💡First, I need to determine the number of eyes for each family member.The mom has 1 eye.The dad has 3 eyes.There are 3 kids, and each kid has 4 eyes.To find the total number of eyes for the kids, I multiply the number of kids by the number of eyes each kid has: 3 kids * 4 eyes = 12 eyes.Next, I add the eyes of the mom, dad, and kids together: 1 (mom) + 3 (dad) + 12 (kids) = 16 eyes.Therefore, the whole monster family has a total of 16 eyes.

💡Okay, so I have this problem about an isosceles trapezoid, and I need to find the length of segment AC. Hmm, let me try to visualize it first. The Asymptote code shows that the trapezoid has points A, B, C, D with coordinates A=(0,0), B=(24,0), C=(17,10), and D=(7,10). The sides are labeled with lengths: CD is 10, DA is 12, AB is 24, and BC is 12. Alright, so it's an isosceles trapezoid, which means the non-parallel sides (the legs) are equal in length. In this case, DA and BC are both 12 units, so that checks out. The bases are AB and CD, with AB being 24 units and CD being 10 units. I remember that in an isosceles trapezoid, the legs are congruent, and the base angles are equal. Also, the diagonals are congruent, which might come in handy. But I need to find the length of AC. Let me think about how to approach this. Since it's a trapezoid, I can maybe drop some perpendiculars from the top base CD down to the bottom base AB to form some right triangles. That might help me find the height of the trapezoid, which could then be used to find the length of AC.So, if I imagine dropping a perpendicular from point C to AB, let's call the foot of that perpendicular E. Similarly, drop a perpendicular from D to AB, and call that foot F. Since it's an isosceles trapezoid, these two perpendiculars should be equal in length, and the segments AE and FB should also be equal. Given that AB is 24 units and CD is 10 units, the difference in the lengths of the bases is 24 - 10 = 14 units. Since the trapezoid is isosceles, this extra length is split equally on both sides. So, each of the segments AE and FB should be 14 / 2 = 7 units. So, AE is 7 units, and since AB is 24 units, that means the segment EF, which is directly under CD, is 24 - 2*7 = 10 units, which matches the length of CD. That makes sense.Now, I can consider triangle AEC, which is a right triangle with legs AE = 7 units and CE (the height of the trapezoid) unknown. The hypotenuse of this triangle is AC, which is the segment we need to find. Wait, but I don't know the height yet. How can I find CE? Well, I know that the legs of the trapezoid (DA and BC) are 12 units each. If I consider triangle DAF, which is also a right triangle, with DF (the height) and AF (which is 7 units) as legs, and DA as the hypotenuse. So, in triangle DAF, DA is 12 units, AF is 7 units, so I can use the Pythagorean theorem to find DF. Let me write that out:DA² = AF² + DF² 12² = 7² + DF² 144 = 49 + DF² DF² = 144 - 49 DF² = 95 DF = √95So, the height of the trapezoid is √95 units. Wait, but in the initial solution, it was mentioned that CE = DF = 10. Hmm, that doesn't match with my calculation here. Did I make a mistake? Let me check.Wait, in the Asymptote code, point C is at (17,10), so the y-coordinate is 10, which is the height. So, actually, the height is 10 units, not √95. Hmm, so maybe my approach is wrong.Wait, hold on. If point C is at (17,10), then the height is indeed 10 units. So, perhaps I confused something earlier. Let me re-examine.If the height is 10 units, then in triangle DAF, DA is 12 units, AF is 7 units, and DF is 10 units. Let me verify that:DA² = AF² + DF² 12² = 7² + 10² 144 = 49 + 100 144 = 149Wait, that's not correct. 49 + 100 is 149, which is not equal to 144. So, that can't be right. There must be a mistake here.Hmm, maybe I misapplied the Pythagorean theorem. Let me think again. If the height is 10 units, then DF is 10 units, AF is 7 units, so DA should be sqrt(7² + 10²) = sqrt(49 + 100) = sqrt(149). But DA is given as 12 units. So, sqrt(149) is approximately 12.20655, which is not exactly 12. So, that's a problem.Wait, but in the Asymptote code, the coordinates are given as A=(0,0), B=(24,0), C=(17,10), D=(7,10). So, the height is indeed 10 units because the y-coordinate of C and D is 10. So, maybe the side lengths are not exactly matching? Or perhaps I'm miscalculating something.Wait, let's calculate the length of DA. Point D is at (7,10), and point A is at (0,0). So, the distance between D and A is sqrt((7-0)^2 + (10-0)^2) = sqrt(49 + 100) = sqrt(149), which is approximately 12.20655, not exactly 12. But in the problem, DA is labeled as 12. So, that seems inconsistent.Similarly, point C is at (17,10), and point B is at (24,0). So, the distance between C and B is sqrt((24-17)^2 + (0-10)^2) = sqrt(49 + 100) = sqrt(149), which is also approximately 12.20655, not exactly 12. So, this is confusing because the problem states that DA and BC are 12 units, but according to the coordinates, they are sqrt(149).Wait, maybe the Asymptote code is just for illustration, and the actual problem has DA and BC as 12 units, and AB as 24, CD as 10. So, perhaps I should ignore the coordinates and just go by the given side lengths.So, let's start over. We have an isosceles trapezoid with bases AB = 24 and CD = 10, and legs DA and BC = 12 each. We need to find the length of diagonal AC.Since it's an isosceles trapezoid, the legs are equal, and the base angles are equal. If I drop perpendiculars from C and D to AB, let's call the feet E and F respectively. Then, EF = CD = 10. The remaining parts of AB are AF and EB, each equal to (AB - EF)/2 = (24 - 10)/2 = 7 units.So, AF = 7, FB = 7, EF = 10. Now, the height of the trapezoid can be found using the Pythagorean theorem on triangle DAF. DA is 12, AF is 7, so DF (the height) is sqrt(12² - 7²) = sqrt(144 - 49) = sqrt(95). So, the height is sqrt(95).Wait, but in the Asymptote code, the height was 10, but according to this, it's sqrt(95). Hmm, sqrt(95) is approximately 9.746, which is close to 10, but not exactly. Maybe the Asymptote code was just approximate.So, moving on. Now, to find AC, which is the diagonal from A to C. If I consider triangle AEC, where E is the foot of the perpendicular from C to AB. Then, AE is AF + FE, which is 7 + 10 = 17 units. Wait, no. AE is just AF, which is 7 units, because E is the foot from C, which is above F. Wait, no, actually, E is the foot from C, which is 17 units from A, as point C is at (17,10). Hmm, this is getting confusing.Wait, maybe I should use coordinates to find AC. If point A is at (0,0) and point C is at (17,10), then AC is the distance between these two points. So, AC = sqrt((17 - 0)^2 + (10 - 0)^2) = sqrt(289 + 100) = sqrt(389). Hmm, sqrt(389) is approximately 19.723, which is close to 19.6469 mentioned in the initial solution, but not exactly.Wait, but in the initial solution, it was mentioned that AC is 19√2, which is approximately 26.87, which is way off. So, that can't be right. So, maybe the initial solution had a mistake.Wait, hold on. Let me clarify. If the trapezoid is isosceles with AB = 24, CD = 10, and legs DA = BC = 12, then the coordinates might not be exactly as given in the Asymptote code. Because in the Asymptote code, the height is 10, but according to the side lengths, the height should be sqrt(95), which is approximately 9.746.So, perhaps the Asymptote code is just for illustration, and the exact coordinates are not to scale. Therefore, I should solve it based on the given side lengths, not the coordinates.So, let's proceed. We have AB = 24, CD = 10, DA = BC = 12. We drop perpendiculars from C and D to AB, meeting AB at E and F respectively. Then, EF = CD = 10, and AF = FB = (AB - EF)/2 = (24 - 10)/2 = 7 units.So, AF = 7, FB = 7, EF = 10. Now, the height h can be found using triangle DAF: DA² = AF² + h² => 12² = 7² + h² => 144 = 49 + h² => h² = 95 => h = sqrt(95).Now, to find AC, which is the diagonal from A to C. Point C is above point E, which is 17 units from A (since AF = 7, and FE = 10, so AE = AF + FE = 7 + 10 = 17). Wait, no, that's not correct. Because E is the foot from C, so AE is actually the horizontal distance from A to E, which is AF + FE. But AF is 7, and FE is the projection of CD onto AB, which is 10 units. So, AE = AF + FE = 7 + 10 = 17 units.Wait, but in reality, since CD is 10 units, and EF is 10 units, then E is 7 units away from F, which is 7 units from A. So, E is at 7 + 10 = 17 units from A. So, AE = 17 units, and the height is sqrt(95). Therefore, AC is the hypotenuse of triangle AEC, where AE = 17 and CE = sqrt(95). So, AC² = 17² + (sqrt(95))² = 289 + 95 = 384. Therefore, AC = sqrt(384).Simplify sqrt(384): 384 = 64 * 6, so sqrt(384) = 8*sqrt(6). So, AC = 8√6 units.Wait, but in the initial solution, it was mentioned that AC is 19.6469, which is approximately sqrt(384) ≈ 19.5959, which is close. But the exact value was given as 19√2, which is about 26.87, which is way off. So, the initial solution might have made a mistake.Wait, let me double-check my calculations. If AF = 7, then AE = AF + FE = 7 + 10 = 17. CE is the height, which is sqrt(95). So, AC² = 17² + (sqrt(95))² = 289 + 95 = 384. So, AC = sqrt(384) = 8√6. That seems correct.Alternatively, maybe I can use the formula for the diagonal of an isosceles trapezoid. The formula is: diagonal = sqrt(a² + b² - c²), where a and b are the lengths of the bases, and c is the length of the legs. Wait, no, that doesn't seem right.Wait, actually, the formula for the diagonal in an isosceles trapezoid can be derived using the Pythagorean theorem as I did earlier. So, I think my approach is correct.Wait, another way to think about it is using coordinates. If I set point A at (0,0), then point B is at (24,0). Since it's an isosceles trapezoid, the legs DA and BC are equal and symmetric. So, point D would be at (x, h) and point C at (24 - x, h), where h is the height.Given that CD = 10, the distance between D and C is 10 units. So, the distance between (x, h) and (24 - x, h) is |(24 - x) - x| = |24 - 2x| = 10. So, 24 - 2x = 10 => 2x = 14 => x = 7. So, point D is at (7, h) and point C is at (17, h).Now, the length of DA is 12 units. DA is the distance from A(0,0) to D(7,h). So, sqrt((7 - 0)^2 + (h - 0)^2) = sqrt(49 + h²) = 12. So, 49 + h² = 144 => h² = 95 => h = sqrt(95).Therefore, point C is at (17, sqrt(95)). So, AC is the distance from A(0,0) to C(17, sqrt(95)). So, AC = sqrt((17)^2 + (sqrt(95))^2) = sqrt(289 + 95) = sqrt(384) = 8√6.So, that confirms my earlier calculation. Therefore, the length of AC is 8√6 units.Wait, but in the initial solution, it was mentioned that AC is 19√2, which is about 26.87, but 8√6 is approximately 19.5959, which is close to 19.6469 mentioned. So, perhaps the initial solution had a rounding error or a miscalculation.Alternatively, maybe the initial solution considered the height as 10 units, which would make AC = sqrt(17² + 10²) = sqrt(289 + 100) = sqrt(389) ≈ 19.723, which is close to 19.6469. But in reality, the height is sqrt(95) ≈ 9.746, so AC is sqrt(17² + 95) = sqrt(384) ≈ 19.5959.So, the exact value is 8√6, which is approximately 19.5959. Therefore, the initial solution might have approximated it as 19.6469, but the exact value is 8√6.Wait, but in the initial solution, it was written as 19√2, which is about 26.87, which is incorrect. So, that must be a mistake.So, to summarize, the correct approach is:1. Recognize it's an isosceles trapezoid with AB = 24, CD = 10, and legs DA = BC = 12.2. Drop perpendiculars from C and D to AB, creating right triangles.3. Calculate the horizontal segments AF and FB as (24 - 10)/2 = 7 units each.4. Use Pythagorean theorem on triangle DAF to find the height h = sqrt(12² - 7²) = sqrt(95).5. Determine the horizontal distance from A to E (foot of perpendicular from C) as AE = AF + FE = 7 + 10 = 17 units.6. Use Pythagorean theorem on triangle AEC to find AC = sqrt(AE² + CE²) = sqrt(17² + (sqrt(95))²) = sqrt(289 + 95) = sqrt(384) = 8√6.Therefore, the exact length of AC is 8√6 units.**Final Answer**The length of segment AC is boxed{8sqrt{6}} units.

💡Let ( x ) represent the total initial number of marbles Pat has.1. **Marbles given to best friend:** [ frac{1}{2}x ]2. **Marbles remaining after giving to best friend:** [ x - frac{1}{2}x = frac{1}{2}x ]3. **Marbles given to sister:** [ frac{1}{3} left( frac{1}{2}x right) = frac{1}{6}x ]4. **Given that sister receives 9 marbles:** [ frac{1}{6}x = 9 ]5. **Solving for ( x ):** [ x = 9 times 6 = 54 ]6. **Marbles Pat keeps:** [ frac{1}{3}x = frac{1}{3} times 54 = 18 ]

💡Alright, so I've got this problem here about three points on a parabola, and I need to figure out the relationship between their y-values. Let me try to break it down step by step.First, the parabola is given by the equation y = -x² + 6x + c. Okay, so it's a quadratic equation, and since the coefficient of x² is negative (-1), I know the parabola opens downward. That means it has a maximum point at its vertex.Now, the points given are P₁(1, y₁), P₂(3, y₂), and P₃(4, y₃). All these points lie on the parabola, so their y-values must satisfy the equation y = -x² + 6x + c. I need to find out how y₁, y₂, and y₃ relate to each other—like which one is bigger or smaller.I remember that for a parabola, the vertex is the highest point if it opens downward. The vertex form of a parabola is y = a(x - h)² + k, where (h, k) is the vertex. But our equation is in standard form, so maybe I can find the vertex using the formula for the x-coordinate of the vertex, which is -b/(2a). In this equation, a is -1 and b is 6.Let me calculate that: x = -6/(2*(-1)) = -6/(-2) = 3. So the vertex is at x = 3. That means the point P₂(3, y₂) is actually the vertex of the parabola. Since the parabola opens downward, this vertex is the highest point on the graph. So y₂ must be the maximum y-value among the three points.Now, let's think about the other points. P₁ is at x = 1, and P₃ is at x = 4. Since the vertex is at x = 3, which is the axis of symmetry, points equidistant from the vertex on either side will have the same y-value. But in this case, P₁ is 2 units to the left of the vertex, and P₃ is 1 unit to the right. So they aren't symmetric.Since the parabola is symmetric around x = 3, the further a point is from the vertex, the smaller its y-value will be. So, P₁ is 2 units away, and P₃ is 1 unit away. That means P₁ is further from the vertex than P₃ is. Therefore, y₁ should be less than y₃ because it's further away from the maximum point.Wait, let me make sure. Since the parabola opens downward, as we move away from the vertex, the y-values decrease. So, moving from the vertex at x = 3 to x = 4, the y-value decreases a little because it's only 1 unit away. Moving to x = 1, which is 2 units away, the y-value decreases more. So, y₁ < y₃ < y₂.Let me plug in the x-values into the equation to confirm.For P₁(1, y₁):y₁ = -(1)² + 6*(1) + c = -1 + 6 + c = 5 + c.For P₂(3, y₂):y₂ = -(3)² + 6*(3) + c = -9 + 18 + c = 9 + c.For P₃(4, y₃):y₃ = -(4)² + 6*(4) + c = -16 + 24 + c = 8 + c.So, y₁ = 5 + c, y₂ = 9 + c, and y₃ = 8 + c.Comparing these, y₁ (5 + c) is less than y₃ (8 + c), which is less than y₂ (9 + c). So, the order is y₁ < y₃ < y₂.That matches what I thought earlier. So, the correct relationship is y₁ < y₃ < y₂, which is option A.I think that makes sense. The vertex is at x = 3, so y₂ is the highest. Then, moving away from the vertex, y decreases. Since P₁ is further away from the vertex than P₃, y₁ is less than y₃. So, y₁ < y₃ < y₂.Yeah, I'm pretty confident now. I don't think I made any mistakes in the calculations or the reasoning. It all lines up.

💡First, I need to determine the percentage of pencils Jeff donated. Let's denote this percentage as ( p ) (in decimal form).Jeff initially had 300 pencils. If he donated ( p ) percent of them, the number of pencils he donated is ( 300p ). Therefore, the number of pencils Jeff has left is ( 300 - 300p ).Vicki had twice as many pencils as Jeff, which means she had ( 2 times 300 = 600 ) pencils. She donated ( frac{3}{4} ) of her pencils, so she donated ( frac{3}{4} times 600 = 450 ) pencils. Consequently, Vicki has ( 600 - 450 = 150 ) pencils remaining.The total number of pencils remaining between Jeff and Vicki is 360. This gives us the equation:[(300 - 300p) + 150 = 360]Simplifying the equation:[450 - 300p = 360]Subtracting 450 from both sides:[-300p = -90]Dividing both sides by -300:[p = frac{90}{300} = 0.3]Therefore, Jeff donated 30% of his pencils.

💡Alright, so I have this problem where the projection of an equilateral triangle ABC onto a plane α results in a triangle with side lengths 2, 3, and 2√3. I need to find the side length of the original equilateral triangle ABC. The options given are A. 2√3, B. √13, C. √15, and D. 4.First, I need to understand what it means to project a triangle onto a plane. Projection, in this context, means that each vertex of the triangle is dropped perpendicularly onto the plane α, forming a new triangle. The side lengths of this projected triangle are given as 2, 3, and 2√3. Since the original triangle is equilateral, all its sides are equal, but when projected, the sides can become unequal because of the angle at which the triangle is inclined relative to the plane α.Let me denote the original equilateral triangle as ABC with side length 's'. When projected onto plane α, the triangle becomes A'B'C' with side lengths 2, 3, and 2√3. The key here is to relate the side lengths of the projected triangle to the original triangle.I recall that when a figure is projected onto a plane, the length of each side in the projection is equal to the original length multiplied by the cosine of the angle between the side and the plane. However, since the triangle is equilateral, all sides make the same angle with the plane, right? Wait, no, that might not necessarily be true because the projection could cause different sides to be foreshortened differently depending on their orientation relative to the plane.Hmm, maybe I need to think in terms of vectors and coordinates. Let me assign coordinates to the triangle ABC and its projection A'B'C'. Suppose I place the plane α as the xy-plane for simplicity. Then, the projection of any point (x, y, z) onto the plane α would be (x, y, 0). So, the z-coordinates of the original triangle's vertices will determine how much each side is foreshortened.Let me denote the coordinates of A, B, and C as (x₁, y₁, z₁), (x₂, y₂, z₂), and (x₃, y₃, z₃) respectively. Their projections A', B', and C' will then be (x₁, y₁, 0), (x₂, y₂, 0), and (x₃, y₃, 0). The side lengths of the projected triangle A'B'C' are given, so I can write equations based on the distances between these projected points.The distance between A' and B' is 2, so:√[(x₂ - x₁)² + (y₂ - y₁)²] = 2Similarly, the distance between B' and C' is 3:√[(x₃ - x₂)² + (y₃ - y₂)²] = 3And the distance between A' and C' is 2√3:√[(x₃ - x₁)² + (y₃ - y₁)²] = 2√3Now, the original triangle ABC is equilateral with side length 's', so the distances between A, B, and C should all be equal to 's'. Therefore, the distance between A and B is:√[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²] = sSimilarly for the other sides:√[(x₃ - x₂)² + (y₃ - y₂)² + (z₃ - z₂)²] = s√[(x₃ - x₁)² + (y₃ - y₁)² + (z₃ - z₁)²] = sSo, if I denote the differences in x, y, and z coordinates between points as Δx, Δy, and Δz, then for each side, the projected length squared plus the square of the z-difference equals s².Let me write that down for each side:For side AB:(Δx_AB)² + (Δy_AB)² = 2² = 4(Δx_AB)² + (Δy_AB)² + (Δz_AB)² = s²So, 4 + (Δz_AB)² = s²For side BC:(Δx_BC)² + (Δy_BC)² = 3² = 9(Δx_BC)² + (Δy_BC)² + (Δz_BC)² = s²So, 9 + (Δz_BC)² = s²For side AC:(Δx_AC)² + (Δy_AC)² = (2√3)² = 12(Δx_AC)² + (Δy_AC)² + (Δz_AC)² = s²So, 12 + (Δz_AC)² = s²Now, since all three expressions equal s², I can set them equal to each other:4 + (Δz_AB)² = 9 + (Δz_BC)² = 12 + (Δz_AC)²So, from the first two:4 + (Δz_AB)² = 9 + (Δz_BC)²Which simplifies to:(Δz_AB)² - (Δz_BC)² = 5Similarly, from the first and third:4 + (Δz_AB)² = 12 + (Δz_AC)²Which simplifies to:(Δz_AB)² - (Δz_AC)² = 8Now, I have two equations:1. (Δz_AB)² - (Δz_BC)² = 52. (Δz_AB)² - (Δz_AC)² = 8Let me denote (Δz_AB)² as a, (Δz_BC)² as b, and (Δz_AC)² as c.So, equation 1: a - b = 5Equation 2: a - c = 8From equation 1: b = a - 5From equation 2: c = a - 8Now, since the triangle is equilateral, the differences in z-coordinates should satisfy some geometric constraints. Also, the projection is a triangle with sides 2, 3, 2√3, which isn't equilateral, so the original triangle is inclined at some angle to the plane.I need another equation to relate a, b, and c. Perhaps using the fact that the original triangle is equilateral, so the vectors AB, BC, and AC should have the same magnitude and satisfy certain vector relationships.Alternatively, maybe I can use the fact that in the projected triangle, the sides are 2, 3, 2√3, so it's a triangle with sides of different lengths. Let me check if this projected triangle is right-angled or something.Let me compute 2² + 3² = 4 + 9 = 13, and (2√3)² = 12. Since 13 ≠ 12, it's not a right-angled triangle. Hmm.Alternatively, maybe I can use the coordinates approach. Let me assign coordinates to the projected triangle A'B'C'.Let me place point A' at (0, 0, 0) for simplicity. Then, point B' is at (2, 0, 0), since the distance between A' and B' is 2. Now, point C' is somewhere in the plane such that the distance from B' to C' is 3 and from A' to C' is 2√3.So, coordinates of C' can be found using the distance formula. Let me denote C' as (x, y, 0). Then:Distance from A' to C':√(x² + y²) = 2√3 ⇒ x² + y² = 12Distance from B' to C':√((x - 2)² + y²) = 3 ⇒ (x - 2)² + y² = 9Subtracting the first equation from the second:(x - 2)² + y² - (x² + y²) = 9 - 12Expanding (x - 2)²: x² - 4x + 4 + y² - x² - y² = -3Simplify: -4x + 4 = -3 ⇒ -4x = -7 ⇒ x = 7/4Now, plug x = 7/4 into x² + y² = 12:(49/16) + y² = 12 ⇒ y² = 12 - 49/16 = (192/16 - 49/16) = 143/16 ⇒ y = ±√(143)/4So, coordinates of C' are (7/4, √143/4, 0) or (7/4, -√143/4, 0). Since the projection is a triangle, the orientation doesn't matter, so I can take the positive y-coordinate.Now, I have coordinates for A', B', and C':A' = (0, 0, 0)B' = (2, 0, 0)C' = (7/4, √143/4, 0)Now, the original triangle ABC has coordinates A = (0, 0, z_A), B = (2, 0, z_B), and C = (7/4, √143/4, z_C). Wait, no, actually, the projection only affects the z-coordinate, so if A' is (0,0,0), then A is (0,0,z_A), similarly for B and C.But actually, the projection is orthogonal, so the x and y coordinates remain the same, only the z-coordinate is dropped. So, if A' is (0,0,0), then A is (0,0,z_A). Similarly, B is (2,0,z_B), and C is (7/4, √143/4, z_C).Now, the distances between A, B, and C should all be equal to 's'.So, distance AB:√[(2 - 0)² + (0 - 0)² + (z_B - z_A)²] = √[4 + (z_B - z_A)²] = sDistance BC:√[(7/4 - 2)² + (√143/4 - 0)² + (z_C - z_B)²] = √[( -1/4)² + (√143/4)² + (z_C - z_B)²] = √[(1/16) + (143/16) + (z_C - z_B)²] = √[(144/16) + (z_C - z_B)²] = √[9 + (z_C - z_B)²] = sDistance AC:√[(7/4 - 0)² + (√143/4 - 0)² + (z_C - z_A)²] = √[(49/16) + (143/16) + (z_C - z_A)²] = √[(192/16) + (z_C - z_A)²] = √[12 + (z_C - z_A)²] = sSo, now I have three equations:1. √[4 + (z_B - z_A)²] = s ⇒ 4 + (z_B - z_A)² = s²2. √[9 + (z_C - z_B)²] = s ⇒ 9 + (z_C - z_B)² = s²3. √[12 + (z_C - z_A)²] = s ⇒ 12 + (z_C - z_A)² = s²Now, since all three equal s², I can set them equal to each other:From 1 and 2:4 + (z_B - z_A)² = 9 + (z_C - z_B)²Which simplifies to:(z_B - z_A)² - (z_C - z_B)² = 5From 1 and 3:4 + (z_B - z_A)² = 12 + (z_C - z_A)²Which simplifies to:(z_B - z_A)² - (z_C - z_A)² = 8Now, let me denote:Let a = z_B - z_ALet b = z_C - z_BLet c = z_C - z_ANote that c = a + b, since z_C - z_A = (z_C - z_B) + (z_B - z_A) = b + aSo, substituting c = a + b into the equations:From the first equation:a² - b² = 5From the second equation:a² - (a + b)² = 8Let me expand the second equation:a² - (a² + 2ab + b²) = 8Simplify:a² - a² - 2ab - b² = 8 ⇒ -2ab - b² = 8 ⇒ 2ab + b² = -8Now, from the first equation:a² - b² = 5 ⇒ a² = b² + 5Let me substitute a² from the first equation into the second equation:2ab + b² = -8But I need to express a in terms of b or vice versa. From a² = b² + 5, I can write a = √(b² + 5). But that might complicate things. Alternatively, let me express a in terms of b from the second equation.From 2ab + b² = -8, let me factor out b:b(2a + b) = -8But I also have a² = b² + 5. Maybe I can solve for a in terms of b and substitute.Let me rearrange the second equation:2ab = -8 - b² ⇒ a = (-8 - b²)/(2b)Now, substitute this into a² = b² + 5:[(-8 - b²)/(2b)]² = b² + 5Let me compute the left side:[( -8 - b² )² ] / (4b²) = (64 + 16b² + b⁴) / (4b²) = (64 + 16b² + b⁴) / (4b²)Set equal to b² + 5:(64 + 16b² + b⁴) / (4b²) = b² + 5Multiply both sides by 4b²:64 + 16b² + b⁴ = 4b⁴ + 20b²Bring all terms to one side:64 + 16b² + b⁴ - 4b⁴ - 20b² = 0 ⇒ -3b⁴ - 4b² + 64 = 0Multiply both sides by -1:3b⁴ + 4b² - 64 = 0Let me let u = b², so the equation becomes:3u² + 4u - 64 = 0Now, solve for u using quadratic formula:u = [-4 ± √(16 + 768)] / (2*3) = [-4 ± √784]/6 = [-4 ± 28]/6So, two solutions:u = (-4 + 28)/6 = 24/6 = 4u = (-4 - 28)/6 = -32/6 = -16/3Since u = b² must be non-negative, we discard the negative solution. So, u = 4 ⇒ b² = 4 ⇒ b = ±2Now, let's find a using a = (-8 - b²)/(2b)Case 1: b = 2a = (-8 - 4)/(4) = (-12)/4 = -3Case 2: b = -2a = (-8 - 4)/(-4) = (-12)/(-4) = 3So, we have two possibilities:1. a = -3, b = 22. a = 3, b = -2Now, let's check which one makes sense in terms of the original triangle.Recall that a = z_B - z_A and b = z_C - z_B.Case 1: a = -3, b = 2So, z_B = z_A - 3z_C = z_B + 2 = z_A - 3 + 2 = z_A - 1Case 2: a = 3, b = -2So, z_B = z_A + 3z_C = z_B - 2 = z_A + 3 - 2 = z_A + 1Now, let's compute c = a + b in both cases.Case 1: c = -3 + 2 = -1Case 2: c = 3 - 2 = 1Now, let's compute s² from each equation.From equation 1: s² = 4 + a² = 4 + 9 = 13From equation 2: s² = 9 + b² = 9 + 4 = 13From equation 3: s² = 12 + c²In Case 1: c = -1 ⇒ c² = 1 ⇒ s² = 12 + 1 = 13In Case 2: c = 1 ⇒ c² = 1 ⇒ s² = 12 + 1 = 13So, in both cases, s² = 13 ⇒ s = √13Therefore, the side length of the original equilateral triangle ABC is √13.