Question

Evaluate \((1 / 4) \sum_{k=1} k(6-k)\) and deduce the number of points of intersection for the diagonals of an octagon if no three diagonals meet at a point.

Short answer

The expression evaluates to 8.75, and the octagon's diagonals intersect at 70 points.

Step-by-step solution

Understanding the Summation

The expression \((1 / 4) \sum_{k=1} k(6-k)\) is a summation that needs to be evaluated. The value 'k' in the summation ranges from 1 to 6. The expression inside the summation is \(k(6-k)\), which needs to be calculated first for each value of 'k'. After that, we can evaluate the full summation.

Expanding the Expression

First, expand the expression for each value of 'k' from 1 to 6. - When \(k = 1\), the expression is \(1(6-1) = 5\).- When \(k = 2\), the expression is \(2(6-2) = 8\).- When \(k = 3\), the expression is \(3(6-3) = 9\).- When \(k = 4\), the expression is \(4(6-4) = 8\).- When \(k = 5\), the expression is \(5(6-5) = 5\).- When \(k = 6\), the expression is \(6(6-6) = 0\).

Calculating the Sum

Add up all the calculated values from the expressions: \(5 + 8 + 9 + 8 + 5 + 0 = 35\). Hence, the sum \(\sum_{k=1} k(6-k) = 35\).

Evaluating the Full Expression

Now, multiply the sum by \(1/4\): \((1/4) \times 35 = 8.75\). This is the result of the first part of the exercise.

Understanding Octagon Diagonals

An octagon has 8 vertices. The formula to calculate the number of diagonals in a polygon is \(\frac{n(n-3)}{2}\), where \(n\) is the number of sides. For an octagon, \(n = 8\).

Calculating the Total Number of Diagonals

Plug \(n = 8\) into the diagonal formula: \(\frac{8(8-3)}{2} = \frac{8 \times 5}{2} = 20\) diagonals.

Calculating Diagonal Intersection Points

Using the combination formula \(\binom{n}{4}\) to find the intersection points for the diagonals: \(\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70\). Hence, there are 70 unique intersection points where the diagonals of the octagon meet.

Key concepts

Summation Evaluation

When dealing with summations, it's important to break down the expression step by step. In this exercise, we evaluated the summation \( (1 / 4) \sum_{k=1}^6 k(6-k) \).

First, we defined the range for \(k\), which is from 1 to 6. The expression \( k(6-k) \) was calculated for every integer within this range:

• For \( k = 1 \), we calculated \( 1(6-1) = 5 \).
• For \( k = 2 \), it's \( 2(6-2) = 8 \).
• Continue similarly for \(k = 3, 4, \,\text{and}\, 5\).
• When \( k = 6 \), the result is zero since \( 6 - 6 = 0 \).

Next, sum these results to obtain \( 5 + 8 + 9 + 8 + 5 + 0 = 35 \). The final step is to multiply this sum by \( 1/4 \), as noted: \( (1/4) \times 35 = 8.75 \).

Summation evaluation allows for complex series to be simplified, which is a powerful tool in mathematics.

Polygon Diagonal Formula

Diagonals are line segments that connect two non-adjacent vertices in a polygon. To find the total number of diagonals in a polygon, we use the formula: \( \frac{n(n-3)}{2} \), where \(n\) represents the number of sides.

In this case, for an octagon \( (n = 8) \):

• Calculate \( 8(n-3) \): \( 8 \times (8-3) = 40 \).
• Divide this by 2: \( \frac{40}{2} = 20 \).

Thus, an octagon has 20 diagonals.

Understanding this formula is essential for determining the structure of more complex polygons, and it provides a clear way to count and visualize the internal connections of regular and irregular polygons.

Intersection Points Calculation

To find where the diagonals of an octagon intersect within the shape, we use combinations. The key here is to determine how many combinations of diagonals can intersect simultaneously at a point.

For an octagon, this is computed by selecting four vertices out of the eight, as four vertices determine a unique intersection point among the diagonals. This is done using the combination formula \( \binom{n}{4} \):

• Calculate \( \binom{8}{4} \): \( \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \).

This tells us that there are 70 distinct points where the diagonals of an octagon intersect.

Combinatorial methods such as this are important in geometrical problems, helping not only in counting solutions but also in understanding spatial relationships.