answer:Given that M is a moving point on the circle C: (x-1)^2 + y^2 = 4, we can denote the coordinates of M as M(a,b). The radius of the circle centered at M is equal to the distance from M to the origin O, which is |OM| = sqrt{a^2 + b^2}. Thus, the equation of the circle centered at M with radius |OM| can be written as: (x-a)^2 + (y-b)^2 = a^2 + b^2. Given the equation of circle C is: (x-1)^2 + y^2 = 4, we subtract the equation of circle M from the equation of circle C to find the equation of line AB: begin{align*} [(x-1)^2 + y^2] - [(x-a)^2 + (y-b)^2] &= 4 - (a^2 + b^2) (x^2 - 2x + 1 + y^2) - (x^2 - 2ax + a^2 + y^2 - 2by + b^2) &= 0 (2a-2)x + 2by - (a^2 - 2a + 1 + b^2 - 4) &= 0 (2a-2)x + 2by - 3 &= 0, end{align*} which simplifies to the equation of line AB: (2a-2)x + 2by - 3 = 0. For a point (m,n) on line AB, we have: (2a-2)m + 2bn - 3 = 0. Rearranging gives: 2ma + 2nb - (2m+3) = 0. Since M lies on circle C, we have: (a-1)^2 + b^2 = 4. The distance d from the center of circle C, (1,0), to the line 2ma + 2nb - (2m+3) = 0 must be less than or equal to the radius of circle C, which is 2. This gives us: frac{|2m+0-(2m+3)|}{sqrt{4m^2+4n^2}} leq 2, which simplifies to: m^2 + n^2 geq frac{9}{16}. Checking each point against this inequality: - For point A, left(frac{4}{5}right)^2 + 0^2 = frac{16}{25} > frac{9}{16}. - For point B, left(frac{1}{2}right)^2 + left(frac{1}{2}right)^2 = frac{1}{4} + frac{1}{4} = frac{1}{2} < frac{9}{16}. - For point C, left(frac{1}{2}right)^2 + left(frac{sqrt{2}}{2}right)^2 = frac{1}{4} + frac{2}{4} = frac{3}{4} > frac{9}{16}. - For point D, 0^2 + left(frac{5}{4}right)^2 = frac{25}{16} > frac{9}{16}. Therefore, the line AB definitely does not pass through point B, ({frac{1}{2}, frac{1}{2}}). Hence, the correct answer is: boxed{B}.

answer:I. Solution: Let's consider the problem solved and draw the circumscribed circle, the diagonal (BD), and let (M) be the intersection point of the diagonals (see Fig. 1).  **Fig. 1** According to the Inscribed Angle Theorem, we have: [ angle BDC = alpha_1 quad text{and} quad angle DBA = gamma_1 ] Therefore, in the triangles (ABM) and (CDM), we know one side and the angle adjacent to it. Due to the equality of their angles, these two triangles are similar, however, with opposite orientations. Based on this, the construction can proceed as follows. 1. **Construct the angle ( alpha_1 ) at point ( A )**: Measure the angle (alpha_1) at point (A) along segment (AB). 2. **Construct the angle ( gamma_1 ) at point ( B )**: Measure the angle (gamma_1) at point (B) along segment (AB). 3. **Find point ( M )**: The intersection of the rays passing through these angles gives point (M). Next steps involve creating parallel lines to use these properties: 4. **Measure (DC) from ( A ) towards ( B ) to point ( E )**. 5. **Draw a line through (E) parallel to (AM) and intersect it with ( BM ) at point ( F )**. 6. **Draw a line through (F) parallel to (AB ) and intersect it with (AM) at point ( G )**. 7. **Find points ( C ) and ( D )**: Extend (AM) and (BM) beyond (M) and measure (MF) and (MG) respectively to find (C) and (D). By constructing parallel lines based on these elements: - (MGF) is similar to (triangle MAB) with identical orientation. - (DC = AE = GF) means that (triangle MDC) is congruent with the opposite travel direction. This construction is executable provided that (alpha_1 + gamma_1 < 180^circ). II. Solution: We can bring the sides (AB) and (CD) next to each other by reflecting triangle (ACD) over a line (k) perpendicular to diagonal (AC) (see Fig. 1 still). With reflection: [ D_1 text{ is the reflection of } D text{ such that } D_1A = DC quad text{and} quad angle D_1AC = angle DCA = gamma_1 ] Hence: [ angle D_1AB = alpha_1 + gamma_1 ] In triangle (D_1AB), we now know two sides and the included angle: 1. **Construct the angle ( alpha_1 ) at point ( A )** along ( AB ). 2. **Construct ( gamma_1 ) from the other ray** of ( alpha_1 ), extending (CD) from ( A ) to point ( D_1 ). 3. **Construct the circumcircle of ( triangle ABD_1 )** and this intersects the second ray of ( alpha_1 ) at ( C ). Finally, reflect (D_1) back over the perpendicular bisector of ( AC ): - (D) is found by intersecting the circle centered at (C) radius (CD), giving (D). This construction is valid if (alpha_1 + gamma_1 < 180^circ). III. Solution: We can also use translation to get a helper triangle. Assuming (alpha_1 geq gamma_1) (which can always be achieved by suitable relabeling), we have (C) at least as far from (AB) as (D). 1. **Translate edge (AB)** with (A) to point (D), placing (B) at (B_1). 2. **Find (N)**: Construct segment (DB_1) and intersect it with (AC) at point (N) (see Fig. 2).  **Fig. 2** (angle B_1DC = alpha_1 - gamma_1), thus enabling us to construct the (DBC) triangle: 3. **Construct rays (e text{ and } f )**: Using diagonals (CA) and (DB). 4. **Place segment (AB) between these rays** parallel to (DB_1). Move segment (AB = N B_2) parallel: - Through point (B_2) draw line parallel to (e), intersect (f) at (B), - Finally, intersect line (BA) parallel to (DB) at point (A). Special case: If (gamma_1 = alpha_1), then (C) lies on line (DB_1) and coincides with (N). (boxed{text{Solution complete}})

answer:To determine which option simplifies to a negative number, we evaluate each option step by step: **Option A**: -left(-5right) - First, we simplify inside the parentheses, which is -5. - Then, we apply the negative sign outside, which negates the -5 to 5. - Therefore, -left(-5right) = 5, which is not a negative number. **Option B**: +|-5| - First, we find the absolute value of -5, which is 5. - Then, we apply the positive sign, which keeps it as 5. - Therefore, +|-5| = 5, which is not a negative number. **Option C**: -left(-5right)^{2} - First, we calculate the square of -5, which is 25. - Then, we apply the negative sign, making it -25. - Therefore, -left(-5right)^{2} = -25, which is a negative number. **Option D**: (-5)times left(-5) - We multiply -5 by -5, which equals 25. - Therefore, (-5)times left(-5) = 25, which is not a negative number. Given the evaluations, only **Option C** simplifies to a negative number. Therefore, the answer is: boxed{C}.

answer:1. **Determine the exterior angle of the original pentagon**: The exterior angle of a regular polygon with n sides can be computed using: [ text{Exterior angle} = frac{360^circ}{n} ] For a pentagon, n=5, giving: [ text{Exterior angle} = frac{360^circ}{5} = 72^circ ] 2. **Angle between tangents at each vertex of the pentagon**: When two tangent lines meet at the vertex, they each form an angle of 90^circ with the radius at the point of tangency. Since each side is tangent in a regular pentagon, the angle where tangents meet at each external intersection is: [ text{Angle between tangents} = 180^circ - 2 times text{Angle between radius and tangent} ] [ text{Angle between tangents} = 180^circ - 2 times 90^circ = 0^circ ] However, in the context of the pentagon's extensions, since the tangents do not overlap directly but diverge towards the corresponding extensions, the relevant angular measure is actually the supplementary angle, which is: [ text{External intersection angle} = 360^circ - 72^circ = 288^circ ] Conclusion: The angle at each external intersection where two tangent lines meet is 288^circ. The final answer is boxed{C) 288°}