HMMT 二月 2001 · 几何 · 第 7 题
HMMT February 2001 — Geometry — Problem 7
题目详情
英文原题
- Equilateral triangle ABC with side length 1 is drawn. A square is drawn such that itsvertex at A is opposite to its vertex at the midpoint of BC . Find the area enclosed within
√ √
2( 3+1)
the intersection of the insides of the triangle and square. Hint: sin 75 = . 4
解析
英文解析
- Equilateral triangle ABC with side length 1 is drawn. A square is drawn such that itsvertex at A is opposite to its vertex at the midpoint of BC . Find the area enclosed within
√ √
2( 3+1)
the intersection of the insides of the triangle and square. Hint: sin 75 = .
Solution: Let D be the midpoint of BC , F 6 = A be the point of intersection of the 4
square and triangle lying on AC , b be the length of F C , x be the side length of the triangle,
2 sin 75 sin 45
and y be the length of AD . By the law of sines on triangle CDF , we have = , sox b
√
x sin 45 2 x 1
b = = . The area of the desired figure can easily be seen to be ( x − b ) y since it
2 sin 75 4 sin 75 2
( )
√
x − b 1 2
can be seen as two triangles of width y and height . This reduces to 1 − xy .
2 2 4 sin 75
√
( )
x 32
2 2
Then by the Pythagorean theorem on triangle ABD , x = + y , so y = x , and the
2 2
( ) ( )
√ √ √ √
3 2 3 2 3
√2
area becomes 1 − x = 1 − = .
4 4 sin 75 4 4 sin 75
4( 3+1)