HMMT 二月 2008 · 几何 · 第 5 题
HMMT February 2008 — Geometry — Problem 5
题目详情
英文原题
- [ 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.
When the piece of paper is unfolded, the resulting hole is a polygon. Let O be one of its vertices.Á
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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).
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解析
英文解析
- [ 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.
When the piece of paper is unfolded, the resulting hole is a polygon. Let O be one of its vertices.Á
°
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 shownbelow. The fold was made on AC , and then BE and DE . Since DC was folded onto DA , we have
∠ ADE = ∠ CDE .
A CEB
° °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 ,
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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 .
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