China | A nice Math Olympiad Radical Simplification Problem | Calculators NOT allowed |

China | A nice Math Olympiad Radical Simplification Problem | Calculators NOT allowed
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[@bpark10001]

Rather than all the substitutions, simplify as you did to [(√3 + 1)/2]^9, then apply cubic formula (a + b)^3 = a³ + 3a²b + 3ab² + b³ twice. Algebra is much simpler & more straightforward.

[@YAWTon]

Let x = (√6 + √2)/(√8) = (1+√3 )/2. x^3 = (5+3√3)/4 by simple application of binomial theorem. Therefore x^9=( (5+3√3)/4)^3. Use binomial theorem once more to obtain x^9 = (265 + 153√3)/32.

[@vacuumcarexpo]

The given expression can be written as (√2cos 15°)^9.
(cos θ)^9=((e^(iθ)+e^(-iθ))/2)^9=(1/2^8)(cos 9θ+C(9,1)cos 7θ+C(9,2)cos 5θ+C(9,3)cos 3θ+C(9,4)cos θ)
Using this identity, the given expression is (16√2/256)(cos 135°+9cos 105°+36cos 75°+84cos 45°+126cos 15°)=(265+153√3)/32.

[@superacademy247]

This is powerful 💪💪💪

[@antoinegrassi3796]

Parfaite méthode, bien que un peu... Complexe. Lol 👍😉
Une fois écrite sous la forme r.e^(i.teta) sa puissance 9 peut directement s'écrire r^9.e^(i.9.teta). Dans le cadre d'une Olympiade on doit pouvoir économiser le temps nécessaire pour détailler l'expression trigonométrique de 9.teta.

[@halidelharbe0787]

If you go to exam, you will miss everything 😅

[@yanfisher2639]

Too much unnecessary writing and talks

[@antoinegrassi3796]

J'espère que tu me pardonneras, c'est le mot OLYMPIADE qui me fait réagir, car il y a beaucoup plus simple et surtout beaucoup plus court. Employer TA MÉTHODE , c'est être RECALÉ aux OLYMPIADES. Désolé mais je ne peux pas laisser passer ça.
Posons Y = (sqr(6) + sqr(2))/sqr(8) = (1+sqr(3))/2 après une simplification évidente par sqr(2), puis calculons successivement Y^2, puis Y^4, puis Y^8 en élevant au carré et enfin Y^9 = Y^8.Y.
À chaque étape on obtiendra un résultat de la forme
a +b.sqr(3) qui se calcule très facilement.
Je détaille le premier pour montrer que ces calculs sont très simples Y^2 = (1+2sqr(2) + 3) / 4 = (4 + 2sqr(3)) / 4
donc Y^2 = (2+sqr(3)) / 2.
On trouvera ensuite Y^4 = (Y^2)^2 = (7 + 4sqr(3)) / 4
puis Y^8 = (97 + 56sqr(3)) / 16
Et enfin Y^9 = (265 + 153sqr(3)) / 32.
L'utilisation de ton équation en X² est astucieuse, mais elle presente deux gros inconvénients:
1- elle est beaucoup trop longue et inutilement compliquée ce qui est incompatible avec une Olympiade.
2- elle montre que tu ignores un résultat connu, à savoir que les nombres de la forme a +b.sqr(3), forment un SOUS-CORPS de Q, ce qui permet de prévoir qu'à chaque étape on aura un résultat simple de la forme a+b.sqr(3).
Sans rancune et bon courage.

[@robisonluiz5826]

Parabéns pela excelente explicação.

[@superacademy247]

Thanks you so much

[@walterwen2975]

A nice Math Olympiad Radical Simplification Problem: [(√6 + √2)/(√8)]⁹ = ?
(√6 + √2)/(√8) = [(√3 + 1)(√2)]/(2√2) = (√3 + 1)/2, Let: a = (1/2)(1 +√3)
a² = [(1/2)²][(1 + √3)²] = (1/4)(1 + 3 + 2√3) = (1/2)(2 + √3)
a³ = a(a²) = (1/4)(1 + √3)(2 + √3) = (1/4)(2 + 3√3 + 3) = (1/4)(5 + 3√3)
a⁶ = (a³)² = [(1/4)²][(5 + 3√3)²] = (1/16)(25 + 27 + 30√3) = (1/8)(26 + 15√3)
a⁹ = (a³)(a⁶) = (1/32)(5 + 3√3)(26 + 15√3) = (1/32)(130 + 135 + 153√3)
= (1/32)(265 + 153√3) = (265 + 153√3)/32
[(√6 + √2)/(√8)]⁹ = a⁹ = (265 + 153√3)/32

[@prime423]

The sq root of 3 is approximately 1.73.Continuous multiplying will get answer much more quickly!!

[@antoinegrassi3796]

On recherche le RÉSULTAT EXACT et sans calculatrice. Sinon c'est sans intérêt. Désolé.