Five Tangent Circles

One should know by now that by drawing 6 circles externally tangent it forms an equilateral hexagon using the tangent farthest away from the original circle. Each surrounding circle will form an equilateral triangle with an angle of 60 degrees with any ray drawn hitting a circle. This is the same question but with 4 circles. Thus it will cover $\frac{4}{6} \Rightarrow \frac{2}{3}$ of the circle.

Legit way is to construct one surrounding circle. To find the angle, draw a line connecting the center from the original to the surrounding and tangent to the circle, which will be half the angle. Then make a right triangle with hypotenuse 2 and one length of 1. Thus it is a 30 degree angle and each circle covers $\frac{30*2}{360}\ \Rightarrow \frac{1}{6}$ for the rays

Let gamma be the given circle with center O and radius 1. Let alpha be one of the circle with center P and radius 1, and touching the circle gamma externally. Therefore |OP|=1+1=2. Consider a half ray from O and intersecting the circle alpha. The limiting point of this half ray to intersect the circle alpha is that the half ray is a tangent to the circle alpha. Let this ray be OA and OB (since there can be two tangents drawn from a point to a circle). Therefore |OA|=|OB|=1 (since radius is 1). Therefore angle(AOP)=angle(BOP)=30 (since radius and tangents are perpendicular). Therefore angle(AOB)=60. Since there are four circles touching externally, the total angle through which the half ray is intersecting the circles = 4*60 =240. The total angle possible is 360. Therefore the probability of the ray to touch the circles =240/360 = 2/3. (which is in the form a/b). Therefore a+b=5.

consider 5 circles such that 4 of them are tangential to a central circle now if one were to draw a tangent from the central circle as we know that the point of tangency and the centres of the 2 circles lie on the same line the distance between the centers is 2

so the length of the tangent is $\sqrt{3}$

so the sides of the triangle are 1,2, $\sqrt{3}$ .we recognize that

cos 30 = $\frac {\sqrt{3}}{2}$ so the angle between the tangent and line joining the centers is 30 degrees.

as 2 tangents can be drawn from a single point to a circle the second tangent makes an angle of 30 in the other direction.

so any line in between these tangents intersect this circle.

so for a line to intersect this circle it has to be in this arc of 60 degrees

the above logic holds for all the other 3 circles as well. as the 4 circles

as the circles don't intersect the total angle that the half ray should be in is $4\times 60$ = 240 degrees

so if the half ray is in the arc of the the remaining 120 degrees (360-240) it doesn't intersect any circle .

now divide this 120 degrees into 2 equal parts of 60 degrees each

So out of these 6 arc each subtending an angle of 60 degrees at the center one has to choose 4 arcs such that the half ray intersects one of the four circles

so 4/6=2/3

thus 2+3=5

A Circle can be tangent to exactly 6 other circles equal in size to the original, therefore each circle cover 1/6 of cycle. No matter how the four circles are drown, the probability that a half ray intersects one of the four circles is 4/6

Focus on one of the four circles first. Let the original circle be Circle O and the othe circle be Circle 1 . The region where which a half-ray from the center of Circle O can intersect Circle 1 is bounded by two tangent lines. Construct a line l from the center of Circle O to the center of Circle 1 . Then, two right triangles are formed, sharing a common hypotenuse with length 1+1=2. Also, the shorter leg of both triangles is a radius of Circle 1 . Then, if \theta is the acute angle between the hypotenuse and the longer side, sin(\theta)=1/2 or simply \theta=30 degrees. The central angle of Circle O bounded by the tangent lines is therefore 30+30=60. Because there are four externally tangent circles to Circle O , then the probability of a half-ray intersecting one of these circles is (240/360)=2/3.

To make it simple, we start with the original circle n circle number 1, let say C 1. So now we draw the 2 circles and 2 lines, both pass the centre of the original circle, O, and tangent to C 1 at point A and B. For triangle OAM, where M is the centre of C 1, AM = 1, OM = 2 and \angle OAM = \frac {pi}{2}. Thus \angle AOM = \frac {pi}{6}. Since triangle OAM is congruent to triangle OBM, the probability of the half ray to intersects C 1 = \frac {2 \times \frac {pi}{6}}{2pi} = \frac {1}{6}. And since all the other four circles do not intersect each other, \frac {a}{b} = 4 \times \frac {1}{6} = \frac {2}{3} and hence a + b = 5

For solving this problem easier, let's imagine a circle of radius 1 is drawn at $O(0,0)$ and one of the four non-overlapping circles is drawn at $A(2,0)$ , externally tangential at $(1,0)$ .

Let us note that the largest half ray drawn from the centre of the original circle, intersects the other circles as their tangent line. Let us denote the point of tangent line from point $O$ to circle A as $P$ and $P'$ .

Note that $OAP$ is a right triangle as $\angle OAP = 90 ^ \circ$ [Tangent line of a triangle].

OA = distance between two circles' center = 2.

AP = radius of circle A = 1.

Hence, $\sin \angle AOP = \frac {\text AP}{\text OA} = \frac {1}{2}$

$\angle AOP = 30 ^\circ$

At the other hand, $OAP'$ is also a right triangle which holds same property as $OAP$ . Thus, $\angle AOP' = 30 ^\circ$ also holds.

P( half ray intersects one circle ) = $\frac {\angle AOP + \angle AOP'}{360} = \frac {60}{360} = \frac{1}{6}$

As all of the circles are defined as non-overlapping:

P( half ray intersects one of the other four circle ) = 4 * P( half ray intersects one circle ) = $\frac {4}{6} = \frac {2}{3}$

The value of $a = 2$ , $b = 3$ , and $a + b = 2 + 3 = 5$ .

Call the center of the original circle $A$ . First note that the problem does not specify where the circles are located relative to circle $A$ , so we keep this in mind while proceeding through the solution. Consider a simpler figure where we only have one of the four externally tangent circles drawn in, and let the center of this circle be $B$ . The half ray drawn from the center of the original circle will hit this circle iff the ray lies between the two tangents emanating from $A$ . Let one of these tangents hit the second circle at point $C$ . Since $\overline{BC}$ is a radius of the circle, it must have length $1$ . A somewhat similar argument can be used to show that $AB=2$ . Finally, since $\overline{BC}$ is the radius drawn to a tangent at the point of tangency, we must have $\angle ACB=90^\circ$ . Combining all of this information means that $\angle BAC=30^\circ$ , so by symmetry the angle between the two tangents is $60^\circ$ . As a result, since there are four circles in the original diagram, the set of possible values of $\gamma$ in $[0,2\pi)$ (or $[0,360^\circ)$ ) consists of four disjoint intervals (since the circles don't necessarily have to touch each other) with a total length of $60\times 4=240^\circ$ . Hence the probability that the half ray intersects a circle is $\dfrac{240}{360}=\dfrac{2}{3}$ , and $a+b=2+3=5$ .

Draw two externally tangent circles each with radius 1 with centers A and B. Let circle A represent the original circle in the problem and let circle B represent one of the four non-overlapping tangential circles to the original. Draw the two tangents from point A to circle B, then draw the perpendiculars from point B to the points of tangency of the two previous lines. Call these two points C and D. Next, connect the centers of A and B.

Segment AB obviously has length 2 (sum of radii), and segments BC and BD both have length 2 (because they're radii of circle B). Angles BCA and BDA are both right angles, so ABD and ABC are clearly 30-60-90 triangles because of the ratio of 1:2 between side length and hypotenuse. Following the rules of 30-60-90 triangles, angles CAB and DAB are both clearly 30 degrees, so angle CAD is 60 degrees.

This is significant because it means that there is a 60 degree interval for each of the four circles in which a randomly chosen half ray will intersect a circle. In total, that makes 240 degrees out of 360 in which a half ray will intersect a circle, which simplifies to 2/3.

a+b = 2+3 = 5

let O be the centre of the original circle and P be the centre of a circle named X touching it. Let OA and OB be tangents to circle X. So OP=2 and AP=1 and thus $\angle PAO = 30^ \circ$ . So $\angle AOB = 60^ \circ$ . So 4 circles occupy $\ 240^ \circ$ . So probability = 240/360 = 2/3. so 5 is the answer.

Draw the tangents from the center of the original circle to one of the four external circle and call A,B the points of tangency and C,D the centers of the original circle and of the choosen circle respectively. Now consider the triangle ACD. We have that CD=2 (the sum of the two rays), AD=1 (ray) and $\angle{CAD}=90$ (the angle formed by a tangent and a ray); so we have $CDsin{\angle{ACD}}=AD$ . From this reletion follows that $\angle{ACB}=2\angle{ACD}=60$ . Hence, the event $a$ is $a=60\cdot 4$ and the sample space $b$ is $b=360$ . In conclusion, the probability is $\frac{a}{b}=\frac{2}{3}$ , so $a+b=5$ .

DEFINITIONS: "radial area": The degrees in which a half-ray intersects an object. For example, the "radial area" of a half-ray crossing through a point is $0$ .

Notice that $6$ circles of the same size can be put around the original circle without overlap. That means that each circle occupies a "radial area" of $360 \div 6 = 60$ degrees. We are using $4$ circles, so the total "radial area" is $60 \times 4 = 240$ . The total possible places the half-ray can be drawn is in the interval $[0, 2\pi)$ , or in degrees: $[0, 360)$ , so this means that the half-ray can be drawn anywhere: $360$ degrees. That means the probability of intersection with a circle is $\frac{240}{360} = \frac{2}{3}$ , so the answer is $2+3=\boxed{5}$

Since γ is a continuous random variable uniformly distributed , the probability that it assumes a value in the range

γ0≤γ≤γ0+α is equal to α2π, we must find the value of the angles subtended by the four external circles respect to the centre O of the original circle.

Now let’s consider one of these four circles and its centre O1 and the two rays drawn from O tangent to it, at points T and T′. The triangle OTO1 , being a rectangle, has side OO1 long 2 radiuses (hypotenuse), while O1T measures 1 radius (cathetus). Let angle in O be α, then

2sinα=1,

from whom we deduce that α=π6. All the same we can say for the other tangent OT′. Thus we conclude that the angle subtended by the circle is 2α or π3, who multiplied for four circles, gives the value of the sector occupied by the circles. The searched probability P will be given by

P=(frac{¾\pi}{2π}=\frac 23}

We have one circle in the centre, tangent to four other circles. If we take the tangent lines through the centre of the middle circle to one of the other circles, then our line will intersect this circle if it is between this pair of lines. If we can determine the angle between these tangents, we will be able to calculate the desired probability.

Let $C_1$ be the centre of the middle circle, $C_2$ be the centre of one of the other circles, and $T$ the point of tangency of a line through $C_1$ intersecting the circle centred at $C_2$ . Triangle $C_1 C_2T$ is right angled at $T$ . $C_1C_2 = 2$ and $C_2T$ has length one. This is one of our special triangles, so we know that the angle $C_2C_1T$ is $\frac{\pi}{6}$ . That means the angle between the two tangents is $\frac{\pi}{3}$ and over all 4 circles this gives a total of $\frac{4\pi}{3}$ . Expressing this as a fraction of the total possible set of angles, we have $\frac{4\pi}{6\pi} = \frac{2}{3}$ . So $a + b = 2 + 3 = 5$ .

firstly the original circle should be considered in such a way that it is tangential to two cirles(1&2) ,which are tengential to each other already,on one side and in the similar way with other two circles(3&4) in the opposite side. now the first circle starts at an angle of 60 degrees and the second circle ends at 150 degrees.similarly the third circle starts at 210 degrees and the fourth circle ends at 330 degrees. the half ray will not intersect the intervals of (0-30),(150-210),(330-360) degrees. Therefore the total angle in which it meets one of the circles=120+120=240. The totol angle around a point(centre)=360 degrees. Therefore probability = 240/360=2/3. Therefore answer=2+3=5.