Maximize an Almost-Factorable Expression

What does $(bc+ ad)$ and $(bd - ac)$ remind you of? It has to be Fermat's Two Square Identity, which states that

$(a^2 + b^2) ( c^2 + d^2) = ( bc + ad) ^2 + (bd - ac) ^2 !$

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For the solution, applying Cauchy Schwarz, we have

$\left[ 20 (bc + ad) + 14 (bd - ac) \right]^ 2 \leq \left[ ( bc+ ad)^2 + (bd-ac)^2 \right] \left[ 20^2 + 14^2 \right ] = ( a^2 + b^2) ^2 \times 596$

This thus gives us

$\left| \frac{ 20 (bc+ad) + 14 (bd - ac) } { a^2 + b^2 + c^2 + d^2 } \right| \leq \sqrt{ 149 }.$

It remains to verify that equality in Cauchy Schwarz can exist. This is done by setting $a = 0, b = \sqrt{596}, c = 20, d = 14$ . Hence, the maximum is $\sqrt{149}$ .

Off the bat, seeing $a^2+b^2=c^2+d^2$ makes us think of the identity $\sin^2\theta+\cos^2\theta=1$ . But will substituting simple variables as complicated trigonometric functions help? Let's find out.

First, to make the substitution, let's define a variable $x$ such that $a^2+b^2=c^2+d^2=x^2$ . Thus, we can substitute $a=x\sin\alpha$ , $b=x\cos\alpha$ , $c=x\sin\beta$ , and $d=x\cos\beta$ . Cool, we reduced it to only three variables to juggle around!

Substituting these values in the expression we want to maximize, along with substituting $a^2+b^2=x^2$ and $c^2+d^2=x^2$ , we obtain that the expression is equivalent to

$\dfrac{20x^2\cos\alpha\sin\beta + 14x^2\cos\alpha\cos\beta + 20x^2\sin\alpha\cos\beta - 14x^2 \sin\alpha\sin\beta}{2x^2}$

Hey look, the $2x^2$ cancel out! We know know that we are definitely heading the right direction, with only two variables to handle now. We are left with the expression

$10\cos\alpha\sin\beta + 7\cos\alpha\cos\beta + 10\sin\alpha\cos\beta - 7 \sin\alpha\sin\beta$

Seeing the coefficients, we should group the pairs together. Grouping gives

$10(\sin\alpha\cos\beta + \sin\beta\cos\alpha)+7(\cos\alpha\cos\beta-\sin\alpha\sin\beta)$

Incredible! We recognize the expressions inside the parentheses as two of the four angle sum/difference identities! Applying the two applicable identities in reverse, we obtain

$10\sin(\alpha+\beta)+7\cos(\alpha+\beta)$

We can now substitute even further by letting $\theta=\alpha+\beta$ , leaving us with a one-variable expression to maximize. We want to maximize

$10\sin\theta+7\cos\theta$

However, it isn't immediate how to maximize this. We can't use $\sin^2\theta+\cos^2\theta=1$ because the coefficients are different (even if they are the same, we actually still cannot obtain the maximum anyways), nor can we use $\sin\theta=\cos(90-\theta)$ because that would make the angle value of the two expression be different...

Well, what if $10=\cos\varphi$ and $7=\sin\varphi$ for some variable $\varphi$ ? We could use the sine angle sum identity one more time to reduce it to one trig function! However, it's clear that there isn't any real $\varphi$ that satisfies either equation.

However, if we scale the entire expression down, maybe there will exist a $\varphi$ ! So let's scale the expression down by a factor of $S$ to obtain

$S\left(\dfrac{1}{S}\cdot 10\sin\theta+\dfrac{1}{S}\cdot 7\cos\theta\right)$

Now, let' set $\cos\varphi = \dfrac{1}{S}\cdot 10$ and $\sin\varphi = \dfrac{1}{S}\cdot 7$ .

Now we can use $\sin^2\varphi + \cos^2\varphi=1$ to find the value of $S$ : $\begin{aligned}\left(\dfrac{1}{S}\cdot 10\right)^2+\left(\dfrac{1}{S}\cdot 7\right)^2&=1\\ \dfrac{100}{S^2}+\dfrac{49}{S^2}&=1\\ 149&=S^2\\ S&=\sqrt{149} \end{aligned}$

Thus, our expression turns into $\sqrt{149}(\sin\theta\cos\varphi+\sin\varphi\cos\theta)$ which is simply $\sqrt{149}\sin(\theta+\varphi)$

The maximum value of this expression is clearly $\sqrt{149}$ , as $\sin(\theta+\varphi) \le 1$ . Thus, the maximum of our original expression is also $\sqrt{149}$ and our final answer is $\boxed{149}$ . $\Box$

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Problem Maker's Note: Using this method, we can also very easily find the minimum value of our expression. Because $\sin(\theta+\varphi)\ge -1$ , the minimum value is simply $-\sqrt{149}$ !