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Points A , B, and C are drawn, with the following:

 

AB = 6

AC = 16

BC = 10

 

Point B is on AC

 

Semicircles are constructed on AB, AC, and BC, and another circle is constructed such that it is tangent to all three semicircles. Find the radius of the circle.

 

 Jun 4, 2024
 #1
avatar+1418 
-1

Let's label the center of the circle O. We can see that triangle ABC is a right triangle with angle B being the right angle.

First, let's find the area of triangle ABC using Heron's formula:
s = (AB + BC + CA) / 2
s = (6 + 10 + 16) / 2
s = 16

Area of triangle ABC = sqrt (s(s-AB)(s-AC)(s-BC))
Area of triangle ABC = sqrt (16(16-6)(16-16)(16-10))
Area of triangle ABC = sqrt (16 * 10 * 6 * 6)
Area of triangle ABC = sqrt (5760)
Area of triangle ABC = 24

Since the radius of the circle is the same as the height of triangle ABC drawn from point O, and the area of a triangle is equal to 0.5 * base * height, we can find the height by dividing the area by the base:

Height = 2 * Area / AB
Height = 2 * 24 / 6
Height = 8

Now, we have a right triangle OBC with OB as the radius of the circle, BC as the base, and height as 8. Using the Pythagorean theorem:

OB^2 = BC^2 + Height^2
OB^2 = 10^2 + 8^2
OB^2 = 100 + 64
OB^2 = 164

 

Therefore, the radius of the semicircle is sqrt(164) = 2*sqrt(41).

 Jun 4, 2024
 #2
avatar+29 
0

Sorry, but I think you missed the fact that B is on AC. Perhaps you could try again? 

dplostthegame  Jun 4, 2024
 #3
avatar+677 
0

Since B is on segment AC, we can divide triangle ABC into two right triangles: right triangle ABI and right triangle BCI.

 

Find the lengths of AI and CI:

 

We know AB = 6 and BC = 10. Since the semi-circle is constructed on AB, diameter AB is twice the radius of the semi-circle, so the radius of the semi-circle on AB is 6 / 2 = 3. This radius (3) is also the height of right triangle ABI with hypotenuse AB (6) and base AI. Using the Pythagorean Theorem on right triangle ABI:

 

AI^2 + 3^2 = 6^2 AI^2 = 27 AI = √27 = 3√3

 

Similarly, the radius of the semi-circle on BC is 10 / 2 = 5. This is the height of right triangle BCI with hypotenuse BC (10) and base CI. Using the Pythagorean Theorem:

 

CI^2 + 5^2 = 10^2 CI^2 = 75 CI = √75 = 5√3

 

Radius of the Tangent Circle:

 

Since the circle is tangent to all three semi-circles, the center of this circle must be at the intersection point of the angle bisectors of triangle ABC. In a right triangle, the angle bisector of the right angle coincides with the median drawn to the hypotenuse, which divides the hypotenuse into two segments with a 1:1 ratio. Therefore, point I, the intersection of AI and CI, is the center of the tangent circle.

 

We now have a right triangle with legs AI = 3√3 and CI = 5√3, and the hypotenuse is the diameter of the tangent circle (which is twice the radius). Using the Pythagorean Theorem:

 

(diameter of tangent circle)^2 = (3√3)^2 + (5√3)^2 (diameter of tangent circle)^2 = 48 diameter of tangent circle = √48 = 4√3

 

Since the diameter is twice the radius, the radius of the tangent circle is:

 

radius = diameter / 2 = (4√3) / 2 = 2√3

 

Therefore, the radius of the circle tangent to all three semi-circles is 2√3.

 Jun 4, 2024
 #4
avatar+129741 
+1

Let the center of the circle  = O

Let the midpoint of the large semi-circle = F

Let the midpoint of the  the right semi-circle = D

Let the center of the semi-circle on the left = E

 

Form triangles EOF   and  DOF

DO =  (5 + r)

FO = (8 - r)

EO = (3 + r)\

DF = 3

EF = 5

 

By the Law of Cosines  we have two equations

 

EO^2  = EF^2  + FO^2  - 2(EF * FO) cos (EFO)

DO^2  =  DF^2 + FO^2 - 2(DF * FO) cos ( DFO)

 

Because  angles EFO  and DFO  are supplementary   cos (DFO) = - cos (EFO)

 

So re-writing....we have

 

EO^2  = EF^2  + FO^2  - 2(EF * FO) cos (EFO)

DO^2  =  DF^2 + FO^2 - 2(DF * FO) -cos ( EFO)

 

EO^2  = EF^2  + FO^2  - 2(EF * FO) cos (EFO)

DO^2  =  DF^2 + FO^2 + 2(DF * FO) cos ( EFO)

 

(3+r)^2  = 5^2 +(8-r)^2  - 2 (5 * (8-r)) cos (EFO)

(5+ r)^2  = 3^2 + (8-r)^2  + 2( 3 (8-r)) cos EFO

 

cos EFO  =   [ (3 + r)^2 - 25 - (8-r)^2] /[ -2 * (40 - 5r) ]

cos EFO  = [ (5 + r)^2 - 9 - (8-r)^2] / [ 2 *(24 - 3r) ]

 

Then

 

  [ (3 + r)^2 - 25 - (8-r)^2] /[ -2 * (40 - 5r) ] =  [ (5 + r)^2 - 9 - (8-r)^2] / [2 *(24 - 3r) ] 

 

This is a little sticky to simplify, but WolframAlpha  helps 

 

We get that

 

272 / (r-8)  + 49 = 0

 

272 / (r -8)  = -49

 

272  = -49 ( r -8)

 

272 = -49r + 392

 

-120  =-49r

 

r =  120 / 49  ≈ 2.449

 

cool cool cool

 Jun 4, 2024
 #5
avatar+29 
0

Thanks so much for your effort, even to those who failed!

 Jun 4, 2024

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