HMMT 二月 2016 · 几何 · 第 6 题

HMMT February 2016 — Geometry — Problem 6

Let ABC be a triangle with incenter I , incircle γ and circumcircle Γ. Let M , N , P be the midpoints of sides BC , CA , AB and let E , F be the tangency points of γ with CA and AB , respectively. Let U , V be the intersections of line EF with line M N and line M P , respectively, and let X be the midpoint ◦ ̂ of arc BAC of Γ. Given that AB = 5, AC = 8, and ∠ A = 60 , compute the area of triangle XU V .

解析

Let ABC be a triangle with incenter I , incircle γ and circumcircle Γ. Let M , N , P be the midpoints of sides BC , CA , AB and let E , F be the tangency points of γ with CA and AB , respectively. Let U , V be the intersections of line EF with line M N and line M P , respectively, and let X be the midpoint ◦ ̂ of arc BAC of Γ. Given that AB = 5, AC = 8, and ∠ A = 60 , compute the area of triangle XU V . Proposed by: Evan Chen √ 21 3 Answer: 8 Let segments AI and EF meet at K . Extending AK to meet the circumcircle again at Y , we see that X and Y are diametrically opposite, and it follows that AX and EF are parallel. Therefore the height from X to U V is merely AK . Observe that AE = AF , so 4 AEF is equilateral; since M N, M P are parallel to AF, AE respectively, it follows that 4 M V U, 4 U EN, 4 F P V are equilateral as well. Then 1 1 1 1 1 M V = M P − P V = AC − F P = AC − AF + AP = AC − AF + AB = BC , since E, F are the 2 2 2 2 2 1 tangency points of the incircle. Since 4 M V U is equilateral, we have U V = M U = M V = BC . 2 7 Now we can compute BC = 7, whence U V = and 2 √ AB + AC − BC 3 3 ◦ AK = · cos 30 = . 2 2 √ 21 3 Hence, the answer is . 8