tomtom

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 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.