqrt 2)x + 1) = 0\)

Again, the quadratic factor has negative discriminant. The only real root in this case is \(x = 3- 2 \sqrt 2\).

Hence, the real roots you are looking for are \(x = 0\), \(x = 3\pm 2 \sqrt 2\), and \(x = 3 \pm \sqrt 7\).

I could not figure out a nice way to calculate the number of "balanced" numbers, but I wrote a Python program to count them.

The output of the program tells me that there are 615 "balanced" numbers. Hence, the required probability is \(\dfrac{615}{9000} = \dfrac{41}{600}\).

Would love to see a combinatorial approach to this problem.

\(\frac n2\) is an integer implies n is even. The smallest four-digit base 9 integer is \((1000)_9\), which is \(9^3 = 729\), but this is not even, so \(\frac n2\) is at least 729 + 1 = 730. This gives the lower bound \(n \geq 1460\).

While n/2 is a four-digit base 9 integer, 3n cannot exceed the maximum value of four-digit base 9 integers. that value is \((8888)_9\), which is \(8(9^3 + 9^2 + 9 + 1) = 6560\), but it is not divisible by 3. That means \(3n \leq 6558\), i.e., \(n \leq 2186\) is the upper bound.

To ensure that n/2 is an integer, the answer is the number of even numbers n in the range \(1460 \leq n \leq 2186\), i.e., \(\dfrac{2186 - 1460}2 + 1 = \boxed{364}\).

Thank you, long time no see