answer:We are given an isosceles triangle with a base of 48 units and two equal sides of 30 units each. Our goal is to find the radius of the inscribed circle ( r ), the radius of the circumscribed circle ( R ), and the distance between their centers. 1. **Find the height of the triangle**: Since the triangle is isosceles, we can drop a perpendicular from the vertex angle to the midpoint of the base. This perpendicular bisects the base into two equal segments, each of length: [ frac{48}{2} = 24 text{ units} ] The height ( h ) of the triangle can then be found using the Pythagorean theorem in one of the right triangles formed: [ h^2 + 24^2 = 30^2 ] Substituting the given values: [ h^2 = 30^2 - 24^2 = 900 - 576 = 324 ] Taking the square root of both sides: [ h = sqrt{324} = 18 text{ units} ] 2. **Radius ( r ) of the inscribed circle**: The radius ( r ) of the inscribed circle can be calculated using the formula for the area of the triangle and its semiperimeter. The area ( A ) of the triangle is: [ A = frac{1}{2} times text{base} times text{height} = frac{1}{2} times 48 times 18 = 432 text{ square units} ] The semiperimeter ( s ) of the triangle is: [ s = frac{30 + 30 + 48}{2} = 54 text{ units} ] The radius ( r ) of the inscribed circle is given by: [ r = frac{A}{s} = frac{432}{54} = 8 text{ units} ] 3. **Radius ( R ) of the circumscribed circle**: The radius ( R ) of the circumscribed circle for an isosceles triangle can be calculated using the formula: [ R = frac{a}{2 sin C} ] where ( a ) is the length of one of the equal sides and ( C ) is the angle at the base. In an isosceles triangle, the height divides the base into two equal parts and the triangle into two right triangles, making ( sin C ) as: [ sin C = frac{text{height}}{text{side}} = frac{18}{30} = frac{3}{5} ] Substituting the value into the formula: [ R = frac{30}{2 times frac{3}{5}} = frac{30 times 5}{6} = 25 text{ units} ] 4. **Calculate the distance between the centers of the circles**: The centers of the inscribed circle and the circumscribed circle are known as the incenter and circumcenter. Since the circumcenter is outside the isosceles triangle and located along the perpendicular bisector of the base, the distance ( d ) between the centers can be calculated as follows: [ d = R - (h - r) = 25 - (18 - 8) = 25 - 10 = 15 text{ units} ] # Conclusion The radius of the inscribed circle is ( 8 ) units, the radius of the circumscribed circle is ( 25 ) units, and the distance between the centers of the inscribed and circumscribed circles is ( 15 ) units. [ boxed{8 , 25 , 15} ]

answer:1. **Formulate the equation of the tangent line**: We start by determining the tangent line to the function y = 5 + sqrt{4 - x} at the point with the abscissa x_0. 2. **Find the derivative**: The first step involves finding the derivative of y with respect to x: [ y' = frac{d}{dx}(5 + sqrt{4 - x}) = -frac{1}{2 sqrt{4 - x}} ] 3. **Equation of the tangent line**: Using the point-slope form of the equation of a line, the tangent at (x_0, y(x_0)) is given by: [ y_{text{kac}} = y'(x_0) (x - x_0) + y(x_0) ] Substituting y'(x_0) and y(x_0): [ y_{text{kac}} = -frac{1}{2 sqrt{4 - x_0}} (x - x_0) + 5 + sqrt{4 - x_0} ] Simplify to: [ y_{text{kac}} = -frac{1}{2 sqrt{4 - x_0}} x + frac{x_0}{2 sqrt{4 - x_0}} + 5 + sqrt{4 - x_0} ] 4. **Determine the figure**: The region of interest is bounded vertically by x = -26 and x = 2, below by the x-axis, and above by this tangent line. As a result, the shape formed is a trapezoid. 5. **Calculate the heights at x = -26 and x = 2**: Evaluate the tangent line at these points. [ y_{text{kac}}(-26) = -frac{1}{2 sqrt{4 - x_0}}(-26 - x_0) + 5 + sqrt{4 - x_0} = frac{26 + x_0}{2 sqrt{4 - x_0}} + 5 + sqrt{4 - x_0} ] [ y_{text{kac}}(2) = -frac{1}{2 sqrt{4 - x_0}}(2 - x_0) + 5 + sqrt{4 - x_0} = -frac{2 - x_0}{2 sqrt{4 - x_0}} + 5 + sqrt{4 - x_0} ] 6. **Calculate the sum of heights**: [ text{Sum of heights} = y_{text{kac}}(-26) + y_{text{kac}}(2) ] [ = left(frac{26 + x_0}{2 sqrt{4 - x_0}} + 5 + sqrt{4 - x_0}right) + left(-frac{2 - x_0}{2 sqrt{4 - x_0}} + 5 + sqrt{4 - x_0}right) ] [ = frac{26 + x_0 - (2 - x_0)}{2 sqrt{4 - x_0}} + 10 + 2sqrt{4 - x_0} ] Simplifying: [ = frac{24 + 2 x_0}{2 sqrt{4 - x_0}} + 10 + 2sqrt{4 - x_0} ] [ = frac{12 + x_0}{sqrt{4 - x_0}} + 10 + 2sqrt{4 - x_0} ] 7. **Evaluate the area of the trapezoid**: Using the formula for the area of a trapezoid: [ A = frac{1}{2} times text{height} times (text{base}_1 + text{base}_2) ] Here, the height is 28: [ A = 14 left(frac{12 + x_0}{sqrt{4 - x_0}} + 10 + 2sqrt{4 - x_0}right) ] 8. **Finding the minimum area**: Differentiate A with respect to x_0 and set the derivative equal to 0 to find the critical points: [ A = 14 left(frac{12 + x_0}{sqrt{4 - x_0}} + 10 + 2sqrt{4 - x_0}right) ] Differentiate: [ A' = 14 left(frac{d}{dx} left(frac{12 + x_0}{sqrt{4 - x_0}} + 10 + 2sqrt{4 - x_0}right)right) ] [ = 14 left(frac{2(4 - x_0) - (12 + x_0) cdot (-1/2)}{(4 - x_0)^{3/2}}right) ] Set A' = 0 to find x_0. After calculations, we find x_0 = -12. 9. **Substitute back the value of x_0 = -12** to find the minimum area: [ A_{text{min}} = 14 left(frac{12 - 12}{sqrt{16}} + 10 + 2sqrt{16 + 12}right) = 14 left(0 + 10 + 2 sqrt{20}right) = 14 times 36 = 504 ] # Conclusion: [ boxed{504} ]

answer:To solve the problem, we need to determine the minimum number of colors ( k ) required to color an infinite planar grid such that no ( n )-gun can cover two squares of the same color. An ( n )-gun is a ( 2n )-mino formed by placing a ( 1 times n ) grid and an ( n times 1 ) grid side by side, sharing a corner. 1. **Understanding the ( n )-gun:** - An ( n )-gun consists of ( 2n ) unit squares. - It can be visualized as an L-shaped figure with ( n ) squares in one direction and ( n ) squares in the perpendicular direction. 2. **Constructing the Coloring Pattern:** - Consider a ((n+1) times (n+1)) square with one corner cut off. This shape will be referred to as a "shield." - The shield contains ((n+1)^2 - 1) squares. 3. **Coloring the Shield:** - We need to color the shield such that no two squares within the shield that can be covered by an ( n )-gun share the same color. - Since any two squares in the shield can be covered by an ( n )-gun, each square must have a unique color. - Therefore, the shield requires ((n+1)^2 - 1) distinct colors. 4. **Tiling the Plane:** - The infinite plane can be tiled with these shields. - Each shield will be colored using the same ((n+1)^2 - 1) colors. - Due to the symmetry and the structure of the shields, no ( n )-gun can cover two squares of the same color. 5. **Proof of Minimum Colors:** - Assume there are fewer than ((n+1)^2 - 1) colors. - Consider the lower-right corner of the shield, which is friendly with all other squares in the shield except the upper-left corner. - If there are fewer colors, the lower-right corner must share a color with the upper-left corner, leading to a contradiction. - Therefore, at least ((n+1)^2 - 1) colors are necessary. 6. **Conclusion:** - The minimum number of colors required to color the infinite planar grid such that no ( n )-gun covers two squares of the same color is ( n(n+2) ). The final answer is ( boxed{n(n+2)} ).

answer:Part (a): 1. **Understanding the Problem**: We have a sequence of 2015 digits where each digit is randomly selected from 0 to 9 and is independent of the others. We perform an operation where if multiple identical digits appear consecutively, they are replaced by a single instance of that digit. We need to calculate the probability that the sequence shortens exactly by one digit. 2. **Identifying the Process**: The first digit remains unchanged, and the following digits each represent a Bernoulli trial with two possible outcomes: - With probability (0.1) the digit disappears (i.e., it is identical to the preceding digit, resulting in a reduction). - With probability (0.9) the digit remains unchanged (i.e., it is different from the preceding digit). 3. **Calculating the Probability**: We want the sequence to shorten exactly by one digit, which means exactly one Bernoulli trial out of 2014 must result in a success (i.e., a digit removal). This is a binomial trial scenario with the number of trials (n = 2014), the number of successes (k = 1), and the probability of success (p = 0.1). The binomial probability formula is: [ P(X = k) = binom{n}{k} p^k (1-p)^{n-k} ] Substituting the values: [ P(X = 1) = binom{2014}{1} (0.1)^1 (0.9)^{2014-1} ] 4. **Calculations**: [ P(X = 1) = 2014 cdot 0.1 cdot 0.9^{2013} ] Using a calculator for the approximation of (0.9^{2013}): [ 0.9^{2013} approx 1.564 times 10^{-90} ] 5. **Combining the Results**: [ P(X = 1) approx 2014 cdot 0.1 cdot 1.564 times 10^{-90} = 201.4 times 1.564 times 10^{-90} approx 1.564 times 10^{-90} ] Therefore, the probability that the sequence shortens by exactly one digit is approximately: [ boxed{1.564 times 10^{-90}} ] Part (b): 1. **Understanding the Problem**: We need to find the expected length of the new sequence after performing the shortening operation. 2. **Expectation of Success**: The number of successes (digits removed) follows a binomial distribution with parameters (n = 2014) and (p = 0.1). The expected number of digits removed (expected number of successes) (E[X]) in a binomial distribution is given by (np): [ E[X] = 2014 times 0.1 = 201.4 ] 3. **Calculating the Expected Length**: Initially, the sequence has 2015 digits. After removing the expected number of 201.4 digits, the expected length becomes: [ text{Expected length} = 2015 - 201.4 = 1813.6 ] 4. **Conclusion**: [ boxed{1813.6} ] Thus, the expected length of the new sequence is ( boxed{1813.6} ).