answer：First, I need to simplify each term modulo 17 to make the calculation easier.Starting with 245 modulo 17:- I divide 245 by 17, which gives a quotient of 14 and a remainder of 7.- Therefore, 245 is congruent to 7 modulo 17.Next, for the term 18 modulo 17:- Dividing 18 by 17 gives a quotient of 1 and a remainder of 1.- So, 18 is congruent to 1 modulo 17.Now, I substitute these simplified values back into the original expression:- The expression becomes (7 * 15 - 1 * 8 + 5) modulo 17.Calculating 7 multiplied by 15:- 7 * 15 equals 105.- Dividing 105 by 17 gives a quotient of 6 and a remainder of 3.- Therefore, 105 is congruent to 3 modulo 17.Finally, I compute the entire expression:- 3 (from 7 * 15) minus 8 (from 18 * 8) plus 5 equals 0.- Thus, the entire expression is congruent to 0 modulo 17.

answer：Alright, so I have this problem here about points on a parabola and some perpendicular lines. Let me try to unpack it step by step. First, the problem says that O is the origin, and points A and B are moving on the parabola y² = 4x. Also, OA is perpendicular to OB. Then, there's this line OH, which is perpendicular to AB and passes through O, with H being the foot of the perpendicular. The question is asking which of the given points has a fixed distance to H. The options are (1,0), (2,0), (1,2), and (2,1).Okay, so I need to figure out the relationship between these points and H. Since A and B are on the parabola y² = 4x, their coordinates can be represented in terms of parameters. Maybe I can let A be (a², 2a) and B be (b², 2b) because for the parabola y² = 4x, the parametric equations are x = t² and y = 2t.Now, OA is perpendicular to OB. That means the vectors OA and OB are perpendicular. The dot product of OA and OB should be zero. So, OA · OB = 0.Calculating the vectors OA and OB, OA is (a², 2a) and OB is (b², 2b). The dot product is (a²)(b²) + (2a)(2b) = a²b² + 4ab. Setting this equal to zero: a²b² + 4ab = 0.Hmm, that's a bit complicated. Maybe I can factor it. Let's see: ab(ab + 4) = 0. So either ab = 0 or ab = -4. But if ab = 0, then either a or b is zero, which would mean either A or B is at the origin, but they are moving points on the parabola, so they can't be at the origin. So, ab must be -4. Therefore, ab = -4.Alright, so the product of the parameters a and b is -4. That might be useful later.Next, I need to find the foot of the perpendicular from O to AB, which is point H. To find H, I need the equation of line AB first.Given points A(a², 2a) and B(b², 2b), the slope of AB is (2b - 2a)/(b² - a²) = 2(b - a)/[(b - a)(b + a)] = 2/(b + a). So, the slope m of AB is 2/(a + b).Since OH is perpendicular to AB, the slope of OH is the negative reciprocal of m. So, the slope of OH is -(a + b)/2.But OH passes through the origin, so its equation is y = [-(a + b)/2]x.Now, to find point H, which is the intersection of OH and AB.First, let's write the equation of AB. We know the slope is 2/(a + b), and it passes through point A(a², 2a). So, the equation is y - 2a = [2/(a + b)](x - a²).Now, set this equal to the equation of OH: y = [-(a + b)/2]x.So, [-(a + b)/2]x = [2/(a + b)](x - a²) + 2a.Let me solve for x:Multiply both sides by 2(a + b) to eliminate denominators:- (a + b)² x = 4(x - a²) + 4a(a + b)Expand:- (a² + 2ab + b²)x = 4x - 4a² + 4a² + 4abSimplify the right side:4x + 4abSo, left side: -a²x - 2abx - b²xBring all terms to left:- a²x - 2abx - b²x - 4x - 4ab = 0Factor x:x(-a² - 2ab - b² - 4) - 4ab = 0Hmm, this is getting messy. Maybe there's a better way.Alternatively, since H is the foot of the perpendicular from O to AB, we can use the formula for the foot of the perpendicular from a point to a line.Given a line in the form Ax + By + C = 0, the foot of the perpendicular from (x0, y0) is given by:(x', y') = (x0 - A(Ax0 + By0 + C)/(A² + B²), y0 - B(Ax0 + By0 + C)/(A² + B²))But in our case, the line AB passes through A(a², 2a) and B(b², 2b). Let me write the equation of AB in standard form.The slope of AB is 2/(a + b), so the equation is y = [2/(a + b)]x + c.To find c, plug in point A:2a = [2/(a + b)]a² + c => c = 2a - [2a²/(a + b)] = [2a(a + b) - 2a²]/(a + b) = [2ab]/(a + b)So, the equation of AB is y = [2/(a + b)]x + [2ab/(a + b)]Multiplying both sides by (a + b):(a + b)y = 2x + 2abBring all terms to left:2x - (a + b)y + 2ab = 0So, A = 2, B = -(a + b), C = 2abNow, the foot of the perpendicular from O(0,0) to AB is:x' = 0 - 2(2*0 + (-(a + b))*0 + 2ab)/(2² + (-(a + b))²) = -2*(2ab)/(4 + (a + b)²) = -4ab/(4 + (a + b)²)Similarly, y' = 0 - (-(a + b))(2*0 + (-(a + b))*0 + 2ab)/(4 + (a + b)²) = (a + b)*(2ab)/(4 + (a + b)²) = 2ab(a + b)/(4 + (a + b)²)So, H has coordinates:H = ( -4ab/(4 + (a + b)²), 2ab(a + b)/(4 + (a + b)²) )Now, since we know from earlier that ab = -4, let's substitute ab = -4.So, H becomes:x' = -4*(-4)/(4 + (a + b)²) = 16/(4 + (a + b)²)y' = 2*(-4)(a + b)/(4 + (a + b)²) = -8(a + b)/(4 + (a + b)²)So, H = (16/(4 + (a + b)²), -8(a + b)/(4 + (a + b)²))Hmm, now I need to see if there's a fixed point among the options that has a fixed distance to H.Let me consider the options:A: (1,0)B: (2,0)C: (1,2)D: (2,1)I need to compute the distance from each of these points to H and see if it's constant.But H is expressed in terms of a + b. Let me see if I can express a + b in terms of something else.From earlier, we have ab = -4.Also, from the equation of AB, we had that the sum of the roots y1 + y2 = 4m, but I'm not sure if that helps here.Wait, maybe I can find a relation between a + b and ab.Since ab = -4, and a + b is some variable, let's denote s = a + b.So, s = a + b, and ab = -4.So, H becomes:x' = 16/(4 + s²)y' = -8s/(4 + s²)So, H = (16/(4 + s²), -8s/(4 + s²))Now, I need to compute the distance from H to each of the points A, B, C, D and see if it's constant.Let's start with point B: (2,0)Distance squared from H to (2,0):(16/(4 + s²) - 2)² + (-8s/(4 + s²) - 0)²Simplify:( (16 - 2(4 + s²))/(4 + s²) )² + ( -8s/(4 + s²) )²= ( (16 - 8 - 2s²)/(4 + s²) )² + (64s²)/(4 + s²)²= ( (8 - 2s²)/(4 + s²) )² + (64s²)/(4 + s²)²Factor out 1/(4 + s²)²:[ (8 - 2s²)² + 64s² ] / (4 + s²)²Expand (8 - 2s²)²:64 - 32s² + 4s⁴So, numerator:64 - 32s² + 4s⁴ + 64s² = 64 + 32s² + 4s⁴Factor numerator:4s⁴ + 32s² + 64 = 4(s⁴ + 8s² + 16) = 4(s² + 4)²So, distance squared:4(s² + 4)² / (4 + s²)² = 4Therefore, distance is sqrt(4) = 2.Wow, so the distance from H to (2,0) is always 2, regardless of s.Let me check with another point to see if it's fixed.Take point A: (1,0)Distance squared from H to (1,0):(16/(4 + s²) - 1)² + (-8s/(4 + s²) - 0)²= ( (16 - (4 + s²))/(4 + s²) )² + (64s²)/(4 + s²)²= ( (12 - s²)/(4 + s²) )² + (64s²)/(4 + s²)²Factor out 1/(4 + s²)²:[ (12 - s²)² + 64s² ] / (4 + s²)²Expand (12 - s²)²:144 - 24s² + s⁴Numerator:144 - 24s² + s⁴ + 64s² = 144 + 40s² + s⁴This doesn't seem to factor nicely, and it's not a perfect square. So, the distance from H to (1,0) varies with s.Similarly, let's check point C: (1,2)Distance squared:(16/(4 + s²) - 1)² + (-8s/(4 + s²) - 2)²This will involve more terms, but it's likely not constant.Point D: (2,1)Distance squared:(16/(4 + s²) - 2)² + (-8s/(4 + s²) - 1)²Again, this seems complicated and likely not constant.So, from the calculations, only point B: (2,0) has a fixed distance of 2 to H, regardless of the values of a and b (as long as OA is perpendicular to OB).Therefore, the answer should be B: (2,0).

answer：Alright, so I have this problem about the function ( f(x) = sin(omega x + frac{pi}{4}) ) where ( omega > 0 ) and it's defined on the interval ([0, pi]). The problem states that this function has exactly 3 axes of symmetry. I need to figure out which of the statements A, B, C, D are correct based on this information.First, let me recall what an axis of symmetry means for a sine function. For a sine function, an axis of symmetry occurs where the function reaches a maximum or minimum, or crosses the midline. Specifically, for ( sin(theta) ), the axes of symmetry are at the points where ( theta = frac{pi}{2} + kpi ) for integer ( k ). These are the points where the sine function reaches its maximum or minimum.So, for the function ( f(x) = sin(omega x + frac{pi}{4}) ), the axes of symmetry will occur where ( omega x + frac{pi}{4} = frac{pi}{2} + kpi ) for some integer ( k ). Solving for ( x ), we get:[omega x + frac{pi}{4} = frac{pi}{2} + kpi omega x = frac{pi}{2} - frac{pi}{4} + kpi omega x = frac{pi}{4} + kpi x = frac{pi}{4omega} + frac{kpi}{omega}]So, the axes of symmetry are at ( x = frac{pi}{4omega} + frac{kpi}{omega} ) for integer ( k ). Now, since the function is defined on the interval ([0, pi]), we need to find how many such ( x ) values lie within this interval.Given that there are exactly 3 axes of symmetry, there must be 3 integer values of ( k ) such that ( x ) is within ([0, pi]). Let's denote these values as ( k = 0, 1, 2 ). So, the corresponding ( x ) values are:1. For ( k = 0 ): ( x = frac{pi}{4omega} )2. For ( k = 1 ): ( x = frac{pi}{4omega} + frac{pi}{omega} = frac{5pi}{4omega} )3. For ( k = 2 ): ( x = frac{pi}{4omega} + frac{2pi}{omega} = frac{9pi}{4omega} )These three ( x ) values must all lie within ([0, pi]). Therefore, the largest ( x ) value, which is ( frac{9pi}{4omega} ), must be less than or equal to ( pi ). So:[frac{9pi}{4omega} leq pi frac{9}{4omega} leq 1 frac{9}{4} leq omega]Similarly, to ensure that the next axis of symmetry (for ( k = 3 )) doesn't fall within the interval, we need:[frac{13pi}{4omega} > pi frac{13}{4omega} > 1 frac{13}{4} > omega]So, combining these two inequalities, we get:[frac{9}{4} leq omega < frac{13}{4}]This gives us the range of ( omega ). So, statement C says the range is ([ frac{9}{4}, frac{13}{4} ]), but our analysis shows it's actually ([ frac{9}{4}, frac{13}{4} )). So, statement C is incorrect because it includes ( frac{13}{4} ), which is not allowed.Moving on to statement A: ( f(x) ) has exactly 3 distinct zeros on the interval ( (0, pi) ). Let's think about the zeros of ( f(x) ). The zeros occur where ( sin(omega x + frac{pi}{4}) = 0 ), which happens when ( omega x + frac{pi}{4} = kpi ) for integer ( k ). Solving for ( x ):[omega x + frac{pi}{4} = kpi omega x = kpi - frac{pi}{4} x = frac{kpi - frac{pi}{4}}{omega}]We need to find how many such ( x ) lie within ( (0, pi) ). Let's consider the range of ( omega ) we found earlier: ( frac{9}{4} leq omega < frac{13}{4} ).The number of zeros depends on how many times the sine function crosses the x-axis within the interval. Given the range of ( omega ), the argument ( omega x + frac{pi}{4} ) will range from ( frac{pi}{4} ) to ( omega pi + frac{pi}{4} ).For ( omega = frac{9}{4} ), the upper limit is ( frac{9}{4}pi + frac{pi}{4} = frac{10}{4}pi = frac{5}{2}pi ). For ( omega = frac{13}{4} ), it's ( frac{13}{4}pi + frac{pi}{4} = frac{14}{4}pi = frac{7}{2}pi ).So, the argument ranges from ( frac{pi}{4} ) to between ( frac{5}{2}pi ) and ( frac{7}{2}pi ). The sine function crosses zero at multiples of ( pi ). So, starting from ( frac{pi}{4} ), the first zero is at ( pi ), then at ( 2pi ), and so on.For ( omega = frac{9}{4} ), the upper limit is ( frac{5}{2}pi ), so the zeros are at ( pi ) and ( 2pi ). But ( 2pi ) is beyond ( frac{5}{2}pi ), so only one zero at ( pi ). Wait, that doesn't seem right.Wait, actually, let's think again. The zeros occur at ( x = frac{kpi - frac{pi}{4}}{omega} ). So, for ( k = 1 ):[x = frac{pi - frac{pi}{4}}{omega} = frac{3pi/4}{omega}]For ( k = 2 ):[x = frac{2pi - frac{pi}{4}}{omega} = frac{7pi/4}{omega}]For ( k = 3 ):[x = frac{3pi - frac{pi}{4}}{omega} = frac{11pi/4}{omega}]Now, we need these ( x ) values to be within ( (0, pi) ).So, for ( k = 1 ):[x = frac{3pi/4}{omega} < pi implies frac{3}{4} < omega]Which is true since ( omega geq frac{9}{4} ).For ( k = 2 ):[x = frac{7pi/4}{omega} < pi implies frac{7}{4} < omega]Again, ( omega geq frac{9}{4} ), so this is true.For ( k = 3 ):[x = frac{11pi/4}{omega} < pi implies frac{11}{4} < omega]But ( omega < frac{13}{4} ), so ( frac{11}{4} < omega < frac{13}{4} ). Therefore, for ( omega ) in ( [frac{9}{4}, frac{11}{4}) ), the zero at ( k = 3 ) is outside the interval, so only two zeros. For ( omega ) in ( [frac{11}{4}, frac{13}{4}) ), the zero at ( k = 3 ) is inside the interval, so three zeros.Therefore, depending on ( omega ), there can be either two or three zeros in ( (0, pi) ). Since the problem states that there are exactly three axes of symmetry, which implies ( omega ) is in ( [frac{9}{4}, frac{13}{4}) ), but the number of zeros can be two or three. So, statement A says there are exactly three zeros, but it's not always true; it depends on ( omega ). Therefore, statement A is incorrect.Moving on to statement B: The minimum positive period of ( f(x) ) could be ( frac{2pi}{3} ). The period of ( f(x) ) is ( T = frac{2pi}{omega} ). The minimum period corresponds to the maximum ( omega ). Since ( omega < frac{13}{4} ), the minimum period is greater than ( frac{2pi}{13/4} = frac{8pi}{13} ). The maximum ( omega ) is approaching ( frac{13}{4} ), so the period approaches ( frac{8pi}{13} ). However, ( frac{2pi}{3} ) is approximately ( 2.094 ), while ( frac{8pi}{13} ) is approximately ( 1.932 ). So, ( frac{2pi}{3} ) is larger than the minimum period. Therefore, the minimum period cannot be ( frac{2pi}{3} ); it's actually smaller. So, statement B is incorrect.Wait, hold on. I think I made a mistake here. The period is ( frac{2pi}{omega} ). The minimum period occurs when ( omega ) is maximum. Since ( omega < frac{13}{4} ), the period is greater than ( frac{2pi}{13/4} = frac{8pi}{13} approx 1.932 ). So, the minimum period is ( frac{8pi}{13} ), which is less than ( frac{2pi}{3} approx 2.094 ). Therefore, ( frac{2pi}{3} ) is larger than the minimum period, so it's possible for the period to be ( frac{2pi}{3} ) if ( omega = 3 ). But wait, ( omega ) is in ( [frac{9}{4}, frac{13}{4}) ), which is approximately ( [2.25, 3.25) ). So, ( omega = 3 ) is within this range. Therefore, the period ( frac{2pi}{3} ) is achievable when ( omega = 3 ). So, statement B is correct.Wait, but earlier I thought the minimum period is ( frac{8pi}{13} ), but actually, the minimum period is the smallest period, which occurs at the maximum ( omega ). Since ( omega ) can be up to just below ( frac{13}{4} ), the period can be just above ( frac{8pi}{13} ). However, ( frac{2pi}{3} ) is a specific period that occurs when ( omega = 3 ), which is within the range ( [frac{9}{4}, frac{13}{4}) ). So, yes, the period could be ( frac{2pi}{3} ). Therefore, statement B is correct.Now, statement D: ( f(x) ) is monotonically increasing on the interval ( (0, frac{pi}{15}) ). To check this, we need to look at the derivative of ( f(x) ):[f'(x) = omega cos(omega x + frac{pi}{4})]For ( f(x) ) to be monotonically increasing, ( f'(x) > 0 ) on ( (0, frac{pi}{15}) ). So, we need:[cos(omega x + frac{pi}{4}) > 0]for all ( x in (0, frac{pi}{15}) ).Let's find the range of ( omega x + frac{pi}{4} ) when ( x in (0, frac{pi}{15}) ):[omega x + frac{pi}{4} in left( frac{pi}{4}, omega cdot frac{pi}{15} + frac{pi}{4} right)]Given ( omega geq frac{9}{4} ), the upper bound is:[frac{9}{4} cdot frac{pi}{15} + frac{pi}{4} = frac{3pi}{20} + frac{5pi}{20} = frac{8pi}{20} = frac{2pi}{5} approx 1.2566]And since ( omega < frac{13}{4} ), the upper bound is less than:[frac{13}{4} cdot frac{pi}{15} + frac{pi}{4} = frac{13pi}{60} + frac{15pi}{60} = frac{28pi}{60} = frac{7pi}{15} approx 1.466]So, ( omega x + frac{pi}{4} ) ranges from ( frac{pi}{4} ) to less than ( frac{7pi}{15} ). The cosine function is positive in the interval ( (-frac{pi}{2}, frac{pi}{2}) ) and negative in ( (frac{pi}{2}, frac{3pi}{2}) ), etc. So, ( frac{pi}{4} ) is approximately 0.785, and ( frac{7pi}{15} ) is approximately 1.466. The cosine function is positive from ( -frac{pi}{2} ) to ( frac{pi}{2} ), which is approximately -1.5708 to 1.5708. Since our interval ( (frac{pi}{4}, frac{7pi}{15}) ) is entirely within ( (-frac{pi}{2}, frac{pi}{2}) ) shifted by ( frac{pi}{4} ), but wait, actually, ( frac{pi}{4} ) is about 0.785, and ( frac{7pi}{15} ) is about 1.466, which is still less than ( frac{pi}{2} ) (which is about 1.5708). Therefore, the entire interval ( (frac{pi}{4}, frac{7pi}{15}) ) is within ( (-frac{pi}{2}, frac{pi}{2}) ) shifted by ( frac{pi}{4} ), but actually, since ( frac{7pi}{15} < frac{pi}{2} ), the cosine function is positive throughout this interval. Therefore, ( f'(x) > 0 ) on ( (0, frac{pi}{15}) ), so ( f(x) ) is monotonically increasing there. Therefore, statement D is correct.So, summarizing:- Statement A: Incorrect- Statement B: Correct- Statement C: Incorrect- Statement D: CorrectTherefore, the correct statements are B and D.

answer：First, I need to determine the time each person has been walking when B overtakes A. A starts walking first and has a 0.5-hour head start. B starts after half an hour and overtakes A after 1 hour and 48 minutes, which is 1.8 hours. Therefore, A has been walking for a total of 2.3 hours when B overtakes.Next, I'll set up the equation for the distance each person has traveled when B overtakes A. Since they start from the same place and walk in the same direction, the distances they cover must be equal at the point of overtaking.The distance covered by A is the product of A's speed and the total time A has been walking. Similarly, the distance covered by B is the product of B's speed and the time B has been walking.I'll use the given speed of B to solve for A's speed. By rearranging the equation, I can isolate A's speed and calculate it using the known values.Finally, I'll perform the calculations to find the exact value of A's speed.

answer：Okay, so I've got this triangle ABC, and the sides opposite angles A, B, and C are a, b, and c respectively. The problem gives me an equation: (c cos A)/(a cos C) - c/(2b - c) = 0. I need to find the measure of angle A, and then if a is 2, find the maximum value of the altitude from BC.Alright, let's start with part (1). I need to find angle A. The equation given is (c cos A)/(a cos C) - c/(2b - c) = 0. Hmm, that looks a bit complicated, but maybe I can simplify it.First, I notice that both terms have a 'c' in them. Maybe I can factor that out or cancel it somehow. Let me write the equation again:(c cos A)/(a cos C) - c/(2b - c) = 0.If I factor out 'c' from both terms, I get:c [ (cos A)/(a cos C) - 1/(2b - c) ] = 0.Since c is a side length of a triangle, it can't be zero, so the expression inside the brackets must be zero:(cos A)/(a cos C) - 1/(2b - c) = 0.So, (cos A)/(a cos C) = 1/(2b - c).Hmm, okay. Maybe I can use the Law of Sines here. The Law of Sines says that a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle. Maybe I can express a and c in terms of sine functions.Let me denote 2R as a common value, so a = 2R sin A, b = 2R sin B, c = 2R sin C.Substituting these into the equation:(cos A)/( (2R sin A) cos C ) = 1/(2*(2R sin B) - 2R sin C )Simplify numerator and denominator:Left side: (cos A)/(2R sin A cos C) = (cos A)/(2R sin A cos C)Right side: 1/(4R sin B - 2R sin C) = 1/(2R(2 sin B - sin C))So, we have:(cos A)/(2R sin A cos C) = 1/(2R(2 sin B - sin C))I can cancel out the 2R from both sides:(cos A)/(sin A cos C) = 1/(2 sin B - sin C)So, (cos A)/(sin A cos C) = 1/(2 sin B - sin C)Hmm, maybe I can cross-multiply:cos A * (2 sin B - sin C) = sin A cos CLet me expand the left side:2 sin B cos A - sin C cos A = sin A cos CHmm, let's see. Maybe I can use some trigonometric identities here. I know that sin(A + C) = sin B, because in a triangle, A + B + C = π, so B = π - A - C, and sin(π - x) = sin x. So, sin B = sin(A + C).Similarly, maybe I can express sin(A + C) as sin A cos C + cos A sin C.So, sin B = sin(A + C) = sin A cos C + cos A sin C.So, sin B = sin A cos C + cos A sin C.Hmm, so in the equation above, we have:2 sin B cos A - sin C cos A = sin A cos CLet me substitute sin B with sin A cos C + cos A sin C:2(sin A cos C + cos A sin C) cos A - sin C cos A = sin A cos CLet me expand this:2 sin A cos C cos A + 2 cos A sin C cos A - sin C cos A = sin A cos CSimplify term by term:First term: 2 sin A cos C cos ASecond term: 2 cos^2 A sin CThird term: - sin C cos ASo, putting it all together:2 sin A cos C cos A + 2 cos^2 A sin C - sin C cos A = sin A cos CHmm, let's see. Let me factor terms where possible.Looking at the first term: 2 sin A cos A cos CSecond term: 2 cos^2 A sin CThird term: - sin C cos ARight side: sin A cos CMaybe I can bring everything to one side:2 sin A cos A cos C + 2 cos^2 A sin C - sin C cos A - sin A cos C = 0Let me factor terms:Looking at the first and last terms: 2 sin A cos A cos C - sin A cos CFactor sin A cos C:sin A cos C (2 cos A - 1)Similarly, the second and third terms: 2 cos^2 A sin C - sin C cos AFactor sin C cos A:sin C cos A (2 cos A - 1)So, combining these:sin A cos C (2 cos A - 1) + sin C cos A (2 cos A - 1) = 0Factor out (2 cos A - 1):(2 cos A - 1)(sin A cos C + sin C cos A) = 0Now, sin A cos C + sin C cos A is equal to sin(A + C), which is sin(π - B) = sin B.So, we have:(2 cos A - 1) sin B = 0In a triangle, sin B is not zero because B is between 0 and π, so sin B ≠ 0. Therefore, 2 cos A - 1 = 0.So, 2 cos A - 1 = 0 => cos A = 1/2Therefore, angle A is arccos(1/2). The arccos of 1/2 is π/3 or 60 degrees.So, angle A is π/3 radians.Alright, that seems solid. So, part (1) is angle A = π/3.Now, moving on to part (2). If a = 2, find the maximum value of the altitude from BC.Wait, altitude from BC. Hmm, in triangle ABC, the altitude from BC would be the height corresponding to side BC, which is side a. So, the altitude from vertex A to side BC is the height h_a.Wait, but the problem says "the altitude from BC". Hmm, maybe I need to clarify. In triangle ABC, the altitude from BC would be the altitude drawn from vertex A to side BC. So, yes, h_a.But let me just confirm: in triangle ABC, side BC is opposite angle A, so the altitude from BC would be the same as the altitude from A to BC.Yes, that makes sense.So, the altitude h from A to BC is given by the formula:h = (2 * area of triangle) / aSince area = (1/2) * base * height, so height = (2 * area) / base.So, h = (2 * area) / a.Given that a = 2, h = (2 * area) / 2 = area.So, h = area of the triangle.Therefore, to maximize h, we need to maximize the area of the triangle.So, the problem reduces to finding the maximum area of triangle ABC with side a = 2 and angle A = π/3.So, how can we maximize the area? The area of a triangle can be given by several formulas. One of them is:Area = (1/2) * b * c * sin ASince angle A is fixed at π/3, sin A = sin(π/3) = √3 / 2.Therefore, Area = (1/2) * b * c * (√3 / 2) = (√3 / 4) * b * c.So, to maximize the area, we need to maximize the product b * c.Given that a = 2 and angle A = π/3, we can use the Law of Cosines to relate sides b and c.Law of Cosines says:a² = b² + c² - 2bc cos AGiven that a = 2 and A = π/3, cos A = 1/2.So,2² = b² + c² - 2bc*(1/2)Simplify:4 = b² + c² - bcSo, we have the equation:b² + c² - bc = 4We need to maximize b * c given this constraint.Hmm, okay. So, we have b² + c² - bc = 4, and we need to maximize bc.This is an optimization problem with constraint. Maybe I can use the method of Lagrange multipliers, but since it's a simple quadratic, perhaps I can express it in terms of one variable.Alternatively, I can use the AM-GM inequality or some other inequality.Wait, let's think about this.We have b² + c² - bc = 4.Let me denote x = b and y = c.So, x² + y² - xy = 4.We need to maximize xy.Hmm, perhaps I can write this as x² + y² - xy = 4.Let me consider that x² + y² - xy is equal to (x - y/2)^2 + (3/4)y².Wait, maybe not. Alternatively, perhaps I can write it as:x² - xy + y² = 4Which is a quadratic in terms of x and y.Alternatively, maybe I can use substitution.Let me consider that x² - xy + y² = 4.Let me set t = x/y, assuming y ≠ 0.Then, x = t y.Substitute into the equation:(t y)² - (t y)(y) + y² = 4Simplify:t² y² - t y² + y² = 4Factor y²:y² (t² - t + 1) = 4So, y² = 4 / (t² - t + 1)We need to maximize xy = t y * y = t y².So, xy = t * y² = t * [4 / (t² - t + 1)] = 4t / (t² - t + 1)So, we need to maximize the function f(t) = 4t / (t² - t + 1) for t > 0.Hmm, okay. So, let's find the maximum of f(t) = 4t / (t² - t + 1).To find the maximum, take the derivative and set it to zero.Let me compute f'(t):f'(t) = [4*(t² - t + 1) - 4t*(2t - 1)] / (t² - t + 1)^2Simplify numerator:4(t² - t + 1) - 4t(2t - 1) = 4t² - 4t + 4 - 8t² + 4tCombine like terms:(4t² - 8t²) + (-4t + 4t) + 4 = (-4t²) + 0 + 4 = -4t² + 4Set numerator equal to zero:-4t² + 4 = 0 => -4t² = -4 => t² = 1 => t = 1 or t = -1But since t = x/y and x, y are sides of a triangle, t must be positive. So, t = 1.So, the maximum occurs at t = 1.Therefore, when t = 1, x = y, so b = c.So, when b = c, we have maximum product bc.Therefore, substituting t = 1 back into y²:y² = 4 / (1 - 1 + 1) = 4 / 1 = 4 => y = 2Therefore, b = c = 2.So, the maximum value of bc is 2 * 2 = 4.Therefore, the maximum area is (√3 / 4) * 4 = √3.Since h = area, the maximum altitude h is √3.Wait, hold on. Earlier, I thought h = area, but actually, h = (2 * area) / a, and a = 2, so h = area.Therefore, the maximum altitude is √3.Wait, but let me double-check.Area = (√3 / 4) * b * cWhen b = c = 2, Area = (√3 / 4) * 4 = √3.So, h = area = √3.Alternatively, using the formula h = (2 * area) / a, since a = 2, h = (2 * √3) / 2 = √3.Yes, that's consistent.Therefore, the maximum altitude from BC is √3.So, summarizing:(1) Angle A is π/3.(2) The maximum altitude from BC is √3.**Final Answer**(1) The measure of angle ( A ) is (boxed{dfrac{pi}{3}}).(2) The maximum value of the altitude from ( BC ) is (boxed{sqrt{3}}).

answer：Okay, so I have this problem where the first, second, and third terms of a geometric sequence are equal to the second, fourth, and seventh terms of an arithmetic sequence, respectively. Also, the product of these three numbers is 64. I need to find the first term of the geometric sequence.Hmm, let's break this down. First, let me recall what a geometric sequence is. It's a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio, usually denoted as ( q ). So, if the first term is ( b ), then the terms are ( b ), ( bq ), ( bq^2 ), and so on.On the other hand, an arithmetic sequence is a sequence where each term after the first is obtained by adding a constant difference, denoted as ( d ), to the previous term. So, if the first term is ( a ), the terms are ( a ), ( a + d ), ( a + 2d ), etc.The problem states that the first, second, and third terms of the geometric sequence are equal to the second, fourth, and seventh terms of the arithmetic sequence. So, translating that into equations:1. First term of geometric sequence ( b ) equals the second term of arithmetic sequence ( a + d ).2. Second term of geometric sequence ( bq ) equals the fourth term of arithmetic sequence ( a + 3d ).3. Third term of geometric sequence ( bq^2 ) equals the seventh term of arithmetic sequence ( a + 6d ).So, writing these as equations:1. ( b = a + d )2. ( bq = a + 3d )3. ( bq^2 = a + 6d )Additionally, the product of these three terms is 64. So,( b times bq times bq^2 = 64 )Simplifying that:( b^3 q^3 = 64 )Which can be written as:( (bq)^3 = 64 )Taking the cube root of both sides:( bq = 4 )So, ( bq = 4 ). That gives us a relationship between ( b ) and ( q ). Maybe we can express ( b ) in terms of ( q ):( b = frac{4}{q} )Okay, so now I can substitute this back into the earlier equations to find relationships between ( a ) and ( d ).From equation 1:( frac{4}{q} = a + d )From equation 2:( frac{4}{q} times q = a + 3d )Simplifying equation 2:( 4 = a + 3d )From equation 3:( frac{4}{q} times q^2 = a + 6d )Simplifying equation 3:( 4q = a + 6d )So now, I have three equations:1. ( frac{4}{q} = a + d ) (Equation 1)2. ( 4 = a + 3d ) (Equation 2)3. ( 4q = a + 6d ) (Equation 3)I can use these equations to solve for ( a ), ( d ), and ( q ). Let me see.From Equation 1: ( a = frac{4}{q} - d )Let me substitute this expression for ( a ) into Equation 2 and Equation 3.Substituting into Equation 2:( 4 = left( frac{4}{q} - d right) + 3d )Simplify:( 4 = frac{4}{q} + 2d )Similarly, substituting into Equation 3:( 4q = left( frac{4}{q} - d right) + 6d )Simplify:( 4q = frac{4}{q} + 5d )So now, I have two equations:1. ( 4 = frac{4}{q} + 2d ) (Equation 4)2. ( 4q = frac{4}{q} + 5d ) (Equation 5)Let me solve Equation 4 for ( d ):( 2d = 4 - frac{4}{q} )( d = 2 - frac{2}{q} )Now, substitute this expression for ( d ) into Equation 5:( 4q = frac{4}{q} + 5 left( 2 - frac{2}{q} right) )Simplify the right side:( 4q = frac{4}{q} + 10 - frac{10}{q} )Combine like terms:( 4q = 10 - frac{6}{q} )Multiply both sides by ( q ) to eliminate the denominator:( 4q^2 = 10q - 6 )Bring all terms to one side:( 4q^2 - 10q + 6 = 0 )Simplify this quadratic equation. Let's see if I can factor it or if I need to use the quadratic formula.First, let me check if it can be factored. The equation is:( 4q^2 - 10q + 6 = 0 )Looking for two numbers that multiply to ( 4 times 6 = 24 ) and add up to -10. Hmm, -6 and -4 multiply to 24 and add to -10. So, let's split the middle term:( 4q^2 - 6q - 4q + 6 = 0 )Grouping terms:( (4q^2 - 6q) + (-4q + 6) = 0 )Factor out common terms:( 2q(2q - 3) - 2(2q - 3) = 0 )Now, factor out ( (2q - 3) ):( (2q - 3)(2q - 2) = 0 )So, either ( 2q - 3 = 0 ) or ( 2q - 2 = 0 )Solving these:1. ( 2q - 3 = 0 implies q = frac{3}{2} )2. ( 2q - 2 = 0 implies q = 1 )Wait, but if ( q = 1 ), then the geometric sequence would have all terms equal, which contradicts the problem statement that the terms are distinct. So, ( q = 1 ) is not acceptable. Therefore, ( q = frac{3}{2} ).Now, let's find ( d ) using the expression we had earlier:( d = 2 - frac{2}{q} )Substituting ( q = frac{3}{2} ):( d = 2 - frac{2}{frac{3}{2}} = 2 - frac{4}{3} = frac{6}{3} - frac{4}{3} = frac{2}{3} )So, ( d = frac{2}{3} ).Now, let's find ( a ) using Equation 1:( a = frac{4}{q} - d )Substituting ( q = frac{3}{2} ) and ( d = frac{2}{3} ):( a = frac{4}{frac{3}{2}} - frac{2}{3} = frac{8}{3} - frac{2}{3} = frac{6}{3} = 2 )So, ( a = 2 ).Wait, but we were supposed to find the first term of the geometric sequence, which is ( b ). Earlier, we had:( b = frac{4}{q} )Substituting ( q = frac{3}{2} ):( b = frac{4}{frac{3}{2}} = frac{8}{3} )So, ( b = frac{8}{3} ).Let me verify if this satisfies all the conditions.First, the geometric sequence is:1. ( b = frac{8}{3} )2. ( bq = frac{8}{3} times frac{3}{2} = 4 )3. ( bq^2 = 4 times frac{3}{2} = 6 )So, the geometric sequence terms are ( frac{8}{3} ), 4, 6.Now, the arithmetic sequence has first term ( a = 2 ) and common difference ( d = frac{2}{3} ). So, the terms are:1. ( 2 )2. ( 2 + frac{2}{3} = frac{8}{3} )3. ( 2 + frac{4}{3} = frac{10}{3} )4. ( 2 + 2 = 4 )5. ( 2 + frac{8}{3} = frac{14}{3} )6. ( 2 + frac{10}{3} = frac{16}{3} )7. ( 2 + 4 = 6 )So, the second term is ( frac{8}{3} ), the fourth term is 4, and the seventh term is 6. These match the geometric sequence terms. Also, the product ( frac{8}{3} times 4 times 6 = frac{8}{3} times 24 = 64 ), which satisfies the given condition.Therefore, the first term of the geometric sequence is ( frac{8}{3} ).**Final Answer**The first term of the geometric sequence is boxed{dfrac{8}{3}}.