Question

a choir director is planning how she will arrange the choir members for an upcoming performance. she arranges them in rows that form a triangle and alternates the shirt color the members will wear between red and blue, to maximize sound and visual effect. a red shirt will always be in the front row. design 1 design 2 design 3 the choir director has 21 choir members to arrange. what color shirt will be in the back row? how many rows will there be? select the correct answer from each dropdown menu. shirts will be in the back row. there will be total rows. the director discovers she only has 20 blue shirts. what is the greatest arrangement design the director

Answer:

To solve the problem, we analyze the designs:

Step 1: Identify the pattern

• Design 1: Front row (red) = 1, Back row (blue) = 0? Wait, no—looking at the images:
• Design 2: Front (red) = 1, Back (blue) = 2
• Design 3: Front (red) = 2, Back (blue) = 3? Wait, no, let’s count total members:

Design 2: 1 (red) + 2 (blue) = 3
Design 3: 3 (red) + 3 (blue)? No, image shows 3 red and 2 blue? Wait, maybe the pattern is triangular numbers (sum of first \( n \) integers: \( S = \frac{n(n+1)}{2} \)).

Wait, the choir has 21 members. Let’s find \( n \) where \( \frac{n(n+1)}{2} = 21 \). Solve \( n^2 + n - 42 = 0 \). Factoring: \( (n+7)(n-6)=0 \), so \( n=6 \). Wait, no—wait, maybe the front row is red, and each row increases by 1. Wait, let's re-express:

Wait, the problem says "triangle" arrangement, so total members \( T = \frac{r(r+1)}{2} \), where \( r \) is number of rows. Wait, no—maybe front row (red) has 1, next (blue) has 2, next (red) has 3, etc., alternating. But the key is the back row (last row) color and number of shirts.

Wait, total members = 21. Let's find the number of rows \( r \) such that the sum of the first \( r \) integers is 21? No, \( \frac{r(r+1)}{2} = 21 \) → \( r^2 + r - 42 = 0 \) → \( r = \frac{-1 \pm \sqrt{1 + 168}}{2} = \frac{-1 \pm 13}{2} \), so \( r=6 \) (since \( r>0 \)). Wait, \( 6 \times 7 / 2 = 21 \). So 6 rows? Wait, no—maybe the front row (red) is row 1, then blue row 2, red row 3, etc. Wait, the problem says "a red shirt will always be in the front row" (so front row is red). Let's list rows:

• Row 1 (front, red): 1 shirt
• Row 2 (blue): 2 shirts
• Row 3 (red): 3 shirts
• Row 4 (blue): 4 shirts
• Row 5 (red): 5 shirts
• Row 6 (blue): 6 shirts

Wait, sum: \( 1 + 2 + 3 + 4 + 5 + 6 = 21 \). Yes! So total 6 rows. Now, the back row is row 6 (last row), which is blue (since row 2,4,6 are blue). Number of shirts in back row (row 6) is 6.

Now, for the second part: director has 20 blue shirts. Let's find the maximum \( r \) where blue shirts used \( \leq 20 \). Blue rows are row 2,4,6,... (even rows). Let's denote number of blue rows as \( k \). Each blue row \( i \) (even) has \( i \) shirts. So sum of blue shirts: \( 2 + 4 + 6 + \dots + 2k = 2(1 + 2 + 3 + \dots + k) = 2 \times \frac{k(k+1)}{2} = k(k+1) \). We need \( k(k+1) \leq 20 \).

Test \( k=4 \): \( 4 \times 5 = 20 \). Perfect! So blue rows: 4 rows (row 2,4,6,8? Wait no—wait, first blue row is row 2 (2 shirts), row 4 (4), row 6 (6), row 8 (8)? Wait no, sum \( 2 + 4 + 6 + 8 = 20 \). Then total rows: red rows (row 1,3,5,7) and blue rows (row 2,4,6,8). Wait, red rows: row 1 (1), row 3 (3), row 5 (5), row 7 (7). Sum of red shirts: \( 1 + 3 + 5 + 7 = 16 \). Total members: \( 20 + 16 = 36 \)? Wait, no—wait, the pattern is alternating red (odd rows) and blue (even rows), with row \( n \) having \( n \) shirts. So total members for \( r \) rows: \( \frac{r(r+1)}{2} \). But we need blue shirts \( \leq 20 \). Blue shirts are sum of even rows: \( 2 + 4 + 6 + \dots + 2m = m(m+1) \) (where \( 2m \) is the last even row). So \( m(m+1) \leq 20 \). \( m=4 \): \( 4 \times 5 = 20 \), so last even row is 8 (row 8, 8 shirts). Then total rows: 8 (since row 8 is the last row). Let's check total members: \( \frac{8 \times 9}{2} = 36 \). Red shirts: sum of odd rows (1,3,5,7) = \( 1 + 3 + 5 + 7 = 16 \). Blue shirts: 20, which matches. So the greatest arrangement design (maximum rows) is when blue shirts are 20, so total rows = 8? Wait, no—wait, the first part: back row color is blue, number of shirts is 6 (for 21 members). For the second part, with 20 blue shirts, the maximum \( m \) where \( m(m+1) = 20 \) is \( m=4 \) (since \( 4 \times 5 = 20 \)), so blue rows are 4 (rows 2,4,6,8), each with 2,4,6,8 shirts. Total blue shirts: 20. Red rows: 4 (rows 1,3,5,7), each with 1,3,5,7 shirts. Total members: 16 + 20 = 36. So total rows: 8.

Final Answers

1. Back row color: Blue; Back row shirts: 6.
2. Greatest arrangement: Total rows = 8 (with blue shirts 20, red shirts 16, total 36 members, arranged in 8 rows: 1,2,3,4,5,6,7,8 shirts, alternating red, blue, red, blue, etc.).

(Note: The first part’s back row for 21 members: row 6 (blue), 6 shirts. The second part: maximum rows with 20 blue shirts is 8 rows, as shown.)