Akhain1

Understanding the Problem

Franklin starts at (0, 0).

He can move up, down, left, or right one unit at each step.

We need to find the total number of unique points he can reach after 12 steps.

Solution Approach

This problem can be solved using a combinatorial approach.

Number of steps in each direction: Since Franklin can move in four directions, we can think of distributing 12 steps among these four directions.

Combinations: We need to find the number of ways to distribute 12 identical objects (steps) into 4 distinct boxes (directions). This is a classic stars and bars problem.

Solving the Problem

Using the stars and bars formula, the number of ways to distribute n identical objects into k distinct boxes is:

C(n + k - 1, k - 1)

In our case, n = 12 (steps) and k = 4 (directions).

So, the number of ways to distribute the steps is:

C(12 + 4 - 1, 4 - 1) = C(15, 3) = 455

Therefore, Franklin the fly can end up at 455 different points after 12 steps.

We can divide the problem into two triangles: △ABC and △ACE. Since we are given three side lengths for each triangle, we can use the [Law of Cosines] to solve for DE.

For △ABC, we have

\begin{align*} BC^2 &= AB^2 + AC^2 - 2 \cdot AB \cdot AC \cos ( \angle A) \ 8^2 &= 3^2 + 7^2 - 2 \cdot 3 \cdot 7 \cos ( \angle A) \ \cos ( \angle A) &= \frac{38}{42} = \dfrac{19}{21}. \end{align*}

Similarly, for △ACE, we obtain

\begin{align*} AC^2 &= AE^2 + CE^2 - 2 \cdot AE \cdot CE \cos ( \angle A) \ 7^2 &= (AE + DE)^2 + 2^2 - 2 \cdot (AE + DE) \cdot 2 \cos ( \angle A) \ 7^2 &= AE^2 + DE^2 + 4AE + 4 - 4(AE + DE) \cdot \dfrac{19}{21} \ 0 &= AE^2 + DE^2 - \dfrac{32}{7} AE - \dfrac{32}{7} DE + 20. \end{align*}

We can complete the square on both the AE2 and DE2 terms. To do this, we take half of the coefficient of our AE term, square it, and add it to both sides of the equation. Similarly, we do this for the DE term.

This gives us

\begin{align*} 0 &= (AE^2 - \dfrac{16}{7} AE + \dfrac{64}{49}) + (DE^2 - \dfrac{16}{7} DE + \dfrac{64}{49}) + 20 - \dfrac{64}{7} (AE + DE) \ &= (AE - \dfrac{8}{7})^2 + (DE - \dfrac{8}{7})^2 + 20 - \dfrac{64}{7} (AE + DE). \end{align*}

Let x=AE−78​ and y=DE−78​. Then AE=x+78​ and DE=y+78​, so

\begin{align*} 0 &= (x)^2 + (y)^2 + 20 - \dfrac{64}{7} ((x + \dfrac{8}{7}) + (y + \dfrac{8}{7})) \ 0 &= x^2 + y^2 + 20 - \dfrac{64}{7} x - \dfrac{64}{7} y - 16. \ 0 &= x^2 + y^2 - \dfrac{64}{7} x - \dfrac{64}{7} y + 4. \end{align*}

Now we can complete the square again, this time on the entire right side of the equation. Taking half of the coefficient of our x term, squaring it, and adding it to both sides gives

\begin{align*} 0 &= (x^2 - \dfrac{64}{7} x + \dfrac{16^2}{49}) + (y^2 - \dfrac{64}{7} y + \dfrac{16^2}{49}) + 4 - \dfrac{16^2}{49}. \ &= (x - \dfrac{32}{7})^2 + (y - \dfrac{32}{7})^2 - \dfrac{225}{49}. \end{align*}

Since the square of a real number is never negative, the only way for the right side of this equation to equal zero is if both (x-32/7​)2 and (y−32/7​)2 are zero. Thus, x = 32/7 and y = 32/7​, which means AE=8 and DE=8. 

Therefore, DE=8​.

Noam pays 13584.5 in tax, and Betty pays 18719.5 in tax.

The difference is 18719.5 - 13584.5 = 5135.

The area of the triangle is 7*sqrt(2)/2.

You get the largest D when you take a = -4, b = 4, and c = 4.  This gives you D = 80.