PUMaC 2014 · 团队赛 · 第 13 题

PUMaC 2014 — Team Round — Problem 13

[ 9 ] There is a right triangle 4 ABC , in which \ A is the right angle. On side AB , there are three points X, Y, and Z that satisfy \ ACX = \ XCY = \ Y CZ = a \ ZCB and BZ = 2 AX . The smallest angle of 4 ABC is degrees, where a, b b are positive integers such that GCD ( a, b ) = 1. Find a + b .

解析

[ 9 ] There is a right triangle 4 ABC , in which ∠ A is the right angle. On side AB , there are three points X, Y, and Z that satisfy ∠ ACX = ∠ XCY = ∠ Y CZ = ∠ ZCB and BZ = 2 AX . The a smallest angle of 4 ABC is degrees, where a, b are positive integers such that GCD ( a, b ) = 1. b Find a + b . Solution: Without loss of generality, let AX = 1. Let XY = x , Y Z = y , CA = h, CX = p, CY = q, CZ = r, CB = s . Reflect the triangle ABC (and all additional points defined on its perime- ter) across line AC to get a triangle BCD , with the perpendicular foot from C to BD being A . By angle bisector theorem, x : y = p : r = 2 : x + y 4 and y : 2 = q : s = x + 2 : x + y + 2 From these, one can get two equations: 2y = x(x + y) and 2(x + 2) = y(x + y + 2). Solving, 3 2 x we get x − 4 x − 4 + 8 = 0. 2 Again, by angle bisector theorem, 1 : x = h : q . Apply Pythagorean Theorem to get ( hx ) = 2 2 2 2 1+ x q = h + (1 + x ) . So h = . 1 − x 2 2 I claim that p = x + y + 2. Using Pythagorean theorem, one gets p = h + 1. Applying this, 2 2 2 it is easy to show that proving p = ( x + y + 2) reduces to proving x (2 − x ) = 8(1 − x ), which 3 2 x when simplified becomes x − 4 x − 4 + 8 = 0, which we have already shown to be true. So, 3 p = x + y + 2. So, CX = XB . So, XBC is a isosceles triangle, so ∠ B = ∠ XCB = ∠ C . Then 4 3 270 the smallest angle is ∠ B , and from the above relationship, one gets that ∠ B = · 90 = . 7 7 So the answer is 277 .