HMMT 二月 2019 · 冲刺赛 · 第 31 题
HMMT February 2019 — Guts Round — Problem 31
题目详情
- [ 20 ] Let ABC be a triangle with AB = 6, AC = 7, BC = 8. Let I be the incenter of ABC . Points Z and Y lie on the interior of segments AB and AC respectively such that Y Z is tangent to the incircle. Given point P such that ◦ ∠ ZP C = ∠ Y P B = 90 , find the length of IP .
解析
- [ 20 ] Let ABC be a triangle with AB = 6, AC = 7, BC = 8. Let I be the incenter of ABC . Points Z and Y lie on the interior of segments AB and AC respectively such that Y Z is tangent to the incircle. Given point P such that ◦ ∠ ZP C = ∠ Y P B = 90 , find the length of IP . Proposed by: Zack Chroman √ 30 Answer: 2 Solution 1. Let P U, P V tangent from P to the incircle. We will invoke the dual of the Desargues Involution Theorem , which states the following: Given a point P in the plane and four lines ` , ` , ` , ` , consider the set of conics tangent to all four 1 2 3 4 lines. Then we define a function on the pencil of lines through P by mapping one tangent from P to each conic to the other. This map is well defined and is a projective involution, and in particular maps P A → P D, P B → P E, P C → P F , where ABCDEF is the complete quadrilateral given by the pairwise intersections of ` , ` , ` , ` . 1 2 3 4 An overview of the projective background behind the (Dual) Desargues Involution Theorem can be found here: https://www.scribd.com/document/384321704/Desargues-Involution-Theorem , and a proof can be found at https://www2.washjeff.edu/users/mwoltermann/Dorrie/63.pdf . Now, we apply this to the point P and the lines AB, AC, BC, Y Z , to get that the pairs ( P U, P V ) , ( P Y, P B ) , ( P Z, P C ) are swapped by some involution. But we know that the involution on lines through P which rotates ◦ by 90 swaps the latter two pairs, thus it must also swap the first one and ∠ U P V = 90. It follows √ by equal tangents that IU P V is a square, thus IP = r 2 where r is the inradius of ABC . Since √ √ √ 21 15 / 2 2 K 15 30 r = = = , we have IP = . a + b + c 21 2 2 Solution 2. Let H be the orthocenter of ABC . 2 2 2 Lemma. HI = 2 r − 4 R cos( A ) cos( B ) cos( C ), where r is the inradius and R is the circumradius. Proof. This follows from barycentric coordinates or the general result that for a point X in the plane, 2 2 2 2 2 2 2 aXA + bXB + cXC = ( a + b + c ) XI + aAI + bBI + cCI , which itself is a fact about vectors that follows from barycentric coordinates. This can also be computed directly using trigonometry. Let E = BH ∩ AC, F = CH ∩ AB , then note that B, P, E, Y are concyclic on the circle of diameter BY , and C, P, F, Z are concyclic on the circle of diameter CZ . Let Q be the second intersection of these circles. Since BCY Z is a tangential quadrilateral, the midpoints of BY and CZ are collinear with I (this is known as Newton’s theorem ), which implies that IP = IQ by symmetry. Note that as BH · HE = CH · HF , H lies on the radical axis of the two circles, which is P Q . Thus, if IP = IQ = x , BH · HE is the power of H with respect to the circle centered at I with radius x , which implies 2 2 BH · HE = x − HI . √ As with the first solution, we claim that x = r 2, which by the lemma is equivalent to BH · HE = 2 4 R cos( A ) cos( B ) cos( C ). Then note that BH · HE = BH · CH cos( A ) = (2 R cos( B ))(2 R cos( C )) cos( A ) , so our claim holds and we finish as with the first solution. Note. Under the assumption that the problem is well-posed (the answer does not depend on the choice √ of Y, Z , or P ), then here is an alternative method to obtain IP = r 2 by making convenient choices. Let U be the point where Y Z is tangent to the incircle, and choose U so that IU ‖ BC (and therefore Y Z ⊥ BC ). Note that Y Z ∩ BC is a valid choice for P , so assume that P is the foot from U to BC . √ If D is the point where BC is tangent to the incircle, then IU P D is a square so IP = r 2. (This disregards the condition that Y and Z are in the interior of segments AC and AB , but there is no reason to expect that this condition is important.)