China Math Olympiad Problem | A Very Nice Geometry Challenge

[@prossvay8744]

Radius of the Large semicircle=5 Radius of the small semicircle=2.5 Connect P to Q (P and Q are ) In ∆ ABC (CQ is radius of the circle) PQ ^2 =CQ^2+PC^2 (2.5+R)^3=(2.5+x)^2+R^2 R^2=x^2+5x (1) R^2+x^2=(5-R)^2 (2) So (1)&(2) X=5/3 (2): R^2+(5/3)^2=(5-R)^2 So R=20/9.❤❤❤

[@marioalb9726]

x² = (R-r)² - r² = (5-r)² - r² x² = (25 -10r+r²) - r² x² = 25 - 10r (x + r₁)² = (r + r₁)² - r² (x + 2,5)² = (r +2,5)² - r² x²+5x+6,25 = (r²+5r+6,25)-r² x²+5x = 5r Replacing: 2x² + 10x = 10r = 25 - x² 3x² + 10x - 25 = 0 x = 5/3 cm R = 20/9 cm ( Solved √ )

[@michaeldoerr5810]

The answer is that the radius is 20/9 units. By golly, I have been seeing videos from this channel and @PreMath channel and this has become such a familiar and easy to understand problem. I better start practicing!!!

[@quigonkenny]

Let P be the center of the radius R circle and S the center of the diameter 5 semicircle. Let T be the point of tangency between circle P and semicircle Q. Let M be the point of tangency between circle P and the circumference of semicircle O, and N the point of tangency between circle P and diameter AB. By observation we can see that the radius of semicircle O is 5, and as the radius of semicircle O is the diameter of semicircle Q, the radius of semicircle Q is 5/2. Let ON = x. As OQ = 5/2, QN = OQ+ON = 5/2+x. As the point of tangency between two circles is always collinear with their centers, T is in-line with PQ, and P is in-line with OM. As QT = 5/2 and PT = R, PQ = QT+PT = 5/2+R. As OM = 5 and PM = R, OP = OM-PM = 5-R. Triangle ∆ONP: ON² + PN² = OP² x² + R² = (5-R)² x² + R² = 25 - 10R + R² x² = 25-10R --- [1] Triangle ∆QNP: QN² + PN² = QP² (5/2+x)² + R² = (5/2+R)² 25/4 + 5x + x² + R² = 25/4 + 5R + R² 5x + x² = 5R 5x + (25 - 10R) = 5R

[@santiagoarosam430]

Potencia de O respecto a la circunferencia =ON²=d²=(5+2r)5= 25-10r=d² → Potencia de P =[(5/2)+d]²=(5/2)[(5/2)+2r]→ 5+d=3r → AN=3r → Potencia de P =[3r-(5/2)]²=(5/2)[(5/2)+2r]→ r=20/9. Gracias y un saludo cordial.

[@marcelowanderleycorreia8876]

Excelente questão de colinearidade. 👍

[@giuseppemalaguti435]

Teorema del coseno triangolo SOR...(2,5+R)^2=(2,5)^2+(5-R)^2-2*2,5*(5-R)cos(180-arcsinR/5-R)...R=20/9...già visto da qualche parte

[@batmunkhenkhbaatar9061]

very nice

[@RealQinnMalloryu4]

(5)^2=25 90°AO/25=3.15 3.3^5 1.3^1 3^1(AO ➖ 3AO+1).