Library / /211 of 740

, 2018

Geometry Difficulty 4.8 AIME Prove it United States

Problem:
A square in the $x y$ -plane has area $A$ , and three of its vertices have $x$ -coordinates $2$ , $0$ , and $18$ in some order. Find the sum of all possible values of $A$ .

Solution

Solution:
More generally, suppose three vertices of the square lie on lines $y = y_{1}$ , $y = y_{2}$ , $y = y_{3}$ . One of these vertices must be adjacent to two others. If that vertex is on $y = y_{1}$ and the other two are on $y = y_{2}$ and $y = y_{3}$ , then we can use the Pythagorean theorem to get that the square of the side length is $(y_{2} - y_{1})^{2} + (y_{3} - y_{1})^{2}$ .

For $(y_{1}, y_{2}, y_{3}) = (2, 0, 18)$ , the possibilities are $2^{2} + 16^{2}$ , $2^{2} + 18^{2}$ , $16^{2} + 18^{2}$ , so the sum is $2(2^{2} + 16^{2} + 18^{2}) = 2(4 + 256 + 324) = 1168$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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