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The area of an equilateral triangle is given by \(\frac{\sqrt{3}}{4}s^2\), where \(s\) is the side length of the equilateral triangle. Consequently, setting this formula and expression equal to \(64\sqrt{3}\), we get \(s^2=256\), and \(s=16\) centimeters. Since each side of the triangle decreases by four centimeters, the new side length of the equilateral triangle is twelve centimeters. Thus, the new area is \(36\sqrt{3}\) centimeters, and it has decreased by \(64\sqrt{3}-36\sqrt{3}=\boxed{28\sqrt{3}}\)centimeters.

-tertre

\(V = IZ\\ 1-i = I(1+3i)\\ I = \dfrac{1-i}{1+3i}\\ I = \dfrac{(1-i)(1-3i)}{(1+3i)(1-3i)}\\ I = \dfrac{-2-4i}{10}\\ I = -\dfrac{1}{5} -\dfrac{2}{5}i\)

We can solve this problem using dynamic programming, a technique taught in computer science, probably in discrete math also.

We can construct a 2-dimensional array called dp. each entry of the array is called dp_i,j, where (i,j) is the coordinates of the bug, and dp_i,j is the number of ways that the bug can reach (i,j).

First, an important observation is that \(dp_{i,j} = dp_{i-1,j} + dp_{i,j-1}\), for \((i,j) \neq (2,3)\).

This is the state transition formula of the dynamic programming.

And, there is only one way to go to (0, n) or (n, 0) for any n, which means \(dp_{0,n} = dp_{n,0} = 1\).

This is the base case of the dynamic programming.

Also, dp_2,3 = 0, as any attempt to go to (2,3) will fail due to the existence of the spider... :(

Now we can construct the array by hand... :)

First, list the base cases.

(Each entry of the matrix aligns with a lattice grid. Numbers are in relative position of the coordinates, which means bottom left corner is dp_0,0 and top right corner is dp_5,7.)

\(dp = \begin{pmatrix} 1&?&?&?&?&?\\ 1&?&?&?&?&?\\ 1&?&?&?&?&?\\ 1&?&?&?&?&?\\ 1&?&0&?&?&?\\ 1&?&?&?&?&?\\ 1&?&?&?&?&?\\ 1&1&1&1&1&1\\ \end{pmatrix}\)

Then use the transition formula to work out the remaining numbers:

\(dp = \begin{pmatrix} 1&8&26&70&180&342\\ 1&7&18&44&110&262\\ 1&6&11&26&66&152\\ 1&5&5&15&40&86\\ 1&4&0&10&25&46\\ 1&3&6&10&15&21\\ 1&2&3&4&5&6\\ 1&1&1&1&1&1\\ \end{pmatrix}\)

Therefore, there are 342 ways to go to (5,7).

The best thing to do here is to check if the determinant is equal to or not equal to zero in the three systems of equations.

Thus, we have the matrix \(\begin{bmatrix} 1 & 1 & 2\\ -2 & 4 & 2\\ 0 & 1 & 1 \end{bmatrix}\), and it's determinant, which is \(D\), is thus \(1\begin{vmatrix} 4 & 2\\ 1 & 1 \end{vmatrix} - 1\begin{vmatrix} -2 & 2\\ 0 & 1 \end{vmatrix} + 2\begin{vmatrix} -2 & 4 \\ 0 & 1 \end{vmatrix}\), and this is equal to \(0\), so the three systems of equations have no solutions.

-tertre

Quite a challenging problem.

By exhaustion of the denominator, it is found that \(\dfrac{5}{6}\) has the smallest denominator :)

Simplify, try all fractions with denominator 1, then try all fractions with denominator 2, then 3 then 4, so on.

Maximum Value: A=B=C=60 degrees, 1/2+1/2+1/2=3/2.

\(\text{A} = 2(1\cdot 1) + 4(16\cdot 1) = 66 \text { square units}\\ \text{B} = 2(8\cdot 2) + 2(8\cdot 1) + 2(2\cdot 1) = 52 \text { square units}\\ |\text{A} - \text{B}| = 66 - 52 = 14\text{ square units}\)

\(\dfrac{3}4 x + \dfrac{5}8 = 4x\\ 6x + 5 = 32x\\ 26x = 5\\ x = \dfrac{5}{26}\)

\(\phantom{= }\sqrt 5 + \dfrac{1}{\sqrt5} + \sqrt 7 + \dfrac{1}{\sqrt7}\\ = \dfrac{6\sqrt5}{5} + \dfrac{8\sqrt7}{7}\\ = \dfrac{42\sqrt5}{35} + \dfrac{40\sqrt7}{35}\\ = \dfrac{42\sqrt5 + 40\sqrt7}{35}\\ a + b + c = 42 + 40 + 35 = 117\)

\(\dfrac{1}{r^3 + 7} - 7 = \dfrac{-r^3}{r^3 + 7}\\ 1 - 7(r^3 + 7) = -r^3, r \neq \sqrt[3]{7}\\ 1 - 7r^3 - 49 = -r^3, r \neq \sqrt[3]{7}\\ 6r^3 = -48, r \neq \sqrt[3]{7}\\ r^3 = -8\\ r = -2, -2\omega, -2\omega^2\text{ where }\omega\text{ and }\omega^2\text{ denotes the cube roots of unity.}\)

   123456789²  -  2 * 123456789 * 123456787  +  123456787²

= (123456789 - 123456787)²

= 2²

= 4

7)

\(\quad 2231_4\\ = 10101101_2\)

6)

\(35 = 2^5 + 2^1 + 2^0\\ 35 = 100011_2\)

5)

\(\quad 999_{16}\\ = 9(16^2 + 16 + 1)\\ = 2457\)

4)

\(\quad 6032_8\\ = 6\cdot 8^3 + 3\cdot 8 + 2\\ = 3098\)

3) 

\(\quad0.04\overline{55}\\ = 0.04 + 0.00\overline{5}\\ = \dfrac{4}{100} + \dfrac{0.\overline{5}}{100}\\ = \dfrac{4}{100} + \dfrac{5}{900}\\ = \dfrac{41}{900}\)

Hmm ok. I will do the searching myself.

2) \(\dfrac{12}{13} = \dfrac{923076}{999999} = 0.\overline{923076}\)

If A is fixed, max value occurs when B = C,

If B is fixed, max value occurs when A = C.

If C is fixed, max value occurs when A = B.

So the maximum will occur when A = B = C

Max value = 3 cos (pi / 3) = 3/2

I got the answer, but why does the max value occur when B = C if A is fixed?

1) \(\dfrac{10}{27} = \dfrac{370}{999} = 0.\overline{370} = 0.370\;370\;370\;3...\)