HMMT 二月 2023 · 几何 · 第 3 题

HMMT February 2023 — Geometry — Problem 3

Suppose ABCD is a rectangle whose diagonals meet at E . The perimeter of triangle ABE is 10 π and the perimeter of triangle ADE is n . Compute the number of possible integer values of n .

解析

Suppose ABCD is a rectangle whose diagonals meet at E . The perimeter of triangle ABE is 10 π and the perimeter of triangle ADE is n . Compute the number of possible integer values of n . Proposed by: Luke Robitaille Answer: 47 Solution: For each triangle T , we let p ( T ) to denote the perimeter of T . 1 First, we claim that p ( △ ABE ) < p ( △ ADE ) < 2 p ( △ ABE ). To see why, observe that 2 p ( △ ADE ) = EA + ED + AD < 2( EA + ED ) = 2( EA + EB ) < 2 p ( △ ABE ) , Similarly, one can show that p ( △ ABE ) < 2 p ( △ ADE ), proving the desired inequality. This inequality limits the possibility of n to only those in (5 π, 20 π ) ⊂ (15 . 7 , 62 . 9), so n could only range from 16 , 17 , 18 , . . . , 62, giving 47 values. These values are all achievable because • when AD approaches zero, we have p ( △ ADE ) → 2 EA and p ( △ ABE ) → 4 EA , implying that 1 p ( △ ADE ) → p ( △ ABE ) = 5 π ; 2 • similarly, when AB approaches zero, we have p ( △ ADE ) → 2 p ( △ ABE ) = 20 π ; and • by continuously rotating segments AC and BD about E , we have that p ( △ ADE ) can reach any value between (5 π, 20 π ). Hence, the answer is 47.