A Nice Math Olympiad Geometry Challenge | Circle inside a semicircle | Find the radius of the circle 

Wow! I get every step. The challenge is knowing where to start!!

1.5

E se fizessemos: (5-2R)²= (3-R)² + (4-R)², estaria correto? 🤔🤔🤔

Bu şekil ve değerler gerçekle uyuşmuyor. Küçük çemberin çapı 3 ise büyük çemberin merkezi; D noktası ile küçük çemberin AB doğrusuna teğet olduğu noktanın tam ortası (O)olmalı. Bu durumda ise büyük çemberin OA ve OB uzunlukları eşit olmuyor.

Dear, NAMASTHE ! You are great. CONGRATULATIONS !

Late to the party... but... 1. Complete the circle 2. Extend CD to become a chord. 3. Use Intersecting Chords to establish AD = 9/4. 4. Use #3 to establish point O (center of semicircle) lying on AB, and R (radius of semicircle) as 24/8. 5. Designate point T as the tangent point of the small circle with the semicircle. 6. Connect points O and T to create segment OT, which passes through point o, the center of the small circle (I forget the theorem that makes this so.) 7. Drop a line segment from point o perpendicular to segment OB, intersecting at E, creating right triangle △OEo. 8. The two legs of that right triangle, OE and Eo are (r - 7/8) and r, respectively (where 'r' is the radius of the small circle). The hypotenuse is the semicircle's radius (25/8, see #4) minus r, the small circle's radius. 9. Then, by Pythagoras, we get: (r - 7/8)² + r² = (25/8 - r)² 10. Simplifying, yields: r² + (9/2)r - 9 = 0 11. Use the Quadratic Formula to arrive at: r = 3/2 Ok, now to watch the video and see if I got it right.

Good job.

Brilliant!