Math Olympiad Geometry Problem | Find the length X in the semicircle

[@jimlocke9320]

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.

[@Ray3-d4v]

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

[@quigonkenny]

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 ]

[@rupeshmishra6739]

I am witnessing your success for years and very glad to see it because your videos helped me to improve my mathematical skill.

[@marioalb9726]

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 √ )

[@soli9mana-soli4953]

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)

[@RAG981]

Nice problem, well solved.

[@michaeldoerr5810]

That is a literal example of easier than it looks. And there is only one positive real solution and that solution is x = 4.

[@giuseppemalaguti435]

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

[@marioalb9726]

Why PC = √(61-x²) ?????

[@Grizzly01-vr4pn]

All fine up to 9:09 which is where I got to, then you just start guessing. Unimpressive.

[@RealQinnMalloryu4]

(5)^2 (6)^2={25 +36}=45 180ABCPX/45=4ABCPX 2^2 (ABCPX ➖ 2ABCPX+2).

[@brettgbarnes]

Lame.