HMMT 十一月 2015 · THM 赛 · 第 3 题
HMMT November 2015 — THM Round — Problem 3
题目详情
- Consider a 3 × 3 grid of squares. A circle is inscribed in the lower left corner, the middle square of the top row, and the rightmost square of the middle row, and a circle O with radius r is drawn such that O is externally tangent to each of the three inscribed circles. If the side length of each square is 1, compute r .
解析
- Consider a 3 × 3 grid of squares. A circle is inscribed in the lower left corner, the middle square of the top row, and the rightmost square of the middle row, and a circle O with radius r is drawn such that O is externally tangent to each of the three inscribed circles. If the side length of each square is 1, compute r . Proposed by: Sam Korsky √ 5 2 − 3 Answer: 6 Let A be the center of the square in the lower left corner, let B be the center of the square in the middle of the top row, and let C be the center of the rightmost square in the middle row. It’s clear that O is the circumcenter of triangle ABC - hence, the desired radius is merely the circumradius of √ √ 1 triangle ABC minus . Now note that by the Pythagorean theorem, BC = 2 and AB = AC = 5 so 2 √ 3 2 we easily find that the altitude from A in triangle ABC has length . Therefore the area of triangle 2 3 ABC is . Hence the circumradius of triangle ABC is given by 2 √ BC · CA · AB 5 2 = 3 6 4 · 2 √ √ 5 2 − 3 5 2 1 and so the answer is − = . 6 2 6