HMMT 二月 2001 · 几何 · 第 7 题

HMMT February 2001 — Geometry — Problem 7

Equilateral triangle ABC with side length 1 is drawn. A square is drawn such that its vertex 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

Equilateral triangle ABC with side length 1 is drawn. A square is drawn such that its vertex 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 Solution: Let D be the midpoint of BC , F 6 = A be the point of intersection of the 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 = , so x 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 √ ( ) 2 x 3 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)