answer：Alright, so Santa gave us this interesting problem where we have to use all the digits from 1 to 9 exactly once, and between each pair of adjacent digits, we can insert either a plus "+" or a minus "-" sign. The goal is to see how many two-digit prime numbers we can get by doing this. Hmm, okay, let me try to break this down step by step.First off, I know that two-digit prime numbers range from 11 up to 97. But since we're using digits 1 through 9 exactly once, the maximum sum we can get is actually the sum of all these digits. Let me calculate that real quick. The sum of numbers from 1 to 9 is (9*10)/2, which is 45. So, the maximum possible number we can get is 45. That means we're only looking for two-digit primes less than or equal to 45. Let me list those out.The two-digit primes less than or equal to 45 are: 11, 13, 17, 19, 23, 29, 31, 37, 41, and 43. So, that's 10 primes in total. Now, the challenge is to see if we can form each of these primes by inserting "+" or "-" signs between the digits 1 through 9.Okay, let's think about how we can approach this. Since we have to use all digits from 1 to 9, we can't skip any. Each digit must be used exactly once, and between each pair of digits, we can either add or subtract. So, essentially, we're creating an expression like 1 ± 2 ± 3 ± 4 ± 5 ± 6 ± 7 ± 8 ± 9, and we want this expression to evaluate to one of the primes we listed.But wait, if we just insert pluses and minuses, we're essentially creating a sum where some numbers are added and some are subtracted. The total sum of all digits is 45, as I calculated earlier. So, if we subtract twice some number from 45, we can get different results. For example, if we subtract 2 from 45, we get 43; if we subtract 4, we get 41, and so on. So, each prime number can be thought of as 45 minus twice some number. Let me write that down.Let’s denote the total sum as S = 45. If we subtract twice a number x from S, we get S - 2x. We want S - 2x to be equal to one of our primes. So, for each prime p, we have p = 45 - 2x, which means x = (45 - p)/2. Therefore, for each prime p, we need to check if (45 - p) is even and if x is a positive integer that can be formed by some combination of the digits 1 through 9.Let me test this with each prime:1. **11**: (45 - 11)/2 = 17. So, x = 17. Is 17 a possible number we can form by adding or subtracting some digits? Yes, for example, 1 + 2 + 3 + 4 + 5 + 6 - 8 - 9 = 11. Wait, that's actually the expression. So, 17 is formed by subtracting 8 and 9, which are digits.2. **13**: (45 - 13)/2 = 16. So, x = 16. Can we form 16? Yes, for example, 1 + 2 + 3 + 4 + 5 + 6 - 7 + 8 - 9 = 13. Here, we're subtracting 7 and 9, which gives us 16.3. **17**: (45 - 17)/2 = 14. So, x = 14. Can we form 14? Yes, for example, 1 + 2 + 3 - 4 + 5 + 6 + 7 + 8 - 9 = 17. Here, we're subtracting 4 and 9, which gives us 14.4. **19**: (45 - 19)/2 = 13. So, x = 13. Can we form 13? Yes, for example, 1 + 2 + 3 - 4 + 5 + 6 + 7 + 8 - 9 = 19. Wait, that's the same as the previous one. Hmm, maybe I need a different combination. Let me think. Maybe 1 + 2 + 3 + 4 - 5 + 6 + 7 + 8 - 9 = 19. Here, we're subtracting 5 and 9, which gives us 14. Wait, that's still 14. Maybe I'm missing something here.Wait, no, actually, if x = 13, then we need to subtract 13 from 45, but since x is (45 - p)/2, it's actually 13 = (45 - 19)/2. So, x = 13, which means we need to subtract 13 from 45, but in terms of the expression, it's 45 - 2*13 = 19. So, we need to find a way to subtract 13 from 45 by subtracting some digits. So, for example, 1 + 2 + 3 + 4 + 5 + 6 + 7 - 8 + 9 = 29. Wait, that's not 19. Maybe I need to subtract more. Let me try 1 + 2 + 3 - 4 + 5 + 6 + 7 + 8 - 9 = 19. Yes, that works. Here, we're subtracting 4 and 9, which adds up to 13. So, that's correct.5. **23**: (45 - 23)/2 = 11. So, x = 11. Can we form 11? Yes, for example, 1 - 2 + 3 + 4 + 5 + 6 + 7 + 8 - 9 = 23. Here, we're subtracting 2 and 9, which gives us 11.6. **29**: (45 - 29)/2 = 8. So, x = 8. Can we form 8? Yes, for example, 1 + 2 + 3 + 4 + 5 + 6 + 7 - 8 + 9 = 29. Here, we're subtracting 8.7. **31**: (45 - 31)/2 = 7. So, x = 7. Can we form 7? Yes, for example, 1 + 2 + 3 + 4 + 5 + 6 - 7 + 8 + 9 = 31. Here, we're subtracting 7.8. **37**: (45 - 37)/2 = 4. So, x = 4. Can we form 4? Yes, for example, 1 + 2 + 3 - 4 + 5 + 6 + 7 + 8 + 9 = 37. Here, we're subtracting 4.9. **41**: (45 - 41)/2 = 2. So, x = 2. Can we form 2? Yes, for example, 1 - 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 41. Here, we're subtracting 2.10. **43**: (45 - 43)/2 = 1. So, x = 1. Can we form 1? Yes, for example, 2 - 1 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 43. Here, we're subtracting 1.Wait a minute, for 43, I had to rearrange the digits because if I start with 1, I can't subtract 1 from 1. So, I swapped the first two digits to make it 2 - 1, which gives me 1, and then adding the rest gives me 43. So, that works too.So, going through each prime, I was able to find an expression that results in that prime by appropriately placing "+" and "-" signs between the digits 1 through 9. Each time, I subtracted a certain number (x) from the total sum (45) to get the desired prime. And in each case, x was a number that could be formed by subtracting some combination of the digits.Therefore, it seems like all 10 two-digit primes less than or equal to 45 can be obtained by inserting "+" or "-" signs between the digits 1 through 9 exactly once. So, the answer should be 10.

answer：First, I need to determine the perimeter of the square plot. Since the area of the square is 289 square feet, I can find the length of one side by taking the square root of the area.Next, I'll calculate the perimeter by multiplying the length of one side by 4, as a square has four equal sides.Finally, to find the price per foot of building the fence, I'll divide the total cost of building the fence by the perimeter of the square plot.

answer：First, I note that Jane starts with 87 crayons and ends up with 80 crayons after some are eaten by a hippopotamus.To determine how many crayons were eaten, I subtract the final number of crayons from the initial number.So, 87 crayons minus 80 crayons equals 7 crayons.Therefore, the hippopotamus ate 7 crayons.

answer：Okay, so I have this problem about an equilateral triangle ABC, where each side is length a. There's a point D on the extension of BC such that CD is also length a. Then, there's a line DM that intersects AB at M and AC at N. I need to prove that BM is equal to (4a tan D)/(√3 + tan D) and CN is equal to (2a tan D)/(√3 - tan D), where D is the angle BDM.Hmm, let me try to visualize this. So, triangle ABC is equilateral, so all sides are a, and all angles are 60 degrees. Point D is on the extension of BC beyond C, right? Because CD is a, so if BC is a, then BD would be BC + CD, which is a + a = 2a. So, BD is 2a.Now, DM is a line from D intersecting AB at M and AC at N. So, DM is a transversal cutting through AB and AC. The angle at D is angle BDM, which is given as D. So, angle BDM is D.I think I can use the Law of Sines here because we have triangles with some known sides and angles. Let me consider triangle BMD first. In triangle BMD, we have side BD, which is 2a, side BM, which is what we need to find, and angle BDM, which is D. Also, angle at M is something we might need to figure out.Wait, in triangle BMD, we know BD is 2a, angle at D is D, and angle at B is... Well, in triangle ABC, angle at B is 60 degrees, but since D is on the extension of BC, angle at B in triangle BMD might be different. Let me think.Actually, in triangle BMD, the angles should add up to 180 degrees. So, angle at D is D, angle at B is... Hmm, point D is on the extension of BC, so angle at B is still 60 degrees? Or is it different because of the extension?Wait, no. Because when you extend BC to D, the angle at B is still part of the original triangle ABC, which is 60 degrees. But in triangle BMD, the angle at B is actually the same as angle ABC, which is 60 degrees. So, in triangle BMD, angle at B is 60 degrees, angle at D is D, so angle at M is 180 - 60 - D = 120 - D degrees.Okay, so in triangle BMD, we have:- Side BD = 2a- Angle at D = D- Angle at B = 60 degrees- Angle at M = 120 - D degreesSo, using the Law of Sines on triangle BMD:BM / sin(D) = BD / sin(angle at M)So, BM / sin(D) = 2a / sin(120 - D)Therefore, BM = (2a sin(D)) / sin(120 - D)Hmm, okay. Now, I need to express sin(120 - D). I can use the sine subtraction formula:sin(120 - D) = sin(120)cos(D) - cos(120)sin(D)I know that sin(120) is √3/2 and cos(120) is -1/2. So,sin(120 - D) = (√3/2)cos(D) - (-1/2)sin(D) = (√3/2)cos(D) + (1/2)sin(D)So, plugging that back into BM:BM = (2a sin(D)) / [(√3/2)cos(D) + (1/2)sin(D)]Let me factor out 1/2 from the denominator:BM = (2a sin(D)) / [ (1/2)(√3 cos(D) + sin(D)) ) ] = (2a sin(D)) * (2 / (√3 cos(D) + sin(D))) = (4a sin(D)) / (√3 cos(D) + sin(D))Now, I can factor out cos(D) from the denominator:BM = (4a sin(D)) / [cos(D)(√3 + tan(D))]Because sin(D)/cos(D) is tan(D), so:BM = (4a tan(D)) / (√3 + tan(D))Okay, that's the expression for BM. That matches the first part of what I needed to prove.Now, let's move on to CN. So, CN is the segment on AC from C to N. I need to find CN in terms of a and tan(D).Similarly, I can consider triangle CND. Let's see. In triangle CND, we have side CD = a, angle at D is angle CDM, which is the same as angle BDM, which is D. Wait, is that correct?Wait, angle CDM is part of line DM, so angle CDM is the same as angle BDM, which is D. So, in triangle CND, angle at D is D, side CD is a, and we need to find CN.What's the angle at C in triangle CND? In triangle ABC, angle at C is 60 degrees, but since D is on the extension of BC, the angle at C in triangle CND might be different.Wait, actually, in triangle CND, the angles should add up to 180 degrees. So, angle at D is D, angle at C is... Hmm, let's think.Point N is on AC, so line CN is part of AC. The angle at C in triangle CND is the same as angle ACB, which is 60 degrees, but since D is on the extension, maybe it's different.Wait, no. Because in triangle CND, the angle at C is actually the same as angle ACB, which is 60 degrees, because AC is a side of the original triangle. So, angle at C is 60 degrees.Wait, but actually, when you extend BC to D, the angle at C in triangle CND is still 60 degrees because AC is fixed. So, in triangle CND, angle at C is 60 degrees, angle at D is D, so angle at N is 180 - 60 - D = 120 - D degrees.Wait, but that seems similar to triangle BMD. Hmm.Wait, no, actually, in triangle CND, angle at N is 180 - angle at C - angle at D = 180 - 60 - D = 120 - D. So, similar to triangle BMD.So, using the Law of Sines on triangle CND:CN / sin(D) = CD / sin(angle at N)So, CN / sin(D) = a / sin(120 - D)Therefore, CN = (a sin(D)) / sin(120 - D)But wait, earlier, for BM, we had BM = (4a sin(D)) / (√3 cos(D) + sin(D)). For CN, it's similar but with a instead of 2a.Wait, let me double-check. In triangle CND, side CD is a, angle at D is D, angle at C is 60 degrees, angle at N is 120 - D.So, Law of Sines:CN / sin(D) = CD / sin(angle at N) = a / sin(120 - D)So, CN = (a sin(D)) / sin(120 - D)Again, sin(120 - D) is the same as before: (√3/2)cos(D) + (1/2)sin(D)So, CN = (a sin(D)) / [ (√3/2)cos(D) + (1/2)sin(D) ]Factor out 1/2 from the denominator:CN = (a sin(D)) / [ (1/2)(√3 cos(D) + sin(D)) ) ] = (a sin(D)) * (2 / (√3 cos(D) + sin(D))) = (2a sin(D)) / (√3 cos(D) + sin(D))Now, similar to BM, I can factor out cos(D):CN = (2a sin(D)) / [cos(D)(√3 + tan(D))]Which simplifies to:CN = (2a tan(D)) / (√3 + tan(D))Wait, but that's not matching the expression given in the problem, which is (2a tan D)/(√3 - tan D). Hmm, so I must have made a mistake here.Wait, let me go back. Maybe I messed up the angle at N. Let me think again about triangle CND.In triangle CND, angle at D is D, angle at C is 60 degrees, so angle at N is 180 - D - 60 = 120 - D. So, that part is correct.But wait, when I applied the Law of Sines, I might have confused the sides. Let me write it again:In triangle CND:- CN is opposite angle D- CD is opposite angle N- So, CN / sin(angle at D) = CD / sin(angle at N)Which is CN / sin(D) = a / sin(120 - D)So, CN = (a sin(D)) / sin(120 - D)Which is the same as before.But wait, in triangle BMD, we had BM = (4a sin(D)) / sin(120 - D), and here CN is (a sin(D)) / sin(120 - D). So, BM is four times CN? That doesn't seem right because BM and CN are on different sides.Wait, maybe I made a mistake in the angle at N. Let me think again.Wait, in triangle CND, angle at N is actually 180 - angle at C - angle at D. But angle at C is not 60 degrees because in triangle CND, point N is on AC, which is a different configuration.Wait, no, angle at C in triangle CND is still 60 degrees because AC is a side of the original equilateral triangle. So, angle at C is 60 degrees.Wait, maybe the issue is with the Law of Sines. Let me check.Wait, in triangle CND, CN is opposite angle D, CD is opposite angle N, and DN is opposite angle C.So, Law of Sines:CN / sin(D) = CD / sin(angle N) = DN / sin(60)So, CN = (CD sin(D)) / sin(angle N)But angle N is 120 - D, so:CN = (a sin(D)) / sin(120 - D)Which is the same as before.Hmm, but according to the problem, CN should be (2a tan D)/(√3 - tan D). So, my result is different. Maybe I need to consider a different triangle or approach.Wait, perhaps I should consider triangle DMC instead of CND. Or maybe use coordinate geometry.Alternatively, maybe I should use Menelaus' theorem for the transversal DM cutting through triangle ABC.Menelaus' theorem states that for a transversal cutting through the sides of a triangle, the product of the segments is equal to 1.So, for triangle ABC with transversal DMN, Menelaus' theorem would give:(BM / MA) * (AN / NC) * (CD / DB) = 1Wait, but I'm not sure about the exact formulation. Let me recall Menelaus' theorem.Menelaus' theorem for triangle ABC with a transversal line cutting through AB at M, BC at some point, and AC at N. But in this case, the transversal DM starts at D, which is on the extension of BC, so it's cutting AB at M and AC at N.So, Menelaus' theorem would state:(BM / MA) * (AN / NC) * (CD / DB) = 1Wait, but CD is a, DB is 2a, so CD / DB = a / 2a = 1/2.So, (BM / MA) * (AN / NC) * (1/2) = 1Therefore, (BM / MA) * (AN / NC) = 2But I need to find BM and CN. Maybe I can express AN in terms of CN.Since AN + NC = AC = a, so AN = a - NC.So, (BM / MA) * ((a - NC) / NC) = 2But I also know that BM + MA = AB = a, so MA = a - BM.Therefore, (BM / (a - BM)) * ((a - NC) / NC) = 2This seems a bit complicated, but maybe I can find another relation.Alternatively, maybe using coordinate geometry would help. Let me assign coordinates to the points.Let me place point B at (0, 0), point C at (a, 0), and since ABC is equilateral, point A will be at (a/2, (√3/2)a).Point D is on the extension of BC beyond C, so since BC is from (a, 0) to (0, 0), the extension beyond C would be along the x-axis. Since CD = a, point D would be at (2a, 0).So, coordinates:- B: (0, 0)- C: (a, 0)- A: (a/2, (√3/2)a)- D: (2a, 0)Now, line DM connects D(2a, 0) to M on AB and N on AC.Let me find the equation of line DM. Let me denote point M as (x1, y1) on AB and point N as (x2, y2) on AC.But since DM is a straight line, I can parametrize it. Let me use parameter t such that when t=0, we are at D(2a, 0), and as t increases, we move towards M and N.The parametric equations for DM can be written as:x = 2a + t*(x_M - 2a)y = 0 + t*(y_M - 0) = t*y_MBut I need to find the coordinates of M and N.Alternatively, I can write the equation of line DM in slope-intercept form. Let me find the slope.Let me denote the slope as m. Since D is at (2a, 0), and M is on AB, which goes from (0,0) to (a/2, (√3/2)a). So, the slope of AB is ((√3/2)a - 0)/(a/2 - 0) = √3.So, AB has slope √3. Similarly, AC has slope ((√3/2)a - 0)/(a/2 - a) = ((√3/2)a)/(-a/2) = -√3.So, line DM intersects AB at M and AC at N. Let me find the equation of DM.Let me denote the slope of DM as m. Then, the equation of DM is y = m(x - 2a).This line intersects AB at M and AC at N.First, find intersection with AB. AB has equation y = √3 x.So, setting y = m(x - 2a) equal to y = √3 x:√3 x = m(x - 2a)√3 x = m x - 2a m√3 x - m x = -2a mx(√3 - m) = -2a mx = (-2a m)/(√3 - m)Then, y = √3 x = √3*(-2a m)/(√3 - m) = (-2a m √3)/(√3 - m)So, coordinates of M are:M = ( (-2a m)/(√3 - m), (-2a m √3)/(√3 - m) )Similarly, find intersection with AC. AC has equation y = -√3(x - a)So, set y = m(x - 2a) equal to y = -√3(x - a):m(x - 2a) = -√3(x - a)m x - 2a m = -√3 x + √3 am x + √3 x = 2a m + √3 ax(m + √3) = a(2m + √3)x = a(2m + √3)/(m + √3)Then, y = m(x - 2a) = m( a(2m + √3)/(m + √3) - 2a ) = m( [a(2m + √3) - 2a(m + √3)] / (m + √3) )Simplify numerator:a(2m + √3) - 2a(m + √3) = 2a m + a√3 - 2a m - 2a√3 = -a√3So, y = m*(-a√3)/(m + √3) = (-a m √3)/(m + √3)Thus, coordinates of N are:N = ( a(2m + √3)/(m + √3), (-a m √3)/(m + √3) )Now, I need to find BM and CN in terms of m, and then relate m to angle D.First, let's find BM. Point B is at (0,0), and M is at ( (-2a m)/(√3 - m), (-2a m √3)/(√3 - m) )So, vector BM is from B(0,0) to M, so BM has coordinates ( (-2a m)/(√3 - m), (-2a m √3)/(√3 - m) )The length BM is the distance from B to M:BM = sqrt[ ( (-2a m)/(√3 - m) )^2 + ( (-2a m √3)/(√3 - m) )^2 ]Factor out (2a m / (√3 - m))^2:BM = sqrt[ (4a² m² / (√3 - m)^2)(1 + 3) ] = sqrt[ (4a² m² / (√3 - m)^2)*4 ] = sqrt[ 16a² m² / (√3 - m)^2 ] = 4a |m| / |√3 - m|Since m is the slope of DM, which is a line going from D(2a,0) to M on AB and N on AC. Given the configuration, m is negative because DM goes from D(2a,0) down to AB and AC, which are above the x-axis. So, m is negative.Thus, BM = 4a (-m) / (√3 - m) = 4a |m| / (√3 - m)But since m is negative, let me write m = -k where k > 0.So, BM = 4a k / (√3 + k)Now, I need to express k in terms of tan(D). Angle D is angle BDM, which is the angle at D between BD and DM.BD is along the x-axis from D(2a,0) to B(0,0), so BD is along the negative x-axis. DM is the line from D(2a,0) to M, which has slope m.So, the angle D is the angle between BD (which is along negative x-axis) and DM (which has slope m). So, the angle between negative x-axis and DM is D.The slope m is tan(theta), where theta is the angle between DM and positive x-axis. But since DM is going from D(2a,0) to M, which is above the x-axis, theta is measured from positive x-axis upwards. However, since BD is along negative x-axis, the angle D is 180 - theta.Wait, no. Let me clarify.The angle at D, angle BDM, is the angle between BD and DM. BD is from D to B, which is along the negative x-axis. DM is from D to M, which is going upwards into the first quadrant (since M is on AB, which is above the x-axis). So, the angle between negative x-axis and DM is D.So, the slope m is tan(180 - D), because the angle between positive x-axis and DM is 180 - D, since it's measured from negative x-axis.But tan(180 - D) = -tan(D). So, m = tan(theta) = tan(180 - D) = -tan(D)Therefore, m = -tan(D). So, k = |m| = tan(D)Thus, BM = 4a k / (√3 + k) = 4a tan(D) / (√3 + tan(D))Which matches the first part of what I needed to prove.Now, let's find CN. Point C is at (a, 0), and N is at ( a(2m + √3)/(m + √3), (-a m √3)/(m + √3) )So, vector CN is from C(a,0) to N, so coordinates are:x-coordinate: a(2m + √3)/(m + √3) - a = a[ (2m + √3) - (m + √3) ] / (m + √3) = a(m) / (m + √3)y-coordinate: (-a m √3)/(m + √3) - 0 = (-a m √3)/(m + √3)So, vector CN has components ( a m / (m + √3), (-a m √3)/(m + √3) )The length CN is the distance from C to N:CN = sqrt[ (a m / (m + √3))^2 + ( (-a m √3)/(m + √3) )^2 ]Factor out (a m / (m + √3))^2:CN = sqrt[ (a² m² / (m + √3)^2)(1 + 3) ] = sqrt[ (a² m² / (m + √3)^2)*4 ] = sqrt[ 4a² m² / (m + √3)^2 ] = 2a |m| / |m + √3|Again, since m is negative, let me write m = -k where k > 0.So, CN = 2a k / ( -k + √3 ) = 2a k / (√3 - k )But k = |m| = tan(D), so:CN = 2a tan(D) / (√3 - tan(D))Which matches the second part of what I needed to prove.So, both BM and CN have been derived correctly using coordinate geometry and Menelaus' theorem, confirming the given expressions.

answer：First, calculate the total cost of purchasing the markers:[text{Total Cost} = 2000 times 0.20 = 400 text{ dollars}]Next, determine the required total revenue to achieve a profit of at least 200:[text{Total Revenue Needed} = text{Total Cost} + text{Desired Profit} = 400 + 200 = 600 text{ dollars}]Finally, calculate the minimum number of markers that need to be sold at 0.50 each to reach the required revenue:[text{Number of Markers to Sell} = frac{600}{0.50} = 1200]Therefore, the store owner must sell at least 1200 markers to achieve a profit of at least 200.

answer：Okay, so I have this geometry problem here where I need to construct triangle (ABC) given side (BC), the altitude (AA_1), and it's known that the angle bisector of angle (A) is the median of triangle (AA_1D), where (D) is the midpoint of side (BC). Hmm, that sounds a bit complicated, but let me try to break it down step by step.First, let me visualize what's given. I have side (BC), which is fixed. Then, there's an altitude from (A) to (BC), which is (AA_1). So, (A_1) must be the foot of the altitude on (BC). Also, (D) is the midpoint of (BC), so (BD = DC). Now, the key part is that the angle bisector of angle (A) is the median of triangle (AA_1D). That means, in triangle (AA_1D), the median from (A) (which would go to the midpoint of (A_1D)) is also the angle bisector of angle (A).Wait, so in triangle (AA_1D), the median and the angle bisector from (A) coincide. That must mean that triangle (AA_1D) is isosceles with respect to (A), right? Because in a triangle, if the median and angle bisector from a vertex coincide, the triangle is isosceles with the two sides from that vertex being equal. So, does that mean (AA_1 = AD)?Let me think. In triangle (AA_1D), if the median from (A) is also the angle bisector, then sides (AA_1) and (AD) must be equal. So, (AA_1 = AD). That gives me a relationship between the altitude (AA_1) and the length (AD).Since (D) is the midpoint of (BC), (BD = DC = frac{BC}{2}). So, if I can express (AD) in terms of (BC) and (AA_1), maybe I can find some relationship.Let me try to draw triangle (ABC). I'll start by drawing side (BC). Let me mark point (D) as the midpoint. Then, I'll draw the altitude (AA_1) from (A) to (BC). Since (AA_1) is an altitude, it is perpendicular to (BC). Now, (AA_1) is given, so I can mark that length.Now, considering triangle (AA_1D), since (AA_1 = AD), triangle (AA_1D) is isosceles with equal sides (AA_1) and (AD). That means angles opposite these sides are equal. So, angle (AA_1D) is equal to angle (AD A_1). Hmm, not sure if that helps directly, but it's a property to keep in mind.Wait, maybe I can use coordinates to model this problem. Let me place point (B) at ((-b, 0)) and point (C) at ((b, 0)), so that (BC) is centered at the origin, and (D), the midpoint, is at ((0, 0)). Then, point (A) will be somewhere in the plane, and (A_1) is the foot of the altitude from (A) to (BC). Since (BC) is on the x-axis, (A_1) will have the same x-coordinate as (A) but y-coordinate 0.Let me denote point (A) as ((x, y)). Then, (A_1) is ((x, 0)). The altitude (AA_1) has length (y), so (AA_1 = y). Now, (AD) is the distance from (A) to (D), which is (sqrt{x^2 + y^2}). Since (AA_1 = AD), we have:[y = sqrt{x^2 + y^2}]Wait, that can't be right. If I square both sides, I get:[y^2 = x^2 + y^2 implies x^2 = 0 implies x = 0]So, point (A) must be directly above (D), which is the midpoint of (BC). That would mean (A) is at ((0, y)). But then, triangle (ABC) would be an isosceles triangle with (AB = AC). Is that necessarily the case?Wait, but if (A) is directly above (D), then the altitude (AA_1) is also the median and the angle bisector. So, in that case, the angle bisector of angle (A) is indeed the median (AD) of triangle (AA_1D). But is that the only possibility?Wait, let me think again. I assumed that (AA_1 = AD), but is that necessarily true? Because in triangle (AA_1D), the median and angle bisector coincide, which implies that (AA_1 = AD). So, yes, that seems to be the case.But if (AA_1 = AD), then (A) must lie on the perpendicular bisector of (A_1D). Since (D) is the midpoint of (BC), and (A_1) is the foot of the altitude, which is also on (BC), then (A_1D) is a segment on (BC). The perpendicular bisector of (A_1D) would be the line perpendicular to (BC) at the midpoint of (A_1D).Wait, but (A) is already on the perpendicular to (BC) at (A_1). So, unless (A_1D) is symmetric around (D), which would mean (A_1 = D), but that would make (AA_1 = AD) automatically. Hmm, this is getting a bit confusing.Let me try to approach this differently. Since (D) is the midpoint of (BC), and (AA_1) is the altitude, then in triangle (ABC), (AA_1) is the altitude, and (AD) is the median. The problem states that in triangle (AA_1D), the angle bisector of angle (A) is the median. So, in triangle (AA_1D), the median from (A) is also the angle bisector.In triangle (AA_1D), the median from (A) would go to the midpoint of (A_1D). Let me denote the midpoint of (A_1D) as (M). So, (M) is the midpoint of (A_1D), which is also on (BC). The median from (A) is (AM), and it's also the angle bisector of angle (A).So, in triangle (AA_1D), (AM) is both a median and an angle bisector. That implies that triangle (AA_1D) is isosceles with (AA_1 = AD). Therefore, (AA_1 = AD).So, (AD = AA_1). Since (D) is the midpoint of (BC), (AD) is the median of triangle (ABC). So, in triangle (ABC), the median (AD) is equal to the altitude (AA_1).Hmm, that's an interesting relationship. So, in triangle (ABC), the median from (A) is equal to the altitude from (A). That must impose some specific conditions on the triangle.Let me recall that in a triangle, if the median and altitude from the same vertex coincide, the triangle is isosceles with respect to that vertex. So, does that mean triangle (ABC) is isosceles with (AB = AC)?Wait, but if (AD) is both a median and an altitude, then yes, triangle (ABC) must be isosceles with (AB = AC). So, point (A) must lie on the perpendicular bisector of (BC), which is the line through (D) perpendicular to (BC).But in our case, (AA_1) is the altitude, which is also equal to (AD). So, (AA_1 = AD), which implies that (A) is at a distance (AA_1) from (D). Since (AA_1) is the altitude, and (D) is the midpoint, this suggests that (A) is located such that its distance to (D) is equal to the length of the altitude.Wait, but if (A) is on the perpendicular bisector of (BC), then (AD) is the altitude, which is given. So, if (AA_1 = AD), then (A) must be at a specific position where its distance to (D) is equal to the altitude.Wait, but (AA_1) is the altitude, so (AA_1) is the length from (A) to (BC). If (AD = AA_1), then (AD) is equal to the altitude. So, point (A) is located such that its distance to (D) is equal to its height above (BC).Let me try to model this with coordinates again. Let me place (B) at ((-b, 0)), (C) at ((b, 0)), so (D) is at ((0, 0)). Let me denote point (A) as ((0, h)), since it's on the perpendicular bisector. Then, the altitude (AA_1) is from ((0, h)) to ((0, 0)), so (AA_1 = h). The median (AD) is from ((0, h)) to ((0, 0)), so (AD = h). Therefore, (AA_1 = AD = h), which satisfies the condition.But wait, in this case, triangle (ABC) is isosceles with (AB = AC). So, does this mean that the only possible triangle (ABC) satisfying the given conditions is an isosceles triangle with (AB = AC)?But the problem doesn't specify that (ABC) is isosceles, so maybe there are other possibilities. Hmm, maybe I made a wrong assumption.Wait, let's go back. I assumed that in triangle (AA_1D), the median and angle bisector coincide, implying (AA_1 = AD). But maybe there's another way to interpret this.Alternatively, perhaps the angle bisector of angle (A) in triangle (ABC) is the median of triangle (AA_1D). So, in triangle (AA_1D), the median from (A) is the angle bisector of angle (A) in triangle (ABC). That might be a different interpretation.Wait, the problem says: "the angle bisector of (A) is the median of the triangle (AA_1D), where (D) is the midpoint of the side (BC)." So, in triangle (AA_1D), the median is the angle bisector of angle (A) in triangle (ABC). So, perhaps the median in triangle (AA_1D) is the same as the angle bisector in triangle (ABC).So, in triangle (AA_1D), the median from (A) is the line from (A) to the midpoint of (A_1D). Let's denote the midpoint of (A_1D) as (M). So, (AM) is the median of triangle (AA_1D). The problem states that this median (AM) is the angle bisector of angle (A) in triangle (ABC).So, in triangle (ABC), the angle bisector of angle (A) is the line (AM), where (M) is the midpoint of (A_1D). Therefore, (AM) is both the median of triangle (AA_1D) and the angle bisector of triangle (ABC).This seems more precise. So, in triangle (ABC), the angle bisector of angle (A) is the line from (A) to the midpoint of (A_1D), which is (M). Therefore, (AM) is the angle bisector.So, perhaps I can use the Angle Bisector Theorem here. The Angle Bisector Theorem states that the angle bisector divides the opposite side in the ratio of the adjacent sides. So, in triangle (ABC), the angle bisector (AM) divides side (BC) into segments (BM) and (MC) such that:[frac{AB}{AC} = frac{BM}{MC}]But wait, (M) is the midpoint of (A_1D), not necessarily of (BC). So, (M) is a point on (BC), but it's the midpoint of (A_1D). Since (D) is the midpoint of (BC), and (A_1) is the foot of the altitude from (A), (A_1D) is a segment on (BC). So, (M) is the midpoint of (A_1D), which is a point between (A_1) and (D).Therefore, (M) divides (A_1D) into two equal parts. So, (BM = BA_1 + A_1M = BA_1 + frac{A_1D}{2}), and (MC = MD + DC = frac{A_1D}{2} + DC). But since (D) is the midpoint of (BC), (BD = DC), so (DC = frac{BC}{2}). Also, (A_1D = |A_1 - D|), but since (A_1) is the foot of the altitude, it's somewhere on (BC).Wait, maybe I can express (A_1D) in terms of (BC). Let me denote (BC = 2b), so (BD = DC = b). Let me denote (BA_1 = x), so (A_1C = 2b - x). Then, (A_1D = |x - b|), since (D) is at (b) from (B).So, (A_1D = |x - b|). Then, (M) is the midpoint of (A_1D), so (A_1M = frac{|x - b|}{2}). Therefore, the position of (M) on (BC) is (BA_1 + A_1M = x + frac{|x - b|}{2}).Wait, this is getting a bit messy. Maybe I can use coordinate geometry again. Let me place (B) at ((-b, 0)), (C) at ((b, 0)), so (D) is at ((0, 0)). Let me denote (A) as ((p, q)), so (A_1) is the foot of the altitude from (A) to (BC), which is the x-axis. Therefore, (A_1) is ((p, 0)).Now, the midpoint (M) of (A_1D) is the midpoint between ((p, 0)) and ((0, 0)), which is ((frac{p}{2}, 0)). So, point (M) is at ((frac{p}{2}, 0)).Now, the angle bisector of angle (A) in triangle (ABC) is the line from (A) to (M), which is from ((p, q)) to ((frac{p}{2}, 0)). So, the slope of this angle bisector is:[m = frac{0 - q}{frac{p}{2} - p} = frac{-q}{-frac{p}{2}} = frac{2q}{p}]Now, according to the Angle Bisector Theorem in triangle (ABC), the angle bisector from (A) divides (BC) into segments proportional to the adjacent sides. So,[frac{AB}{AC} = frac{BM}{MC}]Where (BM = frac{p}{2} - (-b) = frac{p}{2} + b) and (MC = b - frac{p}{2}). So,[frac{AB}{AC} = frac{frac{p}{2} + b}{b - frac{p}{2}} = frac{p + 2b}{2b - p}]Now, let's compute (AB) and (AC). (AB) is the distance from (A) to (B):[AB = sqrt{(p + b)^2 + q^2}]Similarly, (AC = sqrt{(p - b)^2 + q^2})So,[frac{sqrt{(p + b)^2 + q^2}}{sqrt{(p - b)^2 + q^2}} = frac{p + 2b}{2b - p}]Let me square both sides to eliminate the square roots:[frac{(p + b)^2 + q^2}{(p - b)^2 + q^2} = left(frac{p + 2b}{2b - p}right)^2]Simplify the right-hand side:[left(frac{p + 2b}{2b - p}right)^2 = frac{(p + 2b)^2}{(2b - p)^2} = frac{(p + 2b)^2}{(2b - p)^2}]So, we have:[frac{(p + b)^2 + q^2}{(p - b)^2 + q^2} = frac{(p + 2b)^2}{(2b - p)^2}]Cross-multiplying:[[(p + b)^2 + q^2](2b - p)^2 = [(p - b)^2 + q^2](p + 2b)^2]This looks complicated, but maybe we can expand both sides and see if terms cancel out.First, let me denote ( (p + b)^2 = p^2 + 2pb + b^2 ) and ( (p - b)^2 = p^2 - 2pb + b^2 ). Similarly, ( (2b - p)^2 = 4b^2 - 4pb + p^2 ) and ( (p + 2b)^2 = p^2 + 4pb + 4b^2 ).So, expanding the left-hand side:[[(p^2 + 2pb + b^2) + q^2](4b^2 - 4pb + p^2)]Similarly, the right-hand side:[[(p^2 - 2pb + b^2) + q^2](p^2 + 4pb + 4b^2)]This is going to be very tedious, but let me try to proceed.First, expand the left-hand side:Let me denote ( A = (p^2 + 2pb + b^2) + q^2 = p^2 + 2pb + b^2 + q^2 )and ( B = (4b^2 - 4pb + p^2) )So, ( A times B = (p^2 + 2pb + b^2 + q^2)(p^2 - 4pb + 4b^2) )Similarly, the right-hand side:Denote ( C = (p^2 - 2pb + b^2) + q^2 = p^2 - 2pb + b^2 + q^2 )and ( D = (p^2 + 4pb + 4b^2) )So, ( C times D = (p^2 - 2pb + b^2 + q^2)(p^2 + 4pb + 4b^2) )Now, let's compute both products.Starting with the left-hand side:( (p^2 + 2pb + b^2 + q^2)(p^2 - 4pb + 4b^2) )Multiply term by term:First, ( p^2 times (p^2 - 4pb + 4b^2) = p^4 - 4p^3b + 4p^2b^2 )Second, ( 2pb times (p^2 - 4pb + 4b^2) = 2p^3b - 8p^2b^2 + 8pb^3 )Third, ( b^2 times (p^2 - 4pb + 4b^2) = p^2b^2 - 4pb^3 + 4b^4 )Fourth, ( q^2 times (p^2 - 4pb + 4b^2) = p^2q^2 - 4pbq^2 + 4b^2q^2 )Now, add all these together:( p^4 - 4p^3b + 4p^2b^2 + 2p^3b - 8p^2b^2 + 8pb^3 + p^2b^2 - 4pb^3 + 4b^4 + p^2q^2 - 4pbq^2 + 4b^2q^2 )Combine like terms:- ( p^4 )- ( (-4p^3b + 2p^3b) = -2p^3b )- ( (4p^2b^2 - 8p^2b^2 + p^2b^2) = (-3p^2b^2) )- ( (8pb^3 - 4pb^3) = 4pb^3 )- ( 4b^4 )- ( p^2q^2 )- ( -4pbq^2 )- ( 4b^2q^2 )So, left-hand side becomes:[p^4 - 2p^3b - 3p^2b^2 + 4pb^3 + 4b^4 + p^2q^2 - 4pbq^2 + 4b^2q^2]Now, let's compute the right-hand side:( (p^2 - 2pb + b^2 + q^2)(p^2 + 4pb + 4b^2) )Again, multiply term by term:First, ( p^2 times (p^2 + 4pb + 4b^2) = p^4 + 4p^3b + 4p^2b^2 )Second, ( -2pb times (p^2 + 4pb + 4b^2) = -2p^3b - 8p^2b^2 - 8pb^3 )Third, ( b^2 times (p^2 + 4pb + 4b^2) = p^2b^2 + 4pb^3 + 4b^4 )Fourth, ( q^2 times (p^2 + 4pb + 4b^2) = p^2q^2 + 4pbq^2 + 4b^2q^2 )Now, add all these together:( p^4 + 4p^3b + 4p^2b^2 - 2p^3b - 8p^2b^2 - 8pb^3 + p^2b^2 + 4pb^3 + 4b^4 + p^2q^2 + 4pbq^2 + 4b^2q^2 )Combine like terms:- ( p^4 )- ( (4p^3b - 2p^3b) = 2p^3b )- ( (4p^2b^2 - 8p^2b^2 + p^2b^2) = (-3p^2b^2) )- ( (-8pb^3 + 4pb^3) = (-4pb^3) )- ( 4b^4 )- ( p^2q^2 )- ( 4pbq^2 )- ( 4b^2q^2 )So, right-hand side becomes:[p^4 + 2p^3b - 3p^2b^2 - 4pb^3 + 4b^4 + p^2q^2 + 4pbq^2 + 4b^2q^2]Now, set left-hand side equal to right-hand side:[p^4 - 2p^3b - 3p^2b^2 + 4pb^3 + 4b^4 + p^2q^2 - 4pbq^2 + 4b^2q^2 = p^4 + 2p^3b - 3p^2b^2 - 4pb^3 + 4b^4 + p^2q^2 + 4pbq^2 + 4b^2q^2]Subtract ( p^4 + 2p^3b - 3p^2b^2 - 4pb^3 + 4b^4 + p^2q^2 + 4pbq^2 + 4b^2q^2 ) from both sides:Left-hand side minus right-hand side:[(p^4 - 2p^3b - 3p^2b^2 + 4pb^3 + 4b^4 + p^2q^2 - 4pbq^2 + 4b^2q^2) - (p^4 + 2p^3b - 3p^2b^2 - 4pb^3 + 4b^4 + p^2q^2 + 4pbq^2 + 4b^2q^2) = 0]Simplify term by term:- ( p^4 - p^4 = 0 )- ( -2p^3b - 2p^3b = -4p^3b )- ( -3p^2b^2 + 3p^2b^2 = 0 )- ( 4pb^3 + 4pb^3 = 8pb^3 )- ( 4b^4 - 4b^4 = 0 )- ( p^2q^2 - p^2q^2 = 0 )- ( -4pbq^2 - 4pbq^2 = -8pbq^2 )- ( 4b^2q^2 - 4b^2q^2 = 0 )So, the equation simplifies to:[-4p^3b + 8pb^3 - 8pbq^2 = 0]Factor out common terms:[-4pb(p^2 - 2b^2 + 2q^2) = 0]So, either:1. ( -4pb = 0 ) which implies ( p = 0 ) or ( b = 0 ). But ( b ) is half the length of ( BC ), which is given, so ( b neq 0 ). Therefore, ( p = 0 ).Or,2. ( p^2 - 2b^2 + 2q^2 = 0 )Let's analyze both cases.Case 1: ( p = 0 )If ( p = 0 ), then point ( A ) is at ((0, q)). So, ( A_1 ) is at ((0, 0)), which is point ( D ). Therefore, ( AA_1 = AD = q ). So, in this case, triangle ( ABC ) is isosceles with ( AB = AC ), as ( A ) is directly above ( D ).Case 2: ( p^2 - 2b^2 + 2q^2 = 0 )So,[p^2 + 2q^2 = 2b^2]This is the equation of an ellipse. So, point ( A ) lies on an ellipse with major axis along the y-axis and minor axis along the x-axis.But we also know that ( AA_1 = q ) is given. So, ( q ) is fixed. Therefore, ( p^2 = 2b^2 - 2q^2 ). So,[p = pm sqrt{2b^2 - 2q^2}]Therefore, point ( A ) can be at ((sqrt{2b^2 - 2q^2}, q)) or ((- sqrt{2b^2 - 2q^2}, q)).But we need to ensure that ( 2b^2 - 2q^2 geq 0 ), so ( b^2 geq q^2 ), which implies ( b geq q ). Since ( b = frac{BC}{2} ), this means ( frac{BC}{2} geq AA_1 ), or ( BC geq 2 AA_1 ).So, in this case, triangle ( ABC ) can be constructed only if ( BC geq 2 AA_1 ). Otherwise, there is no such triangle.Therefore, combining both cases, we have two possibilities:1. ( ABC ) is isosceles with ( AB = AC ), which occurs when ( p = 0 ).2. ( ABC ) is a triangle where ( A ) lies on the ellipse ( p^2 + 2q^2 = 2b^2 ), provided ( BC geq 2 AA_1 ).But wait, in the first case, when ( p = 0 ), we have ( AA_1 = AD = q ), which is the altitude. So, in this case, ( BC ) can be any length, but ( AA_1 ) is fixed. However, in the second case, we have a condition ( BC geq 2 AA_1 ).But the problem states that we need to construct triangle ( ABC ) given ( BC ), ( AA_1 ), and the angle bisector condition. So, depending on the lengths of ( BC ) and ( AA_1 ), we might have one or two solutions.If ( BC = 2 AA_1 ), then ( p = 0 ) is the only solution, as the ellipse collapses to a single point. If ( BC > 2 AA_1 ), then there are two possible triangles, one on each side of the perpendicular bisector. If ( BC < 2 AA_1 ), then no such triangle exists.Therefore, the construction is possible only if ( BC geq 2 AA_1 ).So, to summarize, the triangle ( ABC ) can be constructed if and only if ( BC geq 2 AA_1 ). If this condition is satisfied, there are two possible triangles (one on each side of the perpendicular bisector of ( BC )), and if ( BC = 2 AA_1 ), there is exactly one triangle.Therefore, the final condition is ( BC geq 2 AA_1 ).