Given a pair of concentric circles, chords AB,BC,CD,... of the outer circle are drawn such that they all touch the inner circle. If ∠ABC = 75°;how many chords can be drawn before returning to the starting point?

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Thought to start by joining points of contact with the centre,but couldn't got any further idea. Any help is appreciated.

Question Source: IOQM Olympiad

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    $\begingroup$ Wouldn't ∠$BCD$ and each successive angle also be 75°...? $\endgroup$ Aug 16, 2022 at 6:58
  • $\begingroup$ @Greg Yes. I have proved it by using congruency in adjacent triangles formed by centre and the two points on each circle. Please tell how to proceed further. $\endgroup$ Aug 16, 2022 at 7:17
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    $\begingroup$ Imagine travelling along the chord AB. When you reach the end you have to turn by 180-75=105 degrees to start travelling along the next chord BC. When you reach the end of that you turn again 105 degrees to head for D, and so on. How many such turns would you need to do before you are facing in the same direction as when you started? $\endgroup$ Aug 16, 2022 at 7:40
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    $\begingroup$ @An_Elephant 105 degrees is the angle between the vector AB and the vector BC. Direction matters. All the chords are tangent to the inner circle. The only other chord parallel to AB would be one on the opposite side of the inner circle. However, that one would correspond to a net turning angle of 180, where you are travelling from right to left (as opposed to left to right from A to B). The central circle is always on your left as you travel along the chords, and your heading angle uniquely identifies which point on the inner circle is touched by the chord, so uniquely identifies the chord. $\endgroup$ Aug 16, 2022 at 9:18
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    $\begingroup$ Let touchpoint of AB is E and touchpoint of BC is F. Angle EOF is 180°–75°=105°. $105=15\cdot 7$, $360=15\cdot 24$, $LCM(105,360)=15\cdot 7 \cdot 24$. Then there is 24 chords to return in the same touch point, then there is 24 chords to return in the same point of outer circle. $\endgroup$ Aug 16, 2022 at 11:25

1 Answer 1


Method using modular arithmetic

Notice that $\triangle AOB$ is isosceles and therefore $\angle AOB = 105^\circ$.

Each successive chord adds $105^\circ$ to this angle so we want to know when this would be a multiple of $360^\circ$. In other words, we want

$$ 105n \equiv 0 \pmod {360} $$

This implies

$$ 7n \equiv 0 \pmod { 24} $$

Since $\gcd{(7, 24)} = 1$, the minimum $n$ that satisfies this equation is $\boxed{n = 24}$

Method using Complex Numbers

Without loss of generality, assume that the outer circle has a radius of $1$. We can rotate the initial diagram such that the point $A$ corresponds to the point $(1, 0)$ on the coordinate plane.

Every time we draw a chord, the angle increases by $105^\circ$, so essentially we are rotating the complex number by $105^\circ$ each time.

Let $z = e^{i \cdot 0}$. The next chord point (in our case $B$) will occour at $z = e^{i \cdot \frac{7\pi}{12}}$ and $C$ will occour at $z = e^{i \cdot \frac{7\pi}{6}}$ and so on.

We want to know, when $z = e^{i \cdot \frac{7\pi}{12} \cdot n}$ will return to the point $A$ at $(1, 0)$. Clearly, this will happen when:

$$ \frac{7\pi}{12}\cdot n = 2\pi k $$

In other words, for some integer $n$ we want:

$$ 7n = 24k $$

The above simply corresponds to the modular equation:

$$ 7n \equiv 0 \pmod {24} $$

Since $\gcd(7,24)=1$, the minimum n that satisfies this equation is $\boxed{n=24}$.


If we use complex numbers, we actually know the coordinates of each point. For example,

$$ e^{i \cdot \frac{7\pi}{12}} \implies (\cos(105^\circ), \ \sin(105^\circ)) $$

Using this we can be sure that the only solution occurs when $n = 24$.

  • $\begingroup$ Thanks but How can we be sure that in the 24th turn I will be exactly at A, instead of being on any other point from which the 25th chord is parallel to AB. Please tell. $\endgroup$ Aug 16, 2022 at 8:32
  • $\begingroup$ Read the verification section! $\endgroup$
    – Manu Anish
    Aug 16, 2022 at 8:55
  • $\begingroup$ Instead of rotating, aren't we translating the complex number also? I mean that both the ends of the affix are not fixed? Is it same as we shift parallel vectors keeping them same? Thanks $\endgroup$ Aug 16, 2022 at 9:56
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    $\begingroup$ That is correct, however, note that the radius of the circle was intentionally chosen to be 1, therefore when we perform a vector multiplication the end is essentially rotated (and not scaled). $\endgroup$
    – Manu Anish
    Aug 16, 2022 at 10:25
  • $\begingroup$ Why is 105 used instead of 75 here. I couldn't find a reason that I could comprehend $\endgroup$ Sep 11 at 3:13

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