MHB Find Lowest Value for A: a1, a2, a3 & 4 | Arithmetic Progression

AI Thread Summary
The discussion focuses on finding the lowest value of A = a1a2 + a2a3 + a3a1 for an arithmetic progression defined by a1, a2, a3, and 4. A participant expresses uncertainty about their approach, having derived A = 3x^2 + 6xd + 2d^2. Another user suggests that the fourth term of the progression can simplify A into a single variable, which would facilitate finding the minimum value. The original poster acknowledges the oversight and appreciates the guidance provided. The conversation emphasizes the importance of utilizing all given terms in mathematical problems for simplification.
mitaka90
Messages
9
Reaction score
0
a1, a2, a3 and 4 make an arithmetic progression with difference d. For which values of d, A = a1a2 + a2a3 + a3a1 has the lowest value?I don't know if I went with the right approach, but I managed to get this : A=3x2 +6xd + 2d2 for a1= x, a2 = x + d, etc... But I don't know what else to do.
 
Mathematics news on Phys.org
mitaka90 said:
a1, a2, a3 and 4 make an arithmetic progression with difference d. For which values of d, A = a1a2 + a2a3 + a3a1 has the lowest value?I don't know if I went with the right approach, but I managed to get this : A=3x2 +6xd + 2d2 for a1= x, a2 = x + d, etc... But I don't know what else to do.
Hi mitaka90!

It seems to me you haven't made good use of the given fourth term in that arithmetic progression...:) the fourth term would help you to simplify your $A$ in terms of only one variable and when you have the quadratic equation in terms of one variable, I believe you could handle from there...
 
anemone said:
Hi mitaka90!

It seems to me you haven't made good use of the given fourth term in that arithmetic progression...:) the fourth term would help you to simplify your $A$ in terms of only one variable and when you have the quadratic equation in terms of one variable, I believe you could handle from there...

Omg, I'm such a moron. I hate it when I do the hard work and then the easiest and most noticable thing just slips from my sight. Thank you sincerely, I guess that little tip is what I needed.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Back
Top