Akhain1

To find the number of ways to choose 3 circles from 12 arranged in a circle such that no two circles are adjacent, we can use a combinatorial approach.

First, since the circles are arranged in a circle, we'll convert the problem into a linear arrangement. We can fix one circle as a reference point, which effectively reduces our problem to a line of 11 circles (with the circle we fixed being considered as already chosen).

Once we fix a circle, we need to select 2 more circles from the remaining 11 while ensuring that no two chosen circles are adjacent. To achieve this, we can represent the chosen circles and the unchosen circles as follows:

If we choose two additional circles, there will be at least one unchosen circle between any two chosen circles. Therefore, we can denote chosen circles as 'C' and unchosen circles as 'U'. So, if we denote our arrangement with chosen circles and a required space, it will look something like this:

- C - U - C - U - C

This configuration illustrates that for each chosen circle, 1 unchosen circle must be placed between them, leading us to count the total number of positions left for unchosen circles after accounting for the chosen ones.

### Create the remaining sequence for unchosen circles:

Given:

- We have chosen 3 'C's, thus requiring 2 'U's to separate them.

- This means we use up 3 + 2 = 5 spots out of the 11 available, leaving us with 6 unchosen circles to place.

### Transform into a stars and bars problem:

Now we need to distribute the remaining unchosen circles among available spots. We can think of placing the 6 remaining unchosen circles in gaps created by the two boundaries (left and right of 'C's):

If we denote the gaps as follows (where '|' represents gaps):

- | U | U | C | U | C | U | C | U | |

We have potential gaps before the first 'C', between the 'C's, and after the last 'C'. In our case, we have fixed the first circle, which means we currently have two gaps to the left and one gap to the right (which connects back to the fixed circle in circular arrangement).

- Let G - denotes the gap available, thus a sequence could be modeled as:

- G (pre gap), C, U, C, U, C (for the already occupied spaces) plus additional G's around.

### Fill spaces with the left-over circles:

Thus, after fixing the first chosen circle, thus we have 8 available spots (in gaps) to place our remaining circles, as constructed above, calculating the arrangement with stars and bars theorem.

Using the formula for distributing \( n \) indistinguishable objects (stars) into \( k \) distinguishable boxes (spaces):

\[ \text{total ways} = \binom{n+k-1}{k-1} \]

where \( n = 6 \) (remaining unchosen circles) and \( k = 3 \) (gaps). Substitute this into the equation:

\[
\text{total ways} = \binom{6 + 3 - 1}{3 - 1} = \binom{8}{2} = 28
\]

### Final Answer:

Thus, the total number of ways to choose the circles such that no two are adjacent is \( \boxed{220} \) ways.

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.