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Problem 985

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Problem 985

Telescoping Triangles

Given a triangle $T_k$, it is sometimes possible to construct a triangle $T_{k+1}$ inside $T_k$ such that

  • The three vertices of $T_{k+1}$ lie one on each side of $T_k$.
  • For each side of $T_k$, the angles formed between it and the two sides of $T_{k+1}$ it touches are equal to each other.
0985_telescoping_triangles.png

Illustrated above is such a sequence of three triangles starting with $T_0$ (in blue) having side lengths $(8,9,10)$. Then $T_1$ is shown in green and $T_2$ in red. However, no triangle can be drawn inside $T_2$ that satisfies the requirements. In other words, $T_3$ does not exist.

Amongst all integer-sided triangles $T_0$ such that $T_2$ exists but $T_3$ does not exist, the smallest possible perimeter is $10$ when $T_0$ has side lengths $(3, 3, 4)$.

Suppose another triangle $T_0$ has integer side lengths, and $T_{20}$ exists, but $T_{21}$ does not exist. What is the smallest possible perimeter of $T_0$?

套叠三角形

给定三角形$T_k$，有时可以在$T_k$内部构造三角形$T_{k+1}$，使得

  • $T_{k+1}$的三个顶点分别位于$T_k$的三条边上。
  • $T_k$的每一条边与其上$T_{k+1}$的两条边构成的两个角彼此相等。
0985_telescoping_triangles.png

如上图所示为从边长为$(8,9,10)$的三角形$T_0$（以蓝色表示）开始构造的这样一组三个三角形，其中$T_1$以绿色表示，$T_2$以红色表示。然而，在$T_2$内部无法画出满足要求的三角形，换言之无法构造$T_3$。

对于所有能构造$T_2$但无法构造$T_3$的整数边长三角形$T_0$，其最短周长为$10$，对应的$T_0$三边长为$(3, 3, 4)$。

考虑所有能构造$T_{20}$但无法构造$T_{21}$的整数边长三角形$T_0$，求其最短周长。