HMMT 二月 2008 · 几何 · 第 5 题

HMMT February 2008 — Geometry — Problem 5

[ 5 ] A piece of paper is folded in half. A second fold is made at an angle φ (0 < φ < 90 ) to the first, and a cut is made as shown below. Á When the piece of paper is unfolded, the resulting hole is a polygon. Let O be one of its vertices. ◦ Suppose that all the other vertices of the hole lie on a circle centered at O , and also that ∠ XOY = 144 , where X and Y are the the vertices of the hole adjacent to O . Find the value(s) of φ (in degrees). ◦

解析

[ 5 ] A piece of paper is folded in half. A second fold is made such that the angle marked below has ◦ ◦ measure φ (0 < φ < 90 ), and a cut is made as shown below. Á When the piece of paper is unfolded, the resulting hole is a polygon. Let O be one of its vertices. ◦ Suppose that all the other vertices of the hole lie on a circle centered at O , and also that ∠ XOY = 144 , where X and Y are the the vertices of the hole adjacent to O . Find the value(s) of φ (in degrees). ◦ Answer: 81 Try actually folding a piece of paper. We see that the cut out area is a kite, as shown below. The fold was made on AC , and then BE and DE . Since DC was folded onto DA , we have ∠ ADE = ∠ CDE . B E A C D ◦ ◦ Either A or C is the center of the circle. If it’s A , then ∠ BAD = 144 , so ∠ CAD = 72 . Using ◦ ◦ ◦ ◦ ◦ CA = DA , we see that ∠ ACD = ∠ ADC = 54 . So ∠ EDA = 27 , and thus φ = 72 + 27 = 99 , ◦ which is inadmissible, as φ < 90 . ◦ ◦ ◦ ◦ ◦ So C is the center of the circle. Then, ∠ CAD = ∠ CDA = 54 , ∠ ADE = 27 , and φ = 54 +27 = 81 . ◦