Problem Analysis #48

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Problem Statement

Consider the parabola with focus $F(p,0)$ $(p > 0)$ and directrix $x = -p$ , as shown in the figure. Among the points on this parabola, let $A$ be a point in the first quadrant, and let $H$ be the foot of the perpendicular from $A$ to the directrix. Consider the ellipse whose foci lie on the $x$ -axis and which passes through the three points $F$ , $A$ , and $H$ . Let $B$ be the focus of this ellipse with positive $x$ -coordinate. Suppose that triangle $AHB$ has perimeter $p + 27$ and area $2p + 12$ . Let the length of segment $\overline{HF}$ be $k$ . Compute $k^2$ . [4 points]

Ground Truth

360

Execution Trace

Reasoning Process

[THOUGHT] Let me analyze this problem step by step.

The parabola has focus F(p,0) and directrix x = -p, so its equation is y² = 4px.
Point A is on the parabola in the first quadrant, so A = (a, 2√(pa)) for some a > 0.
H is the foot of perpendicular from A to the directrix x = -p, so H = (-p, 2√(pa)).
By the definition of parabola, |AF| = |AH| = a + p.
The ellipse passes through F, A, H and has foci on the x-axis. Let the ellipse have semi-major axis 'a_e' and foci at (-c, 0) and (c, 0).
B is the focus with positive x-coordinate, so B = (c, 0).
For the ellipse, the sum of distances from any point to the two foci equals 2a_e.

Let me set up the equations:

F = (p, 0), A = (t², 2pt/p) where I'll parameterize differently
Actually, let's say A = (x_A, y_A) where y_A² = 4p·x_A

Let me use coordinates:

H = (-p, y_A) where y_A is the y-coordinate of A
A = (x_A, y_A) where y_A² = 4p·x_A
|AH| = x_A - (-p) = x_A + p

For the ellipse with foci at F₁ = (-c, 0) and F₂ = B = (c, 0):

Distance from F(p,0) to F₁ + distance to F₂ = 2a_e → (p + c) + |p - c| = 2a_e
Distance from H(-p, y_A) to both foci = 2a_e
Distance from A(x_A, y_A) to both foci = 2a_e

Triangle AHB has:

Perimeter = |AH| + |HB| + |AB| = p + 27
Area = 2p + 12

Let me work with the ellipse property and set up equations. [/THOUGHT] [PYTHON] from sympy import symbols, solve, sqrt, Eq, simplify, Abs, Rational

Let's define variables

p, x_A, y_A, c, a_e = symbols('p x_A y_A c a_e', positive=True, real=True)

A is on parabola: y_A^2 = 4px_A

H = (-p, y_A)

F = (p, 0)

B = (c, 0) - positive focus of ellipse

|AH| = x_A + p (horizontal distance)

AH = x_A + p

For ellipse, sum of distances to foci = 2*a_e

Foci are at (-c, 0) and (c, 0)

Point F(p, 0) on ellipse:

|F to (-c,0)| + |F to (c,0)| = 2*a_e

(p + c) + |p - c| = 2*a_e

If c > p: (p + c) + (c - p) = 2c = 2*a_e, so a_e = c

If c < p: (p + c) + (p - c) = 2p = 2*a_e, so a_e = p

If c = p: 2p = 2*a_e, so a_e = p

Point H(-p, y_A) on ellipse:

sqrt((−p + c)^2 + y_A^2) + sqrt((−p − c)^2 + y_A^2) = 2*a_e

sqrt((c - p)^2 + y_A^2) + sqrt((c + p)^2 + y_A^2) = 2*a_e

Point A(x_A, y_A) on ellipse:

sqrt((x_A + c)^2 + y_A^2) + sqrt((x_A - c)^2 + y_A^2) = 2*a_e

Let's try case c >= p, so a_e = c

Then for H: sqrt((c-p)^2 + y_A^2) + sqrt((c+p)^2 + y_A^2) = 2c

Triangle AHB:

A = (x_A, y_A), H = (-p, y_A), B = (c, 0)

|AH| = x_A + p

|HB| = sqrt((c + p)^2 + y_A^2)

|AB| = sqrt((x_A - c)^2 + y_A^2)

Perimeter = |AH| + |HB| + |AB| = p + 27

Area = (1/2) * base * height = (1/2) * |AH| * |y_B - y_A| where we need to compute properly

Actually, A and H have same y-coordinate, so AH is horizontal

Area of triangle AHB = (1/2) * |AH| * |y_A - 0| = (1/2) * (x_A + p) * y_A

print("Setting up equations...")

Let's use numerical approach

import numpy as np from scipy.optimize import fsolve

def equations(vars): p_val, x_A_val, y_A_val, c_val = vars

# Parabola constraint
eq1 = y_A_val**2 - 4*p_val*x_A_val

# Ellipse: a_e = c (assuming c >= p)
a_e_val = c_val

# H on ellipse
HF1 = np.sqrt((c_val - p_val)**2 + y_A_val**2)
HF2 = np.sqrt((c_val + p_val)**2 + y_A_val**2)
eq2 = HF1 + HF2 - 2*a_e_val

# A on ellipse
AF1 = np.sqrt((x_A_val + c_val)**2 + y_A_val**2)
AF2 = np.sqrt((x_A_val - c_val)**2 + y_A_val**2)
eq3 = AF1 + AF2 - 2*a_e_val

# Triangle perimeter
AH = x_A_val + p_val
HB = np.sqrt((c_val + p_val)**2 + y_A_val**2)
AB = np.sqrt((x_A_val - c_val)**2 + y_A_val**2)
eq4 = AH + HB + AB - (p_val + 27)

# Triangle area
area = 0.5 * AH * y_A_val
eq5 = area - (2*p_val + 12)

return [eq1, eq2, eq3, eq4]

We have 5 equations but 4 unknowns - let me reconsider

Actually we have: parabola, H on ellipse, A on ellipse, perimeter, area

That's 5 equations for 4 unknowns (p, x_A, y_A, c)

def equations_full(vars): p_val, x_A_val, y_A_val, c_val = vars

if p_val <= 0 or x_A_val <= 0 or y_A_val <= 0 or c_val <= 0:
    return [1e10, 1e10, 1e10, 1e10]

# Parabola constraint
eq1 = y_A_val**2 - 4*p_val*x_A_val

# Ellipse: a_e = c (assuming c >= p)
a_e_val = c_val

# H on ellipse
HF1 = np.sqrt((c_val - p_val)**2 + y_A_val**2)
HF2 = np.sqrt((c_val + p_val)**2 + y_A_val**2)
eq2 = HF1 + HF2 - 2*a_e_val

# A on ellipse
AF1 = np.sqrt((x_A_val + c_val)**2 + y_A_val**2)
AF2 = np.sqrt((x_A_val - c_val)**2 + y_A_val**2)
eq3 = AF1 + AF2 - 2*a_e_val

# Triangle perimeter
AH = x_A_val + p_val
HB = np.sqrt((c_val + p_val)**2 + y_A_val**2)
AB = np.sqrt((x_A_val - c_val)**2 + y_A_val**2)
perimeter = AH + HB + AB
eq4 = perimeter - (p_val + 27)

# Triangle area
area = 0.5 * AH * y_A_val
eq5 = area - (2*p_val + 12)

return [eq1, eq2, eq3, eq5]  # Using 4 equations

Try different initial guesses

for guess in [(3, 4, 6, 5), (2, 3, 4, 4), (5, 6, 8, 7), (1, 2, 3, 3)]: try: sol = fsolve(equations_full, guess, full_output=True) x = sol[0] info = sol[1] if np.allclose(equations_full(x), [0,0,0,0], atol=1e-6): p_val, x_A_val, y_A_val, c_val = x print(f"Solution found: p={p_val}, x_A={x_A_val}, y_A={y_A_val}, c={c_val}")

        # Verify perimeter
        AH = x_A_val + p_val
        HB = np.sqrt((c_val + p_val)**2 + y_A_val**2)
        AB = np.sqrt((x_A_val - c_val)**2 + y_A_val**2)
        print(f"Perimeter: {AH + HB + AB}, expected: {p_val + 27}")
        
        # HF distance
        HF = np.sqrt((p_val - (-p_val))**2 + (0 - y_A_val)**2)
        HF = np.sqrt((2*p_val)**2 + y_A_val**2)
        print(f"HF = {HF}, HF^2 = {HF**2}")
except:
    pass

[/PYTHON]

Tool Output

[PYTHON OUTPUT] Setting up equations... [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] The code didn't find a solution with those initial guesses. Let me run more code to properly solve this problem. [/THOUGHT] [PYTHON] import numpy as np from scipy.optimize import fsolve, root

def equations_full(vars): p_val, x_A_val, y_A_val, c_val = vars

# Parabola constraint
eq1 = y_A_val**2 - 4*p_val*x_A_val

# Ellipse: a_e = c (assuming c >= p)
a_e_val = c_val

# H on ellipse
HF1 = np.sqrt((c_val - p_val)**2 + y_A_val**2)
HF2 = np.sqrt((c_val + p_val)**2 + y_A_val**2)
eq2 = HF1 + HF2 - 2*a_e_val

# A on ellipse
AF1 = np.sqrt((x_A_val + c_val)**2 + y_A_val**2)
AF2 = np.sqrt((x_A_val - c_val)**2 + y_A_val**2)
eq3 = AF1 + AF2 - 2*a_e_val

# Triangle perimeter
AH = x_A_val + p_val
HB = np.sqrt((c_val + p_val)**2 + y_A_val**2)
AB = np.sqrt((x_A_val - c_val)**2 + y_A_val**2)
perimeter = AH + HB + AB
eq4 = perimeter - (p_val + 27)

# Triangle area
area = 0.5 * AH * y_A_val
eq5 = area - (2*p_val + 12)

return [eq1, eq2, eq3, eq5]

Try many different initial guesses

found = False for p0 in np.linspace(1, 20, 20): for x0 in np.linspace(1, 20, 20): for y0 in np.linspace(1, 20, 20): for c0 in np.linspace(1, 20, 20): try: sol = fsolve(equations_full, [p0, x0, y0, c0], full_output=True) x = sol[0] residual = equations_full(x) if all(v > 0 for v in x) and np.allclose(residual, [0,0,0,0], atol=1e-8): p_val, x_A_val, y_A_val, c_val = x

                    # Verify perimeter
                    AH = x_A_val + p_val
                    HB = np.sqrt((c_val + p_val)**2 + y_A_val**2)
                    AB = np.sqrt((x_A_val - c_val)**2 + y_A_val**2)
                    perimeter = AH + HB + AB
                    
                    if abs(perimeter - (p_val + 27)) < 0.01:
                        print(f"Solution: p={p_val:.6f}, x_A={x_A_val:.6f}, y_A={y_A_val:.6f}, c={c_val:.6f}")
                        print(f"Perimeter check: {perimeter:.6f} vs {p_val + 27:.6f}")
                        
                        # HF distance: H=(-p, y_A), F=(p, 0)
                        HF_sq = (2*p_val)**2 + y_A_val**2
                        print(f"HF^2 = {HF_sq:.6f}")
                        found = True
                        break
            except:
                pass
        if found:
            break
    if found:
        break
if found:
    break

if not found: print("No solution found in grid search") [/PYTHON]

Tool Output

[PYTHON OUTPUT] Timeout: Python execution exceeded 30 seconds [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] The grid search is too slow. Let me use a smarter approach with symbolic computation or a more targeted numerical search. [/THOUGHT] [PYTHON] import numpy as np from scipy.optimize import fsolve

def equations_full(vars): p_val, x_A_val, y_A_val, c_val = vars

# Parabola constraint
eq1 = y_A_val**2 - 4*p_val*x_A_val

# Ellipse: a_e = c (assuming c >= p)
a_e_val = c_val

# H on ellipse
HF1 = np.sqrt((c_val - p_val)**2 + y_A_val**2)
HF2 = np.sqrt((c_val + p_val)**2 + y_A_val**2)
eq2 = HF1 + HF2 - 2*a_e_val

# A on ellipse
AF1 = np.sqrt((x_A_val + c_val)**2 + y_A_val**2)
AF2 = np.sqrt((x_A_val - c_val)**2 + y_A_val**2)
eq3 = AF1 + AF2 - 2*a_e_val

# Triangle area
AH = x_A_val + p_val
area = 0.5 * AH * y_A_val
eq4 = area - (2*p_val + 12)

return [eq1, eq2, eq3, eq4]

Try specific guesses

guesses = [ (3, 3, 6, 5), (3, 4, 6, 6), (4, 4, 8, 6), (2, 4, 4, 5), (6, 9, 12, 10), (3, 9, 6, 8), (4, 8, 8, 9), (5, 5, 10, 8), (3, 6, 6, 7), (2, 2, 4, 4), (1, 1, 2, 3), (6, 6, 12, 9), (4, 4, 8, 7), (3, 3, 6, 6), (5, 10, 10, 12), ]

for guess in guesses: try: sol = fsolve(equations_full, guess, full_output=True) x = sol[0] residual = equations_full(x) if all(v > 0.1 for v in x) and np.allclose(residual, [0,0,0,0], atol=1e-6): p_val, x_A_val, y_A_val, c_val = x

        # Verify perimeter
        AH = x_A_val + p_val
        HB = np.sqrt((c_val + p_val)**2 + y_A_val**2)
        AB = np.sqrt((x_A_val - c_val)**2 + y_A_val**2)
        perimeter = AH + HB + AB
        expected_perimeter = p_val + 27
        
        print(f"Solution: p={p_val:.6f}, x_A={x_A_val:.6f}, y_A={y_A_val:.6f}, c={c_val:.6f}")
        print(f"Perimeter: {perimeter:.6f}, expected: {expected_perimeter:.6f}")
        print(f"Residuals: {residual}")
        
        # HF distance: H=(-p, y_A), F=(p, 0)
        HF_sq = (2*p_val)**2 + y_A_val**2
        print(f"HF^2 = {HF_sq:.6f}")
except Exception as e:
    pass

[/PYTHON]

Tool Output

[PYTHON OUTPUT] Solution: p=3.000002, x_A=2.999999, y_A=6.000001, c=764326059.401721 Perimeter: 1528652124.803447, expected: 30.000002 Residuals: [np.float64(0.0), np.float64(0.0), np.float64(0.0), np.float64(0.0)] HF^2 = 72.000057 Solution: p=2.999999, x_A=3.000000, y_A=6.000000, c=240787556.374795 Perimeter: 481575118.749590, expected: 29.999999 Residuals: [np.float64(3.694822225952521e-13), np.float64(1.1920928955078125e-07), np.float64(1.1920928955078125e-07), np.float64(-1.4779288903810084e-12)] HF^2 = 71.999981 Solution: p=3.000000, x_A=3.000000, y_A=6.000000, c=263419502.830505 Perimeter: 526839011.661010, expected: 30.000000 Residuals: [np.float64(1.2278178473934531e-11), np.float64(1.1920928955078125e-07), np.float64(1.1920928955078125e-07), np.float64(-1.9414514440541097e-10)] HF^2 = 71.999996 Solution: p=3.000000, x_A=3.000000, y_A=6.000000, c=1618021178.694430 Perimeter: 3236042363.388861, expected: 30.000000 Residuals: [np.float64(0.0), np.float64(0.0), np.float64(0.0), np.float64(0.0)] HF^2 = 72.000009 Solution: p=2.999999, x_A=3.000000, y_A=6.000000, c=469883302.076019 Perimeter: 939766610.152036, expected: 29.999999 Residuals: [np.float64(0.0), np.float64(0.0), np.float64(0.0), np.float64(3.552713678800501e-15)] HF^2 = 71.999983 Solution: p=3.000000, x_A=3.000000, y_A=6.000000, c=439024990.558566 Perimeter: 878049987.117133, expected: 30.000000 Residuals: [np.float64(0.0), np.float64(0.0), np.float64(0.0), np.float64(0.0)] HF^2 = 71.999999 [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] The numerical solver is finding solutions where c is extremely large, which doesn't satisfy the perimeter constraint. The perimeter constraint is not being enforced properly. Let me reconsider - I need to include the perimeter constraint as one of the equations. Let me run a corrected version. [/THOUGHT] [PYTHON] import numpy as np from scipy.optimize import fsolve

def equations_with_perimeter(vars): p_val, x_A_val, y_A_val, c_val = vars

# Parabola constraint
eq1 = y_A_val**2 - 4*p_val*x_A_val

# Ellipse: a_e = c (assuming c >= p)
a_e_val = c_val

# H on ellipse
HF1 = np.sqrt((c_val - p_val)**2 + y_A_val**2)
HF2 = np.sqrt((c_val + p_val)**2 + y_A_val**2)
eq2 = HF1 + HF2 - 2*a_e_val

# A on ellipse
AF1 = np.sqrt((x_A_val + c_val)**2 + y_A_val**2)
AF2 = np.sqrt((x_A_val - c_val)**2 + y_A_val**2)
eq3 = AF1 + AF2 - 2*a_e_val

# Triangle perimeter
AH = x_A_val + p_val
HB = np.sqrt((c_val + p_val)**2 + y_A_val**2)
AB = np.sqrt((x_A_val - c_val)**2 + y_A_val**2)
perimeter = AH + HB + AB
eq4 = perimeter - (p_val + 27)

# Triangle area
area = 0.5 * AH * y_A_val
eq5 = area - (2*p_val + 12)

return [eq1, eq2, eq4, eq5]  # Using perimeter instead of eq3

Try specific guesses

for guess in guesses: try: sol = fsolve(equations_with_perimeter, guess, full_output=True) x = sol[0] residual = equations_with_perimeter(x) if all(v > 0.1 for v in x) and np.allclose(residual, [0,0,0,0], atol=1e-6): p_val, x_A_val, y_A_val, c_val = x

        # Verify all constraints
        AH = x_A_val + p_val
        HB = np.sqrt((c_val + p_val)**2 + y_A_val**2)
        AB = np.sqrt((x_A_val - c_val)**2 + y_A_val**2)
        perimeter = AH + HB + AB
        area = 0.5 * AH * y_A_val
        
        # Check A on ellipse
        AF1 = np.sqrt((x_A_val + c_val)**2 + y_A_val**2)
        AF2 = np.sqrt((x_A_val - c_val)**2 + y_A_val**2)
        
        print(f"Solution: p={p_val:.6f}, x_A={x_A_val:.6f}, y_A={y_A_val:.6f}, c={c_val:.6f}")
        print(f"Perimeter: {perimeter:.6f}, expected: {p_val + 27:.6f}")
        print(f"Area: {area:.6f}, expected: {2*p_val + 12:.6f}")
        print(f"A on ellipse: AF1+AF2={AF1+AF2:.6f}, 2c={2*c_val:.6f}")
        
        # HF distance: H=(-p, y_A), F=(p, 0)
        HF_sq = (2*p_val)**2 + y_A_val**2
        print(f"HF^2 = {HF_sq:.6f}")
        print()
except Exception as e:
    pass

[/PYTHON]

Tool Output

[PYTHON OUTPUT] No output [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] No solution found with those guesses. Let me try a different approach - maybe the other focus of the ellipse is at a different location. Let me reconsider the problem setup and try with the left focus being at a specific point. [/THOUGHT] [PYTHON] import numpy as np from scipy.optimize import fsolve

Let me reconsider: The ellipse passes through F, A, H

F = (p, 0), A = (x_A, y_A), H = (-p, y_A)

The ellipse has foci on x-axis. Let foci be at (f1, 0) and (f2, 0) with f1 < f2

B is the focus with positive x-coordinate

For any point P on ellipse: |PF1| + |PF2| = 2a (constant)

Let's try: one focus at origin, i.e., f1 = 0, f2 = 2c for some c > 0

Or more generally, foci at (-d, 0) and (d, 0)

Actually, let me think about this differently.

The problem says B is the focus with positive x-coordinate.

So B = (f2, 0) where f2 > 0.

Let foci be at (-c, 0) and (c, 0), so B = (c, 0)

For F(p, 0) on ellipse: sqrt((p+c)^2) + sqrt((p-c)^2) = 2a

If p > c: (p+c) + (p-c) = 2p = 2a, so a = p

If p < c: (p+c) + (c-p) = 2c = 2a, so a = c

If p = c: 2p = 2a, so a = p

Let's try case a = p (i.e., p >= c)

def equations_case1(vars): p_val, x_A_val, y_A_val, c_val = vars

a_e = p_val  # semi-major axis

# Parabola constraint
eq1 = y_A_val**2 - 4*p_val*x_A_val

# H(-p, y_A) on ellipse: distances to foci sum to 2a
HF1 = np.sqrt(((-p_val) + c_val)**2 + y_A_val**2)
HF2 = np.sqrt(((-p_val) - c_val)**2 + y_A_val**2)
eq2 = HF1 + HF2 - 2*a_e

# A(x_A, y_A) on ellipse
AF1 = np.sqrt((x_A_val + c_val)**2 + y_A_val**2)
AF2 = np.sqrt((x_A_val - c_val)**2 + y_A_val**2)
eq3 = AF1 + AF2 - 2*a_e

# Triangle AHB perimeter
AH = x_A_val + p_val
HB = np.sqrt((c_val + p_val)**2 + y_A_val**2)
AB = np.sqrt((x_A_val - c_val)**2 + y_A_val**2)
eq4 = AH + HB + AB - (p_val + 27)

# Triangle area
area = 0.5 * AH * y_A_val
eq5 = area - (2*p_val + 12)

return [eq1, eq2, eq3, eq4]

We have 5 equations but only 4 unknowns, so system is overdetermined

Let's use 4 equations and check the 5th

def equations_v2(vars): p_val, x_A_val, y_A_val, c_val = vars

a_e = p_val  # semi-major axis (case p >= c)

# Parabola constraint
eq1 = y_A_val**2 - 4*p_val*x_A_val

# H(-p, y_A) on ellipse
HF1 = np.sqrt((c_val - p_val)**2 + y_A_val**2)
HF2 = np.sqrt((c_val + p_val)**2 + y_A_val**2)
eq2 = HF1 + HF2 - 2*a_e

# Triangle AHB perimeter
AH = x_A_val + p_val
HB = np.sqrt((c_val + p_val)**2 + y_A_val**2)
AB = np.sqrt((x_A_val - c_val)**2 + y_A_val**2)
eq3 = AH + HB + AB - (p_val + 27)

# Triangle area
area = 0.5 * AH * y_A_val
eq4 = area - (2*p_val + 12)

return [eq1, eq2, eq3, eq4]

Try many guesses

found_solutions = [] for p0 in [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]: for x0 in [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]: for y0 in [2, 4, 6, 8, 10, 12]: for c0 in [1, 2, 3, 4, 5, 6, 7, 8]: try: sol = fsolve(equations_v2, [p0, x0, y0, c0], full_output=True) x = sol[0] residual = equations_v2(x) if all(v > 0.1 for v in x) and np.allclose(residual, [0,0,0,0], atol=1e-6): p_val, x_A_val, y_A_val, c_val = x

                    # Check if A is on ellipse
                    a_e = p_val
                    AF1 = np.sqrt((x_A_val + c_val)**2 + y_A_val**2)
                    AF2 = np.sqrt((x_A_val - c_val)**2 + y_A_val**2)
                    A_check = abs(AF1 + AF2 - 2*a_e)
                    
                    if A_check < 0.01:  # A is also on ellipse
                        key = (round(p_val, 2), round(x_A_val, 2), round(y_A_val, 2), round(c_val, 2))
                        if key not in found_solutions:
                            found_solutions.append(key)
                            print(f"Solution: p={p_val:.4f}, x_A={x_A_val:.4f}, y_A={y_A_val:.4f}, c={c_val:.4f}")
                            
                            # Verify
                            AH = x_A_val + p_val
                            HB = np.sqrt((c_val + p_val)**2 + y_A_val**2)
                            AB = np.sqrt((x_A_val - c_val)**2 + y_A_val**2)
                            perimeter = AH + HB + AB
                            area = 0.5 * AH * y_A_val
                            
                            print(f"  Perimeter: {perimeter:.4f}, expected: {p_val + 27:.4f}")
                            print(f"  Area: {area:.4f}, expected: {2*p_val + 12:.4f}")
                            print(f"  A on ellipse check: {A_check:.6f}")
                            
                            # HF distance
                            HF_sq = (2*p_val)**2 + y_A_val**2
                            print(f"  HF^2 = {HF_sq:.4f}")
            except:
                pass

if not found_solutions: print("No solutions found with case a=p") [/PYTHON]

Tool Output

[PYTHON OUTPUT] No solutions found with case a=p [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] Let me try a different approach - maybe the left focus of the ellipse is at a different position, not symmetric about the origin. Let me reconsider the problem more carefully. [/THOUGHT] [PYTHON] import numpy as np from scipy.optimize import fsolve

Let the ellipse have foci at (f1, 0) and (f2, 0) with f1 < f2

B is the focus with positive x-coordinate, so B = (f2, 0) if f2 > 0

The ellipse passes through F(p,0), A(x_A, y_A), H(-p, y_A)

For ellipse: |PF1| + |PF2| = 2a for all points P on ellipse

Let's parameterize: f1 and f2 are the foci x-coordinates

5 unknowns: p, x_A, y_A, f1, f2

5 equations: parabola, F on ellipse, A on ellipse, H on ellipse, perimeter, area

That's 6 equations for 5 unknowns - overdetermined

But for ellipse, if F, A, H are on it, we can use:

|FF1| + |FF2| = |AF1| + |AF2| = |HF1| + |HF2| = 2a

def equations_general(vars): p_val, x_A_val, y_A_val, f1, f2 = vars

# Parabola constraint
eq1 = y_A_val**2 - 4*p_val*x_A_val

# F(p, 0) on ellipse
FF1 = abs(p_val - f1)
FF2 = abs(p_val - f2)
sum_F = FF1 + FF2

# H(-p, y_A) on ellipse
HF1 = np.sqrt((-p_val - f1)**2 + y_A_val**2)
HF2 = np.sqrt((-p_val - f2)**2 + y_A_val**2)
sum_H = HF1 + HF2

# A(x_A, y_A) on ellipse
AF1 = np.sqrt((x_A_val - f1)**2 + y_A_val**2)
AF2 = np.sqrt((x_A_val - f2)**2 + y_A_val**2)
sum_A = AF1 + AF2

# All sums equal (ellipse property)
eq2 = sum_F - sum_H
eq3 = sum_F - sum_A

# B = (f2, 0) since f2 > 0 (positive x-coordinate focus)
# Triangle AHB: A=(x_A, y_A), H=(-p, y_A), B=(f2, 0)
AH = x_A_val + p_val
HB = np.sqrt((f2 + p_val)**2 + y_A_val**2)
AB = np.sqrt((x_A_val - f2)**2 + y_A_val**2)

# Perimeter
eq4 = AH + HB + AB - (p_val + 27)

# Area
area = 0.5 * AH * y_A_val
eq5 = area - (2*p_val + 12)

return [eq1, eq2, eq3, eq4, eq5]

Try many guesses

found_solutions = [] for p0 in [2, 3, 4, 5, 6]: for x0 in [2, 3, 4, 5, 6, 8, 10]: for y0 in [3, 4, 6, 8, 10]: for f1_0 in [-5, -3, -2, -1, 0, 1]: for f2_0 in [3, 5, 7, 9, 11]: try: sol = fsolve(equations_general, [p0, x0, y0, f1_0, f2_0], full_output=True) x = sol[0] residual = equations_general(x) p_val, x_A_val, y_A_val, f1, f2 = x

                    if p_val > 0 and x_A_val > 0 and y_A_val > 0 and f2 > 0 and f1 < f2:
                        if np.allclose(residual, [0,0,0,0,0], atol=1e-6):
                            key = (round(p_val, 2), round(x_A_val, 2), round(y_A_val, 2))
                            if key not in [s[0] for s in found_solutions]:
                                found_solutions.append((key, x))
                                print(f"Solution: p={p_val:.4f}, x_A={x_A_val:.4f}, y_A={y_A_val:.4f}, f1={f1:.4f}, f2={f2:.4f}")
                                
                                # HF distance: H=(-p, y_A), F=(p, 0)
                                HF_sq = (2*p_val)**2 + y_A_val**2
                                print(f"  HF^2 = {HF_sq:.4f}")
                except:
                    pass

if not found_solutions: print("No solutions found") [/PYTHON]

Tool Output

[PYTHON OUTPUT] Solution: p=9.0000, x_A=1.0000, y_A=6.0000, f1=-15.2583, f2=7.2583 HF^2 = 360.0000 [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] The numerical solver found a solution with p=9, x_A=1, y_A=6, and the foci at f1=-15.2583 and f2=7.2583.

The HF distance is calculated as:

H = (-p, y_A) = (-9, 6)
F = (p, 0) = (9, 0)
HF² = (2p)² + y_A² = (18)² + 6² = 324 + 36 = 360

So k² = 360. [/THOUGHT]

\boxed{360}

Final Model Output

360

Comparative Analysis (P48) (Average across runs)

Avg Time Taken

Claude-Opus-4.5

581.5

Deepseek-V3.2

288.6

Gemini-3-Pro-Preview

268.3

GPT-5.1 (high)

225.0

Grok-4.1-fast

149.7

Solar-Pro-2 (31B)(high)

107.3

Llama-VARCO-8B-Instruct

81.1

A.X-4.0 (72B)

60.2

EXAONE-4.0.1-32B (high)

34.9

HCX-007(high)

29.0

Avg Token Usage

Claude-Opus-4.5

59392.0

Grok-4.1-fast

23504.0

Deepseek-V3.2

22811.0

EXAONE-4.0.1-32B (high)

14729.0

Solar-Pro-2 (31B)(high)

14626.0

Gemini-3-Pro-Preview

13540.0

GPT-5.1 (high)

13286.0

A.X-4.0 (72B)

7884.0

HCX-007(high)

4314.0

Llama-VARCO-8B-Instruct

2849.0