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, 2024

Algebra Difficulty 5.4 AIME, harder Prove it United States

Problem:
Let $a$ , $b$ , and $c$ be real numbers such that
$\begin{aligned} a + b + c &= 100 \\ ab + bc + ca &= 20, \text{ and } \\ (a + b)(a + c) &= 24 \end{aligned}$
Compute all possible values of $bc$ .

Solution

Solution:
We first expand the left-hand side of the third equation to get $(a + b)(a + c) = a^2 + ac + ab + bc = 24$ . From this, we subtract the second equation to obtain $a^2 = 4$ , so $a = \pm 2$ .

If $a = 2$ , plugging into the first equation gives us $b + c = 98$ and plugging into the second equation gives us $2(b + c) + bc = 20 \Rightarrow 2(98) + bc = 20 \Rightarrow bc = -176$ .

Then, if $a = -2$ , plugging into the first equation gives us $b + c = 102$ , and plugging into the second equation gives us $-2(b + c) + bc = 20 \Rightarrow -2(102) + bc = 20 \Rightarrow bc = 224$ .

Therefore, the possible values of $bc$ are $224, -176$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.