Problem
GEO-B1-M02-P023 The Three Longest Sides
Six segments can be split into two triples, each triple forming a triangle. Order their lengths as \(a_1\ge a_2\ge a_3\ge a_4\ge a_5\ge a_6\). Prove that the segments \(a_1,a_2,a_3\) can always form a triangle. Also show that the analogous statement for the three shortest segments is false.
C. Hint 1. Use the original triangle that contained the longest segment.
D. Hint 2. The longest segment is less than the sum of the other two sides of its original triangle; replace those two sides by no shorter ones.
E. Full solution.
The segment of length \(a_1\) belonged to one of the two original triangles. Let the other two sides of that triangle have lengths \(u\) and \(v\). By the triangle inequality, \(a_1 Among the five segments different from \(a_1\), the two longest have lengths \(a_2\) and \(a_3\). Hence \(a_2+a_3\ge u+v\). Therefore \(a_1 Since \(a_1\ge a_2\ge a_3\), the other two triangle inequalities are immediate: \(a_2 The statement for the three shortest segments is false. For example, take two congruent triangles with side lengths \(1,3,3\). Then the three shortest lengths are \(1,1,3\), which cannot form a triangle.
A. Source analysis. Main objects: six sides of two triangles, ordered by length. The obvious but insufficient approach is to try to reconstruct the original triples. The hidden observation is that one only needs the original triangle containing the longest side. The needed transformation is to replace its other two sides by the two longest remaining segments. Number of key ideas: 2.
F. Difficulty justification. This is Level 6: a regional-style architecture, with one main move, namely applying the triangle inequality to an ordered set.
G. Check. This is not a one-step definition check: one must select the correct original triangle and then build a counterexample for the shortest triple.