You are the best, i told you already. I am italian, but i prefer to follow you that explain in english instead to follow italian youtubers. Are you a math professor? Anyway gratz and thank you for your lessons
I used a slightly different approach. In △ACD, we can use Pythagorean theorem to find CD = √(x² − 1) In △BCD, we use Law of Sines to get: BD/sin(∠BCD) = CD/sin(∠B) 1/sin(30°) = √(x² − 1)/sin(∠B) sin(∠B) = sin(30°) √(x² − 1) sin(∠B) = (1/2) √(x² − 1) In △ABC, we use Law of Sines to get: AB/sin(∠ACB) = AC/sin(∠B) (x+1)/sin(120°) = 1/sin(∠B) sin(∠B) = sin(120°)/(x+1) sin(∠B) = (√3/2) / (x+1) Now we set both values of sin(∠B) equal to each other (1/2) √(x² − 1) = (√3/2) / (x+1) √(x² − 1) = √3 / (x+1) Square both sides: (x² − 1) = 3 / (x² + 2x + 1) (x² − 1) (x² + 2x + 1) = 3 x⁴ + 2x³ + x² − x² − 2x − 1 = 3 x⁴ + 2x³ − 2x − 4 = 0 x³ (x + 2) − 2 (x + 2) = 0 (x + 2) (x³ − 2) = 0 Possible solutions: x = −2 → not possible, since x is a side length x³ = 2 *x = ∛2*
My solution as well. (I wasn't as direct as yours.. Did some extra work that wasn't needed) got the same quartic polynomial, and was pleasantly surprised that it easily factored by grouping. Are you as surprised as I am that out of a simple geometric situation sprang a cube root?! 😅
A fascinating and unexpected result. Because 2^(1/3) is not constructible with compass and straight edge, the entire figure as shown is not constructible.
You can build a drawing that reflects real proportions: 1. Build a DCB angle with infinite beams 2. Lay an AC beam 3. Apply the ruler, constantly monitoring the lengths of AC and DB segments until they are approximately equal.
@@PreMath You're most welcome and thank you for producing the video! Here's another way to the solution after we have found BF = √3/x. From the Pythagorean theorem, CD = √(x²-1), thus CD/BF = (√(x²-1))/(√3/x). From similar triangles, CD/BF = AD/AB and we have AD/AB = x/(x+1), so (√(x²-1))/(√3/x) = x/(x+1). Multiplying both sides by (√3/x), we get √(x²-1) = (√3/x)( x/(x+1)), which simplifies to √(x²-1) = √3/(x+1). Square both sides and we get x²-1 = 3/(x+1)² or x²-1 = 3/(x²+2x+1). Multiply both sides by (x²+2x+1) and simplify: x⁴ +2x³-2x-4= 0. Factor out (x+2) to get (x+2)(x³)-(x+2)(2)= 0 or (x+2)(x³-2)= 0. The 2 solutions are x = -2 and x³ = 2. The value of x must be positive, so x = -2 is invalid and x = ³√2, as PreMath found.
My solution was all based on geometry: First idea how to start solving this was looking at the angle 30 degrees, should I construct some angle of 90 or 60 degrees out of that, which I also did: First I made a rotated copy of triangle CBD, so that the new triangle BEC forms a trapezoid with CBD, and the edge BE is parallel to DC, and CE parallel to AB. Next, looking at angles CAD = theta, and ABE = 90-theta, so that if I extend the lines AC and BE into a new point F, the angle AFB is a right angle. Therefore, the large triangle ABF and the smaller one CEF are uniform having the same angles, and proportional lengths of edges. Next, the triangles BCF is a right angle triangle having also a 60 degrees angle at point C, a triangle having lengths with proportions (1, sqrt(3), 2). The lengths of the large right triangle were AF = 1+1/x, BF = sqrt(3)/x, and the hypothenyse AB = x + 1, so applying the Pythogorean gave (1+1/x)^2 + (sqrt(3)/x)^2 = (x+1)^2 and solving that resulted with x = cuberoot(2). (I probably should have made a video out of this solution, instead of writing this as a text that might be really hard for anyone to get the idea)
You could avoid fractional denominators by using sec (Hypotenuse ÷ Adjacent) and csc (Hypotenuse ÷ Opposite). The Law of Sines can ironically be expressed as a csc(α) = b csc(β).
Feels very difficult. 😗First cos A is 1/x, then BC^2=1+(1+x)^2-2(1+x)/x, DE=sqrt(x^2-1)/2, CE=sqrt(x^2-1)sqrt(3)/2, EB=sqrt(1-(x^2-1)/4)=sqrt((5-x^2)/4), so BC=CE+EB=a clumsy expression of x, so I cannot continue my work.😢
cut out ACD and rotate anticlockrise until AC is the bace line and put it on BDC, notice they have two same side, rotate the line DC from ACD anticlockrise can make the triangle BDC, therefore it should be a formula to get the information of x.
Actually, the other "solution" is x = -2 , which is not allowed because x is a length of a line segment, so it must be true that x > 0 (which it ain't, if x = - 2 ! 😊 ) .
I thought I had solved this, but my reasoning was wrong. I'm not even going to watch the video to find out how I should have done it, though, because you have deliberately tried to fool us with the diagram. If a triangle is not isosceles, it should not be drawn as if it might be. I don't like that sort of trickery. Similarly, if an angle is not a right angle, it should not be drawn in such a way that it is near-indistinguishable from one. The fact is that triangle CDB _looks_ isosceles, yet it can't be, because triangle ACD would then have to be an isosceles right triangle, which, in turn, would be incompatible with triangle CDB if it were isosceles. I know you always say that the diagram may not be 100% to scale, but here have tried to run us off the rails from the start. I hope this is not the beginning of a new trend.
@@WernHerr Yes, I was just angry with myself, really, for not being able to do it. I still can't do, but I can't face watching the video to find out what I'm missing. It will just have to be a question that I _can_ afford to ignore. Be aware, though, that congruent or similar features are _not_ always indicated in these problems. Sometimes, the answer depends on determining that two given lengths are equal, or that two triangles are similar.
@@AnonimityAssuredYou can build a drawing that reflects real proportions: 1. Build a DCB angle 30 grad with infinite beams 2. Lay an AC beam 3. Apply the ruler, constantly monitoring the lengths of AC and DB segments until they are approximately equal.
@@valentinaivanova4806 Thanks. I found a way of constructing a reasonably accurate diagram within the limits of MS Word. I was right that triangle CDB should be significantly more asymmetrical, with CD a bit less than 4/5 of DB. Now, having finally watched the video, I see that I would never have solved the problem. I'll just have to archive it under the label 'Techniques I should know, but never will.'
A me risulta x=rad((1+rad13)/2)...dal teorema dei seni...x+1:sin120=1:sin(120+arccos1/x)...no!ho sbagliato...risulta ,dal teorema dei seni..x^4+2x^3-2x-4=0...che da soluzione positiva x=rad3(2)...spero sia giista
