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Reach The Stars
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Добавлен 14 авг 2021
Q365 | Math Olympiad | Geometry | 2016 AMC 12B Problem 17 | Heron Formula | Angle Bisector Theorem
Q365 | Math Olympiad | Geometry | 2016 AMC 12B Problem 17 | Heron Formula | Angle Bisector Theorem
Просмотров: 2
Видео
Q364 | Math Olympiad | Algebra | Odd Function | Casework
Просмотров 1
Q364 | Math Olympiad | Algebra | Odd Function | Casework
Q363 | Math Olympiad | Algebra | Logarithms | Base Change Property
Q363 | Math Olympiad | Algebra | Logarithms | Base Change Property
Q362 | Math Olympiad | Algebra | Sum of Functions | Period
Просмотров 202 часа назад
Q362 | Math Olympiad | Algebra | Sum of Functions | Period
Q361 | Math Olympiad | Algebra | Function | Period
Просмотров 132 часа назад
Q361 | Math Olympiad | Algebra | Function | Period
Q360 | Math Olympiad | Geometry | 2016 AMC 12A Problem 12 | Angle Bisector Theorem
Просмотров 232 часа назад
Q360 | Math Olympiad | Geometry | 2016 AMC 12A Problem 12 | Angle Bisector Theorem
Q359 | Math Olympiad | Geometry | 2005 AMC 12A Problem 15 | Circle | Pythagorean Theorem | Ratio
Просмотров 424 часа назад
Q359 | Math Olympiad | Geometry | 2005 AMC 12A Problem 15 | Circle | Pythagorean Theorem | Ratio
Q358 | Math Olympiad | Algebra | Function | Absolute Value | Graph | Monotonic Interval
Просмотров 244 часа назад
Q358 | Math Olympiad | Algebra | Function | Absolute Value | Graph | Monotonic Interval
Q357 | Math Olympiad | Algebra | Functions | Period | Substitution
Просмотров 824 часа назад
Q357 | Math Olympiad | Algebra | Functions | Period | Substitution
Q356 | Math Olympiad | Geometry | 2013 AMC 12A Problem 11 | Equilateral Triangle | Perimeter
Просмотров 827 часов назад
Q356 | Math Olympiad | Geometry | 2013 AMC 12A Problem 11 | Equilateral Triangle | Perimeter
Q355 | Math Olympiad | Algebra | 2010 AMC 12A Problem 11 | Exponents | Logarithms
Просмотров 497 часов назад
Q355 | Math Olympiad | Algebra | 2010 AMC 12A Problem 11 | Exponents | Logarithms
Q354 | Precalculus Basics | Matrix | Vector Multiplication | Substitution
Просмотров 257 часов назад
Q354 | Precalculus Basics | Matrix | Vector Multiplication | Substitution
Q353 | Math Olympiad | Algebra | 2010 AMC 12B Problem 12 | Logarithm Properties
Просмотров 569 часов назад
Q353 | Math Olympiad | Algebra | 2010 AMC 12B Problem 12 | Logarithm Properties
Q352 | Math Olympiad | Geometry | 2013 AMC 12B Problem 19 | Cyclic Quadrilateral | Ptolemy Theorem
Просмотров 849 часов назад
Q352 | Math Olympiad | Geometry | 2013 AMC 12B Problem 19 | Cyclic Quadrilateral | Ptolemy Theorem
Q351 | Math Olympiad | Algebra | 2013 AMC 12B Problem 17 | Cauchy Schwarz Inequality
Просмотров 819 часов назад
Q351 | Math Olympiad | Algebra | 2013 AMC 12B Problem 17 | Cauchy Schwarz Inequality
Q350 | Math Olympiad | Geometry | 2012 AMC 12B Problem 21 | Equiangular Hexagon | Square | Sin Cos
Просмотров 9912 часов назад
Q350 | Math Olympiad | Geometry | 2012 AMC 12B Problem 21 | Equiangular Hexagon | Square | Sin Cos
Q349 | Math Olympiad | Algebra | 2020 AMC 12B Problem 13 | Square Root | Logarithms
Просмотров 22012 часов назад
Q349 | Math Olympiad | Algebra | 2020 AMC 12B Problem 13 | Square Root | Logarithms
Q348 | Math Olympiad | Algebra | 2020 AMC 12A Problem 13 | Roots | Exponents
Просмотров 3412 часов назад
Q348 | Math Olympiad | Algebra | 2020 AMC 12A Problem 13 | Roots | Exponents
Q347 | Math Olympiad | Geometry | 2009 AMC 12B Problem 16 | Trapezoid | Similar Triangle
Просмотров 9012 часов назад
Q347 | Math Olympiad | Geometry | 2009 AMC 12B Problem 16 | Trapezoid | Similar Triangle
Q346 | Math Olympiad | Algebra | 2012 AIME II Problem 8 | System of Complex Equations
Просмотров 4212 часов назад
Q346 | Math Olympiad | Algebra | 2012 AIME II Problem 8 | System of Complex Equations
Q345 | Precalculus Basics | Cauchy Schwarz Inequality
Просмотров 2316 часов назад
Q345 | Precalculus Basics | Cauchy Schwarz Inequality
Q344 | Math Olympiad | Geometry | 2009 AMC 12A Problem 20 | Area of Trapezoid
Просмотров 6616 часов назад
Q344 | Math Olympiad | Geometry | 2009 AMC 12A Problem 20 | Area of Trapezoid
Q343 | Math Olympiad | Algebra | 2006 AMC 12A Problem 15 | Cosine
Просмотров 4516 часов назад
Q343 | Math Olympiad | Algebra | 2006 AMC 12A Problem 15 | Cosine
Q342 | Math Olympiad | Geometry | 2004 AMC 12B Problem 14 | Pentagon Area
Просмотров 6919 часов назад
Q342 | Math Olympiad | Geometry | 2004 AMC 12B Problem 14 | Pentagon Area
Q341 | Math Olympiad | Algebra | 2004 AMC 12B Problem 13 | Function | Inverse Function
Просмотров 61419 часов назад
Q341 | Math Olympiad | Algebra | 2004 AMC 12B Problem 13 | Function | Inverse Function
Q340 | Math Olympiad | Algebra | 2007 AMC 12A Problem 17 | Trigonometric Pythagorean Identities
Просмотров 35019 часов назад
Q340 | Math Olympiad | Algebra | 2007 AMC 12A Problem 17 | Trigonometric Pythagorean Identities
Q339 | Math Olympiad | Geometry | 2003 AMC 12A Problem 17 | Coordinate Geometry | Circle Equation
Просмотров 14321 час назад
Q339 | Math Olympiad | Geometry | 2003 AMC 12A Problem 17 | Coordinate Geometry | Circle Equation
Q338 | Math Olympiad | Geometry | 2004 AMC 12A Problem 18 | Two Tangent Theorem
Просмотров 27121 час назад
Q338 | Math Olympiad | Geometry | 2004 AMC 12A Problem 18 | Two Tangent Theorem
Q337 | Math Olympiad | Algebra | Square Root | Function | Domain | Inequality
Просмотров 6021 час назад
Q337 | Math Olympiad | Algebra | Square Root | Function | Domain | Inequality
Q336 | Math Olympiad | Geometry | 2002 AMC 12B Problem 20 | Pythagorean Theorem
Просмотров 90День назад
Q336 | Math Olympiad | Geometry | 2002 AMC 12B Problem 20 | Pythagorean Theorem
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??
it goes amc12 to aime to usamo to imo this is the easiest level of olympiad
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?
On the graph it says 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.
What is a power of point theorem?
Very nice geometric solution. For those of us who struggle more with geometry, here is a trigonometric solution: Start as in video to find that △BAP ≅ △DAP → ∠BAP = ∠DAP = ∠BAD/2 = 90°/2 = 45° Using Law of Sines in △BAP, we get: sin(∠ABP)/AP = sin(∠BAP)/BP sin(∠ABP)/1 = sin(45°)/√2 sin(∠ABP) = (1/√2)/√2 = 1/2 ∠ABP = 30° Now we add angles in △BAP ∠BAP + ∠ABP + ∠APB = 180° 45° + 30° + ∠APB = 180° ∠APB = 180° − 45° − 30° *∠APB = 105°*
ChatGPT 4o got it in two seconds, I uploaded your thumbnail and said: Solve this, use python if you need to, and elaborate on the answer. Show me your steps.
Nice solution. For those unfamiliar with Power of a Point theorem, we can use Pythagorean theorem by constructing right triangles. Drop perpendicular from A to BC at point P. Since △ABC is isosceles (with AB = AC), altitude AP bisects BC. Let AP = h, BP = CP = a, CD = b. Then BD = 2a + b Using Pythagorean theorem in △APC, we get: h² + a² = 2² (1) Using Pythagorean theorem in △APD, we get: h² + (a+b)² = 4² (2) Subtracting (1) from (2) we get: (a+b)² − a² = 16 − 4 a² + 2ab + b² - a² = 12 2ab + b² = 12 (2a+b) * b = 12 BD * CD = 12
You could also get the product by using Stewart's Formula to find the Cevian of a Triangle.
Do the same problem for 20^21. it becomes more interesting
and then for 21^21.
the question did not state x can be written as m/n but logarithmic equation is equal to m/n, it was so confusing!!! the value of the logarithm is irrational so there is no m/n value possible.
I used coordinate geometry. Let, D be origin, DC x axis and DA y axis. With given info easy to find coordinate values of A, E, F, G, C points. Find equation of AG, AC, EF to get P, Q point coordinates. Find length of PQ and EF then divide.
Nice one
BD =2√(3+4√2)
It's wrong, because in ²√x, x can't be negative number. If we replace 8 with xes, it will be (²√49-8²)-(²√25-8²) => (²√49-64)-(²√25-64) => (²√-15)-(²√-39) which is impossible and even if you remove 8 from root it will be -14 and if you add () to each side it will be 2
8 is not the value of x , but the value of a.
Solve the equation for x and plug the result(s) into second equation. You may be surprised :)
@@usiek ok, I'll try it
This answer is fine. The point is that this equation is possible only for x€[-5;5]. a=8 and if you are looking for the x value, you will find x = + or -squareroot(18,75). I am sorry I answer with my phone and I don't know how to make math signs properly in RUclips comments.
@@cxfx4199 use this √, it can help you, copy ut so you will never face up with such a promlems. And about your information, I think I don't watch the video carefully, thank you for your information and help ^_^
You must using complex analysis method sir, if you using it, you can solving this problem easyly
But what does n equal? 🤔🤔🤔