Arbab Qamar

Algebra Level 5

$\begin{cases} a^2+b^2+c^2+d^2+e^2 = 25 \\ a+2b+c+ d \sqrt{3}+e = 13\end{cases}$

If $a,b,c,d$ and $e$ are real numbers satisfying the two equation above, find the sum of the minimum and maximum possible values of $e$ .

If you got your answer as $\dfrac{A}{B}$ , where $A$ and $B$ are coprime positive integers, submit your answer as $A\times B$ .

The answer is 65.

1 solution

Manuel Kahayon
Feb 27, 2016

$a^2+b^2+c^2+d^2 = 25-e^2$

$a+2b+c+ d \sqrt{3}= 13-e$

By Cauchy-schwarz inequality,

$(1^2+2^2+1^2+\sqrt{3}^2)(a^2+b^2+c^2+d^2) \geq (a+2b+c+ d \sqrt{3})^2$

$9(25-e^2) \geq (13-e)^2$

By factoring,

$0 \geq (5e+7)(e-4)$

$4 \geq e \geq \frac{-7}{5}$

So, the maximum is $4$ and minimum is $\frac{-7}{5}$ So, our answer is $4 + \frac{-7}{5} = 13/5$

Then, $13\cdot 5\boxed{65}$

We don't even need to factorize! We can just find the sum of roots of the quadratic in the inequality as $\frac{-b}{a}$ , which even though not necessary here, can be helpful in preventing lengthy factorization/evaluation of roots

Ashish Gupta - 5 years, 1 month ago

Oh, yeah! But, showing the minimum and maximum values of e made me happy, I guess :)

Manuel Kahayon - 5 years, 1 month ago

Your solution is incomplete.

You've just shown that $dfrac{-7}{5}$ is a lower bound and $4$ is a upper bound.

You've to show that the obtained extreme values can indeed be attained.

Aditya Sky - 4 years, 11 months ago

Arbab Qamar

The answer is 65.

1 solution

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