For each natural number k, let Ck denote the circle with radius k centimetres and centre at the orig

Q: Wait, why is the angle covered on every circle exactly 1 radian? Doesn't the radius change?

That is the beauty of the problem! While the radius $r$ increases as $k$ increases, the arc length $s$ is defined to be equal to $k$. Since the angle $\theta = \frac{s}{r}$, we get $\theta = \frac{k}{k} = 1$. The radius and arc length grow in perfect proportion, keeping the angular sweep constant at $1$ radian per circle.

Q: Why do we use $2\pi$ as the threshold? Is it because the particle starts at $(1,0)$?

Exactly! Starting at $(1,0)$ corresponds to an angular position of $0$ radians. To cross the positive x-axis again, the particle must complete a full rotation, which is $2\pi$ radians. Any angle $\theta$ such that $\theta \geq 2\pi$ means the particle has completed at least one full lap.

Q: Why is the interval for the crossing on circle $C_n$ defined as $(n-1) < 2\pi < n$?

Think of the particle's journey as a sequence of segments. After $n-1$ circles, the particle has covered a total angle of $(n-1)$ radians. During its motion on the $n$-th circle, it covers an additional $1$ radian, ending at an angle of $n$ radians. If $2\pi$ falls between these two values, it means the crossing happened while the particle was traversing the $n$-th circle.

Q: Can I just use the circumference formula $2\pi r$ to solve this?

Not directly. The circumference formula tells you the total distance for a full circle, but our particle isn't completing full circles. It is performing a 'staircase' motion where it jumps to a new radius after only $1$ radian of travel. Stick to the angular accumulation method; it is much more robust here.

Q: Is there any chance the particle lands exactly on the x-axis?

In this specific problem, we are looking for the first crossing. Since $2\pi$ is an irrational number and our cumulative angles are integers ($1, 2, 3, ...$), the particle will never land exactly on the positive x-axis. It will 'jump' over it during the $7^{th}$ step, which is why we look for the interval containing $2\pi$.

Question

For each natural number $k$, let $C_k$ denote the circle with radius $k$ centimetres and centre at the origin. On the circle $C_k$, $\alpha$-particle moves $k$ centimetres in the counter-clockwise direction. After completing its motion on $C_k$, the particle moves to $C_{k+1}$ in the radial direction. The motion of the particle continues in this manner. The particle starts at $(1, 0)$. If the particle crosses the positive direction of the x-axis for the first time on the circle $C_n$, then $n = .........$

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Staircase

Imagine you are standing at the origin of a coordinate plane, watching a tiny $\alpha$-particle begin an extraordinary journey. It starts at the point $(1, 0)$, resting on the innermost circle $C_1$.

This is a rhythmic, expanding dance. The particle travels along the circumference of $C_1$, then jumps radially outward to $C_2$, travels along $C_2$, and jumps again to $C_3$. It is a staircase of circles spiraling outward.

### The Constant Rhythm

In circular motion, the relationship between arc length $s$, radius $r$, and the angle $	heta$ (in radians) is governed by the identity:

$$	heta = \frac{s}{r}$$

On the first circle $C_1$, the radius is $r = 1$ cm, and the particle travels an arc length of $s = 1$ cm. Thus, the angle covered is $	heta_1 = \frac{1}{1} = 1$ radian.

Now, consider the $k$-th circle $C_k$. The radius is $r = k$ cm, and the particle travels an arc length of $s = k$ cm. The angle covered is:

$$	heta_k = \frac{k}{k} = 1 	ext{ radian}$$

This is the key insight: regardless of how large the circle becomes, the particle always sweeps out exactly $1$ radian of angle before jumping to the next circle. The rhythm is constant and steady.

### The Cumulative Journey

If the particle covers $1$ radian on every circle, then after completing its motion on $n-1$ circles, the total angle $\Theta$ it has accumulated is the sum of these individual steps:

$$\Theta_{n-1} = \sum_{k=1}^{n-1} 1 = n-1$$

We are looking for the moment the particle crosses the positive x-axis for the first time. Since it starts at an angle of $0$, it will cross the positive x-axis whenever its total angular displacement $\Theta$ reaches or exceeds $2\pi$ radians.

Given that $2\pi \approx 6.283$ radians, we seek the smallest integer $n$ such that the particle crosses the threshold during its traversal of circle $C_n$.

### The Final Leap

As the particle enters circle $C_n$, its starting angle is $(n-1)$ radians. By the time it finishes its motion on $C_n$, it will have reached an angle of $n$ radians.

Therefore, the crossing occurs on circle $C_n$ if the value $2\pi$ lies within the interval $[n-1, n]$.

Let us test our integers:
* For $n = 6$, the interval is $[5, 6]$. Since $6 < 6.283$, the particle has not yet reached the positive x-axis.
* For $n = 7$, the interval is $[6, 7]$. Because $6 < 6.283 < 7$, the value $2\pi$ is trapped inside this interval.

This means that while the particle is traversing the $7^{th}$ circle, it crosses the $2\pi$ threshold. It has completed its first full revolution and is crossing the positive x-axis for the first time.

The beauty of this problem lies in how a complex-sounding motion simplifies into a basic inequality. By focusing on the angular displacement, we find that the particle is simply a clock, ticking one radian at a time.

The final answer is **$n = 7$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Concentric Circles $C_{k}$

The Arc-Angle Relationship

Constant Angle Increment per Circle

Cumulative Angle Covered

Condition for Crossing the Positive X-Axis

Analyzing the Interval on Circle $C_{n}$

Testing Integer Values for $n$

Final Answer and Conclusion

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Geometry of the Staircase

The Constant Rhythm

The Cumulative Journey

The Final Leap

Similar Questions

For a positive integer $n$ , let $f_{n} (θ) = (tan \frac{θ}{2}) (1 + sec θ) (1 + sec 2 θ) (1 + sec 4 θ) \dots (1 + sec 2^{n} θ)$ . Then

Let $A_{1}, A_{2}, \dots, A_{n}$ be the vertices of an $n$ -sided regular polygon such that $\frac{1}{A _{1} A _{2}} = \frac{1}{A _{1} A _{3}} + \frac{1}{A _{1} A _{4}}$ . Find the value of $n$ .

If for $x \in (0, \frac{π}{2})$ , $lo g_{10} sin x + lo g_{10} cos x = - 1$ and $lo g_{10} (sin x + cos x) = \frac{1}{2} (lo g_{10} n - 1)$ , then the value of $n$ is equal to :

If $x = \sum_{n = 0}^{\infty} (- 1)^{n} tan^{2 n} θ$ and $y = \sum_{n = 0}^{\infty} cos^{2 n} θ$ , where $0 < θ < \frac{π}{4}$ , then:

The maximum value of $(cos α_{1}) \cdot (cos α_{2}) \dots (cos α_{n})$ , under the restrictions $0 \leq α_{1}, α_{2}, \dots, α_{n} \leq π /2$ and $(cot α_{1}) \cdot (cot α_{2}) \dots (cot α_{n}) = 1$ is

Let $n$ be a positive integer such that $sin \frac{π}{2 n} + cos \frac{π}{2 n} = \frac{n}{2}$ . Then

The positive integer value of $n > 3$ satisfying the equation $\frac{1}{s i n ( π / n )} = \frac{1}{s i n ( 2 π / n )} + \frac{1}{s i n ( 3 π / n )}$ is ____.

Let $θ \in (0, π /4)$ and $t_{1} = (tan θ)^{t a n θ}, t_{2} = (tan θ)^{c o t θ}, t_{3} = (cot θ)^{t a n θ}$ and $t_{4} = (cot θ)^{c o t θ}$ , then

If $t_{1}$ and $t_{2}$ are the times of flight of two particles having the same initial velocity u and range R on the horizontal, then $t_{1}^{2} + t_{2}^{2}$ is equal to

The number of integral values of $k$ for which the equation $7 cos x + 5 sin x = 2 k + 1$ has a solution is

Enter Numerical Value:

Visualized Solution

Visualizing the Concentric Circles .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ck​

The Arc-Angle Relationship

Constant Angle Increment per Circle

Cumulative Angle Covered

Condition for Crossing the Positive X-Axis

Analyzing the Interval on Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Cn​

Testing Integer Values for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Final Answer and Conclusion

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Geometry of the Staircase

The Constant Rhythm

The Cumulative Journey

The Final Leap

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a positive integer n, let fn​(θ)=(tan2θ​)(1+secθ)(1+sec2θ)(1+sec4θ)…(1+sec2nθ). Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A1​,A2​,…,An​ be the vertices of an n-sided regular polygon such that A1​A2​1​=A1​A3​1​+A1​A4​1​. Find the value of n.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If for x∈(0,2π​), log10​sinx+log10​cosx=−1 and log10​(sinx+cosx)=21​(log10​n−1), then the value of n is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x=∑n=0∞​(−1)ntan2nθ and y=∑n=0∞​cos2nθ, where 0<θ<4π​, then:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The maximum value of (cosα1​)⋅(cosα2​)…(cosαn​), under the restrictions 0≤α1​,α2​,…,αn​≤π/2 and (cotα1​)⋅(cotα2​)…(cotαn​)=1 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let n be a positive integer such that sin2nπ​+cos2nπ​=2n​​. Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The positive integer value of n>3 satisfying the equation sin(π/n)1​=sin(2π/n)1​+sin(3π/n)1​ is ____.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let θ∈(0,π/4) and t1​=(tanθ)tanθ,t2​=(tanθ)cotθ,t3​=(cotθ)tanθ and t4​=(cotθ)cotθ, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If t1​ and t2​ are the times of flight of two particles having the same initial velocity u and range R on the horizontal, then t12​+t22​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of integral values of k for which the equation 7cosx+5sinx=2k+1 has a solution is