At about 10:05, Math Booster appears to pull the solution x = 4 out of the air. Let Δ = x³ + 12x² + 11x - 300 and we search for a positive value of x which will produce Δ = 0. If x = 10, Δ = 2010 and, if x = 0, Δ = -300. We also know that Δ is monotonic (in this case, steadily increasing) over the range x = 0 to x = 10, so there is a zero crossing. So, we write a python program to conduct a half interval search: import math x = 5 dx = 2.5 # define function to compute delta = ∆ def compute_delta(x): delta = math.pow(x,3) + 12*math.pow(x,2) + 11*x - 300 return(delta) for a in range(0,51): # after 50 divisions by 2, dx is sufficiently small delta = compute_delta(x) if (delta > 0): x = x - dx else: x = x + dx dx = dx/2 x = round(x , 12) # round x to 9 decimal places formatted = "{:.12f}".format(x) print(f"x = ",formatted) Output: x = 4.000000000000 === Code Execution Successful === We try x = 4 in Δ = x³ + 12x² + 11x - 300 on our own and find that it produces Δ = 0 exactly.
I like your approach. I might add that since the first derivative of the cubic function is easy to calculate, Newton/Raphson converges very quickly. I got lazy so instead of writing some Python code I used an Excel spreadsheet to do Newton. Using an initial guess of x = 4.9 (from the diagram it must be less than 5) it took just 3 iterations to get a value of x = 4.000000417 f(x) f'(x) x0 = 4.9 159.669 200.63 x1 = 4.10416189 16.40661587 160.0323198 x2 = 4.00164125 0.254458393 155.0787881 x3 = 4.000000417
In this question, to avoid the third degree equation, I used a shortcut, I don't know how correct it is, but I reasoned like this: since the hypotenuse of APQ, AP = 5, I supposed that its sides could be the Pythagorean triple 3,4,5. A couple of tests are enough to realize that it is possible. If x = 3 => PQ = 10/3 and this is not good If x = 4 => PQ = 3 and it is ok Once I obtained these values, I verified that with them the triangles PCQ and ABC were still similar, because we know that they are. And it is easy to see that the values obtained are consistent with the similarity between the two triangles (6:3 = (6+4)/5)
I am witnessing your success for years and very glad to see it because your videos helped me to improve my mathematical skill.
As ⌒AP = ⌒CB, then AP = CB = 5 and PC and AB are parallel. As ∠PCQ and ∠CAB are alternate interior angles, then ∠PCQ and ∠CAB are congruent. As ∠CQP = 90° and, by Thales' Theorem, ∠ACB = 90° (as AB is a diameter and C is on the circumference), then ∆CQP and ∆ACB are similar triangles.
Triangle ∆PQA:
PQ² + QA² = AP²
PQ² + x² = 5²
PQ² = 25 - x²
PQ = √(25-x²)
AC/CQ = CB/QP
(6+x)/6 = 5/√(25-x²)
30 = (6+x)√(25-x²)
30/√(25-x²) = 6 + x
900/(25-x²) = 36 + 12x + x²
900 = (25-x²)(36+12x+x²)
900 = 900 + 300x + 25x² - 36x² - 12x³ - x⁴
x⁴ + 12x³ + 11x² - 300x = 0
x(x³+12x²+11x-300) = 0
x³ + 12x² + 11x - 300 = 0 --- x ≠ 0
By observation, 0 < x < 5:
(1)³ + 12(1)² + 11(1) - 300 =
1 + 12 + 11 - 300 ≠ 0 ❌
(2)³ + 12(2)² + 11(2) - 300 =
8 + 48 + 22 - 300 ≠ 0 ❌
(3)³ + 12(3)² + 11(3) - 300 =
27 + 108 + 33 - 300 ≠ 0 ❌
(4)³ + 12(4)² + 11(4) - 300 =
64 + 192 + 44 - 300 = 0 ✓
x³ + (16x²-4x²) + (75x-64x) - 300 = 0
(x³ - 4x²) + (16x² - 64x) + (75x - 300) = 0
(x-4)x² + (x-4)16x + (x-4)75 = 0
(x-4)(x²+16x+75) = 0
x = 4 | x² + 16x + 75 = 0
√((16)²-4(1)75) = √(256-300) = √(-44)
No additional real solutions
[ x = 4 ]
Similarity of triangles:
y/6 = 5/(6+x) --> y = 30/(6+x)
Pytagorean theorem:
y² + x² = 5²
[30/(6+x)]² + x² = 5²
30² / (6+x)² = 5² - x²
(5²-x²)(6+x)² = 30²
(x²-5²)(x²+12x+6²) = -30²
x⁴+12x³+6²x²-5²x²-12.5²x-30²=-30²
x³+12x²+11x-300=0
x= 4 cm ( Solved √ )
At about 10:05, Math Booster appears to pull the solution x = 4 out of the air. Let Δ = x³ + 12x² + 11x - 300 and we search for a positive value of x which will produce Δ = 0. If x = 10, Δ = 2010 and, if x = 0, Δ = -300. We also know that Δ is monotonic (in this case, steadily increasing) over the range x = 0 to x = 10, so there is a zero crossing. So, we write a python program to conduct a half interval search:
import math
x = 5
dx = 2.5
# define function to compute delta = ∆
def compute_delta(x):
delta = math.pow(x,3) + 12*math.pow(x,2) + 11*x - 300
return(delta)
for a in range(0,51): # after 50 divisions by 2, dx is sufficiently small
delta = compute_delta(x)
if (delta > 0):
x = x - dx
else:
x = x + dx
dx = dx/2
x = round(x , 12) # round x to 9 decimal places
formatted = "{:.12f}".format(x)
print(f"x = ",formatted)
Output:
x = 4.000000000000
=== Code Execution Successful ===
We try x = 4 in Δ = x³ + 12x² + 11x - 300 on our own and find that it produces Δ = 0 exactly.
I like your approach. I might add that since the first derivative of the cubic function is easy to calculate, Newton/Raphson converges very quickly. I got lazy so instead of writing some Python code I used an Excel spreadsheet to do Newton. Using an initial guess of x = 4.9 (from the diagram it must be less than 5) it took just 3 iterations to get a value of
x = 4.000000417
f(x) f'(x)
x0 = 4.9 159.669 200.63
x1 = 4.10416189 16.40661587 160.0323198
x2 = 4.00164125 0.254458393 155.0787881
x3 = 4.000000417
3
In this question, to avoid the third degree equation, I used a shortcut, I don't know how correct it is, but I reasoned like this: since the hypotenuse of APQ, AP = 5, I supposed that its sides could be the Pythagorean triple 3,4,5. A couple of tests are enough to realize that it is possible.
If x = 3 => PQ = 10/3 and this is not good
If x = 4 => PQ = 3 and it is ok
Once I obtained these values, I verified that with them the triangles PCQ and ABC were still similar, because we know that they are. And it is easy to see that the values obtained are consistent with the similarity between the two triangles (6:3 = (6+4)/5)
Nice problem, well solved.
PA=5...PC=√(61-x^2)..gli angoli opposti del trapezio APCB sono supplementari...arcsin(x/5)+arcsin(6/√(61-x^2))+arctg(6+x)/5=180....calcoli x=4
Why PC = √(61-x²) ?????
That is a literal example of easier than it looks. And there is only one positive real solution and that solution is x = 4.
All fine up to 9:09 which is where I got to, then you just start guessing. Unimpressive.
(5)^2 (6)^2={25 +36}=45 180ABCPX/45=4ABCPX 2^2 (ABCPX ➖ 2ABCPX+2).
Lame.