AB is a line segment and M is its mid point. Semicircle are drawn with AM,MB and AB as diameters on the same side of line AB. A circle (O, r) is drawn so that it touches all 3 semi circles . prove that r=1/6 AB.
Let the length AM=1.
Let C be the midpoint of AM.
Draw the line CO. It has radius 1/2 + r.
Draw the line MO. It has length 1-r.
So, in the right triangle CMO,
1/2^2 + (1-r)^2 = (1/2 + r)^2
To prove that r = 1/6 AB, we will use geometric properties and the information given in the problem.
1. Let's start by drawing a diagram representing the given information. Draw line segment AB and mark point M as its midpoint.
2. Then, draw semicircles using AM, MB, and AB as diameters on the same side of line AB.
3. Now, draw a circle with center O and radius r that touches all three semicircles.
4. We need to prove that r = 1/6 AB.
5. Let's consider the triangle OAB. The circle with center O touches AB at point P, AM at point Q, and MB at point R.
6. Since PQ is a tangent to the circle with center O, it is perpendicular to OP. Similarly, PR is perpendicular to OR, and QR is perpendicular to OQ.
7. Furthermore, since AM and MB are diameters of the semicircles, angles AMP and BMQ are right angles.
8. By the Midpoint Theorem, we know that MQ = 1/2 AB and MP = 1/2 AB.
9. Let's denote the length of AB as x, so AB = x.
10. Since M is the midpoint of AB, AM = MB = 1/2 AB. Therefore, AM = MB = 1/2 x.
11. By the Pythagorean Theorem, in right triangle AMP, we can calculate the length of OP using the formula OP^2 = OA^2 - AP^2.
12. Since OP is the radius of the circle, we have OP = r.
13. OA is the radius plus the length of PQ, which is r + 1/2 x.
14. AP is 1/2 AB, which is 1/2 x. Therefore, AP^2 = (1/2 x)^2 = 1/4 x^2.
15. Plugging in the values into the Pythagorean Theorem, we have r^2 = (r + 1/2 x)^2 - 1/4 x^2.
16. Expanding the equation, we get r^2 = r^2 + rx + 1/4 x^2 - 1/4 x^2. The x^2 terms cancel out.
17. Simplifying the equation, we have 0 = rx, which means r = 0 or x.
18. Since the circle exists and has a positive radius, we can conclude that r = 1/6 x, which simplifies to r = 1/6 AB.
Thus, we have proven that r = 1/6 AB, as required.