answer:First, we use the Angle Bisector Theorem to find the ratio of the areas of triangles XYD and XZD: The theorem states that frac{YD}{ZD} = frac{XY}{XZ} = frac{18}{27} = frac{2}{3}. Therefore, the ratio of the area of triangle XYD to the area of triangle XZD is boxed{frac{2}{3}}. Next, to find the length of the altitude from D to YZ, we calculate the area of triangle XYZ using Heron's formula and then use the area to find the altitude. - Step 1: Calculate the semi-perimeter, s = frac{XY + XZ + YZ}{2} = frac{18 + 27 + 30}{2} = 37.5. - Step 2: Use Heron's formula to find the area, A = sqrt{s(s-XY)(s-XZ)(s-YZ)} = sqrt{37.5(37.5-18)(37.5-27)(37.5-30)}. - Step 3: Simplify to find A = sqrt{37.5 cdot 19.5 cdot 10.5 cdot 7.5}. Calculating this gives A approx 243.562 (exact value unnecessary for ratio calculations). - Step 4: To find the altitude h from D to YZ, use the area formula for triangles: A = frac{1}{2} times text{base} times text{height}. Thus, 243.562 = frac{1}{2} times 30 times h Rightarrow h = frac{2 times 243.562}{30}. Calculating the exact value of h: [ h = frac{487.124}{30} approx 16.237 ] So, the length of the altitude from D to YZ is boxed{16.237}.

answer:1. **Initial Setup:** We begin with an 11 times 11 grid, labeled with rows 0, 1, 2, ldots, 10 and columns 0, 1, 2, ldots, 10. The grid is to be filled with integers from 1 to 2^{10} inclusive. We know that the sum of all numbers in row n and in column n must be divisible by 2^n for each n from 0 to 10. 2. **Filling the 10 times 10 Subgrid:** Let's consider filling the 10 times 10 grid that spans rows 1 through 10 and columns 1 through 10. We need to assign an integer from 1 to 2^{10} to each cell in this subgrid. Since there are 100 cells in total: [ text{Number of ways to fill the 10 times 10 subgrid} = (2^{10})^{100} = 2^{1000}. ] 3. **Filling Column 0:** Next, we consider filling column 0. For each row i (i = 0 to 10), the sum of all numbers in row i must be divisible by 2^i. As we already chose numbers for rows 1 to 10 while filling the 10 times 10 subgrid, there’s no constraint on choosing them independently except for the row sums. - In row 10: The number in column 0 can only take 1 possible value since any number is divisible by 2^{10}. - In row 9: The number in column 0 can take 2 possible values based on the sums of the rest of the row being divisible by 2^9. - Progressively, the number in column 0 can take 2^{(10-(i+1))} possible values for row i from 0 to 9. Therefore: [ text{Ways to fill column 0 considering constraints} = 2^1 cdot 2^2 cdot 2^3 cdots 2^{10} = 2^{sum_{i=1}^{10} i} = 2^{55}. ] 4. **Filling Row 0:** Similarly, filling row 0 works under the constraints that the column sums are divisible by 2^i for each column i = 0 to 10. - For column 10: Only one choice. - For column 9: Two choices. - Proceeding in same pattern until column 0: Two choices. Therefore: [ text{Ways to fill row 0 considering constraints} = 2^1 cdot 2^2 cdot 2^3 cdots 2^{10} = 2^{55}. ] 5. **Combining Results:** Combining the independent counts of filling the 10 times 10 subgrid, and then separately filling column 0 and row 0: [ text{Total number of distinct grids} = 2^{1000} cdot 2^{55} cdot 2^{55} = 2^{1000} cdot 2^{110} = 2^{1110}. ] 6. **Conclusion:** Thus, the total number of possible distinct grids is: [ boxed{2^{1110}} ]

answer:We will show that the midpoint of (BC) lies on the circumcircle of (triangle PQR). 1. **Apply Ceva's Theorem and Menelaus' Theorem:** - By Ceva's theorem and Menelaus' theorem, we know that [ frac{AE}{EC} cdot frac{CD}{DB} cdot frac{BF}{FA} = 1 quad text{and} quad frac{AE}{EC} cdot frac{PC}{PB} cdot frac{BF}{FA} = 1 ] - Therefore, by comparing the two expressions, we get [ frac{CD}{DB} = frac{PC}{PB} ] 2. **Define Points and Segments:** - Let (M) be the midpoint of (BC). - Put (PM = x), (DM = y), and let (BM = R) (since (M) is the midpoint, (BM = CM = R)). - Then, the segment lengths can be expressed as: [ PB = x - R, quad PC = x + R, quad DC = y + R, quad BD = R - y ] 3. **Use the Ratio from Step 1:** - From the ratio (frac{CD}{DB} = frac{PC}{PB}), we have: [ frac{y + R}{R - y} = frac{x + R}{x - R} ] - Cross-multiplying yields: [ (y + R)(x - R) = (x + R)(R - y) ] - Expanding and simplifying: [ yx - yR + Rx - R^2 = xR - yx + R^2 - yR ] - Combining like terms: [ 2yx = 2R^2 ] - Hence, we obtain: [ yx = R^2 ] 4. **Prove Cyclic Quadrilateral (RBQC):** - Note that (F, E) are points where altitudes intersect (AC) and (AB). Since (angle BFE = angle BCE): [ angle BFC = angle BEC ] - This implies that (B, F, E, C) are concyclic. - Since (QR) is parallel to (EF), the angles satisfy: [ angle AEF = angle AQR ] - Therefore, the angles (angle RBC) and (angle RQC) satisfy: [ angle RBC = 180^circ - angle B = 180^circ - angle AQR = angle RQC ] indicating that (R, B, Q, C) are concyclic. 5. **Apply the Power of a Point Theorem on (D) and (M):** - Since (RBQC) is cyclic, we use the power of a point: [ BD cdot DC = RD cdot DQ ] - From step 3, we have shown: [ PD cdot DM = BD cdot DC ] - Therefore: [ RD cdot DQ = PD cdot DM ] - Since (RDQ) is cyclic ((RBQC)), (M), the midpoint of (BC), lies on the circumcircle of (triangle PQR) by the above relation. Conclusion: [ boxed{text{The midpoint of (BC) lies on the circumcircle of ( triangle PQR )}} ]

answer:From the given information, we have: S_n = 4n^2 - n + 2 To find the general term a_n, we need to express a_n in terms of S_n and S_{n-1}. For n=1, we calculate the first term directly from S_1: a_1 = S_1 = 4(1)^2 - 1 + 2 = 5 For ngeq 2, we find a_n by taking the difference between the sum of the first n terms and the sum of the first n-1 terms: begin{align*} a_n &= S_n - S_{n-1} &= left[ 4n^2 - n + 2 right] - left[ 4(n-1)^2 - (n-1) + 2 right] &= 4n^2 - n + 2 - left[ 4(n^2 - 2n + 1) - n + 1 + 2 right] &= 4n^2 - n + 2 - (4n^2 - 8n + 4 - n + 1 + 2) &= 4n^2 - n + 2 - 4n^2 + 8n - 7 &= 8n - 5 end{align*} So the general term a_n is 5 when n = 1 and 8n - 5 for n geq 2. Thus, the sequence can be defined as: a_n = begin{cases} 5, & n=1 8n-5, & ngeq 2 end{cases} And the sequence is an element of the natural numbers excluding zero, i.e., nin mathbb{N}^*. The correct option is therefore [ boxed{B: a_n = begin{cases} 5, & n=1 8n-5, & ngeq 2 end{cases} } ]