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van Aubel's theorem. AF/FD= 8/7 + 6/7=2

Why is the point M the circle of center?

why does the 5 cancel the log in base 5 ?

Knew answer in 5 seconds

cool solution

didn't understood a shit

It may not be obvious for some how you got from the x^2(y^2+1)+(y^2+1) step to (x^2+1)(y^2+1). You should also mention that the (y^2+1) multiplier was also factored out from both terms leaving (x^2+1). Just for clarity’s sake.

why does f(-x) the same as -f(x)? i want to learn but i cant grasp this part yet

The thumbnail does not match the video presentation. For those interested in the thumbnail's solution: g(x)=1/(1-x^2) and f(x)=x/(x^2-1)

This function is periodic over the integers with a period of 2. You can start evaluating f(6) to move forward to see a period of 2, and you can also move backward with a little more effort. The assumption is that this function is defined over the integers.

Cos(EBC)=Sin(ABE). S=5^2-0.5*3*4-0.5*3*5*(3/5).

how do you know the period is 6?

The value??

BX=40/sin(60°-a)sina,s=20√3/sin (60°-a) FZ=41(√3-1)-40/sin(60°-a) sina a=-arccos((163-41√3)/2/√(83 24-41*82√3)+arccos(-40/√(8324- 41*82√3)) 11896 285/1487× s=20√3/sin120°=40

{y=X²-4X,0=(x-2)⁴-12(x-2)²-(x-2)+32; (z-2)³-17 3/4(z-2)-27 33/64=0 z=2+(880 1/2+√(1761²-71³*256/2 7)/2)¹/³/4+(..-..)..;2-(880 1/2+√ (1761²-71³*256/27)/2)¹/³/8-(..-..)..±√(3/4((880 1/2+√(1761²-71³*25 6/27)/2)¹/³/4+(..-..)..)²-17,75)i

x³(-√5x+x²+1)=0

you can find the inverse of the first one, f(x) = ax + b => f^-1(x) = (x - b)/a. Then equal to the other inverse, (x - b) / a = bx + a <=> x - b = abx + a^2, you find that: x = abx <=> ab = 1, and a^2 = -b. Furthermore, b = -a^2 and replacing in the other one: a(-a^2) = 1 <=> a = -1. Finally, (-1)^2 = - b, b = - 1, so a + b = -1 + (-1) = -2.

10/√(y²-100)(18-y) x=10√(y²-100)/y 5√(5+400/y²) 3/√(5+400/y²)=1,8√(y ²-100)/(18-y) 0=y⁴+45y³-450y²-18000 у=10,88 DF=16 324/536 x=19 12/17 K=653 (78 10/17)/134 ..=922(~✓)

10/√(y²-100)(18-y) x=10√(y²-100)/y 5√(5+400/y²) 3/√(5+400/y²)=1,8√(y ²-100)/(18-y) 0=y⁴+45y³-450y²-18000

x<log3(1/2) 0=3(3^x)²+2*3^x-1 3^x=(-2±√16)/6=1/3;-1>0 x=-1✓

a=10log5(2) b=10lg(2) 1/10log2(5)-1/10log2(10)=-1/10

S∆=√(12*3*4*5)=12√5 h=3√5 PQ=17/15√5

2log³10+6(log10-1)log²2=2 Ну в этом случае да.

just for fun. let O be center of circle. tan(OCB) = 1/2 thus tan(ECB) = 4/3 thus cos(DCE) = cos(90 - ECB) = sin(ECB) = 4/5 EC = 2 / cos(DCE) = 2 / (4/5) = 5/2

its crazy how the first step is thought of

I found another way to find the solution. Step 1) Find the inverse f^-1(x) of f(x) while assuming that the given f^-1(x) is different to the actual f^-1(x). Let's call the given f^-1(x) from now on g(x). Well, you'll get f^-1(x) = 1/a * x - b/a as the inverse of f(x) . Step 2) Equate the calculated f^-1(x) and g(x), which means that 1/a * x - b/a has to be equal to bx + a. Step 3) That leads to a system of equations with I) 1/a = b and II) -b/a = a, which is a solvable system of equation. You'll get a = -1 and b = -1. Step 4) Calculate a + b = -1 + (-1) = -2 (sry for my bad english)

b(ax+b)+a=x, abx+b²+a=x. ab=1, b²+a=0. a=1/b, b²+1/b=0. b isn't 0 or ∞, so b³+1=0. Meaning b is one of the 3 cube roots of -1. Meanwhile, a, being its reciprocal, is b's conjugate. Either a=b=-1, or a=(1±√(3)i)/2 and b=(1-±√(3)i)/2. Meaning either a+b=-2, or a+b=1. The two solutions are -2 and 1.

Easy ass shiiiiit bruh was this Olympiad considered hard or moderate??

I didn't got the last part, last row.

If a and b are not real, you will get three solutions. Sums would be -2, 1, 1.

Exactly how I did it.

Well done. I got 1/3 as well.

what???????

(2)^2=4 360°ABCD/4 = 90°ABCD 3^30 3^5^6 3^5^13^2 1^1^13^2 3^2 (ABCD ➖ 3ABCD+2)

I did it a completely different way, but I got the same answer. Nice video!

absolutely i loved this question🙏🏼

Use tan I'm guessing after finding out that KLB = LCM = MDN = NAK and notice that its also a square

I would Never find this solution

Wow. Good this was fun

V nice sol.!! Here's another fun but longer solution w. indices. log x(w) = 24 | x^24 = w log y(w) = 40 | y^40 = w log xyz(w) = 12 | (xyz)^12 = w log z(w) = ? | z^? = w therefore: x^24 = w | x^12 = w^(1/2) [1] y^40 = w | (y^12)^(10/3) = w | y^12 = w^(3/10 ) [2] x^12 * y^12 * z^12 = w [3] substituting [1] and [2] into [3]: w^(1/2) * w^(3/10) * z^12 = w w^(4/5) * z^12 = w^1 z^12 = w^(1/5) z^60 = w therefore: log z(w) = 60

ez

Nice explanation 👌

math is mathing

Is this seriously Olympiad level?!

1:06 Shouldn't BG = 2?

this is a romanian baccalaureate mock exam question...

This was hard

Very cool!

Nice synthetic proof 🎉

32 = r * SQRT(2) thus r = 16 * SQRT(2) the area of the triangle is 16*32/2 is 256. Since the triangle is anchored to the center of the Square we would like to know how much space to rotate it . The chord of 90° is SQRT(2) on the unit circle the bisector and halfchords are r SQRT(2)/2. Thus thus is bisector SQRT(2)/2 * 16 *SQRT(2) = 16. While we are here the area between a chord and the origin is bisector * halfchord = 256. That’s rather obvious but it proves our math so far is sound. If we inscribed the square into a circle the the radius is 21/SQRT(2) or 10.5 * SQRT(2) So this is smaller than 16. This means we can rotate the triangle so that its sides are orthogonal to the sides of the square, done. The square is 10.5 x 10.5 = 21^2/4 The shaded area is 256 - 21^2/4 = 21x21/4 = 441/4 = 110.25 256-110.25 = 145.75 This is the complete answer.