How did you arrive at this biquadratic equation? What is its form? I tried to do it for y, x, and for u (the name I gave to the red line), but in all cases, I arrived at very complicated equations due to irrational terms (variables inside square roots). I tried to isolate the irrational terms to square both terms, but it only made the equation look worse… Could you explain?”
Me also here are the equations y²=(x+5)²+3² pythagore c²=a²+b²-2ab cos(angle) 5²=(9+x²)+y²-2 sqrt(9+x²)y (sqrt(2)/2) 5²=(9+x²)+y²- sqrt(9+x²) y sqrt(2) with pythagore equation we obtain 25=(9+x²)+((x+5)²+3²)- sqrt(9+x²) sqrt((x+5)²+3²) sqrt(2) 2x²+10x+18= sqrt(2) sqrt(9+x²) sqrt((x+5)²+9) 4(x²+5x+9)²=2((x+5)²+9) (9+x²) 2((x²)²+10x³+43x²+90x+81)=(x²)²+10x³+43x²+90x+306 (x²)²+10x³+43x²+90x-144=0 x=1 is a solution we divide (x²)²+10x³+43x²+90x-144 by x-1 (x²)²+10x³+43x²+90x-144=(x³+11x²+54x+144)(x-1) is there another solution if there is (x³+11x²+54x+144)=0 or (x³+54x)+(11x²+144))=0 (11x²+144) is always posifif, in order to have a solution we must have x³+54x less the zero x³+54x=x(x²+5) , x³+54x is less then zero if only if x is less then 0, in this case it is not a solution because it is a distance. x=1 is the only solution and y=sqrt(45).
England 🏴 Math Olympiad 2009 - Algebra - Find f(x)??
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solved by cosine rule and biquadratic i.e. full algebra but your method is superior
How did you arrive at this biquadratic equation? What is its form? I tried to do it for y, x, and for u (the name I gave to the red line), but in all cases, I arrived at very complicated equations due to irrational terms (variables inside square roots). I tried to isolate the irrational terms to square both terms, but it only made the equation look worse… Could you explain?”
Me also
here are the equations
y²=(x+5)²+3² pythagore
c²=a²+b²-2ab cos(angle)
5²=(9+x²)+y²-2 sqrt(9+x²)y (sqrt(2)/2)
5²=(9+x²)+y²- sqrt(9+x²) y sqrt(2)
with pythagore equation we obtain
25=(9+x²)+((x+5)²+3²)- sqrt(9+x²) sqrt((x+5)²+3²) sqrt(2)
2x²+10x+18= sqrt(2) sqrt(9+x²) sqrt((x+5)²+9)
4(x²+5x+9)²=2((x+5)²+9) (9+x²)
2((x²)²+10x³+43x²+90x+81)=(x²)²+10x³+43x²+90x+306
(x²)²+10x³+43x²+90x-144=0
x=1 is a solution
we divide (x²)²+10x³+43x²+90x-144 by x-1
(x²)²+10x³+43x²+90x-144=(x³+11x²+54x+144)(x-1)
is there another solution if there is
(x³+11x²+54x+144)=0 or (x³+54x)+(11x²+144))=0
(11x²+144) is always posifif, in order to have a solution we must have x³+54x less the zero
x³+54x=x(x²+5) , x³+54x is less then zero if only if x is less then 0, in this case it is not a solution because it is a distance.
x=1 is the only solution and y=sqrt(45).
What about using tangens to get the lenght of the middle line. With Pythagoras you will get x. And with x you get y
No unique way to solve. Using trigonometric function works as well 👍
@@MathAPlus So this is NOT an Olympiad question if it can be solved by a calculator!
i can't belive it i spent more then an hour and you did it in ten minutes and with a brillant methode, unfortunately i have to leave mathematics.
Thank you. Your channel is gonna help me a lot(from a subscriber)
My pleasure
Solved it using trigonometry btw can you tell me this question was of which class
Great, first stage of high school math olympiad from China , Grade 11 and 12 should solve it
how bro
how did you solve it?
I thought using cosine rule but can’t solve equations
same lol i ended up with a biquadratic could have solved it but x=1 guess and check possible only
I am kinda bad in geometry what should I do
I will post more videos stay tune
How do you know the triangle you made and the other triangle are equal, what if they are similar
Same angles and lengths means both are equal. Just same angles mean similar
fuzzy drawings and dirty fingernails
What does it have to do with his mathematical skills ?