AIME 2011 I · 第 10 题

Solution 1

Inscribe the regular polygon inside a circle. A triangle inside this circle will be obtuse if and only if its three vertices lie on one side of a diameter of the circle. (This is because if an inscribed angle on a circle is obtuse, the arc it spans must be 180 degrees or greater).

Break up the problem into two cases: an even number of sides $2n$, or an odd number of sides $2n-1$. For polygons with $2n$ sides, the circumdiameter has endpoints on $2$ vertices. There are $n-1$ points on one side of a diameter, plus $1$ of the endpoints of the diameter for a total of $n$ points. For polygons with $2n - 1$ points, the circumdiameter has $1$ endpoint on a vertex and $1$ endpoint on the midpoint of the opposite side. There are also $n - 1$ points on one side of the diameter, plus the vertex for a total of $n$ points on one side of the diameter.

Case 1: $2n$-sided polygon. There are clearly $\binom{2n}{3}$ different triangles total. To find triangles that meet the criteria, choose the left-most point. There are obviously $2n$ choices for this point. From there, the other two points must be within the $n-1$ points remaining on the same side of the diameter. So our desired probability is $\frac{2n\binom{n-1}{2}}{\binom{2n}{3}}$ $=\frac{n(n-1)(n-2)}{\frac{2n(2n-1)(2n-2)}{6}}$ $=\frac{6n(n-1)(n-2)}{2n(2n-1)(2n-2)}$ $=\frac{3(n-2)}{2(2n-1)}$

so $\frac{93}{125}=\frac{3(n-2)}{2(2n-1)}$

$186(2n-1)=375(n-2)$.

$372n-186=375n-750$ $3n=564$

$n=188$ and so the polygon has $376$ sides.

Case 2: $2n-1$-sided polygon. Similarly, $\binom{2n-1}{3}$ total triangles. Again choose the leftmost point, with $2n-1$ choices. For the other two points, there are again $\binom{n-1}{2}$ possibilities.

The probability is $\frac{(2n-1)\binom{n-1}{2}}{\binom{2n-1}{3}}$

$=\frac{3(2n-1)(n-1)(n-2)}{(2n-1)(2n-2)(2n-3)}$ $=\frac{3(n-2)}{2(2n-3)}$

so $\frac{93}{125}=\frac{3(n-2)}{2(2n-3)}$

$186(2n-3)=375(n-2)$ $375n-750=372n-558$ $3n=192$

$n=64$ and our polygon has $127$ sides.

Adding, $127+376=\boxed{503}$

Solution 2

We use casework on the locations of the vertices, if we choose the locations of vertices $v_a, v_b, v_c$ on the n-gon (where the vertices of the n-gon are $v_0, v_1, v_2, ... v_{n-1},$ in clockwise order) to be the vertices of triangle ABC, in order, with the restriction that $a.

By symmetry, we can assume WLOG that the location of vertex A is vertex $v_0$.

Now, vertex B can be any of $v_1, v_2, ... v_{n-2}$. We start in on casework.

Case 1: vertex B is at one of the locations $v_{n-2}, v_{n-3}, ... v_{\lfloor n/2 \rfloor +1}$. (The ceiling function is necessary for the cases in which n is odd.)

Now, since the clockwise arc from A to B measures more than 180 degrees; for every location of vertex C we can choose in the above restrictions, angle C will be an obtuse angle.

There are $\lceil n/2 \rceil - 2$ choices for vertex B now (again, the ceiling function is necessary to satisfy both odd and even cases of n). If vertex B is placed at $v_m$, there are $n - m - 1$ possible places for vertex C.

Summing over all these possibilities, we obtain that the number of obtuse triangles obtainable from this case is $\frac{(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil) - 1}{2}$.

Case 2: vertex B is at one of the locations not covered in the first case.

Note that this will result in the same number of obtuse triangles as case 1, but multiplied by 2. This is because fixing vertex B in $v_0$, then counting up the cases for vertices C, and again for vertices C and A, respectively, is combinatorially equivalent to fixing vertex A at $v_0$, then counting cases for vertex B, as every triangle obtained in this way can be rotated in the n-gon to place vertex A at $v_0$, and will not be congruent to any obtuse triangle obtained in case 1, as there will be a different side opposite the obtuse angle in this case.

Therefore, there are $\frac{3(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil - 1)}{2}$ total obtuse triangles obtainable.

The total number of triangles obtainable is $1+2+3+...+(n-2) = \frac{(n-2)(n-1)}{2}$.

The ratio of obtuse triangles obtainable to all triangles obtainable is therefore

$\frac{\frac{3(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil - 1)}{2}}{\frac{(n-2)(n-1)}{2}} = \frac{3(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil - 1)}{(n-2)(n-1)} = \frac{93}{125}$.

So, $\frac{(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil - 1)}{(n-2)(n-1)} = \frac{31}{125}$.

Now, we have that $(n-2)(n-1)$ is divisible by $125 = 5^3$. It is now much easier to perform trial-and-error on possible values of n, because we see that $n \equiv 1,2 \pmod{125}$.

We find that $n = 127$ and $n = 376$ both work, so the final answer is $127 + 376 = \boxed{503}$.

Solution 3

Let the regular $n$-gon be inscribed in a circle, allowing the use of the inscribed angle theorem. We wish to determine the number of three distinct vertices such that they form an obtuse triangle.

WLOG fix vertex $A$ Let $B$ be $k$ edges away from $A$ clockwise $C$ be $j$ edges away from $A$ in the other direction

To ensure $\triangle ABC$ has no conciding vertices the edges between them $k, j, n - (k+j) > 0$ \begin{cases} k > 0\\ j > 0\\ k + j < n\\ \end{cases} The angle of $\triangle ABC$ are $\frac{180^\circ}{n} \cdot (n - k - j), \frac{180^\circ}{n} \cdot j, \frac{180^\circ}{n} \cdot k$ respectively. Solving for the angles $\frac{180^\circ}{n} \cdot x > 90$ for an obtuse, \begin{cases} k > \frac{n}{2}\\ j > \frac{n}{2}\\ k + j < \frac{n}{2} \end{cases} These are the conditions for each angle of $\triangle ABC$ to be obtuse. Hence we wish to find lattice point $(k, j)$ that satisfy the inequalities.

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As expected these are disjoint, as a triangle contain at most one obtuse angle.

We wish to take the union of lattice points within the pink shaded regions. The blue line represents the conditions that make $k$ and $j$ form a non degenerate triangle.

Case $n = 2c$ Focus on of the pink regions. The number of lattice points is $1 + 2 + 3 + \cdots + (c - 2) = \frac{1}{2}(c - 2)(c-1)$ Multiply by $3$ for each of the regions resulting in $\frac{3}{2}(c - 2)(c-1)$

Case $n = 2c - 1$ Proceed similarly and get $\frac{3}{2}(c - 2)(c-1)$ too.

At first the WLOG fixing a single vertex, hence $\times n$ Eac $\triangle ABC$ is counted $3$ times, once for each vertex, hence $\div 3$

$$\frac{3}{2}(c - 2)(c-1) \cdot n \div 3 = \frac{n(c-2)(c-1)}{2}$$ Where $c =\left \lfloor \frac{n}{2} \right \rfloor$

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The total number of ways to choose 3 distinct vertices on the $n$-gon is $\binom{n}{3}$

$$\frac{n(c-2)(c-1)}{2 \dbinom{n}{3} } = \frac{93}{125}$$ \begin{align*} 125 \cdot \frac{n(c-2)(c-1)}{2} &= 93 \cdot \dbinom{n}{3}\\ 125 \cdot \frac{n(c-2)(c-1)}{2} &= 93 \cdot \frac{n(n-1)(n-2)}{6}\\ 125 (c-2)(c-1) &= 31 (n-1)(n-2)\\ 125 (\left \lfloor \frac{n}{2} \right \rfloor-2)(\left \lfloor \frac{n}{2} \right \rfloor-1) &= 31 (n-1)(n-2) \end{align*}

Case $n = 2c$ \begin{align*} 125(c-2)(c-1) &= 31(2c - 1)(2c - 2) \\ &= 31 \cdot 2 \cdot (2c - 1)(c-1)\\ 125(c-2) &= 62(2c-1)\\ c &= 188\\ n &= 2c = 376 \end{align*}

Case $n = 2c - 1$ \begin{align*} 125(c-2)(c-1) &= 31(2c - 2)(2c - 3) \\ &= 31 \cdot 2 \cdot (c - 1)(2c-3)\\ 125(c-2) &= 62(2c-3)\\ c &= 64\\ n &= 2c - 1 = 127 \end{align*}

The final is thus $376 + 127 = \boxed{503}$ ~clod

Video Solution

2011 AIME I #10

MathProblemSolvingSkills.com

Video Solution

https://youtu.be/FyTEjRjW_pQ

~IceMatrix