answer:1. **Identify Key Points and Angles:** - Let the center of the circumcircle of the triangle ABC be O. - Denote the foot of the altitude from A to side BC as P. - Define F as the midpoint of side BC. 2. **Angle Analysis at the Midpoint `F`:** - The angle angle FOC is half of the central angle angle CAB denoted as alpha. Therefore, [ angle FOC = alpha. ] - Since triangle ABC is acute, the angles involving circles and perpendiculars can be simplified. 3. **Relating Angle angle OCP:** - Given that angle OCB = 90^circ - alpha, we have: [ angle OCP = 90^circ - alpha. ] - Our goal is to show: [ angle POC < angle OCP Longrightarrow CP < OP. ] 4. **Reflecting Point A:** - Reflect A over the perpendicular bisector of BC to get point A'. - Since A' lies on the circumcircle and A'CB = ABC, by given condition: [ angle ACA' geq 30^circ. ] - The chord cut by angle ACA' has length at least r, where r is the radius of the circumcircle. 5. **Length Calculation:** - The segment AA', being the chord cut by an angle at least 30^circ, has a length [ AA' geq r. ] - The height split into two equal parts, thus FP = frac{AA'}{2} geq frac{r}{2}. 6. **Comparing Chords and Segments:** - Since triangle ABC is acute, the chord BC is less than the diameter 2r. - Thus, CF, half of BC, is less than r, i.e., [ CF < r. ] 7. **Deriving Final Length Inequality:** - Now, examine the length: [ CP = CF - FP quad text{with} quad FP geq frac{r}{2}. ] - Thus, [ CP = CF - FP < r - frac{r}{2} = frac{r}{2}. ] - Therefore, [ CP < FP. ] 8. **Conclusion:** - Given that CP < FP, it implies: [ PO < OP ] leading to: [ angle POC < angle OCP. ] Thus, [ boxed{angle CAB + angle COP < 90^{circ}} ]

answer:To find the x-intercept, we set y = 0 in the equation 5x - 7y = 35: [ 5x - 7(0) = 35 ] This simplifies to: [ 5x = 35 ] Solving for x, we divide both sides by 5: [ x = frac{35}{5} = 7 ] Thus, the x-intercept is (7, 0). Putting it in the ordered pair notation, the x-intercept is boxed{(7,0)}.

answer:Given the set A={x|(kx-k^2-6)(x-4)>0, xin mathbb{Z}}, Since the equation (kx-k^2-6)(x-4)=0, We find: x_{1}=k+ frac {6}{k}, and x_{2}=4, Therefore, for (kx-k^2-6)(x-4)>0, xin mathbb{Z} When k=0, A=(-infty,4); When k>0, 4<k+ frac {6}{k}, A=(-infty,4) cup (k+ frac {6}{k},+infty); When k<0, k+ frac {6}{k}<4, A=(k+ frac {6}{k},4). Therefore, when kgeq0, the number of elements in set A is infinite; When k<0, k+ frac {6}{k}<4, A=(k+ frac {6}{k},4). The number of elements in set A is finite, and at this time, the number of elements in set A is the least. Thus, we have: begin{cases} k<0 k+ frac {6}{k}<4end{cases}, solving this gives: k<0. Hence, the answer is: boxed{(-infty,0)}. By simplifying set A and discussing based on k, we can find the range of x and thus the answer. This problem examines the distribution of set elements and operational issues, the application of equation thinking, and the transformation of problems within the question. This is a comprehensive problem involving set operations, equations, and inequalities, which is worth students' careful reflection and summarization.

answer:Given: - Initial numbers on the board: (2023, 2023, ldots, 2023) (a total of 2023 numbers) - In each step, two numbers (x) and (y) are chosen and replaced by (frac{x+y}{4}). We aim to prove that the final number left on the board is always greater than 1. 1. **Consider the Arithmetic-Harmonic Mean Inequality:** The Arithmetic-Harmonic Mean Inequality states that for any positive numbers (x) and (y), [ frac{x+y}{2} geq frac{2}{frac{1}{x} + frac{1}{y}} ] From this inequality, we can derive that: [ frac{1}{x} + frac{1}{y} geq frac{4}{x+y} ] 2. **Transformation of the original problem:** In every step, we replace numbers (x) and (y) with (frac{x+y}{4}). Looking at the harmonic mean context, we rewrite this in terms of their reciprocals. Given: - Each initial number is (2023) - We will focus on the reciprocal values: starting with ( frac{1}{2023} ). 3. **Initial sum of reciprocals:** Initially, we have: [ sum_{i=1}^{2023} frac{1}{2023} = 2023 times frac{1}{2023} = 1 ] 4. **Evolution of the sum of reciprocals through each step:** When two numbers (x) and (y) are chosen and replaced by (frac{x+y}{4}), their reciprocals (harmonic means) must be considered: [ frac{1}{frac{x+y}{4}} = frac{4}{x+y} ] From the Arithmetic-Harmonic Mean Inequality: [ frac{1}{x} + frac{1}{y} geq frac{4}{x+y} ] Replacing the two numbers (x) and (y) on the board thus increases or retains the sum of their reciprocals. - If (x) and (y) are equal, (frac{1}{x} + frac{1}{y} = frac{2}{x}). So, the sum of the reciprocals remains the same. - If (x neq y), then (frac{1}{x} + frac{1}{y} geq frac{4}{x+y}), ensuring that the sum of the reciprocals does not decrease. 5. **Consistency across steps:** Since each transformation satisfies the above condition: - The sum of reciprocals remains at least 1. - The transformation ensures that we cannot reach a final single reciprocal value that sums exactly to 1 from non-decreasing intermediate sums, implying the final value itself must be greater than 1. 6. **Conclusion:** Because the initial sum of reciprocals is 1 and each step respects the arithmetic-harmonic mean relationship ensuring non-decrease from this value: [ boxed{ text{The final number left on the board is always greater than 1.} } ]