In a triangle ABC,
BK is an angle bisector. A circle with
radius 5/3
passes through the vertex
B, intersects
AB at a
point L, and is tangent to
AC at
K. It is known that the
length of AC is
3√3, and the ratio of the lengths
|AK| to
|BL| is
6:5. The area of the triangle
ABC can be written
as a√b/c
, where
a and
c are coprime positive integers, and
b is not divisible by the square of any prime. What is the value of a+b+c?
To solve this problem, we need to find the area of triangle ABC using the given information.
First, let's draw a diagram to visualize the triangle and its properties.
Based on the problem statement, we have:
1. Triangle ABC, with angle bisector BK.
2. A circle with radius 5/3 passing through B and intersecting AB at point L.
3. The circle is tangent to AC at point K.
4. The length of AC is 3√3.
5. The ratio of AK to BL is 6:5.
Let's start by finding the lengths of AK and BL.
Given the ratio of AK to BL is 6:5, let's assign a variable to the common ratio:
Let AK = 6x and BL = 5x.
Next, since AK is a linear segment from A to K and BL is a linear segment from B to L, we can use the angle bisector theorem to establish a relationship between the lengths of AK, BK, and their respective sides of the triangle ABC.
According to the angle bisector theorem, we have:
AC / AB = KX / XB
Since we know that AC = 3√3 and AK + KB = AB, we can substitute the values:
3√3 / AB = 5x / (5x + 6x)
√3 / AB = 5x / 11x
√3 / AB = 5 / 11
AB = 11√3 / 5
Now, we can find the lengths of AK and BL:
AK = 6x = 6 * (11√3 / 5) = 66√3 / 5
BL = 5x = 5 * (11√3 / 5) = 11√3
Since the circle is tangent to AC at K, the distance from the center of the circle to AC is equal to the radius of the circle. Let's denote the center as O.
Applying the Pythagorean theorem in triangle OCK, we have:
OK^2 + CK^2 = CO^2
Since OK is equal to the radius, which is 5/3, and CK is equal to AK (as they are tangent to the circle), we can substitute the values:
(5/3)^2 + (66√3 / 5)^2 = CO^2
25/9 + (66^2 * 3) / (5^2 * 9) = CO^2
25/9 + 4356/225 = CO^2
(225 * 25 + 9 * 4356) / (9 * 25) = CO^2
(5625 + 39104) / 225 = CO^2
44729 / 225 = CO^2
CO^2 ≈ 198.57
CO ≈ √198.57
CO ≈ 14.08
Now that we have the length of OC, we can find the length of BC by subtracting BO (which is the radius of the circle) from CO:
BC = CO - BO
BC = 14.08 - (5/3)
BC = 37/3
Now we have all three sides of triangle ABC:
AB = 11√3 / 5
BC = 37/3
AC = 3√3
To find the area of triangle ABC, we can use Heron's formula:
Area = √(s(s-a)(s-b)(s-c))
where s is the semi-perimeter of the triangle, and a, b, and c are the side lengths.
First, calculate the semi-perimeter:
s = (AB + BC + AC) / 2
s = [(11√3 / 5) + (37/3) + (3√3)] / 2
Now, plug in the values and calculate the area:
Area = √[s(s-a)(s-b)(s-c)]
Finally, we can simplify the expression and find the values of a, b, and c to determine the final answer of a+b+c.