2024 AIME I Problems/Problem 7

Let such that . The expression becomes:

Call this complex number . We simplify this expression.

$$\begin{array}{rl}w& =(75+117i)(a+bi)+\frac{96+144i}{a+bi}\\ & =(75a-117b)+(117a+75b)i+48\left(\frac{2+3i}{a+bi}\right)\\ & =(75a-117b)+(116a+75b)i+48\left(\frac{(2+3i)(a-bi)}{(a+bi)(a-bi)}\right)\\ & =(75a-117b)+(116a+75b)i+48\left(\frac{2a+3b+(3a-2b)i}{a^2+b^2}\right)\\ & =(75a-117b)+(116a+75b)i+48\left(\frac{2a+3b+(3a-2b)i}{16}\right)\\ & =(75a-117b)+(116a+75b)i+3(2a+3b+(3a-2b)i)\\ & =(75a-117b)+(116a+75b)i+6a+9b+(9a-6b)i\\ & =(81a-108b)+(125a+69b)i.\end{array}$$

We want to maximize . We can use elementary calculus for this, but to do so, we must put the expression in terms of one variable. Recall that ; thus, . Notice that we have a in the expression; to maximize the expression, we want to be negative so that is positive and thus contributes more to the expression. We thus let . Let . We now know that , and can proceed with normal calculus.

$$\begin{array}{rl}f(a)& =81a+108\sqrt{16-a^2}\\ & =27(3a+4\sqrt{16-a^2})\\ f^{\prime }(a)& =27{(3a+4\sqrt{16-a^2})}^{\prime }\\ & =27(3+4{\left(\sqrt{16-a^2}\right)}^{\prime })\\ & =27(3+4\left(\frac{-2a}{2\sqrt{16-a^2}}\right))\\ & =27(3-4\left(\frac{a}{\sqrt{16-a^2}}\right))\\ & =27(3-\frac{4a}{\sqrt{16-a^2}}).\end{array}$$

We want to be to find the maximum.

$$\begin{array}{rl}0& =27(3-\frac{4a}{\sqrt{16-a^2}})\\ & =3-\frac{4a}{\sqrt{16-a^2}}\\ 3& =\frac{4a}{\sqrt{16-a^2}}\\ 4a& =3\sqrt{16-a^2}\\ 16a^2& =9(16-a^2)\\ 16a^2& =144-9a^2\\ 25a^2& =144\\ a^2& =\frac{144}{25}\\ a& =\frac{12}{5}\\ & =2.4.\end{array}$$

We also find that .

Thus, the expression we wanted to maximize becomes .

~Technodoggo

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.