Library / Annual Harvard-MIT Mathematics Tournament /5 of 72

Annual Harvard-MIT Mathematics Tournament

Geometry Difficulty 4.2 AIME Find the answer United States

Problem:

Square $ABCD$ has side length $2$ , and $X$ is a point outside the square such that $AX = XB = \sqrt{2}$ . What is the length of the longest diagonal of pentagon $AXB C D$ ?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:

Answer: $\sqrt{10}$

Since $AX = XB = \sqrt{2}$ and $AB = 2$ , we have $\angle AXB = 90^{\circ}$ . Hence, the distance from $X$ to $AB$ is $1$ and the distance from $X$ to $CD$ is $3$ . By inspection, the largest diagonals are thus $BX = CX = \sqrt{3^{2} + 1^{2}} = \sqrt{10}$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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