China Math Olympiad | A Very Nice Geometry Problem

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[@santiagoarosam430]

Si ON=d→ En el triángulo ONQ: (5-r)²=r²+d²→ d²=25-10r → En el triángulo QNP: [(5/2)+r]²=r²+[(5/2)+d]²→ 5r=d²+5d→ Sustituyendo d²→ 5r=25-10r+5d→ 3r-5=d→ 3r=d+5=AN → Potencia de P respecto a la circunferencia de centro Q =PN²=PM(PM+2r)→ [3r-(5/2)]²=(5/2)[(5/2)+2r]→ 9r²-20r=0→ r=20/9.
Interesante problema. Gracias y un saludo cordial.

[@murdock5537]

Interesting. φ = 30° → sin⁡(3φ) = 1; ∆ PNQ → PN = PO + NO = 5/2 + x; QN = r; PQ = 5/2 + r
QO = 5 - r = √(r^2 + x^2); QPN = δ; QON = α; NQO = γ; sin⁡(PNQ) = 1 →
sin⁡(α) = cos⁡(γ) → cos⁡(α) = sin⁡(γ); POQ = 6φ - α → sin⁡(6φ - α) = sin⁡(α); OQP = β
sin⁡(γ)/x = sin⁡(α)/r = sin⁡(3φ)/(5 - r); cos⁡(α)/x = sin⁡(α)/r → r/x = tan⁡(α)
sin⁡(α)/r = 1/(5 - r) → sin⁡(α) = r/(5 - r); √(r^2 + x^2) = 5 - r → x = (√5)√(5 - 2r)
cos⁡(δ) = (5/2 + x)/(5/2 + r) = (5 + (2√5)√(5 - 2r))/(5 + 2r); sin⁡(δ) = 2r/(5 + 2r) →
sin^2(δ) + cos^2(δ) = 1 → (5 + (2√5)√(5 - 2r))^2 + 4r^2 = (5 + 2r)^2 →
60r - 100 = (20√5)√(5 - 2r) → 9r = 20 → r = 20/9
btw: sin⁡(δ) = 8/17 → ∆ PNQ = pyth. triple = (5/18)(8 - 15 - 17)
sin⁡(α) = r/(5 - r) = 4/5 → ∆ ONQ = pyth. triple = (5/9)(3 - 4 - 5); x = (√5)√(5 - 2r) = 5/3 🙂

[@quigonkenny]

Let P be the center of the inscribed circle and Q the center of the inscribed semicircle. Let S be the point of tangency between circle P and OB, let V be the point of tangency between the circumferences of circle P and semicircle O, and let T be the point of tangency between circle P and semicircle Q.
As the diameter of semicircle Q (AO) is 5, then the radius of semicircle Q is 5/2. As AO is a radius of semicircle O, the radius of semicircle O is 5.
Draw OV, PQ, and PS. As OV is a radius of semicircle O, OV = 5. As V is the point of tangency between semicircle O and circle P, then it is collinear with points O and P. Therefore OV = OP+PV. As PV is a radius of circle P, PV = R, so OP = 5-R.
As T is the point of tangency between circle P and semicircle Q, it is collinear with Q and P, so QP = QT+PT. As radii of their respective shapes, QT = 5/2 and PT = R, so QP = R+5/2.
As PS is a radius of circle P and OB is tangent to circle P at S, then ∠OSP = 90°. If OS = x, then as OQ = 5/2, then QS = x+5/2.
Triangle ∆OSP:
OS² + PS² = OP²
x² + R² = (5-R)²
x² + R³ = 25 - 10R + R²
x² - 25 = -10R
R = (25-x²)/10
Triangle ∆QSP:
PS² + QS² = QP²
R² + (x+5/2)² = (R+5/2)²
R² + x² + 5x + 25/4 = R² + 5R + 25/4
x² + 5x = 5R
x² + 5x = 5(25-x²)/10 = 25/2 - x²/2
3x²/2 + 5x - 25/2 = 0
3x² + 10x - 25 = 0
3x² + 15x - 5x - 25 = 0
3x(x+5) - 5(x+5) = 0
(x+5)(3x-5) = 0
x = -5 ❌ | x = 5/3
R = (25-x²)/10 = (25-(5/3)²)/10
R = (25-25/9)/10 = ((225-25)/9)/10
R = (200/9)/10 = 20/9 = 2.22… units

[@michaeldoerr5810]

The answer is 20/9. And it looks like a reason practice Pythagorean Theorem twice and then do the substitution one time. Also that is the fifth time I have noticed that nifty simplification at the 13:17 mark. I better use that as practice!!!

[@ritwikgupta3655]

Very nice solving for x.

[@TheAlavini]

Congrat. Nice solution.

[@MathBooster]

Thank you very much 😊

[@RealQinnMalloryu4]

180°ABRO/5 =3 .30ABRO 3^1.5^6ABRO 3^1.5^3^2ABRO 1^1.5^1^3^2 1^13^2 3^2 (ABRO ➖ 3ABRO+2).

[@imetroangola4943]

The madness of mathematics! Only you understand! 😂😂😂😂😂😂😂😂

[@giuseppemalaguti435]

(√((2,5+R)^2-R^2)-2,5)^2+R^2=(5-R)^2...R=20/9

[@prossvay8744]

R=20/9