P is a point outside of circle Γ. The tangent from P to Γ touches at A. A line from P intersects Γ at B and C such that ∠ACP=120∘. If AC=16 and AP=19, then the radius of Γ can be expressed as a√b/c where b is an integer not divisible by the square of any prime and a,c are
coprime positive integers. What is a+b+c?
To find the radius of the circle Γ, we can use the properties of tangents and secants.
First, let's draw the diagram to visualize the situation described:
```
B
/ \
/ \
/ \
/ \
/ \
/ P \
/ | \
/ | \
/ | \
/ | \
/ | \
A-----------O-----------C
```
In the diagram, O represents the center of the circle Γ, and AC is a chord of the circle intersected by the line segment PB.
Since the tangent from P to Γ touches the circle at A, we know that PA is perpendicular to OA. This means that triangle PAO is a right triangle.
Also, since the line segment PB intersects the circle at B and C, we have two secants, PA and PC.
Now, let's use the given information to find the radius of the circle.
We are given that AP = 19 and AC = 16.
From triangle PAC, we can use the Law of Cosines to find the length of PC (let's call it x):
PC^2 = PA^2 + AC^2 - 2 * PA * AC * cos(∠ACP)
x^2 = 19^2 + 16^2 - 2 * 19 * 16 * cos(120°)
Simplifying the equation:
x^2 = 361 + 256 - 2 * 19 * 16 * (-0.5)
x^2 = 361 + 256 + 304
x^2 = 921
Taking the square root of both sides:
x = √921
Now, let's find the length of AO. Since AO is the radius of the circle, let's call it r:
AO = AC - OC
AO = 16 - x
Substituting the value of x:
AO = 16 - √921
So, the radius of the circle Γ is √921.
To express the radius in the form a√b/c, we can simplify √921:
√921 = √(3 * 307) = √3 * √307
Now, our expression for the radius is a√b/c, where a = 1, b = 307, and c = 1.
Therefore, a + b + c = 1 + 307 + 1 = 309.