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Multiply uv together, expand, and you get uv = -1. Since you now know u+v=-1 and uv=-1, u and v are roots of the quadratic x^2+x-1=0. Use the quadratic formula and you will find that |u-v|=sqrt5. Then using your trig expression and the fact that cos2pi/5>cos4pi/5, you know that u-v>0, therefore...

Yeah I did 😭 , it's a rhombus cos a rhombus is (well can be) defined as a four sided shape with equal side lengths and it's the only one with the definition (square is a rhombus) so that's the conclusion right?

Ah I can say that cos our four vectors are z_1, z_2, z_3 and 1, and they all have modulus one, i.e 4 sides are equal, it has to be a rhombus which has parallel sides. so therefore one of them also has to be 1. That works right (hopefully lmao my yr 9 geometry is rusty)?

It's just I graphed the curve she got for x on desmos, and it's symmetrical about the turning point implying that the time it takes to reach max height is the same as it takes to fall to the ground and that's not right right? It should take longer to fall down is my understanding of mechanics...

It's impossible to construct unit vectors that add together to get one if one of the is not one. For example, construct the first one anywhere you want, then consider the second vector as one, if it shifts up or down, the distance between its tip and 1 would change and it no longer 1, so the...

Tbh the polynomial thing for 16d wasn't that necessary. Once you have that they're all magnitude 1, you can use z_1+z_2+z_3=1 to draw a rhombus (since their lengths are the same) and deduce one of them has to be 1 and the other two are of the form e^(itheta) and -e(itheta), then you can use the...

I'm still confused by this solution lmao does anyone know how she's allowed to just use the upwards equation the whole way through and sub in x=0, seems to contradict everything I've learnt in mechanics 😭

Question 9 didn't need circle geo right, it was just pythag with hypotenuse as radius of the sphere in the first set, the radius that we're trying to find and 1/4 the length of AB

Yeah surely that's wrong, the person assumed that the impact speed of impact to the ground is the same as the initial velocity unless I'm missing something normally it's different

I could not do that stupid square question for my life

Well, I'm sure we'll activate our big brains for the nesa overlords 💀. Not much prep I can do today so I'm just gonna lay back for a bit, good luck for tomorrow : )

It's the same assumption but I said that the area of two trapeziums would be greater than just the one trapezium cos it's concave down so that mid point on the graph will be above the line drawn for one trapezium. It was the only thing that came into my mind during the exam, kinda funny how I...

Yeah, I think this does work, I mean in the exam I did it funkily by finding trapezium areas. The suggested probably makes the must intuitive sense with concavity honestly

Ok complex is chill but that vectors one is weird cos in my mind, aren't there an infinite amount of values for R that trace an annulus about the sphere?

This one is not too bad actually, I can do this one. The one that I've attached is the inequality that threw off most people in our yr in our trials and it's what I thought geometric inequalities meant lmao

If I open up Q16 on Thursday and they give us a munted inequality and write prove geometrically I'm gonna ask for a refund

As in draw a triangle to prove Am-Gm for 2? 💀 Got scammed. Normally all the geometrical proofs I see are all functions and areas under the curve tbh

The length of the vector (0,y,z) or like a semi circle equation, idk how that's gonna help though 😅

Yeah the algebraic way is really straightforward, you do Am-Gm for 2 to get x^2+(y^2+z^2)≥2xsqrt(y^2+z^2) and y^2+(x^2+z^2)≥2ysqrt(x^2+z^2) then you just add then together. I don't really know what "geometric" proof you're supposed to provide given that the RHS is not a curve in 3d space we can...

To try find the substitutions the trick is often to try match up the powers. For example, if I was like prove that a^4+b^4+c^4>abc(a+b+c), you see the 4th power so instantly you're thinking along the lines of Am-Gm for four. Then, you see three terms on the LHS and you notice that on the RHS...