Track / Stage 5 / 229 of 400 #829 of 1964

Problem 829

AIME late

Algebra Difficulty 5.5 Find the answer

# Problem 6. (4 points)

Six positive numbers, not exceeding 3, satisfy the equations $a+b+c+d=6$ and $e+f=2$ . What is the smallest value that the expression

$\left(\sqrt{a^{2}+4}+\sqrt{b^{2}+e^{2}}+\sqrt{c^{2}+f^{2}}+\sqrt{d^{2}+4}\right)^{2}$

can take?

Your answer

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

Answer: 72

In the image, there are three rectangles $2 \times (a+b)$ and three rectangles $2 \times (c+d)$ , forming a $6 \times 6$ square, since $a+b+c+d=6$ . The segment of length 2 in the center of the image is divided into segments $e$ and $f$ , which sum to 2, as required by the condition.

The broken line depicted in the image, going from one corner of the square to the other, by the Pythagorean theorem, has a square of its length equal to the expression whose minimum value we need to find. Such a broken line can be constructed for each set of six numbers, so we only need to find the length of the shortest broken line, which is the diagonal of the square. The square of its length is $2 \cdot 6^{2}=$ 72.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.