Solutions to Geometry Puzzles from the 35-th Lomonosov Tournament
I recently posted two geometry problems. Now is the time for solutions:
Problem 1. Is it possible to put positive numbers at the vertices of a triangle so that the sum of two numbers at the ends of each side is equal to the length of the side?
One might guess that the following numbers work: (a+b-c)/2, (b+c-a)/2 and (c+a-b)/2, where a, b, and c are the side lengths. But there exists a geometric solution: Construct the incircle. The tangent points divide each side into two segment, so that the lengths of the segments ending at the same vertex are the same. Assigning this length to the vertex solves the problem. Surprisingly, or not surprisingly, this solution gives the same answer as above.
Problem 2. Prove that it is possible to assign a number to every edge of a tetrahedron so that the sum of the three numbers on the edges of every face is equal to the area of the face.
The problem is under-constrained: there are six sides and four faces. There should be many solutions. But the solution for the first problem suggests a similar idea for the second problem: Construct the inscribed sphere. Connect a tangent point on each face to the three vertices on the same face. This way each face is divided into three triangles. Moreover, the lengths of the segments connecting the tangent points to a vertex are the same. Therefore, two triangles sharing the same edge are congruent and thus have the same area. Assigning this area to each edge solves the problem.
There are many solutions to the second problem. I wonder if for each solution we can find a point on each face, so that the segments connecting these points to vertices divide the faces into three triangles in such a way that triangles sharing an edge are congruent. What would be a geometric meaning of these four points?Share:
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