HMMT 二月 2023 · 冲刺赛 · 第 18 题

HMMT February 2023 — Guts Round — Problem 18

[16] Elisenda has a piece of paper in the shape of a triangle with vertices A , B , and C such that AB = 42. She chooses a point D on segment AC , and she folds the paper along line BD so that A lands at a point E on segment BC . Then, she folds the paper along line DE . When she does this, B lands at the midpoint of segment DC . Compute the perimeter of the original unfolded triangle.

解析

[16] Elisenda has a piece of paper in the shape of a triangle with vertices A , B , and C such that AB = 42. She chooses a point D on segment AC , and she folds the paper along line BD so that A lands at a point E on segment BC . Then, she folds the paper along line DE . When she does this, B lands at the midpoint of segment DC . Compute the perimeter of the original unfolded triangle. Proposed by: Drake Du, Eric Shen √ Answer: 168 + 48 7 Solution: Let F be the midpoint of segment DC . ◦ Evidently ∠ ADB = 60 = ∠ BDE = ∠ EDC . Moreover, we have BD = DF = F C , AD = DE , and AB = BE . Hence angle bisector on BDC gives us that BE = 42, EC = 84, and hence angle bisector on ABC gives us that if AD = x then CD = 3 x . Now this gives BD = 3 x/ 2, so thus the Law of √ Cosines on ADB gives x = 12 7. √ √ √ Hence, BC = 42 + 84 = 126 and AC = 4 x = 48 7. The answer is 42 + 126 + 48 7 = 168 + 48 7.