Math Olympiad Challenge | A Very Nice Geometry Problem

[@quigonkenny]

As ∠BCA = 105° and ∠CAB = 30°, ∠ABC = 180°-(105°+30°) = 45°. As ∠ABD = θ, ∠DBC = 45°-θ. Extend BC to E, where AE is perpendicular to BC. As ∠ABC = 45° and ∠BEA = 90° (by construction), ∠EAB = 90°-45° = 45°, and thus ∆BEA is an isosceles right triangle and BE = EA = a. Triangle ∆BEA: BE² + EA² = AB² a² + a² = AB² AB² = 2a² AB = √(2a²) = √2a As ∠ACE is an exterior angle to ∆ABC at C, ∠ACE = ∠CAB+∠ABC = 30°+45° = 75°. As ∠CEA = 90°, ∠EAC = 90°-75° = 15°. sin(75°) = sin(45°+30°) sin(75°) = sin(45°)cos(30°) + cos(45°)sin(30°) sin(75°) = (1/√2)(√3/2) + (1/√2)(1/2) sin(75°) = (√3+1)/2√2 cos(75°) = cos(45°+30°) cos(75°) = cos(45°)cos(30°) - sin(45°)sin(30°) cos(75°) = (1/√2)(√3/2) - (1/√2)(1/2) cos(75°) = (√3-1)/2√2 sin(75°) = EA/AC AC = EA/sin(75°) = x/((√3+1)/2√2) AC = 2√2x/(√3+1) AC = 2√2x(√3-1)/(√3+1)(√3-1) AC = 2√2x(√3-1)/(3-1) AC = √2x(√3-1) = (√6-√2)x CD = DA = CA/2 CD = DA = (√6-√2)x/2 = (√3-1)x/√2 cos(75°) = CE/AC CE = ACcos(75°) = (√6-√2)x(√3-1)/2√2 CE = x(√3-1)²/2 = x(3+1-2√3)/2 CE = x(4-2√3)/2 = (2-√3)x BE = BC + CE x = BC + (2-√3)x BC = x - (2-√3)x = x + √3x - 2x BC = √3x - x = (√3-1)x Triangle ∆BCD: DB² = BC² + CD² - 2BC(CD)cos(105°) DB² = ((√3-1)x)² + ((√3-1)x/√2)² - 2((√3-1)x)((√3-1)x/√2)(-cos(75°)) DB² = (4-2√3)x² + (2-√3)x² + √2(4-2√3)x²(√3-1)/2√2 DB² = (6-3√3)x² + (2-√3)(√3-1)x² DB² = (6-3√3)x² + (2√3-2-3+√3)x² DB² = (6-3√3)x² + (3√3-5)x² = x² DB = √(x²) = x Triangle ∆ABD: cos(θ) = (BD²+AB²-DA²)/(2BD(AB)) cos(θ) = (x²+(√2x)²-((√3-1)x/√2)²)/2x(√2x) cos(θ) = (x²+2x²-(2-√3)x²)/2√2x² cos(θ) = (3-(2-√3))/2√2 = (√3+1)/2√2 sin(90°-θ) = (√3+1)/2√2 = sin(75°) 90° - θ = 75° θ = 15°

[@ludmilaivanova1603]

@ 2:34 EC is a leg against 30 degrees angle and equal the half of the hypotenuse, no need to involve cos and calculations.

[@jimlocke9320]

Extend AC down and drop a perpendicular to it from point B. Label the intersection as point F. Note that

[@pwmiles56]

I did it by sine rule. AD=CD=x, BD=y. sin theta / x = sin 30 /y sin(45-theta) / x = sin 105 / y sin theta sin 105 = sin(45-theta) sin 30 2 sin theta sin 75 = sqrt2/2 (cos theta - sin theta) 4 sin theta sin(45+30) = sqrt2 (cos theta - sin theta) 2 sqrt2 sin theta (sin 30 + cos 30) = sqrt2 (cos theta - sin theta) sin theta (1+sqrt3) = cos theta - sin theta 1/tan theta - 1 = 1 + sqrt3 tan theta = 1 / (2 + sqrt(3)) theta = 15 deg CHECK: tan theta = tan(45 - 30) = (1 - 1/sqrt(3))/(1+1/sqrt(3)) = (sqrt3 - 1)/(sqrt3+1) = 2 / (4+2 sqrt 3)

[@chandrasekaranpanchakshara6010]

How can the sum of interior angles of triangle be 195degrees? Question seems to be wrongly coined. Procedure makes use of those wrong measures of angles.

[@oscarcastaneda5310]

I only got as far as : Theta = inverse cosine of (1/2)sqrt(sqrt(3) + 2). Where do I go from here ?

[@michaeldoerr5810]

The answer is 15 degrees. Near the end of this video, it looks like the exterior angle theorem is only equivalent to 2theta=30 degrees because of congruent sides. And that all started off with a simple construction that of the 30-60-90 traingle. Come to think of it any triangle that has a 30 degree or 60 degree angle requires that construction and working with that results in knowing which triangles within that triangle DO NOT exist. I hope that that is a good takeway. And goodness I need to pair thia video with videos that require that construction on your channel.

[@giuseppemalaguti435]

ctgθ=1+2√2sin105=2+√3...θ=15

[@Celanish]

Hard and nice !

[@devondevon4366]

15

[@RealQinnMalloryu4]

{30°B+30°A+105°C}=165°BAC {165°BAC+15D}= 180°BACD 3^60 3^30^2 3^3^10^2 1^3^2^5^2 1^3^1^1^2 32(x ➖ 3x+2).