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Obtuse triangle

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Source: IMO LongList 1959-1966 Problem 41

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#1 h Sep 2, 2004, 1:53 AM • 2 Y

Y by Adventure10, Mango247

Given a regular $n$ -gon $A_{1}A_{2}...A_{n}$ (with $n\geq 3$ ) in a plane. How many triangles of the kind $A_{i}A_{j}A_{k}$ are obtuse ?

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#2 h Sep 2, 2004, 10:32 AM • 2 Y

Y by Adventure10, Mango247

- If $n = 2k+1$ is odd :
The triangle $A_1A_iA_j$ is obtuse at $A_1$ , where $i < j$ , if and only if $i \in \{2, \ldots, k+1 \}$ and $j \in \{ k+1+i, \ldots, n \}$ .
Thus, for each $i \in \{2, \ldots, k+1 \}$ there are $k+1-i$ possibles values for $j$ . Summing over $i$ , it leads to $\displaystyle \frac {k(k-1)} 2$ obtuse triangles at $A_1$ .
Similarly, there are as many obtuse triangles at $A_i$ for each $i$ , and non is counted twice, since a triangle cannot have more than one obtuse angle...

Thus, there are a total of $\displaystyle \frac {(2k+1)k(k-1)} 2$ obtuse triangles in that case.

- If $n = 2k$ is even :
The triangle $A_1A_iA_j$ is obtuse at $A_1$ , where $i < j$ , if and only if $i \in \{2, \ldots, k \}$ and $j \in \{ k+1+i, \ldots, n \}$ .
Thus, for each $i \in \{2, \ldots, k+1 \}$ there are $k-i$ possibles values for $j$ . Summing over $i$ , it leads to $\displaystyle \frac {(k-2)(k-1)} 2$ obtuse triangles at $A_1$ .
Similarly as above, it leads to a total of $k(k-2)(k-1)$ obtuse triangles in that case.

Pierre.

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