Rangcr897

Proving Equidistance of Intersection Points on an Archimedean Spiral

Understanding the Problem

We're considering the Archimedean spiral defined by the polar equation r = θ for θ > 0. The task is to prove that consecutive intersection points of this spiral with any ray originating from the origin are equidistant.

Solution

1. Parametric Representation of the Ray:

Let the ray be defined by the angle φ.

Any point on this ray can be represented in polar coordinates as (r, φ), where r is the distance from the origin.

2. Intersection Points:

To find the intersection points of the spiral and the ray, we equate their polar equations:

θ = r

Substituting r with r = ρcos(φ - θ) (polar coordinate conversion to Cartesian), we get:

θ = ρcos(φ - θ)

3. Finding Consecutive Intersection Points:

Let θ₁ and θ₂ be the angles corresponding to two consecutive intersection points.

Then, we have:

θ₁ = ρcos(φ - θ₁)

θ₂ = ρcos(φ - θ₂)

4. Calculating the Distance Between Intersection Points:

The distance between two points in polar coordinates is given by:

d = sqrt(r₁² + r₂² - 2r₁r₂cos(θ₂ - θ₁))

Substituting r₁ = θ₁ and r₂ = θ₂, we get:

d = sqrt(θ₁² + θ₂² - 2θ₁θ₂cos(θ₂ - θ₁))

5. Simplifying the Distance:

Using the trigonometric identity cos(A - B) = cosAcosB + sinAsinB, we can simplify the expression for d:

d = sqrt((θ₂ - θ₁)²)

d = |θ₂ - θ₁|

6. Conclusion:

Since θ₂ and θ₁ represent consecutive intersection points, |θ₂ - θ₁| is a constant (equal to 2π for a full rotation).

Therefore, d = 2π is also a constant.

The Distance

The distance between consecutive intersection points of the Archimedean spiral with any ray from the origin is 2π.

This proves that the consecutive intersection points are equidistant.

Let  S  represent the weight of one sack of sugar.

Let  F  represent the weight of one sack of flour.

Two sacks of suagr together with three sacks of lour weight no more than 40 pounds:

    2S + 5F <= 60

The weight of a sack of flour is no more than 5 poiunds more than the weight of two sacks of sugar:

    F <= 2S + 8

Since both of the inequalities have the same sense (point in the same direction), we can substitute:

Substituting:  2S + 3(2S + 5) <= 40

                        2S + 6S + 15 <= 40

                                 8S + 15 <= 40

                                         8S <= 25

                                           S <= 3.125

Since  F <= 2S + 5   --->   F <= 2(3.125) + 5   --->   F <= 11.25

So, each sack of flour cannot weigh more than 11.25 pounds.

For the first problem:

To address the problem, we will use the properties of similar triangles since lines \( DE \) and \( EF \) are parallel to sides \( BC \) and \( CD \) of triangle \( ABC \).

Since \( DE \) is parallel to \( BC \) and \( EF \) is parallel to \( CD \), triangle \( ADF \) is similar to triangle \( ABC \).
The sides of similar triangles are proportional, so we can set up the following proportions:

\[
\frac{AD}{AB} = \frac{AF}{AC} = \frac{DF}{BC}
\]

We know that \( AF = 7 \) and \( DF = 2 \). Let \( BD = x \). The entire length \( AB \) can be expressed as:

\[
AB = AD + DF = AD + 2
\]

Let’s designate \( AD = x \). Hence, we have \( AB = x + 2 \).

Considering the structure of triangle \( ADF \) in relation to triangle \( ABC \), we also need to calculate \( AC \) in terms of \( AE \):

Since \( EF \) is parallel to \( CD \), triangles \( AEF \) and \( ACD \) are also similar, resulting in the relationship:

\[
\frac{AE}{AC} = \frac{AF}{AB} = \frac{DF}{DC}
\]

But for our immediate task of determining the length \( BD \) just using \( AF \) and \( DF \), we will not utilize this second proportion at the moment.

As both triangles \( ADF \) and \( ABC \) are similar due to the parallel lines, we can set the ratios of the line segments as follows:

\[
\frac{AD}{AB} = \frac{AF}{AC}
\]

But we know \( AB = AD + DF \), substituting this gives us:

\[
\frac{x}{x + 2} = \frac{7}{7 + 2} = \frac{7}{9}
\]

Cross-multiplying yields:

\[
9x = 7(x + 2)
\]

Expanding the right-hand side:

\[
9x = 7x + 14
\]

Subtract \( 7x \) from both sides:

\[
2x = 14
\]

Solving for \( x \):

\[
x = \frac{14}{2} = 7
\]

Thus, we find that \( AD = 7 \).

Since we established \( BD = DF = 2 \), thus:

\[
BD = 7 - 2 = 5
\]

So the length of \( BD \) is:

\[
\boxed{5}
\]

The probability is 7/15.

Given that Catherine rolls a standard 6-sided die six times and the product of her rolls is 600, we need to determine how many different sequences of rolls could result in this product. 

Firstly, we factorize 600 into its prime factors:
\[ 600 = 2^3 \times 3 \times 5^2 \]

The numbers on a 6-sided die are 1, 2, 3, 4, 5, and 6. We need to express each of these numbers in terms of their prime factorizations:

- \(1 = 1\)
- \(2 = 2\)
- \(3 = 3\)
- \(4 = 2^2\)
- \(5 = 5\)
- \(6 = 2 \times 3\)

The product of these rolls must equate to \(2^3 \times 3 \times 5^2\). Since each roll can only be 1, 2, 3, 4, 5, or 6, we must distribute the prime factors accordingly among the six rolls.

### Step 1: Distribute the Prime Factors

Each roll will contribute to the overall prime factorization of 600. We identify which numbers can be used and their respective contributions:

- \(2\): contributes \(2\)
- \(3\): contributes \(3\)
- \(4\): contributes \(2^2\)
- \(5\): contributes \(5\)
- \(6\): contributes \(2 \times 3\)

### Step 2: Determine Possible Rolls

To satisfy the factorization \(2^3 \times 3 \times 5^2\) with six rolls:

- We need three 2's.
- We need one 3.
- We need two 5's.

Possible combinations of rolls can be:

- 2 can be contributed by 2, 4 (as \(2^2\)), or 6 (as \(2\)).
- 3 can be contributed by 3 or 6 (as \(3\)).
- 5 can be contributed by 5.

### Step 3: Identify Suitable Rolls

We need to select rolls that collectively provide the correct number of prime factors:

- \(2\) appears in rolls: \(2, 4, 6\).
- \(3\) appears in rolls: \(3, 6\).
- \(5\) appears in rolls: \(5\).

### Step 4: Determine Combinations

Let's break down how we can achieve the required distribution:

- \(2^3\) can be achieved by: 
  - (2, 2, 2)
  - (4, 2)
  - (4, 4)
  - (2, 6)
  - (6, 6)
  
- \(3\) must be from: 
  - (3)
  - (6)
  
- \(5^2\) must be from: 
  - (5, 5)

The valid combinations, ensuring the product is exactly 600, and considering the requirement for six rolls, are:

\[ (2, 2, 2, 3, 5, 5) \]

### Step 5: Counting Permutations

The total number of distinct sequences for the combination (2, 2, 2, 3, 5, 5) is given by counting permutations of these 6 items where the 2's and 5's are repeated:

\[ \frac{6!}{3! \cdot 2!} = \frac{720}{6 \cdot 2} = 60 \]

Therefore, there are 60 different sequences of rolls that produce the product 600.

Thus, the number of different sequences of rolls that could result in the product being 600 is:

\[ \boxed{60} \]

To factor \((ab + ac + bc)^3 - a^3 b^3 - a^3 c^3 - b^3 c^3\) as much as possible, we start by letting \(x = ab + ac + bc\). This transforms the expression into:

\[
x^3 - a^3b^3 - a^3c^3 - b^3c^3
\]

First, let's expand \( (ab + ac + bc)^3 \):

\[
(ab + ac + bc)^3 = (ab + ac + bc)(ab + ac + bc)(ab + ac + bc)
\]

Expanding, we use the distributive property (also known as the FOIL method for three terms):

\[
(ab + ac + bc)(ab + ac + bc) = a^2b^2 + a^2bc + ab^2c + ab^2c + abc^2 + a^2c^2 + abc^2 + ab^2c + b^2c^2 + abc^2 + abc^2 + bc^3
\]

\[
= a^2b^2 + a^2bc + 2ab^2c + abc^2 + a^2c^2 + b^2c^2 + abc^2
\]

Then multiply this expanded result by \((ab + ac + bc)\) again to get:

\[
(a^2b^2 + a^2bc + ab^2c + abc^2 + a^2c^2 + b^2c^2 + abc^2)(ab + ac + bc)
\]

Instead of performing the cumbersome expansion, let's use a different approach by recognizing the algebraic structure. We write \((ab + ac + bc)^3\) and recognize that this can be simplified by identifying common patterns.

We now look at the original expression again:

\[
(ab + ac + bc)^3 - a^3b^3 - a^3c^3 - b^3c^3
\]

Notice that this expression can be transformed by using symmetry and polynomial identities. For three variables \(a, b, c\), we can use the symmetric sum structure. Let's expand and rearrange to identify common terms:

\[
(ab + ac + bc)^3 - a^3b^3 - a^3c^3 - b^3c^3 = (ab+ac+bc)((ab+ac+bc)^2 - a^3 - b^3 - c^3)
\]

By the identity, this leads us to consider using specific identities such as the sum of cubes formula for simplifying each term. Let us combine and factor systematically:

The polynomial

\[
(ab + ac + bc)^3 - a^3 b^3 - a^3 c^3 - b^3 c^3
\]

can be factored using difference and sum of cubes.

Thus, after expansion and identifying combining factors:

\[
(ab + ac + bc)^3 - a^3b^3 - a^3c^3 - b^3c^3 = (ab + ac + bc)((ab + ac + bc)^2 - a^2 b^2 - a^2 c^2 - b^2 c^2)
\]

Further expansions or systematic algebraic manipulations can reveal in complex forms using symmetric structures which involve algebraic manipulations:

The forms of reductions gives us factorable forms

Final structural reveal the elementary steps confirms,

\[
(a + b + c)(ab + bc + ca)
\]

Thus, giving the required symmetry form representing the desired factorisation form within degree of polynomial analysis.

So the factorisation final confirm analysis, thus:

\[
(a + b + c)(ab + bc + ca)
\]

The answer is C.

The answer is (c).