A Very Nice Geometry Challenge | Maths Olympiad | 2 Different Methods to Solve

[@quigonkenny]

As ∠DBC = 15° and ∠BCD = 30°, ∠CBD = 180-45 = 135°. As CA is a straight angle, ∠BDA = 180-135 = 45°. Drop a perpendicular from D to E on BC. As ∠ECD = 30°, ∆DEC is a 30-60-90 special right triangle and DE = CD/2 = 5/2. Triangle ∆BED: sin(15°) = DE/DB sin(45-30) = (5/2)/DB sin(45)cos(30) - cos(45)sin(30) = 5/2DB (1/√2)(√3/2) - (1/√2)(1/2) = 5/2DB √3/2√2 - 1/2√2) = 5/2DB (√3-1)/2√2 = 5/2DB 10√2 = 2DB(√3-1) DB = 5√2/(√3-1) DB = 5√2(√3+1)/(√3-1)(√3+1) DB = 5√2(√3+1)/(3-1) DB = 5√2(√3+1)/2 AB² = BD² + DA² - 2(BD)DAcos(45°) x² = (5√2(√3+1)/2)² + 5² - 2(5√2(√3+1)/2)(5)(1/√2) x² = 50(√3+1)²/4 + 25 - 25(√3+1) x² = 25(3+2√3+1)/2 + 25 - 25√3 - 25 x² = (100 + 50√3)/2 - 25√3 x² = 50 + 25√3 - 25√3 = 50 x = √50 = 5√2

[@johnlee6304]

1. Draw a line DE such that DE = EC 2. ==> AED is a equilateral triangle 3. AE=BE=5 ==> AB = SQRT(50)

[@russcooper9628]

Calculate AE as in Method 1 Then note that EC is (AE * sqrt(3)) Note that angle BDC is 180 - 15 - 30 = 135 degrees Use Law of Sines to calculate BC BE = BC - EC Use Pythagorean Theorem to find X from BE and AE

[@MarieAnne.]

That's what I did. Uses the simplest steps of Method 1 and Method 2 shown in video.

[@jimlocke9320]

Drop perpendiculars from A to BC, label the intersection M, and from D to BC, intersection N. Note that ΔAMC and ΔDNC are special 30°-60°-90° right triangles, so, from ratios of sides, lengths DN = 2.5, AM = 5, CN = (2.5)√3 and CM = 5√3. ΔBDN is a 15°-75°-90° triangle with long side BN having length (2 + √3) times DN, = 5 + (2.5)√3. BC = BN + CN = 5 + (2.5)√3 + (2.5)√3 = 5 + 5√3. MB = BC - CM = 5 + 5√3 - 5√3 = 5. Therefore, in right ΔBMA, sides AM and MB are both length 5, so hypotenuse X = length AB = 5√2, as Math Booster also found.

[@RobertHering-tq7bn]

Hello, here is another way to find the length of x. We can add a further edge with length h from B to a new point E on the line thru A and C with angle DBE being 45°. As angl e BDA is 45° as well, the length of EDB is h, too. Angle CBE is 15°+45° = 60° and so angle DEB is 90°. From that we know that the length of BC is 2h. Now... (h+5)² + h² = (2h)² 2h² - 10h - 25 = 0 h² - 5h - 25/2 = 0 h² - 2h*(5/2) + (5/2)² - 25/4 - 50/4 = 0 (h-5/2) - 75/4 = 0 (h-5/2-(5/2)*Sqrt(3))(h-5/2+(5/2)*Sqrt(3)) ==> h = (5/2)(Sqrt(3)+1) This is true, because h cannot be negative. The value is about 6.8. This means also that BC = 5(Sqrt(3)+1). Furthermore... h-5 = (5/2)(Sqrt(3)+1) - (5/2)*2 = (5/2)(Sqrt(3)-1) and so x² = h² + (h-5)² = (25/4)(1+2*Sqrt(3)+3+3-2*Sqrt(3)+1) = (25/4)*8 = 50 and x = Sqrt(50) = 5*Sqrt(2) .

[@AndreasPfizenmaier-y7w]

AE=5 and the quicker Solution tan 30 degrees = BE:5, BE= 2,88 and x=5,77

[@dnarna8994]

You seem to have taken a complicated route. Why not just use the sine/cosine rules?

[@dotsucks]

I was having trouble getting rid of that sin(15); thanks for the video

[@giuseppemalaguti435]

ABD=α...teorema dei seni...x/sin45=5/sinα...x/sin30=10/sin(α+15)...divido le equazioni...risulta ctgα=(√2-cos15)/sin15...α=30...x=(5√2/2)/sinα=5√2

[@josegiovannygio3679]

Obrigado professor por compartilha tanto conhecimento de geometria, obrigado tambem por contribuir na minha evolução da matemática( sou brasileiro)!!!!

[@nunoalexandre6408]

Love it!!!!!!!!!!!!!!!!

[@someonespadre]

Only a mathematician would think 5 times the square root of 2 is a valid answer. Breaking this down, the line in question is a 5 by 5 which is easy to remember, a 5 by 5 equals Seven oh Seven, a pretty good airplane.

[@misterenter-iz7rz]

Your figure is much out of scale, hardly AEB is liked a isosceles right-angled triangle 45-90-45.😊

[@ai2657]

Sir i found BD and BC and used apollonius theorem

[@nathancc2526]

Same

[@danyetdeco]

gracias !!!!!!!!!!!