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In a regular polygon of $12$ sides , three vertices are selected at random to form a triangle. Then no. of right angle triangles formed.

And also find probability that triangle formed is equilateral given that triangle formed is obtuse.

Solution I try:

Total number of triangles is $\displaystyle \binom{12}{3}$ and number of equilateral triangle is $4$ .how i find right angle triangle and obtuse angle triangle.

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  • $\begingroup$ An obtuse triangle is never equilateral, so that part is easy: probability $0.$ Perhaps it was supposed to be "acute" rather than "obtuse"? $\endgroup$ – David K Feb 16 '18 at 12:39
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    $\begingroup$ By $\dfrac{12}3$, I suppose you mean the binomial coefficient $\dbinom{12}3$? $\endgroup$ – Bernard Feb 16 '18 at 12:45
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For the right triangle question . . .

Pick a vertex $P$, and consider triangle $APB$, where $A,B$ are two other vertices.

Then triangle $APB$ has a right angle at $P$ if and only if $AB$ is a diameter.

For each $P$, there are $5$ diameters which don't meet $P$.

Since there are $12$ choices for $P$, it follows that there are $(12)(5) = 60$ right triangles, hence the probability of getting a right triangle is $$\frac{60}{\binom{12}{3}}=\frac{60}{220}=\frac{3}{11}$$

For the non-obtuse triangle question . . .

Define the vertex distance between two vertices as the least number of vertex steps to get from one vertex to the other.

For distinct vertices $A,P,B$ of the $12$-gon, triangle $APB$ has an obtuse angle at $P$ if and only if $A,B$ are not diametrically opposite, and $P$ is strictly between $A$ and $B$ on minor arc $AB$.

Suppose distinct vertices $A,B$ of the $12$-gon are not diametrically opposite.

Then the vertex distance from $A$ to $B$ is at least $2$, and at most $5$.

For each $k \in \{2,3,4,5\}$, there are $12$ choices for a chord $AB$ for which the vertex distance from $A$ to $B$ is $k$, and $k-1$ choices for the point $P$, hence the number of obtuse triangles is $$\sum_{k=2}^5 (12)(k-1) = 12(1+2+3+4)=120$$

It follows that of the ${\large{\binom{12}{3}}}$ possible triangles, $${\small{\binom{12}{3}}}-120=220-120=100$$ of them are non-obtuse, and of those, $4$ are equilateral, so the probability of getting an equilateral triangle, given that the chosen triangle is non-obtuse is $$\frac{4}{100}=\frac{1}{25}$$

If you want "acute" instead of "non-obtuse", a minor adjustment is needed . . .

Of the ${\large{\binom{12}{3}}}$ possible triangles, $${\small{\binom{12}{3}}}-(60+120)=220-180=40$$ of them are acute, and of those, $4$ are equilateral, so the probability of getting an equilateral triangle, given that the chosen triangle is acute is $$\frac{4}{40}=\frac{1}{10}$$

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Hint:

Denote $O$ the centre of the regular dodecahedron, and $A_1,\dots, A_{12}$ its vertices. Computing the angles in triangles $OA_iA_{i+1}$, you can check the only way to obtain a right triangle $A_iA_jA_k$ is to take three consecutive vertices.

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  • $\begingroup$ I don't think $O$ was meant to be a vertex of the triangle. $\endgroup$ – David K Feb 16 '18 at 12:52
  • $\begingroup$ I didn't say so, but you can solve the problem if you know these angles. $\endgroup$ – Bernard Feb 16 '18 at 12:55
  • $\begingroup$ By my reckoning, three consecutive vertices form a 150-15-15 triangle. There are 12 such triangles, whereas there are 60 right triangles (as shown in the other answer). $\endgroup$ – David K Feb 16 '18 at 15:31
  • $\begingroup$ @DavidK: You're right. I guess I messed some calculations with another problem. I'll update in a moment. $\endgroup$ – Bernard Feb 16 '18 at 17:20

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