Find the radius of the semicircle | A Nice Geometry Problem 

extending the line AC in the upper part until it intersects the perpendicular to AB from B, at the intersection point D, we obtain the right-angled triangle ABD which has sides twice the sides of AOP. Therefore if OP = 3 it follows that DB = 6 Let DC = x and apply the tangent-secant theorem from the external point D 6² = (10+x)*x x² + 10x - 36 = 0 x = √ 61 - 5 AD = 10 + √ 61 - 5 = 5 + √ 61 we apply the Pythagorean theorem on ABD (2r)² = (5 + √ 61)² - 6² r² = (25 + 5√ 61)/2

cosα = R/c = 10/(2R) --> R² = 5c Pytagorean theorem: c² = R² + 3² = 5c + 9 c² - 5c - 9 = 0 --> c= 6,4051cm R = 6,659 cm ( Solved √ )

Segment AC= 'a' ; Segment BC= 'b' Similarity of triangles: b/10=3/R b² = 900/R² Pytagorean theorem: a² + b² = (2R)² b² = 4R² - 10² Equalling : 900/R² = 4R² - 100 4R⁴ - 100R² - 900 = 0 R⁴ - 25R² - 225 = 0 R = 5,659 cm ( Solved √ )

cosα = 10/(2R) ; cos²α = 25/R² Intersecting chords theorem: (R+3cosα)(R-3cosα) = (½10)² R² - 9.cos²α = 5² R² - 9.(25/R²) = 25 R⁴ - 25R² - 225 = 0 R = 5,659 cm ( Solved √ )

cosα= 10/(2R) ; tanα= 3/R R = 5/cosα = 3/tanα 5/3 sinα = cos²α = 1 - sin²α sin²α + 5/3 sinα - 1 = 0 sin α = 0,46837 --> α = 27,929° R = 5,659 cm ( Solved √ )

AP=x. R^2+9=x^2 ∆APO~∆ACB 10/(2R)=R/x R^2=5x x^2-9=5x x=(5+√61)/2 R^2=5x=(25+5√61)/2

Draw BC. By Thales' Theorem, as A and B are ends of a diameter and C is a point on the circumference of semicircle O, then ∠BCA = 90°. As ∠BCA = ∠AOP = 90° and ∠A is common, ∆BCA and ∆AOP are similar triangles. Triangle ∆AOP: OA² + OP² = PA² x² + 3² = PA² PA² = x² + 9 PA = √(x²+9) Triangle ∆BCA: BC² + CA² = AB² BC² + 10² = (2x)² BC² + 100 = 4x² BC² = 4x² - 100 BC = √(4x²-100) BC/AB = OP/PA √(4x²-100)/2x = 3/√(x²+9) √(4x²-100)√(x²+9) = 6x (4x²-100)(x²+9) = 36x² 4x⁴ + 36x² - 100x² - 900 - 36x² = 0 4x⁴ -100x² - 900 = 0 x⁴ - 25x² - 225 = 0 x² = [-(-25)±√(-25)²-4(1)(-225)]/2(1) x² = 25/2 ± √(625-900)/2 x² = 25/2 ± √1525/2 = (25±5√61)/2 x² = (25+5√61)/2 | x² = (25-5√61)/2 ❌ x² > 0 x = √((25+5√61)/2) x = √(50+10√61)/2 ≈ 5.659 units

x/3=10/BC and 4x^2=100+BC^2=100+900/x^2. Thus, 4x^4-100x^2-900=0 or x^4-25x^2-225=0; x^2=(25+5√61)/2 and x=√[(25+5√61)/2]=5.65912... units.

Nice problem In triangle AOP tan A = 3/x In triangle ACB sec A = 2x/10 = x/5 eliminating A x^2/25 - 9/( x^2) = 1 x^4- 25 x^2- 225 =9 rest follows at once.

The answer is x = sqrt((25+5*sqrt(61))/2). Also I really have to say that it would be nice if there was a mnemonic device for HL similarity that involve the 90° and the common angle because I have always noticed that the similarity rules are simply the first two letters of the similar triangles are ratioed and then the first and last letters of the similar triangles that get ratioed.

3/R=BC/10 BC=30/R 100+(30/R)^2=(2R)^2 So R=5.66 units.

cosθ = 10/2r = 5/r (sinθ)/(cosθ) = 3/r = (sinθ)r/5 (sinθ) = 15/r^2 (5/r)^2 + (15/r^2)^2 = 1 25/r^2 + 225/r^4 = 1 r^4 - 25r^2 - 225 = 0 let r^2 = x x^2 - 25x - 225 = 0 x = (25 + 5✓61)/2 r = ✓(50 + 10✓61)/2

(10)^2 (3)^2={100+9}=109 180°ABCPO/109=1.71ABCPO 1.71^1 1.71 (ABCPO ➖ 71ABCPO+1).

2rcosθ=10..3/r=tgθ...rcosθ=5..r^2=25(secθ)^2=25(1+9/r^2)..r^4=25(r^2+9)..r^2=(25+√1525)/2=(25+5√61)/2

A questão é muito boa. Parabéns.