HMMT 二月 2023 · 几何 · 第 9 题
HMMT February 2023 — Geometry — Problem 9
题目详情
- Point Y lies on line segment XZ such that XY = 5 and Y Z = 3. Point G lies on line XZ such that there exists a triangle ABC with centroid G such that X lies on line BC , Y lies on line AC , and Z lies on line AB . Compute the largest possible value of XG .
解析
- Point Y lies on line segment XZ such that XY = 5 and Y Z = 3. Point G lies on line XZ such that there exists a triangle ABC with centroid G such that X lies on line BC , Y lies on line AC , and Z lies on line AB . Compute the largest possible value of XG . Proposed by: Luke Robitaille 20 Answer: 3 1 1 1 Solution: The key claim is that we must have + + = 0 (in directed lengths). XG Y G ZG We present three proofs of this fact. Proof 1: By a suitable affine transformation, we can assume without loss of generality that ABC is equilateral. Now perform an inversion about G with radius GA = GB = GC . Then the im- ′ ′ ′ ages of X, Y, Z (call them X , Y , Z ) lie on ( GBC ) , ( GAC ) , ( GAB ), so they are the feet of the per- pendiculars from A , B , C to line XY Z , where A , B , C are the respective antipodes of G on 1 1 1 1 1 1 ( GBC ) , ( GAC ) , ( GAB ). But now A B C is an equilateral triangle with medial triangle ABC , so its 1 1 1 ′ ′ ′ centroid is G . Now the centroid of (degenerate) triangle X Y Z is the foot of the perpendicular of the ′ ′ ′ centroid of A B C onto the line, so it is G . Thus X G + Y G + Z G = 0, which yields the desired 1 1 1 claim. ■ Proof 2: Let W be the point on line XY Z such that W G = 2 GX (in directed lengths). Now note that ( Y, Z ; G, W ) is a harmonic bundle, since projecting it through A onto BC gives ( B, C ; M , ∞ ). BC BC 1 1 2 By harmonic bundle properties, this yields that + = (in directed lengths), which gives the Y G ZG W G desired. ■ Proof 3: Let P ̸ = G be an arbitrary point on the line XY Z . Now, in directed lengths and signed areas, [ GBP ] [ GCP ] [ GBP ] − [ GCP ] [ GBP ] − [ GCP ] 3([ GBP ] − [ GCP ]) GP GP = = , so = = = . Writing analogous GX [ GBX ] [ GCX ] GX [ GBX ] − [ GCX ] [ GBC ] [ ABC ] GP GP GP GP GP equations for and and summing yields + + = 0, giving the desired. ■ GY GZ GX GY GZ With this lemma, we may now set XG = g and know that 1 1 1
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- = 0 g g − 5 g − 8 20 Solving the quadratic gives the solutions g = 2 and g = ; the latter hence gives the maximum (it is 3 not difficult to construct an example for which XG is indeed 20 / 3).