answer:**Answer:** According to the rules of multiplication for rational numbers, we calculate the result. The final answer is boxed{A: 6}.

answer:1. **Given:** - Triangle (ABC) with (M) as the midpoint of side (AB). - (D) is the foot of the altitude from (C) to (AB). - We need to prove that (angle A = 2angle B) if and only if (AC = 2MD). 2. **Introduce Midpoint (N):** - Let (N) be the midpoint of side (AC). - Since (M) and (N) are midpoints, we have (AM = MB) and (AN = NC). 3. **Isosceles Triangle (DMN):** - Consider triangle (DMN). - Since (M) and (N) are midpoints, (DM) and (DN) are equal in length. - Therefore, triangle (DMN) is isosceles with (DM = DN). 4. **Using the Midpoint Theorem:** - By the midpoint theorem, (MN) is parallel to (BC) and (MN = frac{1}{2}BC). 5. **Angle Relationships:** - Since (D) is the foot of the altitude from (C), (angle CDB = 90^circ). - Let (angle B = beta). Then (angle A = 2beta). 6. **Proving (AC = 2MD):** - In (triangle ABC), using the Law of Sines: [ frac{AC}{sin B} = frac{BC}{sin A} ] Since (angle A = 2beta) and (angle B = beta), we have: [ frac{AC}{sin beta} = frac{BC}{sin 2beta} ] Using the double angle identity (sin 2beta = 2 sin beta cos beta), we get: [ frac{AC}{sin beta} = frac{BC}{2 sin beta cos beta} ] Simplifying, we find: [ AC = frac{BC}{2 cos beta} ] 7. **Relating (MD) to (AC):** - Since (M) is the midpoint of (AB), (AM = MB = frac{AB}{2}). - In (triangle CMD), using the Pythagorean theorem: [ MD = sqrt{CM^2 + CD^2} ] Since (M) is the midpoint, (CM = frac{AC}{2}). 8. **Final Calculation:** - Given (AC = 2MD), we substitute (MD) from the previous step: [ AC = 2 sqrt{left(frac{AC}{2}right)^2 + CD^2} ] Simplifying, we get: [ AC = 2 sqrt{frac{AC^2}{4} + CD^2} ] [ AC = 2 sqrt{frac{AC^2 + 4CD^2}{4}} ] [ AC = sqrt{AC^2 + 4CD^2} ] Squaring both sides, we get: [ AC^2 = AC^2 + 4CD^2 ] [ 0 = 4CD^2 ] [ CD = 0 ] This implies that (D) is the midpoint of (AB), which is a contradiction unless (angle A = 2angle B). Therefore, we have shown that (angle A = 2angle B) if and only if (AC = 2MD). (blacksquare)

answer:Let's denote AC = 8 + 22 = 30 and introduce the variables x = AD = CD and y = BD. Using the Pythagorean Theorem for triangles ABD and BCD: - For triangle ABD, h^2 = y^2 - 8^2, where h is distance from D to line overline{AC}. - For triangle BCD, h^2 = x^2 - 22^2. Equating the two expressions for h^2, gives: y^2 - 64 = x^2 - 484 Longrightarrow x^2 - y^2 = 420 Rewriting, we have: (x + y)(x - y) = 420 Factor pairs of 420 that can form valid (x,y): - (1,420), (2,210), (3,140), (4,105), (5,84), (6,70), (7,60), (10,42), (12,35), (14,30), (15,28), (20,21) Verify integers x,y: - (x,y) that yields integer solutions to x = frac{p+r}{2} and y = frac{p-r}{2} (where p and r are factor pairs) and such that AD and BD are integers. Exclude pairs leading to impossible triangle configurations. Resulting valid cases using some sample factor pairs: [ (71, 349), (106, 314), (72, 348), ] [ x_1 + x_2 + x_3 = 71 + 106 + 72 = 249 ] Sum of all possible perimeters: [ 3 times AC + 2 times (71 + 106 + 72) = 3 times 30 + 2 times 249 = 90 + 498 = boxed{588} ]

answer:# Step-by-Step Solution Part 1: Center and Radius of the Circle, and Relationship with Line l Given the equation of the circle C: x^{2}+y^{2}-2y-4=0, we can rewrite it in the standard form of a circle's equation by completing the square for the y terms. 1. Rewrite the circle's equation: [ x^{2}+(y^{2}-2y+1)-4-1=0 ] [ x^{2}+(y-1)^{2}=5 ] 2. From this, we identify the center and radius of the circle: - Center C(0,1) - Radius r=sqrt{5} 3. The line l: mx-y+1-m=0 can be rearranged as: [ y=mx-m+1 ] [ y-1=m(x-1) ] This shows that the line always passes through the point M(1,1). 4. To determine the relationship between the line l and the circle C, we check the position of point M relative to the circle: [ 1^{2}+(1-1)^{2}=1 lt 5 ] Since the distance squared from M to the center of the circle is less than the radius squared, point M lies inside the circle C. Therefore, the line l must intersect the circle. Part 2: Length of Chord AB 1. The slope of the line l is given as 120^{circ}, which corresponds to a slope of k=tan 120^{circ}=-sqrt{3}. 2. The equation of the line l with m=-sqrt{3} becomes: [ y=-sqrt{3}x+sqrt{3}+1 ] Or rearranged to standard form: [ sqrt{3}x+y-sqrt{3}-1=0 ] 3. To find the distance d from the center C(0,1) to the line l, we use the formula for the distance from a point to a line: [ d=frac{|sqrt{3}cdot0+1cdot1-(sqrt{3}+1)|}{sqrt{(sqrt{3})^{2}+1^{2}}}=frac{|sqrt{3}|}{2} ] 4. The length of chord AB is found using the formula |AB|=2sqrt{r^{2}-d^{2}}: [ |AB|=2sqrt{5-left(frac{sqrt{3}}{2}right)^{2}}=2sqrt{5-frac{3}{4}}=sqrt{17} ] Therefore, the length of chord AB is boxed{sqrt{17}}.