Tedious Chords

Probability Level 4

Suppose that six points which are evenly spaced around the circumference of a circle are chosen. Now, two of the points are chosen and connected to form chords. There will be a total of $6 \choose 2$ $= 15$ chords being drawn. Two of the chords are chosen and the angle formed between them is deemed valid with the following conditions:

The extension of chords to produce intersection is not allowed.
If there are two angles formed between the chords, only the acute angle is considered valid.
If the two chords are perpendicular to each other, there is only one valid angle, $90^\circ$ .
If the two chords are parallel to each other, in this case, there is no valid angle.

All the angles are measured in degrees ( $^\circ$ ).

Denote the average of valid angles by $\deg_{avg}$ , the total number of valid angles by $N$ and the sum of all valid angles by $\sum_{i=1}^{N} \deg_{i}$ .

Given that $\deg_{avg}=\cfrac{\sum_{i=1}^{N} \deg_{i}}{N}$ , compute $10 \deg_{avg}$ in degrees ( $^\circ$ ).

$\textbf{Bonus:}$ Don't use a calculator, use math tricks.

This is part of the set Things Get Harder! .

The answer is 624.

1 solution

Donglin Loo
May 31, 2018

It is rather obvious that the six points are the vertices of a regular hexagon.

We need only consider two main cases:

$\textbf{Case1:}$

$1.$ The valid angles are formed by two chords involving three points only. In other words, one of the points is used twice.

From the image above, notice that three of such valid angles formed are located in one triangle. There are a total of $6 \choose 3$ = $20$ triangles that can be drawn. Therefore, in this case, the sum of valid angles $=180^\circ \cdot 20$ , the number of valid angles $=20\cdot3=60$ .

$\textbf{Case2:}$

$2.$ The valid angles are formed by two chords involving four points.

$\textbf{Sub-case 1}$

The valid angle formed in this sub-case is the acute angle $60^\circ$ formed by the diagonals of a rectangle. This is easily found by knowing that $\angle P_{1}P_{3}P_{6}=30^\circ$ and $\angle P_{3}P_{1}P_{4}=30^\circ$ . There are $3$ such rectangles. So, in this sub-case, the sum of valid angles $=60^\circ \cdot 3$ , the number of valid angles $=3$ .

$\textbf{Sub-case 2}$

The valid angle formed in this sub-case is the acute angle $60^\circ$ formed by the diagonals of a trapezium. This is easily found by knowing that $\angle P_{2}P_{4}P_{3}=30^\circ$ and $\angle P_{5}P_{4}P_{3}=30^\circ$ . There are $6$ such trapezium. So, in this sub-case, the sum of valid angles $=60^\circ \cdot 6$ , the number of valid angles $=6$ .

$\textbf{Sub-case 3}$

The valid angle formed in this sub-case is $90^\circ$ formed by the diagonals of a kite. There are $6$ such kites. So, in this sub-case, the sum of valid angles $=90^\circ \cdot 6$ , the number of valid angles $=6$ .

$\therefore deg_{avg}$ $=\cfrac{180^\circ \cdot 20+60^\circ \cdot 3+60^\circ \cdot 6+90^\circ \cdot 6}{60+3+6+6}$

$=\cfrac{180^\circ \cdot 20+60^\circ \cdot 3+60^\circ \cdot 6+90^\circ \cdot 6}{75}$

$=\cfrac{60^\circ \cdot 60+60^\circ \cdot 3+60^\circ \cdot 6+60^\circ \cdot 9}{75}$

$=\cfrac{60^\circ\cdot(60+3+6+9)}{75}$

$=\cfrac{60^\circ\cdot78}{75}$

$=\cfrac{4^\circ\cdot78}{5}$

$=\cfrac{8^\circ\cdot78}{10}$

$10 deg_{avg}=8^\circ\cdot78=624^\circ$

Tedious Chords

The answer is 624.

1 solution

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