Let ABC be a right triangle with ACB =90, AC =6 , and BC =2. E is the midpoint of AC , and F is the midpoint of AB . If CF and BE intersect at G , then cos(CGB),
in simplest radical form, is ((k square root w)/f) where k , w , and f are positive integers. Find the value of k+w+ f.
Since E and F are midpoints, EF ║ CB, and has length CB/2 = 1
CE = AC/2 = 3
Label angles
CGB = a
EBC = b
FCB = c
so, intriangle CGB, a+b+c = 180 degrees
we want
cos a = cos *180 - (b+c))
= -cos(b+c)
= sinb sinc - cosb cosc
Looking at the right triangles,
EB = √13 and FC = √10, so we have
cos a = 3/√13 * 3/√10 - 2/√13 * 1/√10
= 9/√130 - 2/√130
= 7/√130
= 7√130 / 130
k+w+f = 7+130+130 = 267
To find cos(CGB), we first need to find the lengths of CF, BE, and BG.
We know that E is the midpoint of AC, so AE = EC = 6/2 = 3.
Similarly, F is the midpoint of AB, so AF = FB = BC = 2.
Now, let's find the length of BG.
Using the Midpoint Theorem, we can determine that EF is parallel to AB and EF = (1/2) * AB.
Therefore, AB = 2 * EF = 2 * AF = 2 * 2 = 4.
Since AF = FB = 2, we can use the Pythagorean theorem to find AB:
AB^2 = AF^2 + FB^2
AB^2 = 2^2 + 2^2
AB^2 = 4 + 4
AB^2 = 8
AB = √8 = 2√2 (simplified in radical form)
Now, consider the triangles CEB and CBG. Since AE is parallel to BG (both are medians of triangle ABC), we have:
AE/BG = EC/CG
3/BG = 3/CG
BG = CG
So, BG = CG.
Therefore, in triangle CGB, the angle at G is a right angle.
Now, let's find the length of CF.
Using the Pythagorean theorem in triangle ABC, we have:
AC^2 = AB^2 + BC^2
6^2 = (2√2)^2 + 2^2
36 = 8 + 4
36 = 12
AC = 6
Using the Pythagorean theorem in triangle ACF, we have:
AC^2 = AF^2 + CF^2
6^2 = 2^2 + CF^2
36 = 4 + CF^2
CF^2 = 36 - 4
CF^2 = 32
CF = √32 = 4√2 (simplified in radical form)
Now, we can find cos(CGB) using the cosine ratio in triangle CGB:
cos(CGB) = BG/CF
cos(CGB) = BG/4√2
cos(CGB) = (BG√2)/(4√2)
cos(CGB) = (BG√2)/(4√2) * (√2/√2) [Multiplying top and bottom by √2 to rationalize the denominator]
cos(CGB) = (BG√2)/(4*2)
cos(CGB) = (BG√2)/(8)
cos(CGB) = BG/(8/√2)
cos(CGB) = BG/(8/√2) * (√2/√2) [Multiplying top and bottom by √2 to rationalize the denominator]
cos(CGB) = (BG√2)/(8√2)
cos(CGB) = (BG√2)/(8√2) * (1/√2) [Simplifying √2/√2 to 1]
cos(CGB) = (BG√2)/(8*2)
cos(CGB) = (BG√2)/16
cos(CGB) = (BG/16)√2
Since BG = CG, we can further simplify:
cos(CGB) = (CG/16)√2
Therefore, k = 1, w = 2, and f = 16.
Finally, the value of k + w + f = 1 + 2 + 16 = 19.
To find the value of cos(CGB), we can start by finding the coordinates of points A, B, C, E, F, and G on a coordinate plane.
Let's choose point C as the origin (0, 0) on the coordinate plane. Given that AC = 6 and BC = 2, point A will have coordinates (6, 0) and point B will have coordinates (0, 2).
Since E is the midpoint of AC, its coordinates will be the average of the coordinates of points A and C. Therefore, the coordinates of E will be ((6+0)/2, (0+0)/2), which simplifies to (3, 0).
Similarly, F is the midpoint of AB, so its coordinates will be the average of the coordinates of A and B. Hence, the coordinates of F will be ((6+0)/2, (0+2)/2), which simplifies to (3, 1).
Now, we can find the equation of the line CF. The slope of CF can be calculated as (y coordinate of C - y coordinate of F) divided by (x coordinate of C - x coordinate of F).
Slope of CF = (0 - 1) / (0 - 3) = -1 / -3 = 1/3.
Using the slope-intercept form y = mx + b, where m is the slope and b is the y-intercept, we can write the equation of CF as y = (1/3)x + b.
Since CF passes through point C(0, 0), substituting the coordinates of C into the equation, we have 0 = (1/3)(0) + b.
Simplifying, we find that b = 0.
Therefore, the equation of line CF is y = (1/3)x.
Next, let's find the equation of line BE. We know that E has coordinates (3, 0). Since BE is perpendicular to CF, it will have the negative reciprocal slope of CF. Therefore, the slope of BE is -1/3.
Using the slope-intercept form again and substituting the coordinates of point E, the equation of line BE is y = (-1/3)(x - 3).
Expanding and simplifying, we have y = (-1/3)x + 1.
To find the coordinates of point G, we need to solve the system of equations formed by the equations of CF and BE. Setting the two equations equal to each other, we have:
(1/3)x = (-1/3)x + 1.
Simplifying, we find (2/3)x = 1.
Multiplying both sides by 3/2, we get x = 3/2.
Substituting this value back into either equation (let's use the equation of CF), we find y = (1/3)(3/2) = 1/2.
Therefore, point G has coordinates (3/2, 1/2).
Now, we have all the coordinates we need to find cos(CGB).
Using the distance formula, we can find the lengths of the sides of triangle CGB:
CG = sqrt((3/2 - 0)^2 + (1/2 - 0)^2) = sqrt(9/4 + 1/4) = sqrt(10/4) = sqrt(10)/2.
GB = sqrt(0^2 + 1^2) = sqrt(1) = 1.
CB = sqrt((3/2 - 0)^2 + (1/2 - 0)^2) = sqrt(9/4 + 1/4) = sqrt(10/4) = sqrt(10)/2.
Now, we can use the law of cosines to find cos(CGB):
cos(CGB) = (CG^2 + GB^2 - CB^2) / (2 * CG * GB)
= ( (sqrt(10)/2)^2 + 1^2 - (sqrt(10)/2)^2 ) / (2 * sqrt(10)/2 * 1)
= (10/4 + 1 - 10/4) / (sqrt(10)/2)
= 1 / sqrt(10).
Therefore, cos(CGB) in simplest radical form is 1 / sqrt(10).
The value of k is 1, w is 10, and f is 1. Hence, k + w + f = 1 + 10 + 1 = 12.