Minimum and Maximum #4

Algebra Level 4

x x and y y are nonzero reals such that x y ( x 2 − y 2 ) xy\big(x^2-y^2\big) and x 2 + y 2 x^2+y^2 are the same number.

What is the minimum value of this number to 2 decimal places?


The answer is 4.

Boi (보이) by Boi (보이) , 19, South Korea

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1 solution

Boi (보이)
Jul 26, 2017

Solution 1)

First, since x y > 0 xy>0 and x 2 + y 2 > 0 , x^2+y^2>0, we know that x 2 − y 2 > 0 x^2-y^2>0 and therefore x > y . x>y.

Square both sides and you get

x 2 y 2 ( ( x 2 + y 2 ) 2 − 4 x 2 y 2 ) = ( x 2 + y 2 ) 2 . x^2y^2((x^2+y^2)^2-4x^2y^2)=(x^2+y^2)^2.

Let x 2 y 2 = X , x 2 + y 2 = Y . x^2y^2=X,~x^2+y^2=Y.

X ( Y 2 − 4 X ) = Y 2 4 X 2 − Y 2 X + Y 2 = 0. X(Y^2-4X)=Y^2 \\ 4X^2-Y^2X+Y^2=0.

The discriminant of this equation must be larger than or equal to 0. 0.

Y 4 − 16 Y 2 ≥ 0 Y 2 ( Y − 4 ) ( Y + 4 ) ≥ 0 ∴ Y ≥ 4. Y^4-16Y^2\ge0 \\ Y^2(Y-4)(Y+4)\ge0 \\ \therefore~Y\ge 4.

Therefore the miniimum of k = x 2 + y 2 = Y k=x^2+y^2=Y is 4 , \boxed{4},

with equality achieved when x = 2 + 2 , y = 2 − 2 . ( ∵ x > y ) x=\sqrt{2+\sqrt{2}},~y=\sqrt{2-\sqrt{2}}.~(\because x>y)

Solution 2)

Let x = k cos ⁡ θ , y = k sin ⁡ θ . x=\sqrt{k}\cos\theta,~y=\sqrt{k}\sin\theta.

k 2 sin ⁡ θ cos ⁡ θ ( cos ⁡ 2 θ − sin ⁡ 2 θ ) = k k ⋅ 1 2 sin ⁡ 2 θ cos ⁡ 2 θ = 1 k ⋅ 1 4 sin ⁡ 4 θ = 1 k = 4 sin ⁡ 4 θ ≥ 4. k^2\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)=k \\ k\cdot\dfrac{1}{2}\sin2\theta\cos2\theta=1 \\ k\cdot\dfrac{1}{4}\sin4\theta=1 \\ k=\dfrac{4}{\sin4\theta}\ge4.

Therefore the minimum of k k is 4 . \boxed{4}.

7 Helpful 0 Interesting 0 Brilliant 0 Confused

In solution 1 1 , you need to show that the minimum can be achieved - x = 2 + 2 x = \sqrt{2+\sqrt{2}} , y = 2 − 2 y = \sqrt{2-\sqrt{2}} . In solution 2 2 , it is more straightforward - θ = 1 8 π \theta = \tfrac18\pi .

Mark Hennings - 3 years, 10 months ago

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Will do, sir!

Boi (보이) - 3 years, 10 months ago

Nice solutions. One small thing, you should say x,y are non-zero reals, otherwise the minimum of k would be 0 when both x and y are zero.

Wei Chen - 3 years, 10 months ago

Note: In solution 1, the equation that we obtain from squaring both sides is a necessary, but not (yet) sufficient condition to satisfy the original conditions.

As pointed out by Mark, we had to verify that the minimum can be achieved by non-zero reals that satisfy the original condition.

E.g. x = 2 − 2 , y = 2 + 2 x =\sqrt{ 2 - \sqrt{2} } , y = \sqrt{ 2 + \sqrt{2} } would satisfy the squared condition, but not the original condition.

Calvin Lin Staff - 3 years, 10 months ago

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Yeah. Since A 2 = B 2 A^2=B^2 yields A = B A=B and A = − B , A=-B, x = 2 − 2 , y = 2 + 2 x=\sqrt{2-\sqrt{2}},~y=\sqrt{2+\sqrt{2}} would satisfy x y ( x 2 − y 2 ) = − ( x 2 + y 2 ) . xy(x^2-y^2)=-(x^2+y^2). We know this because x y xy is positive, x 2 + y 2 x^2+y^2 is also positive, but x 2 − y 2 x^2-y^2 is negative.

Boi (보이) - 3 years, 10 months ago

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