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We can divide this problem into two cases:

Case 1: Each row has exactly one child from each family.

Choose one child from each family for the first row: there are 3 choices

.

Arrange the remaining two children in the first row (siblings can't be together): there are 2!=2 ways.

Arrange the second row similarly: 3⋅2=6​ ways total.

Case 2: One row has two children from the same family.

There are two subcases depending on the arrangement of siblings in the rows:

Subcase 2a: The first child in each row is from the same family.

Choose one pair of siblings: 3 choices

.

Arrange the siblings within the pair: 2 ways.

Arrange the remaining 4 children (2 from another pair and 2 from the third pair) in the second row: 4! ways.

Overcount: we've counted the arrangement as if sibling order matters within a pair (which it doesn't) twice (once for each sibling in the first row). So, we divide by 2!⋅2!.

Total: 2!⋅2!3⋅2⋅4!​=36 ways.

Subcase 2b: The first child in the first row is NOT a sibling of the first child in the second row.

Choose one pair to have their children occupy the third seats in each row: 3 choices.

Arrange the remaining 4 children in the first row: 4! ways.

Overcount: similar to subcase 2a, we divide by 2!⋅2! to account for sibling order not mattering.

Total: 2!⋅2!3⋅4!​=36 ways.

Since Cases 1 and 2 are mutually exclusive, to find the total number of arrangements, we simply add the number of arrangements from each case:

6 + 36 + 36 = 78​.

This gives us a final answer of 78.

We can solve this problem by considering the empty spaces between the circles. There are 11 spaces between the circles, since there are 12 circles forming a circle.

Since no two chosen circles can be next to each other, the chosen circles must correspond to every other empty space. There are C(11,3) = 165​ ways to choose 3 empty spaces out of 11.

Problem 2)

Since BD=AD=AC then triangle ABC is isosceles with AB=AC. Therefore, ∠ABC=∠ACB=x (for some angle measure x ).

Since D is on side BC so BD=AD, then line AD is an angle bisector of ∠A. Therefore, ∠BAD=∠CAD=2∠A​.

Since the angles in a triangle sum to 180∘, then in triangle ABC, ∠A+∠B+∠C=180∘. Substituting x for ∠B and ∠C, we get ∠A+x+x=180∘.

Solving for ∠A, we get ∠A=180∘−2x.

We are also given that ∠BAD=∠CAD=2∠A​. Substituting the expression for ∠A that we just found, we get 2180∘−2x​=∠BAD=∠CAD.

Because the sum of the angles in a triangle is 180∘, then in triangle ABD, ∠BAD+∠ABD+∠ADB=180∘.

Substituting 78∘ for ∠ABD (since BD=AD=AC and triangle ABC is isosceles), we get 2180∘−2x​+78∘+90∘=180∘.

Solving for x, we get x=31∘​.

Therefore, angle ACB = 31 degrees.

Problem 1)

Since ∠ABC=78∘ and ∠ACB=47∘, then ∠BAC=180∘−78∘−47∘=55∘.

Let D be the foot of the altitude from B to AC. Then ∠ABD=∠ABC=78∘.

Triangle ABD is a 30-60-90 triangle, so ∠ADB=60∘ and ∠BAD=30∘.

Since I is the incenter, ∠BAI=∠CAI=∠BAC/2=27.5∘. Then ∠ABI=∠BAI+∠BAD=57.5∘.

Triangle ABI is a 30-57.5-92.5$ triangle, so ∠BID = ∠ABI/2 = 46.25∘​.

Since a rotation about point O maps A to B, B to C, and C to D, we know the following:

Angle AOD: This is given as 24 degrees.

Full Rotation: A full rotation around a point is 360 degrees.

Possible Rotations:

There are three possibilities for the rotation that maps A to B, B to C, and C to D, resulting in three possible measures for angle AOB:

Case 1: Single Rotation of 24 Degrees

In this case, the rotation about O maps A to B by 24 degrees clockwise, B to C by 24 degrees clockwise, and C to D by 24 degrees

clockwise, resulting in a total rotation of A to D of 24 + 24 + 24 = 72 degrees.

Since the full rotation is 360 degrees, the remaining angle for AOB must be 360 - 72 = 288 degrees.

However, angle measures are typically represented between 0 and 180 degrees. We can achieve this by subtracting a multiple of 180 (full rotations) from 288:

AOB = 288° - 180 = 108.

Case 2: Double Rotation (360 + 24 Degrees)

Here, the rotation about O maps A to B by a full rotation (360 degrees) followed by a 24-degree clockwise rotation. This effectively brings B back to its original position and then continues the rotation to C and D.

The total rotation for AOD in this case becomes 360 + 24 + 24 = 408 degrees.

Similar to case 1, we can adjust this to fit the 0-180 degree range:

AOB = 408° - 360° = 48°

Case 3: Triple Rotation (2 * 360 + 24 Degrees)

This case involves two full rotations followed by a 24-degree clockwise rotation from A to B. Again, the first two rotations effectively bring B back to its original position.

The total rotation for AOD becomes 2 * 360 + 24 = 744 degrees.

Adjusting for the 0-180 degree range:

AOB = 744° - 2 * 360° = 124°

Summary:

Therefore, the three possible degree measures for angle AOB, considering rotations between 0 and 180 degrees, are:

108° (Case 1)

48° (Case 2)

124° (Case 3)

Understanding the Transformations:

Tasha's Transformation: Moves the point sqrt(2) units to the right.

Richard's Transformation: Rotates the point clockwise about point O by 90 degrees.

Round 1:

A' is sqrt(2) units to the right of A (Tasha's translation).

A" is obtained by rotating A' by 90 degrees clockwise around O (Richard's rotation).

Round 2:

A1 is obtained by rotating A by 90 degrees clockwise around O (Richard's rotation).

A2 is sqrt(2) units to the right of A1 (Tasha's translation).

Finding A"A2:

To find the distance between A" and A2, we can consider the following:

A" and A1 are the same distance away from O (since both are obtained by a 90-degree rotation from A and they share the same original distance from O).

A'A2 forms a right triangle with A"A1 as the hypotenuse (due to the 90-degree rotations).

Pythagorean Theorem:

We can use the Pythagorean theorem to find A"A2:

a^2 + b^2 = c^2

where:

a = A'A2 (we know this is sqrt(2) from Tasha's translation)

b = A"A1 (distance between A" and A1, which is the same as the distance between A and O, which is sqrt(3))

c = A"A2 (what we want to find)

Substituting Values:

(sqrt(2))^2 + (sqrt(3))^2 = (A"A2)^2

2 + 3 = (A"A2)^2

5 = (A"A2)^2

Taking the Square Root (consider both positive and negative):

A"A2 = ±√5

Since distance cannot be negative, we take the positive square root:

A"A2 = √5

Therefore, the distance between A" and A2 is √5 units.

This is a series where the numerator follows the Fibonacci sequence and the denominator is a geometric sequence with common ratio 1/10. To find the sum of such a series, we can use a technique involving manipulation and summation of geometric series.

Here's how to solve it:

Step 1: Splitting the Series

We can represent the series as the sum of two series:

S = (1/10^2 + 1/10^3 + 1/10^4 + ...) + (2/10^4 + 3/10^5 + 5/10^6 + ...)

Step 2: Recognizing Geometric Series

The first series is a geometric series with first term (a = 1/10^2) and common ratio (r = 1/10). The second series is another geometric series with first term (a = 2/10^4) and common ratio (r = 1/10).

Step 3: Find the Sum of Each Series

The formula for the sum (Sn) of a finite geometric series is:

Sn = a(1 - r^n) / (1 - r)

where:

a is the first term

r is the common ratio

n is the number of terms

Step 4: Apply the Formula to Each Series

First Series:

S1 = (1/10^2) * (1 - (1/10)^n) / (1 - 1/10)

Second Series:

S2 = (2/10^4) * (1 - (1/10)^n) / (1 - 1/10)

Step 5: Note on Infinite Series

Since we're dealing with an infinite series (n tends to infinity), both (1/10)^n terms in the numerators approach zero. Therefore, we can simplify the expressions:

S1 ≈ (1/10^2) * (1 - 0) / (1 - 1/10) = 1/100

S2 ≈ (2/10^4) * (1 - 0) / (1 - 1/10) = 1/50

Step 6: Sum the Simplified Series

The total sum (S) of the original series is the sum of S1 and S2:

S = S1 + S2 ≈ 1/100 + 1/50 = 3/100

Answer:

Therefore, the sum of the series is 3/100.

The answer is 1 + 2 + 3 + 4 = 10.

The strategy here is to calculate the probability of the event's complement (i.e., the probability that the coin is tossed ten times or fewer) and subtract this from 1.

Favorable Outcomes:

There are two favorable outcomes:

HHH appears in exactly 10 flips: There are 3 choices for the position of the first heads (leaving 9 flips remaining), then 2 choices for the second heads (leaving 8 flips remaining), and 1 choice for the third heads (leaving 7 flips remaining).

So, there are 3⋅2⋅1=6 successful outcomes where HHH appears in exactly 10 flips.

TTT appears in exactly 10 flips: This follows the same logic as scenario 1, so there are also 6 successful outcomes.

Total Favorable Outcomes:

There are a total of 6 + 6 = 12 successful outcomes where the coin is tossed exactly 10 times.

Total Outcomes:

Since the coin is fair, there are 210=1024 total possible outcomes for 10 flips (heads or tails for each flip).

Probability of Favorable Outcomes:

The probability of needing exactly 10 flips is the number of successful outcomes divided by the total number of outcomes:

P(exactly 10 flips) = 12 / 1024

Complement's Probability:

We want the probability of needing more than 10 flips. This is the complement of the event where the coin is tossed ten times or fewer.

P(more than 10 flips) = 1 - P(exactly 10 flips)

Final Answer:

Substitute the probability of needing exactly 10 flips:

P(more than 10 flips) = 1 - (12 / 1024)

Simplify:

P(more than 10 flips) = (1024 - 12) / 1024 = 1012 / 1024

Both the numerator and denominator have a common divisor of 4, so we can simplify:

P(more than 10 flips) = 253 / 256

Therefore, the probability of needing more than 10 flips is 253/256.

Let's denote the common difference of the arithmetic sequence by \(d\).

For the sum of an arithmetic series, we have the formula:

\[ S = \frac{n}{2} \cdot (a_1 + a_n) \]

Given that \(a_1 + a_3 + a_5 + a_7 + a_9 + a_{11} = -2\), we can write the sum as:

\[ S_1 = \frac{6}{2} \cdot (a_1 + a_{11}) = 3 \cdot (a_1 + (a_1 + 10d)) = 3 \cdot (2a_1 + 10d) \]

\[ S_1 = 6a_1 + 30d \]

Similarly, for \(a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} = 1\), we get:

\[ S_2 = \frac{6}{2} \cdot (a_2 + a_{12}) = 3 \cdot (a_2 + (a_2 + 11d)) = 3 \cdot (2a_2 + 11d) \]

\[ S_2 = 6a_2 + 33d \]

Given that \(a_1 + a_3 + a_5 + a_7 + a_9 + a_{11} = -2\) and \(a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} = 1\), we can write the following system of equations:

\[ 6a_1 + 30d = -2 \]

\[ 6a_2 + 33d = 1 \]

Now, let's solve this system of equations. Subtracting the first equation from the second, we get:

\[ 6a_2 + 33d - (6a_1 + 30d) = 1 - (-2) \]

\[ 6a_2 - 6a_1 + 3d = 3 \]

\[ 6(a_2 - a_1) + 3d = 3 \]

\[ 6d + 3d = 3 \]

\[ 9d = 3 \]

\[ d = \frac{1}{3} \]

Now, substitute \(d = \frac{1}{3}\) into one of the original equations to find \(a_1\).

Let's use the first equation:

\[ 6a_1 + 30 \left(\frac{1}{3}\right) = -2 \]

\[ 6a_1 + 10 = -2 \]

\[ 6a_1 = -12 \]

\[ a_1 = \boxed{-2} \]

So, \(a_1 = -2\).

Diagonal across the base: This can be calculated using the Pythagorean theorem on the right triangle formed by half the width, half the length, and the diagonal across the base.

Applying the formula, we get 25​2+23​2=d12​, where d1​ is the diagonal across the base. Solving for d1​, we get d1​=38​​.

There are two cases to consider:

Case 1: C is on the horizontal line through A or the vertical line through B.

Without loss of generality, suppose C is on the horizontal line through A. If AB is a horizontal segment, then △ABC cannot be isosceles.

Otherwise, AC=AB if and only if C is one of the two points on this horizontal line that are the same distance from A as B is.

The same argument applies if C is on the vertical line through B. Thus, there are at most 2+2=4 such points C.

Case 2: C is not on the horizontal line through A or the vertical line through B.

Then △ABC is isosceles if and only if C lies on the perpendicular bisector of AB. There are 4 such points.

Hence, there are 4+4=8​ such points C.

Combining like terms, we have the inequality 2x2+nx−135<0. We can use the quadratic formula to solve for the roots of 2x2+nx−135=0.

The quadratic formula gives us:

x=2⋅2−n±n2−4⋅2⋅−135​​

x=4−n±n2+1080​​

Since we are given that the inequality has no real solutions, the discriminant n2+1080 must be negative. This gives us the inequality n2<−1080.

Since n is an integer, the only possible values of n are −34, −33, −32, ..., 32, 33, and 34. Therefore, there are 67​ possible values of n.