Can you find lenght of Red line? | (Fun Geometry Problem) |

[@murdock5537]

Very interesting, many thanks, Sir! Here an old fashioned way finding roots (√): φ = 30° → sin⁡(3φ) = 1; ∆ PSQ → PS = 8; QS = 9; QP = 5; QPS = θ ∆ QPR → QP = 5; RP = 7; QR = r = ? QPR = α ∆ RPS → RPS = β = θ - α; PS = 8; RS = 6; RP = 7 → ∆ PSQ → law of cosines → cos⁡(θ) = 1/10 → sin⁡(θ) = 3√11/10 ∆ PSR → law of cosines → cos⁡(β) = 11/16 → sin⁡(β) = 3√15/16 → cos⁡(α) = cos⁡(θ - β) = cos⁡(θ)cos⁡(β) + sin⁡(θ)sin⁡(β) = (1/10)(11/16) + (3√11/10)(3√15/16) = (1/160)(11 + 9√165) → r^2 = 25 + 49 - 2(5)7cos⁡(α) = (9/16)(123 - 7√165) 144 < 165 < 169 → 165 = (13 - x)^2 = 169 + x^2 - 26x ≈ 169 - 26x → x = 2/13 → 13 - 2/13 = (1/13)(169 - 2) = 167/13 ≈ √165 → r^2 = (9/16)(123 - 7√165) = (9/16)(123 - 7(167/13)) = (9/4)(5/2)(43/13) → r = (3/2)√5590 → √5590 = ? → 4900 < 5590 < 6400 → (70 + x)^2 = 5590 ≈ 4900 + 140x → x = 69/14 → 70 + x = 70 + 69/14 = (1/14)(14(70) + 69) = 1049/14 → (1049/14 - x)^2 = 5590 ≈ (1049/14)^2 - 1049x/7 → x = (4761/196)(7/1049) ≈ 162/1000 = 81/500 → 1049/14 - x = 1049/14 - 81/500 = 261683/3500 ≈ 74,76657 ≈ √5590 → r= (3/4)(√430/√13) = (3/52)√5590 = (3/52)(261683/3500) ≈ 4,313456 → 3√(123 - 7√165)/4r = 1,000093882 (looks like a good approximation 🙂)

[@MrPaulc222]

I used right triangles and it became very convoluted. I had perpendiculars from the base (8) up to Q and R from corresponding points on the base. I then set up two sets of equations to find the dimensions of the two right triangles created. I then calculated the difference in heights for the short lengths down from Q and the other length I projected up from the base line as the gap between the two perpendicular points. The redline was then calculated by Pythagoras. It was very complex. I don't doubt your way was better, but I'm not yet at a level where I could understand all of it.

[@soli9mana-soli4953]

I also used this method, but as you told it's very complex due to radicals. However formalizing, being QH and RK the heights of the two triangles perpendicular to base PS, applying pythagorean theorem we have: r² = (8 - PH - KS)² + (QH - RK)²

[@marioalb9726]

Cosine rule: c²=a²+b²-2ab.cosα cosα = (a²+b²-c²)/(2.a.b) Angle PSQ: cosα₁ = (8²+9²-5²)/(2*8*9) α₁ = 33,5573° Angle PSR: cosα₂ = (8²+6²-7²)/(2*8*6) α₂ = 57,9100° Angle QSR: α = α₂ - α₁ = 24,3527° Triangle QSR: r² = 9²+6²-2*9*6*cosα r = 4,3138 cm ( Solved √ )

[@marioalb9726]

Cosine rule: c²=a²+b²-2ab.cosα cosα = (a²+b²-c²)/(2.a.b) Angle RPS: cosα₁ = (8²+7²-6²)/(2*8*7) α₁ = 46,5675° Angle QPS: cosα₂ = (8²+5²-9²)/(2*8*5) α₂ = 84,2608° Angle QPR: α = α₂ - α₁ = 37,6933° Triangle QPR: r² = 5²+7²-2*5*7*cosα r = 4,3138 cm ( Solved √ )

[@AzouzNacir]

We use AI Cachi theorem in triangle PSR and find cosα= (8²+6²-7²)/(2*8*6)=17/32. Then we apply it in triangle PSQ and find cosβ= (8²+9²-5²)/(2*8*9)=5/6. So sinα=√11/6 and sinβ=7√15/32. Hence x²=6²+9²-2*6*9*cos(α-β)=117-108*((5/6*17/32)+(√11/6*7√15/32))= (1107-63√165)/16. Hence x= (3√(123-7√165))/4.

[@santiagoarosam430]

Según fórmula de Heron: Área de PQS =6√11=19,8997---> Altura respecto a PS =3√11/2---> Proyección horizontal de PQ=a=½ ; Área de PRS =21√15/4---> Altura respecto a PS =21√15/16---> Proyección horizontal de SR =b≈5,5602... ---> Proyección horizontal de QR =8-a-b≈ 4,3125. ; Proyección vertical de QR =21√15/16 - 3√11/2≈0,10826---> QR =√(4,3125²+0,10826²) ≈4,3138. Gracias y saludos