HMMT 二月 2019 · 几何 · 第 8 题
HMMT February 2019 — Geometry — Problem 8
题目详情
- In triangle ABC with AB < AC , let H be the orthocenter and O be the circumcenter. Given that the midpoint of OH lies on BC , BC = 1, and the perimeter of ABC is 6, find the area of ABC.
解析
- In triangle ABC with AB < AC , let H be the orthocenter and O be the circumcenter. Given that the midpoint of OH lies on BC , BC = 1, and the perimeter of ABC is 6, find the area of ABC. Proposed by: Andrew Lin 6 Answer: 7 ′ ′ ′ ′ Solution 1. Let A B C be the medial triangle of ABC , where A is the midpoint of BC and so on. Notice that the midpoint of OH , which is the nine-point-center N of triangle ABC , is also the ′ ′ ′ circumcircle of A B C (since the midpoints of the sides of ABC are on the nine-point circle). Thus, if ′ ′ ′ N is on BC , then N A is parallel to B C , so by similarity, we also know that OA is parallel to BC . Next, AB < AC , so B is on the minor arc AC . This means that ∠ OAC = ∠ OCA = ∠ C , so ∠ AOC = 180 − 2 ∠ C . This gives us the other two angles of the triangle in terms of angle C : ∠ B = 90 + ∠ C and ∠ A = 90 − 2 ∠ C . To find the area, we now need to find the height of the triangle from A to BC , and this is easiest by finding the circumradius R of the triangle. We do this by the Extended Law of Sines. Letting AC = x and AB = 5 − x , 1 x 5 − x = = = 2 R, sin(90 − 2 C ) sin(90 + C ) sin C which can be simplified to 1 x 5 − x = = = 2 R. cos 2 C cos C sin C This means that 1 ( x ) + (5 − x ) 5 = = cos 2 C (cos C ) + (sin C ) cos C + sin C and the rest is an easy computation: 2 2 cos C + sin C = 5 cos 2 C = 5(cos C − sin C ) 1 = cos C − sin C 5 Squaring both sides, 1 2 2 = cos C − 2 sin C cos C + sin C = 1 − sin 2 C 25 24 7 1 25 so sin 2 C = , implying that cos 2 C = . Therefore, since = 2 R from above, R = . Finally, 25 25 cos 2 C 14 viewing triangle ABC with BC = 1 as the base, the height is √ ( ) 2 BC 12 2 R − = 2 7 1 12 6 by the Pythagorean Theorem, yielding an area of · 1 · = . 2 7 7 Solution 2. The midpoint of OH is the nine-point center N . We are given N lies on BC , and we also know N lies on the perpendicular bisector of EF , where E is the midpoint of AC and F is the midpoint of AB . The main observation is that N is equidistant from M and F , where M is the midpoint of BC . Translating this into coordinates, we pick B ( − 0 . 5 , 0) and C (0 . 5 , 0), and arbitrarily set A ( a, b ) where a +0 . 5 b a − 0 . 5 b (without loss of generality) b > 0. We get E ( , ), F ( , ), M (0 , 0). Thus N must have 2 2 2 2 a a x -coordinate equal to the average of those of E and F , or . Since N lies on BC , we have N ( , 0). 2 2 2 2 a 1 b 2 2 1 Since M N = EN , we have = + . Thus a = b + . The other equation is AB + AC = 5, 4 16 4 4 which is just √ √ 2 2 2 2 ( a + 0 . 5) + b + ( a − 0 . 5) + b = 5 . This is equivalent to √ √ 2 2 2 a + a + 2 a − a = 5 √ √ 2 2 2 a + a = 5 − 2 a − a √ 2 2 2 2 a + a = 25 − 10 2 a − a + 2 a − a √ 2 25 − 2 a = 10 2 a − a 2 2 625 − 100 a + 4 a = 200 a − 100 a 2 196 a = 625 2 625 2 576 24 12 b 6 Thus a = , so b = . Thus b = = , so [ ABC ] = = . 196 196 14 7 2 7